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First Edition Published 1986
1986
Published 2000
Second Edition Published
Printed in the United States of America
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10113 pt.
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this book
book isis available
availablefrom
from the
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Cataloging in Publication
Publication Data
Data
Library
Aschbacher, Michael,
Michael, 19441944Aschbacher,
Finite group theory / M. Aschbacher. -- 2nd ed.
p.
p.
cm. - (Cambridge
(Cambridge studies
studies in
in advanced
advanced mathematics
mathematics ;;10)
10)
Includes bibliographical references and
and index.
index.
ISBN 0-521-78675-4
0-521-78675-4 (pbk.)
(pbk.)
ISBN 0-521-78145-0 (hb) - ISBN
1. Finite groups.
I. Title. 11.
II, Series.
1.
Series.
QA177.A82
QA177 .A82
2000
512'.2
512'.2 -- dc2l
dc21
99-055693
hardback
ISBN 0 521 78145 0 hardback
ISBN 0 521 78675 4 paperback
To Pam
Contents
Contents
Preface
Preface
11
1
22
33
44
1
Preliminaryresults
results
Preliminary
Elementary group
grouptheory
theory
Elementary
Categories
Categories
Graphs and
andgeometries
geometries
Graphs
Abstractrepresentations
representations
Abstract
Permutationrepresentations
representations
22 Permutation
Permutationrepresentations
representations
55 Permutation
Sylow's Theorem
Theorem
66 Sylow's
Representationsof
ofgroups
groupson
on groups
groups
33 Representations
Normal series
series
77 Normal
Characteristic subgroups
subgroupsand
and commutators
commutators
88 Characteristic
Solvableand
andnilpotent
nilpotentgroups
groups
99 Solvable
10 Semidirect
Semidirectproducts
products
10
Central products
products and
and wreath
wreath products
products
11 Central
11
Linearrepresentations
representations
44 Linear
12 Modules
Modules over
overthe
the group
group ring
ring
12
13
13
The
The general
generallinear
lineargroup
groupand
and special
speciallinear
lineargroup
group
14 The
The dual
dualrepresentation
representation
14
Permutation
Permutationgroups
groups
The
The symmetric
symmetric and
and alternating
alternating groups
groups
16 Rank
Rank3 3permutation
permutationgroups
groups
16
55
15
15
66
17
17
18
18
Extensions
Extensions of
of groups
groups and
and modules
modules
1-cohomology
1-cohomology
Coprime
Coprime action
action
Spaces with
with forms
forms
77 Spaces
19 Bilinear,
Bilinear, sesquilinear,
sesquilinear, and
and quadratic
quadratic forms
19
20 Witt's
Witt's Lemma
Lemma
20
21 Spaces
Spaces over
overfinite
finitefields
fields
21
Theclassical
classicalgroups
groups
22 The
88
p-groups
p-groups
23 Extremal
Extremal p-groups
p-groups
23
24 Coprime action on p-groups
24
Contents
Contents
viii
99
25
25
26
26
27
27
10
10
28
28
29
29
30
30
Change of field of
ofaalinear
linear representation
representation
Tensor
Tensor products
products
Representations
Representations over
over finite
finite fields
fields
Minimal
Minimal polynomials
polynomials
Presentations
Presentationsof
of groups
groups
Free
Free groups
groups
Coxeter groups
Root systems
systems
11 The
Thegeneralized
generalizedFitting
Fittingsubgroup
subgroup
31
31
The
The generalized
generalized Fitting
Fitting subgroup
subgroup
117
117
123
127
138
138
141
148
156
157
32 Thompson
Thompsonfactorization
factorization
Central
33
33 Central extensions
extensions
12 Linear
Linearrepresentations
representationsof
offinite
finite groups
groups
34 Characters
Characters in
incoprime
coprimecharacteristic
characteristic
35 Characters
Characters in
in characteristic
characteristic 00
36 Some
Somespecial
specialactions
actions
162
Transfer
Transferand
andfusion
fusion
Transfer
37 Transfer
Alperin's Fusion
Fusion Theorem
Theorem
38 Alperin's
39 Normal
Normal p-complements
p-complements
40 Semiregular
Semiregular action
action
197
14 The
Thegeometry
geometryofofgroups
groupsofofLie
Lietype
type
41 Complexes
Complexes
42 Buildings
Buildings
43 BN-pairs
BN-pairs and
andTits
Tits systems
systems
209
13
166
177
178
181
192
197
200
202
205
209
215
218
15
44
Signalizer
Signalizerfunctors
functors
Solvable
Solvablesignalizer
signalizerfunctors
functors
229
16
45
46
47
Finite
Finite simple
simple groups
groups
242
Involutions
Involutions in
infinite
finitegroups
groups
Connected
Connected groups
groups
The
The finite
finite simple
simple groups
groups
48 AnAnoutline
outlineofofthe
theClassification
ClassificationTheorem
Theorem
229
243
245
249
260
Appendix
Appendix
269
References
References
297
List
Listof
ofSymbols
Symbols
299
Index
Index
301
Preface
Finite Group
reference
Group Theory
Theory is intended
intended to serve
serve both as a text and as a basic reference
on finite groups. In neither role do II wish
wish the
the book
book to
tobe
beencyclopedic,
encyclopedic,so
so
I've included
I've
included only
only the
the material
material II regard
regard as
asmost
most fundamental.
fundamental. While
While such
such
judgments are subjective, I've been guided by a few basic principles which I
feel are important
important and
and should
should be
be made
made explicit.
explicit.
One unifying notion is that of a group
group representation.
representation. The term
term representarepresentaNamely in this book
book a
tion is used here in a much
much broader sense than usual. Namely
of a group
group G
G in
in aa category
category +?is
6 is a homomorphism
homomorphism of G into the
representation of
automorphism group of some object of -C
6 Among these representations, the
permutation representations,
permutation
representations,the
the linear
linear representations,
representations,and
and the
the representations
representations
fundamental. As a result much of the
of groups on groups seem to be the most fundamental.
book is devoted to these
these three classes
classes of
of representations.
representations.
The first step in investigating
investigating representations of finite
finite groups or
or finite
finite didimensional groups
groups is to break up the
mensional
the representation
representation into
into indecomposable
indecomposable or
irreducible representations.
representations.This
Thisprocess
process focuses
focuses attention
attentionon
on two
two areas
areas of
irreducible
the irreducible
irreducible and indecomposable
indecomposable representations
representations themselves,
study: first on the
and second on the recovery of the general representation
representation from its irreducible
irreducible
constituents. Both areas receive attention
constituents.
attention here.
The irreducible objects in the category of groups are the simple
simple groups. I
regard the finite
finite simple
simple groups
groups and
and their
their irreducible
irreduciblelinear
linearand
andpermutation
permutation
representations as the center of interest
interest in
in finite
finite group
group theory.
theory. This point
point of
of
view above all others has dictated the choice of material. In particular
view
particular I feel
feel
groups are
are best
best answered
answered through
through the
the
many of the deeper questions about finite groups
following
following process. First reduce the question to a question about
about some class of
irreducible
irreducible representations
representations of simple
simple groups
groups or almost
almost simple
simple groups.
groups. Second
Second
of the finite simple groups to conclude the group
appeal to the classification of
is an alternating
alternating group, a group
group of Lie
Lie type,
type, or
or one
one of
of the
the 26
26 sporadic
sporadicsimple
simple
groups.
groups. Finally invoke the irreducible representation theory of these groups.
The book serves
foundation for the proof of the Classification Theorem.
serves as a foundation
Almost all material
turns out
out
material covered plays a role in the classification, but as itit turns
almost
interest outside
outside that framework
framework too. The only major result treated
almost all
all is of interest
here which has
has not found application
application outside
outside of simple
simple group
group theory is the Signalizer
nalizer Functor
Functor Theorem.
Theorem. Signalizer
Signalizerfunctors
functors are
are discussed
discussednear
near the
the end
end of the
book. The
The last
last section
section of the book discusses
discusses the
the classification
classification in
in general
general terms.
terms.
x
Preface
The first edition of the book included
The
included a new proof of the Solvable Signalizer
Signalizer
Functor Theorem, based
based on
on earlier work of
of Helmut
Helmut Bender.
Bender. Bender's
Bender's proof
proof
Functor
was valid only for the prime 2, but it is very short and elegant. I've come to
believe that my extension
extension to arbitrary
arbitrary primes
primes in
in the
the first
first edition
edition is
is so
socomplicomplicated that it obscures the proof, so this edition
edition includes only a proof of the
the
Solvable 2-Signalizer Functor Theorem, which is closer to Bender's original
original
proof. Because of this change, section 36 has also been truncated.
In some sense most of the
the finite
finite simple
simple groups
groups are
are classical
classical linear
linear groups.
groups.
Thus the classical groups serve as the best example
example of finite simple groups.
They are also representative
representative of the groups of Lie type, both classical
classical and
and exexceptional, finite or infinite. A significant fraction of the book is devoted to the
ceptional,
classical groups. The discussion is not restricted to groups
groups over
over finite
finite fields.
classical groups
groups are examined via their representation
representation as the automorphism
automorphism
The classical
forms and their
their representation
representation as the
the automorphism
automorphism groups
groups of spaces of forms
of buildings. The Lie theoretic point of view enters
enters into
into the
the latter
latterrepresentation
representation
and into a discussion
discussion of Coxeter groups
groups and
and root
root systems.
systems.
I assume
assume the reader
reader has been
been exposed
exposed to a first
first course
course in algebra
algebra or
or its
its
equivalent; Herstein's
Herstein's Topics
Topics in
in Algebra
Algebra would
would be
be a representative text
text for
equivalent;
such a course. Occasionally
Occasionally some deeper algebraic
algebraic results are
are also
also needed;
needed;
in such instances the result is quoted and a reference
is
given
for
its
reference
its proof.
proof.
Lang's Algebra is one reference
reference for such results. The group
group theory I assume is
listed explicitly in section 1. There
There isn't
isn't much; for example Sylow's
Sylow's Theorem
is proved in chapter
chapter 2.
As indicated
indicated earlier,
earlier, the
the book
book isis intended
intended to
to serve
serveboth
both as
as aa text
text and
and as a
As
basic reference.
reference. Often these
these objectives are compatible, but when compromise
compromise
is necessary it is usually in favor of the role as aa reference.
reference. Proofs are more
terse than in most
most texts.
texts. Theorems
Theorems are
are usually
usually not
not motivated
motivated or
or illustrated
illustrated
with examples,
exercises are
examples, but there are exercises. Many of the results in the exercises
an appeal
appeal to
to the
the exercises
exercises in
in the
the
interesting in their own right; often there is an
book proper. In this second edition I've added
appendix containing
containing solutions
added an appendix
solutions
to some of the most difficult and/or
andlor important
important exercises.
exercises.
book is used as a text
text the
the instructor
instructor will probably wish to expand many
If the bookis
Here are some
proofs in lecture and omit some of the more difficult sections. Here
suggestions
suggestions about which sections
sections to
to skip
skip or
or postpone.
postpone.
A good basic course in finite group theory would consist of the first eight
chapters, omitting sections 14, 16, and 17 and chapter 7, and adding sections
38, and 39
28, 31, 34, 35, and 37. Time permitting, sections 32, 33,
33,38,
39 could be
added.
The classical groups and some associated Lie theory are treated in chapter
7, sections 29 and 30, chapter 14, and the latter part of section 47. A different
sort of course could be built around
around this
this material.
material.
Preface
xi
concepts in the theory of linear representations
Chapter 9 deals with various concepts
representations
those in
in chapters
chapters 44 and
and 12.
12.Much
Muchof
of
which are somewhat less basic than most of those
interest
for
representations
over
fields
the material in chapter 9 is of principal
principal interest
representations
of prime
prime characteristic.
characteristic. A
A course
course emphasizing
emphasizing representation
representation theory would
probably include
include chapter
chapter 9.
9.
Chapter 15 is the
only of
of
the most
most technical
technical and specialized.
specialized. ItIt is probably
probably only
interest to potential simple groups theorists.
16 discusses
discusses the finite
finite simple
simple groups
groups and
and the
the classification.
classification. The
Chapter 16
latter part of section
section 47
47 builds on chapter
chapter 14,
14, but the rest of chapter 16
16 is pretty
latter
consists of a very brief outline of the
the proof
proof of
of the
the finite
finite
easy reading. Section
Section 48 consists
simple groups makes use of results
results from
from earlier
earlier in the book and thus
thus motivates
motivates
those results by exhibiting
exhibiting applications
applicationsof the
the results.
results.
Each chapter begins with a short introduction describing the major results
in the chapter.
chapter. Most chapters close with a few
few remarks.
remarks. Some remarks acknowledge sources for material covered in the chapter or suggest
suggest references
references
for further
further reading.
reading. Similarly,
Similarly, some
some of
of the
the remarks
remarks place
place certain
certainresults
resultsin
in concontext and hence
hence motivate
motivate those results. Still others warn that some
some section in
the chapter is technical
technical or specialized
specialized and
and suggests
suggests the casual
casual reader
reader skip
skip or
or
postpone the section.
section.
In addition to the introduction and the remarks, there is another good way
to decide which results in aa chapter
chapter are
are of
of most
most interest:
interest: those
those results
results which
which
bear some
some sort
sort of descriptive
descriptive label
label (e.g.
(e.g. Modular Property of Groups,
Groups, Frattini
Frattini
Argument) are
are often
often of
of most
most importance.
importance.
Preliminary results
I assume
assume familiarity
familiaritywith
with material
material from
from aa standard
standard course
course on
on elementary
elementaryalgealgebra. A typical text for such a course is Herstein [He].
[He]. A few deeper algebraic
results are
are also
also needed;
needed; they
they can
can be
be found
foundfor
forexample
examplein
inLang
Lang[La].
[La].Section
Section11
elementary group theoretic results assumed and also contains a list of
lists the elementary
of
notation. Later
Later sections
sections in
in chapter
chapter 11 introduce
introducesome
some terminology
terminology and
and nonobasic notation.
tation
of algebra.
algebra. Deeper
Deeper algebraic
algebraic results
results are introduced
introduced
tation from
from a few other areas of
are needed.
needed.
when they are
The last
last section
section of chapter 11 contains
contains a brief discussion of group
group representarepresentaThe
tions. The
The term representation
representation is
is used here
here in
in aa more
more general
general sense
sense than
than usual.
usual.
tions.
Namely a representation of a group
group G
G will
will be
be understood
understood to
to be
be aa group
group homohomomorphism of G into the group of automorphisms
automorphisms of an object
object X.
X. Standard use
use
of the term representation requires X to be a vector space.
1 Elementary
Elementary group
group theory
Recall that a binary operation on a set
set G
G isisaafunction
functionfrom.
from the set product
product
G
G into G.
Gx G
G. Multiplicative notation will usually be used. Thus the image
image of
a pair (x,
xy. The operation is
(x,y)
y) under
under the
the binary
binary operation will be written xy.
associative if (xy)z ==x(yz)
x(yz)for
forall
allx,x,y,y,zzininG.
G.The
Theoperation
operationisiscommutative
commutative
if xy ==yx
yxfor
forall
allx,x,yyininG.
G.An
Anidentity
identityfor
forthe
the operation
operation is an element 1 in
G such
x 1=
= lx
such that xl
lx==xxfor
forall
allxxininG.G.An
Anoperation
operationpossesses
possessesatatmost
mostone
one
identity. Given
Given an
an operation
operation on
on G possessing an identity 1, an inverse for an
element
in G
G such that
thatxy
x y=
= yx =
=1.1.IfIf our
our operation
operation is
element x of G is an element yy in
associative
is unique
unique and is
is denoted
denoted
associativeand x possesses
possesses an inverse
inverse then that inverse is
in multiplicative
multiplicative notation.
notation.
by x-'
x-1 in
is aa set
setGGtogether
togetherwith
withananassociative
associativebinary
binaryoperation
operation which
which
A group is
possesses an identity and such that each element of G possesses
possesses an
an inverse.
inverse.
abelian ifif its
its operation
operationisis commutative.
commutative. In
In the
the remainder
remainder of
of this
this
The group is abelian
G is
is aa group
group written
written multiplicatively.
multiplicatively.
section G
x" denotes the product of x with
Let xX EE G and n a positive
positive integer. xn
with itself
x_n
x" is aa well-defined
n times. Associativity insures xn
well-defined element of G. Define
Define x-"
(x-1)"
to be (
x-')" and
and x°
x0to
to be
be 1.1.The
The usual
usual rules
rules of
of exponents
exponents can be derived from
2
results
Preliminary results
this definition:
definition:
(1.1) Let G be a group, x EE G,
G, and
and nn and
and m integers. Then
(1.1)
= (xm)(xn).
(xn)(xm)=
xn+m =
(xrn)(x").
(1) (xn)(xm)
= xn+m
(2) (xn)m
(2)
(xn)m =
= xnm.
xnm
subgroup of G is a nonempty
H, xy
xy
A subgroup
nonempty subset
subset H
H of G such that for
for each
eachx,
x, yy E H,
x-' are
are in
in H.
H.This
Thisinsures
insuresthat
that the
the binary
binary operation
operation on G restricts to a
and x-1
which makes
makes H
Hinto
intoaagroup
groupwith
with the
the same
same identity
identity as
as
binary operation on H which
G and the same inverses.
inverses. IIwrite
writeH
H(
<G
H is
G to
to indicate
indicate that H
is a subgroup
subgroup of G.
The intersection
intersectionof
of any
any set
set of
of subgroups
subgroupsof
of G
G isis also
also aa subgroup
subgroupof
of G.
G.
(1.2) The
Let
Let SS C
E GGand
anddefine
define
(S) = n H
scH<G
( S )is
is a subgroup
subgroupof G
G and
and by construction
construction it is the smallest subgroup
By 1.2,
1.2,(S)
By
subgroup of
G containing
containingS.
S. The
The subgroup
subgroup(S)
( S )isiscalled
calledthe
thesubgroup
subgroupof
of GGgenerated
generatedby
byS.
S.
(1.3)
(1.3) Let S C
g G.
G. Then
(S)
-1}.
(S) =
={(sl)E1
{ ( s ~ )...~ '(S,)"':
. . . ( sSi
~ )EE~S,
S
" :, s ~e
& == +1
+ I or
or-1).
x) =
{xn:nn E
Z].
(1.4) Let xX EE G.
(1.4)
G. Then
Then ((x)
= {xn:
c 1].
course 1.4
1.4 is a special
special case of 1.3.
1.3. A group
group G is
is cyclic
cyclic if itit is generated by
Of course
generator of
of G
G and
and by
by 1.4,
1.4,GG
some element x.
x. In that case x is said to be a generator
consists of the powers of x.
consists
The order
order of
of aa group
group G
Gisisthe
thecardinality
cardinalityof
of the
the associated
associated set
setG.
G.Write
WriteIGI
IGl
The
for the
the order
orderof
ofaaset
setGGororaagroup
groupG.G.For
ForXxEEG,G,IxIxI Idenotes
denotesI I(x)
(x)I1 and is
is called
called
for
the order of x.
A
group G into a group
group H is a function a :: G
G4
-*
A group
group homomorphism
homomorphismfrom
from a group
of the
the set
setGGinto
intothe
theset
setHHwhich
whichpreserves
preservesthe
thegroup
groupoperations:
operations: that
H of
is for
for all x, y in
in G,
G, (xy)a
(xy)cr ==xaya.
xaya.Notice
Noticethat
thatI Iusually
usuallywrite
write my
my maps
maps on
on
is
right, particularly
particularly those
those that
thatare
arehomomorphisms.
homomorphisms. The
Thehomomorphism
homomorphism
the right,
anisomorphism
isomorphism if acr isisaabijection.
bijection. In
Inthat
thatcase
case aapossesses
possessesan
aninverse
inverse
aa isisan
function a-':
a`1: H
H-*
+GGand
andititturns
turnsout
outa-1
a-'isisalso
alsoa agroup
grouphomomorphism.
homomorphism.GG
is isomorphic to H
H if there
there exists
exists an
anisomorphism
isomorphismofofGGand
andH.
H.Write
WriteGGZ= H to
indicate that G is isomorphic to H.
indicate
H. Isomorphism is an equivalence relation.
relation. H
H
is
homomorphic image of G
surjective homomorphism
is said
said to be a homomorphic
G if there
there is a surjective
homomorphism
of G onto H.
H.
Elementary group theory
3
subgroup H
H of
of G is normal
if g-'hg
g-lhg c-E H for
H..
A subgroup
normal if
for each
each g EE G
G and
and h cEH
Write
of G. If a:
G to
to indicate
indicate H
H isisaanormal
normal subgroup
subgroup of
a:G
G + X is aa
Write H Ia! G
homomorphismthen
thenthe
thekernel
kernelofofaaisisker(a)
ker(a)=
_ {g EE G: ggaa ==1}
group homomorphism
11and
and itit
of G. Also write
write G
Gaa for
for the
the image
turns out that ker(a) is a normal subgroup of
{ga:
Gin
{ga:g E G}
GI of G
in X.
X. Ga
Gaisisaasubgroup
subgroupof
of X.
X.
Let H <5G.
X EE G
= {hx:
xH =
= {xh:
G.For
Forx
G write
write Hx =
{hx:h EE H}
HI and
andxH
{xh:h E
E H}.
HI. Hx
and xH are cosets of H in
G.
Hx
is
a
right
coset
and
xH
a
left
coset.
in G. Hx is a right coset and xH
coset.
be consistent
consistent I'll
I'll work with right cosets Hx
Hx in
in this
this section.
section. GIH
G/H denotes
denotes
To be
of all
all (right)
(right)cosets
cosetsof
ofHH in
inG.
G. GIH
G/H is the coset space
space of
of H
H in G. Denote
the set of
by IIG
G ::HI
H Ithe
theorder
orderofofthe
thecoset
cosetspace
spaceG/H.
G/H.As
Asthe
themap
maphht+H hx
hx isisaa bijection
bijection
of H with
with Hx,
Hx, all
all cosets
cosets have the same
same order, so
IHJ
(1.5)
(1.5) (Lagrange's
(Lagrange'sTheorem)
Theorem) Let
Let G
G be
be aagroup
groupand
andHHi
< G.
G. Then
ThenIGI
G I=
_H
I
I( G
G:: H 1.1.In
In particular
particular if
G I.
if GGisisfinite
finitethen
thenI H
IHI( divides
dividesI \GI.
If H a5GGthe
thecoset
cosetspace
spaceG/H
GIHisismade
madeinto
intoaagroup
groupby
bydefining
defining multiplication
multiplication
via
(Hx)(Hy) = Hxy
x, y E G
homomorphismnrr:
-* GIH
G/H defined
: GG+
defined
Moreover there is a natural surjective homomorphism
by nrr:: xx t+H Hx.
ker(rr) =
= H.
Hx. Notice
Notice ker(n)
H.Conversely
Conversely if aa:: G -*
+LLisisaasurjective
surjective
homomorphism
ker(a) ==HHthen
i-+ xxaa is
isomorphism
thenthe
themap
map,B:
/3: Hx t+
is an isomorphism
homomorphismwith ker(a)
of G/H
= a.
GIH with
with LL such
such that
that ir,B
np =
a.The
The group
group G/H
GIHisiscalled
calledthe
thefactor group
group of
G by H.
H. Therefore
Thereforethe
the factor
factor groups
groups of G
G over
over its
its various
various normal
normal subgroups
subgroups
are, up to isomorphism,
isomorphism, precisely the homomorphic images of G.
G.
(1.6)
G.
is isa bijection
(1.6) Let H aI!
G.Then
Thenthe
themap
map LL H
t+ L/H
LIH
a bijectionbetween
betweenthe
theset
setof
of all
all
subgroups of
of G containing H
H and the set of all
all subgroups
subgroups of G/H.
GIH. Normal
Normal
subgroups
subgroups correspond
correspond to normal subgroups
subgroups under this bijection.
x, yy E G, set Xxy
= y-lxy.
{xY:
x EEX).
For x,
Y =
y-lxy. For
ForXXCG
g GsetsetXyXY_ =
{xY:x
X).XY
XY isisthe
the
conjugate of X under y. Write
of X
xGfor the set {X8:
{Xg:gg Ec G}
G) of conjugates of
Write XG
under G. Define
NG(X)
= {g E G:Xg = X).
NG(X)={gEG:XB=X).
NG(X)
normalizer in
in G
G oof
X and
and is
is aa subgroup
subgroupof
of G.
G. Indeed
IndeedififXX 5
<G
NG(X) is the normalizer
f X
then NG(X)
NG(X)is the largest subgroup of G in which X is
is normal.
normal. Define
Define
_ {g EE G: xg
xg =
= gx
CG(X) =
for all x EE X}.
X).
CG(X)
X. CG(X)
CG(X) is
is also
also a subgroup of
of G.
CG(X)is the centralizer
centralizer in
in G
G of
of X.
4
results
Preliminary results
define XY
XY =={xy:
{xy:xx EE X,
X,yyEE Y).
Y } The
.Theset
setXY
XY isisthe
the product
product
G define
For
For X, Y C G
of X with
with Y.
Y.
(1.7) Let XX,, Y
Y5
< G.
G. Then
(1)
(1)XYis
XY isaasubgroup
subgroupofofGGififand
andonly
onlyififXY
XY ==YX.
YX.
(2)
XY/X 2
= Y/(Y n X).
(2)IfIfYY<5NG(X)
N G ( Xthen
)thenXY
XY is
is aa subgroup
subgroup of G and XY/X
(3)
IXYI
=
)XHHYl/IX
n
Y.
(3) IXYt = tXIIYIIIX n YI.
(1.8)
subgroupsofofGGwith
withKK5<H.
H.Then
ThenGG/K/H/K
(1.8) Let
Let H and
and K be normal subgroups
/K/H/K G
G/H.
G/H.
Let
LetG1,
G 1 ,...
. . ., ,G.G ,bebeaafinite
finiteset
setof
of groups. The
I xx- ...
. . x Gn
G, =
=
Thedirectproduct
directproduct G
G1
ny=,GiGi ofof thethe groups
groups G1,
G I ,...,
. . .Gn
, G,isisthe
thegroup
groupdefined
definedon
onthe
theset
setproduct
product
l 1n=1
G1
G 1 xx ...... xx Gn
G, by
by the
theoperation
operation
(x1,...,Xn)(y1,...,yn)=(xiyi,...,xnyn) xi,yi c Gi
Let G
G be
be aa group
group and (Gi:
(Gi:1 <
5 ii <5n)n a) afamily
familyofofsubgroups
subgroupsof
of G.
G.Then
Then
(1.9) Let
the
the following
followingare
areequivalent:
equivalent:
(1)
(1)The
Themap
map (xi,
( x l ,...,
. . .xn)
, x,) HI+ x1x...
l . x,
. .x,is an
is an
isomorphismofofGGwith
withG1
GI xx .. . xx
isomorphism
G.
G, .
(2) G
G ==(Gi:
(Gi:11 <5i i<n)
5 nand
) andfor
foreach
eachi,i ,11<5i i<5n,n Gi
, Gi49
andGin
Gi n(GG:
( G j :j
(2)
GGand
# i) = 1.
(3)
GGfor
( 3 ) Gi
Gia9
foreach
eachi,i,11<5i i< 5n,nand
, andeach
eachg gE EGGcan
canbebewritten
writtenuniquely
uniquely as
as
gg ==x1
xi EE G,.
x l ...
...xnx, with
withxi
Gi.
If
If any
any of the equivalent conditions of 1.9 hold, G will be said to be.the
be the direct
product
of
the
subgroups
(Gi:
1
<
i
<
n).
(Gi:
1
5
i
5
n).
product the subgroups
(1.10) Let
Let G
G ==(g)
(g)bebea acyclic
cyclicgroup
groupand
andZZthe
thegroup
groupofofintegers
integersunder
under addiaddi(1.10)
tion.
tion.Then
Then
(1)
(I) IfIfHHisisa anontrivial
nontrivialsubgroup
subgroup of
of ZZ then
then H
H==(n),
( n )where
,wherennisisthe
theleast
least
positive
positive integer in H.
H.
(2)
ma ==g'g"isisa asurjective
(2)The
Themap
mapa:a:71
Z -±
-+ GGdefined
defined by ma
surjectivehomomorphism
homomorphism
with
kernel
(
n
)
,
where
n
=
0
if
g
is
of
infinite
order
and
n
min{m
{rn >>0:0:
with kernel (n), where = 0 if g is of infinite order and n ==min
gm
gm ==11
1 ) ifif gg has
hasfinite
finiteorder.
order.
(3)
ordernnthen
thenGG== {gi:0
(3) Iflf gg has
has finite
finite order
(g" 0 <
5 ii <<n}n )and
andnnisisthe
theleast
least
positive
m with
withgg'm ==1.1.
positive integer rn
(4)
(4)Up
Uptotoisomorphism
isomorphism71
Z is
is the
the unique
unique infinite
infinitecyclic
cyclic group
group and
and for
for each
each
positiveinteger
integern,n,the
thegroup
group71n
Z, of integers
modulo nn isis the
the unique
uniquecyclic
cyclicgroup
group
positive
integers modulo
of
of order
ordern.
n.
Elementary group theory
55
(5) Let
isisthe
(5)
LetIgI
Ig ( ==n.n.Then
Thenfor
foreach
eachdivisor
divisorm
m of n, (g"/')
(gn/m)
theunique
uniquesubgroup
subgroup
of G of order
order m.
m. In
In particular
particular subgroups
subgroups of cyclic groups are cyclic.
cyclic.
(1.11)
(1.11) Each finitely generated abelian
abelian group is the direct
direct product
product of
of cyclic
cyclic
groups.
power of
of p.
p. More
Let p be
be aa prime.
prime. A
A p-group is a group whose order is a power
set of
of primes
primes then
then aa n-group
rr-group is a group G of
of finite
finite order
generally if nTr isis aa set
such that 7r
(G) gC n
jr,, where
(G) denotes
denotes the set of prime divisors
GI. p'
n(G)
where 7r
n(G)
divisors of
of I/GI.
denotes the
the set of
of all primes
primes distinct
distinctfrom
fromp.
p. An
An element
element xx in
in a group
group G
G is a
7r -elementifif (x)
(x) is a n-group.
7r -group.An
Aninvolution
involutionisis an
an element
element of order 2.
n-element
(1.12)
Let 1 #
: GGbe
bean
anabelian
abelianp-group.
p-group. Then
Then G
G isisthe
thedirect
directproduct
product of
of
(1.12) Let
cyclic
subgroupsGi
G,2= Zpe,,
Z pe;1,1
i <n,n,ele 2
l >eze22>. ...
> een>>1. 1.
Moreover
cyclic subgroups
_( <
i _(
. . L.
Moreoverthe
the
(ei: 115< i <5 n)
n)are
areuniquely
uniquely determined
determinedby
by G.
G.
integers n and (e1:
of a finite group G is the
The exponent of
the least
least common
common multiple
multiple of the orders
orders
of the elements of G. An elementary
p-group is an abelian
elementary abelian p-group
abelian p-group
of exponent
by 1.12,
abelian p-group
p-group of
of
exponent p. Notice that by
1.12, G is an
an elementary
elementary abelian
order p"
pnifif and
andonly
onlyififGGisisthe
thedirect
directproduct
productofofnncopies
copiesofofTLp.
Z,. In particular
up to isomorphism there
there is
is aa unique
unique elementary
elementaryabelian
abelianp-group
p-group of
of order
order pn,
p",
Ep,.. The
p-rank
The integer n is the p-rank of Ep».
Epa. The p-rank
which will be denoted by Ep.
of a general finite group G is the maximum p-rank of
of an
an elementary
elementary abelian
abelian
mp(G).
p-subgroup of
of G,
G, and
and is
is denoted
denoted by mp(G).
(1.13)
(1.13) Each
Each group
groupof
of exponent
exponent 22 is
is abelian.
abelian.
setof
ofprimes
primesand
andGGaafinite
finitegroup,
group,write
write0,
0, (G) for the largest normal
If nTrisisaaset
7r-subgroup of
of G,
G, and
and OK(G)
O" (G) for
for the
the smallest
smallest normal
normal subgroup
subgroup H
H of G
n-subgroup
G such
such
0, (G) and
(G) are
that G/H isisaa7r-group.
n-group. O,(G)
and 0'OK(G)
arewell
welldefined
definedby
byExercise
Exercise1.1.
1.1.
(G) and call Z(G)
Z(G) the center
center of G. If
If G is aa p-group
p-group then
then
CG(G)
Define Z(G) ==CG
define
Q,(G)=(xEG:x"
Q,,(G) = (x E G: xp"_1)
= 1)
= (xp":x E G).
Un(G)
Y(G)=(x'':xEG).
For X <
= NG(X)/CG(X)
NI(X)/CG(X) to be the automizer in
in G of
(G
G define
define AUtG(X)
AutG(X) =
For
by Exercise
Exercise 1.3,
1.3,AutG(X)
AutG(X)_(< Aut(X)
Aut(X) and
and indeed
indeedAutc(X)
Autc(X) is the
X. Notice that by
group of automorphisms induced on X in G.
G.
subgroupof
of aa group
group G
G is a proper
A maximal
maximal subgroup
proper subgroup
subgroup of G which
which is
is
properly
properly contained
contained in
in no proper subgroup
subgroup of G.
G. That
That is
is aa maximal
maximal subgroup
subgroup is
66
Preliminary results
results
Preliminary
a maximal member of
of the set of proper subgroups of
of G, partially
partially ordered by
by
inclusion.
inclusion.
If a:
the
a:SS -+
+TTisisaafunction
functionand
and RRCGSSthen
thena)R
~ I denotes
denotes
R
the restriction
restriction of a
to R.
4R:RR+
- *T
T is
is the
the function
function from
from R into T agreeing with a.
R. That
That is
is aalR:
a.
Here's aa little
little result
result that's
that's easy
easyto
toprove
provebut
but useful.
useful.
(ModularProperty
Propertyof
of Groups)
Groups)Let
Let A,
A, B,
B, and
and C
(1.14) (Modular
C be
be subgroups
subgroupsof
of a group
G
with A <5C.
Then AB F) C == A(B
GwithA
C.ThenABnC
A ( BF)
n CC).
).
If
- (1)
If G
G isis aa group
group write G#
G' for the set G (1)ofofnonidentity
nonidentity elements
elementsof
of G.
G. On
On
the
R# =
= R --(O).
the other
other hand if R is a ring define R'
(0).
Denote by C,
C, R,
R, and
and 0Qthe
thecomplex
complexnumbers,
numbers,the
thereals,
reals,and
andthe
therationals,
rationals,
Denote
respectively.
1 will
respectively. Often Z
will denote
denotethe
the integers.
integers.
Given
H of
of G,
G,and
andaa collection
collectionC
C of
of subgroups
subgroupsof
of G,
G,
Given a group G, a subgroup
subgroup H
I'll
I'll often
often write
write cCnnHHfor
forthe
theset
setofofmembers
membersof
ofCCwhich
whichare
aresubgroups
subgroupsof
of H.
H.
I'll
I'll use
use the
the bar
barconvention.
convention. That
That is
is I'll
I'll often
oftendenote
denoteaahomomorphic
homomorphic image
image
Gaof
ofaagroup
groupG
G by
by GG (or
(orG*
G*or
or G)
G )and
andwrite
write gg (or
(or g*
g* or g)
8 ) for ga.
g a .This
This will
will
Ga
be done
done without
without comment.
comment.
Other notation and
and terminology are
are introduced
introducedin
in later
later chapters.
chapters. The
The List
List of
Other
Symbols
introduced and
Symbolsgives
givesthe
the page
page number
numberwhere
whereaa notation
notation is
is first introduced
and defined.
defined.
22 Categories
Categories
willbe
beconvenient
convenient to
to have
have available
availablesome
some of
of the
the elementary
elementary concepts
concepts and
and
ItIt will
languageof
of categories.
categories.For
Foraasomewhat
somewhatmore
moredetailed
detaileddiscussion,
discussion,see
seechapter
chapter11
language
of
of Lang
Lang [La].
[La].
A
A category
category fi'i?consists
consists of
of
(1)
Ob(i) ofofobjects.
objects.
(1) AAcollection
collection Ob(&)
(2) For
Foreach
eachpair
pairA,B
A,Bof
of objects,
objects,aaset
setMor(A,B)
Mor(A,B) of
of morphisms
morphisms from
from A
Ato
to B.
B.
(2)
(3)
ofobjects
objectsaamap
map
(3) For
Foreach
eachtriple
tripleA,
A, B,
B,CCof
Mor(A, B) x Mor(B, C)
Mor(A, C)
called composition.
composition. Write
g) under
under the
the
called
Write ff gg for the image of
of the pair (f, g)
composition
compositionmap.
map.
Moreoverthe
the following
followingthree
threeaxioms
axiomsare
arerequired
requiredtotohold:
hold:
Moreover
Cat
Cat(1)
(1)For
Foreach
eachquadruple
quadrupleA,
A,B,
B,C,C,DDofofobjects,
objects,Mor(A,
Mor(A,B)B)f1nMor(C,
Mor(C,D)
D)isis
empty
unless A
A=
= C and B =
empty unless
=D.
D.
Cat
Cat (2)
(2)Composition
Compositionisisassociative.
associative.
Cat
Mor(A,A)A)possesses
possessesananidentity
identitymorphism
morphism1A
1.4such
such
Cat(3)
(3)For
Foreach
eachobject
objectA,A,Mor(A,
that
and all
all ffininMor(A,
Mor(A,B)
B)and
andggininMor(B,
Mor(B,A),
A),
that for
for all
all objects
objects B and
lAf=fandglA=g.
1.4f = f and g1.4 = g.
Graphs and geometries
77
Almost
categories of sets with
with structure.
structure.
Almost all
all categories
categories considered
consideredhere
here will
will be categories
That is the objects
objects of the category are
are sets
sets together with some
some extra structure,
structure,
Mor(A, B) consists of
of all functions
from the
the set associated
associated to
to A
A to the set
functions from
associated to B which preserve the extra structure,
structure, and composition
composition is ordinary
associated
composition
forced to be the identity
identity
compositionof
of functions.
functions.The
The identity
identity morphism
morphism IA
lAis forced
map on A.
A. Thus
Thus we
we need
need to
toknow
knowthe
theidentity
identitymap
mappreserves
preservesstructure.
structure.We
We
to know
know the
the composition
composition of
of maps
mapswhich
which preserve
preservestructure
structurealso
also
also need to
preserves structure. These facts will usually be obvious in the examples
examples we
consider.
consider.
We'll be
be most
most interested
interested in
in the
the following
following three
three categories,
categories, which are all
all
categories
categories of
of sets
sets with
with structure.
structure.
(1) The category
category of sets
sets and
and functions:
functions: Here
Here the objects are the sets
sets and
and
Mor(A, B)
B) is
is the
the set
set of
of all
all functions
functionsfrom
from the
the set
set A
A into
into the
the set
set B.
B.
(2) The
The category
category of
of groups
groups and
and group
group homomorphisms: The objects are the
groups
groups and
and morphisms
morphisms are
are the
the group
grouphomomorphisms.
homomorphisms.
transformations: Fix a field
field F.
F. The
The
(3) The
Thecategory
categoryof
of vector
vector spaces
spaces and linear transformations:
objects are the vector spaces
spaces over
over F
F and the morphisms
are the
the F-linear
morphisms are
transformations.
transformations.
Let ff be
to an
anobject
objectB.
B.An
Aninverse
inversefor
forf fin
inff?
-'
beaamorphism
morphism from
from an object A to
is a morphism
morphism gg E
e Mor(B,
= ff g and 1l B
B== ggf.
Mor(B, A)
A) such that l1A
A=
f .The
The morphism
ff isisan
anisomorphism
isomorphismifif itit possesses
possessesan
an inverse
inverse in w.
ff?.An
An automorphism
automorphismof
of AAisis
an isomorphism
isomorphism from A to A. Denote
Denote by Aut(A) the set
set of all
all automorphisms
automorphisms
of A and observe Aut(A)
Aut(A) forms
formsaagroup
groupunder
underthe
thecomposition
compositioninin6'.
i.
If aa:: A + Bisanisomorphismdefinea*:Mor(A,
B is an isomorphismdefine a*:Mor(A ,A)
A) + Mor(B,
Mor(B, B)
B) by B +
a`1- lfla
isomorphism of Aut(A) with Aut(B).
Aut(B).
a
~ and
a
andobserve
observea*
a*restricts
restrictsto
to aa group
group isomorphism
Let (Ai:
(Ai: i Ee II)) be aa family
family of objects in a category
category -i'.
ff?. A
A coproduct
coproduct of
of the
the
family is an
an object
objectCCtogether
togetherwith
withmorphisms
morphismsci:
c, Ai
: A;+-- CC,, ii E I,
satisfying
the
I, satisfying
universal property: whenever
whenever X is an object and at:
ai:Al
Ai--+
-+ X
X are
are morphisms,
morphisms,
there exists
exists aaunique
uniquemorphism
morphisma :a:CC+
-+XXwith
withcia
cia=
= ai
ai for each ii eE I.I.As
As aa
consequence
consequenceof
of the
the universal
universal property,
property, the coproduct
coproductof
of a family
family is determined
determined
up to isomorphism,
isomorphism, if
if itit exists.
exists.
The product of the
the family
family is
is defined
defined dually.
dually. That is
is to
to obtain
obtainthe
the definition
definition
of the product, take the definition of the coproduct and reverse the direction of
all arrows.
arrows.
Exercise
Exercise 1.2 gives a description of coproducts and products in the three
three
categories
categories listed above.
above.
3 Graphs
Graphsand
andgeometries
geometries
This section contains a brief
brief discussion
discussion of
of two
two more
more categories
categorieswhich
which will
will
occasional appearances
make occasional
appearances in these notes.
88
Preliminary results
B =(V,
(V,*)*)consists
consistsofofa aset
setVVofofvertices
vertices(or
(orobjects
objectsor
or points)
points) totoA graph fi=
symmetric relation ** called adjacency
adjacency (or incidence
incidenceor
or something
something
gether with a symmetric
else). The
The ordered pairs in the relation are
are called the edges
edges of the graph. I write
else).
u ** vvto
toindicate
indicate two
two vertices
vertices are
are related
related via
via ** and
and say
say uu is
is adjacent
adjacent to v. Apath
A path
of length n from u to vv is
is aa sequence
sequence of
of vertices
vertices u ==uo,
ug,uu1,1, ...
. . .,,u,
U, =
=vvsuch
such
of
u,*u,+ for
foreach
each i.i.Denote
Denoteby
by d(u,
d(u,v)v)the
theminimal
minimallength
lengthof
of aa path
path from
from uu
that ui.ui+l
v . If no
(u , v)
oo.d(u,
d (u ,v)
v)isisthe
the distance
distance from
from u to v.
to v.
no such
suchpath
pathexists
existsset
setdd(u,
v)=
= no.
on V
V defined
defined by
and only
only if d(u, v) < no
oo is an
The relation
The
relation - on
by uu - vv ifif and
called
equivalence relation
relation on V. The equivalence classes of this relation are called
the connected components
components of
of the
the graph.
graph. The
The graph
graph is
is connected ifif itit has
has just
just
one connected component. Equivalently
Equivalently there
there isis aa path
path between
between any
any pair of
vertices.
vertices.
A morphism
morphismaa:: B9 -,i'g'
graphsisisaafunction
functiona:
a:VV-+
+V'V'from
fromthe
thevertex
vertex
ofofgraphs
whichpreserves
preservesadjacency;
adjacency; that
that is if u and
set V of $?to
9 to the vertex set
set V'
V' of
of @''
" which
'.
v are vertices adjacent
adjacentin
in @'then
07thenua
ua is
is adjacent
adjacenttotova
va in
in i?9'.
geometries. In this book I adopt a notion of gegeSo much for graphs; on to geometries.
ometry due to Tits.
Tits. Let
Let II be a finite set. A
A geometry
geometryover
overII is
is a triple
triple ((F,
r, *)
r , t.,
*)
+IIisisaatype
typefunction,
function,and
and ** is
is aa symmetric
symmetric
where F is a set of
of objects,
objects,t.:
r: rr -+
incidence
incidence relation on rl?such
suchthat
thatobjects
objectsuuand
andvv of
of the
the same
sametype
type are
are incident
incident
and only if
=v.v.r(u)
t.(u)isisthe
thetype
typeofofthe
theobject
objectu.u.Notice
Notice (F,
( r ,*)
*)isisaagraph.
graph.
if and
if u =
usually write r
forthe
the geometry
geometry (F,
( r , r,t.,*).
*).
I'll usually
r for
r'ofof geometries
geometries is a function aa:: rF -+
+F'r'ofofthe
theassoassoA morphism
morphisma:
a: rF +
-+ F'
ciated object sets which
which preserves
preserves type
typeand
andincidence;
incidence;that
thatisisififu,u,vvEc rF with
u ** v then
thent.(u)
r(u) =
= ~r'(ua)
' ( u aand
and
) ua *'
*' va.
A
flag of the geometry
geometry rr isis aa set
A$ag
set TTof
ofobjects
objectssuch
suchthat
that each
each pair
pair of
of objects
objects
is incident.
incident. Notice
Notice our
our one
one (weak)
(weak) axiom
axiom insures
insures that aa flag
flag TT possesses
possesses
in TT is
at
function t.
r induces
at most one object of each type, so that the type function
induces an
an injection
injection
into I.I.The
Theimage
image rt.(T)
calledthe
the type
type of T. The rank and corank of T
of T into
(T) isis called
respectively. The residue
residue rT
rTofofthe
theflag
flag TT
are the order of
of t.(T)
r (T) and
and I --r t.(T),
(T ), respectively.
forall
alltt cETT]
regarded as aa geometry
is (v cE Fr --T:T:vv**t tfor
J regarded
geometryover
overI I-- t.(T).
r(T).
The geometry rrisisconnected
connectedififits
itsgraph
graph (I',
( r ,*)
*)isisconnected.
connected. Fr isisresidually
residually
The
connected and
and the
the
residue of every flag of corank
corank at least 2 is connected
connected if the residue
residue of
of every
every flag
flag of corank
corank 11is
residue
is nonempty.
nonempty.
Here's aa way
way to
to associate
associate geometries
geometries to groups.
groups. Let
be aa group
group and
and
Here's
Let G be
9==(Gi:
(G,:ii EEI)I )aafamily
familyofofsubgroups
subgroupsof
of G.
G.Define
Define F(G,
r ( G , F ))to be the geogeoGIG, and
andwith
withobjects
objects
metry whose set of objects of type i is
is the
the coset space G/Gi
metry
Gix
For JJ G
c_II write
write J' for
G,xand
and Gay
G,y incident
incident ifif Gix
G,x fl
n Gay
G, y is nonempty.
nonempty. For
for
and define
G,.Observe
Observe that
that for
for
the complement
complementI I-- JJ of
of JJ in I and
defineGGJ =
j = fl G3.
xEG,Sj,X=(Gjx:jEJ}isaflag
x E G, Sj,, = (GJx: j E J ) is a flag of r ( G , g)
of type J .
A group H of automorphisms
automorphisms of a geometry r
saidto
to be
beflag
transitive if
r isissaid
flag transitive
H isis transitive
transitive on flags
I.
H
flags of
of type
type JJ for each subset
subset JJ of I.
-
-
nJEJ
Abstract representations
Abstract
99
4 Abstract
Abstractrepresentations
representations
Let 6'
B be
be aacategory.
category. A -2-representation
&-representation of a group G is a group
group homomorG -f+Aut(X)
Aut(X)ofofGGinto
intothe
thegroup
groupAut(X)
Aut(X)ofofautomorphisms
automorphismsof
of some
some
phism nn::G
object X in 6.
6'.(Recall
(Recallthe
thedefinition
definitionof
of Aut(X)
Aut(X)in
in section
section2.)
2.) We
We will be most
concerned with the following three classes of representations.
concerned
A permutation representation
representation is
is aa representation in the category of sets and
functions. The
The group
group Aut(X)
Aut(X) of automorphisms
of set
set X is
automorphisms of
is the
the symmetric
symmetric
functions.
Sym(X) is
is the
the group
group of
of all
all permutations
permutations of X
group Sym(X) of X. That is Sym(X)
composition.
under composition.
A linear
linear representation
representationisis aa representation
representationin
in the
the category
category of
of vector
vector spaces
spaces
and linear transformations. Aut(X)
Aut(X) is
is the
the general
general linear group GL(X) of the
vector space X. That is GL(X) is the group
group of all invertible linear transformations of X.
Finally we will of course
course be interested
interested in the category of groups
groups and
and group
group
homomorphisms. Of particular interest
representation of G
homomorphisms.
interest is the representation
G via
via conjugaconjugation on itself (cf.
(cf. Exercise
Exercise 1.3).
1.3).
Two '-representations
= 1,
2, are said to be equiva+Aut(X1
Aut(Xi),),ii =
1,2,
Two
B-representationsnlni:: G -+
lent if there exists an isomorphism a:
a: X1 +
-+X2
X2such
such that
that n2
n2 ==Trla*,
nla*,where
where
a*: Aut(X1)
Aut(X1) -+
a*:
+Aut(X2)
Aut(X2)isis the
the isomorphism
isomorphismdescribed
describedin section
section 2. The map a
is said to
to be
be an
an equivalence
equivalenceof
ofthe
therepresentations.
representations.&-representations
6-representations n,
-+
ni:: Gi +
Aut(Xl) i ==1,1,2,2,are
Aut(Xi)
aresaid
saidto
tobe
bequasiequivalent
quasiequivalentifif there
thereexists
existsaa group
group isomorisomorphismQQ,B:
groups
anda a&-isomorphism
6-isomorphismaa:: X1
X1 +
-+ X2 such that
phism
/3: G2 G2-+
+ G1G1
of of
groups
and
n2
=
/3n1a*.
n2 = N7rla*
Equivalent representations
representationsofofaa group
group G
G are the same for our
Equivalent
our purposes.
purposes.
Quasiequivalent representations
representations are almost the same, differing only by an automorphism of G.
A representation
representation 7r
faithful if n7risisan
aninjection.
injection.In
Inthat
thatevent
eventnn induces
induces
n of G isfaithful
isomorphism of G with the subgroup
an isomorphism
subgroup G7r
G n of Aut(X), so G may be regarded
as a group
group of
of automorphisms
automorphismsofofXXvia
via7Tn .
ni: G -+
= 1,
2, be
Let ni:
+Aut(XZ
Aut(Xi),),ii =
1,2,
be 6-representations.
&-representations.Define
Defineaa G-morphism
G-morphism
a:
X1
-+
X2
to
be
a
morphism
a
Of
X1
to
X2
which
commutes
with
the action
a : X1 + X2 to be a morphism a of X1 X2
commutes
of G in the sense that (gnl)a
(gal)a ==a(g7r2)
a(gnz)for
foreach
eachggEEG.G.Write
WriteMorG(X1,
Morc(X1,X2)
X2)
Notice that
that the composition
composition of
of GGfor the set of G-morphisms
G-morphisms of X1 to X2. Notice
morphisms
morphisms is a G-morphism
G-morphism and the
the identity
identity morphism
morphism is aa G-morphism.
G-morphism.
Similarly define a G-isomorphism
G-isomorphism to
to be
be a G-morphism which is also an isomorphism. Notice the G-isomorphisms
G-isomorphisms are the equivalences of
of representations
representations
morphism.
of G.
One focus
focus of this book is the decomposition
decomposition of
representation 7r
into smaller
n into
smaller
One
of aa representation
representations. Under
representations.
Under suitable
suitablefiniteness
finiteness conditions
conditions (which
(which are always present
here) this process of decomposition
in the representations
representations considered here)
decomposition must
n certain
certain indecomposable
indecomposable
terminate, at which point we
we have
have associated
associated to
to Tr
Preliminary results
10
10
or irreducible
irreducible representations
representations which cannot be broken
broken down
down further.
further. ItIt will
will
develop
develop that the indecomposables
indecomposablesassociated
associated to
to rr
n are
aredetermined
determinedup
up to
to equivequivalence.
alence. Thus
Thus we are
are reduced
reduced to a consideration of indecomposable
indecomposablerepresentarepresentations.
tions.
In general indecomposables
indecomposablesare not irreducible,
irreducible, so an indecomposable
indecomposable representation itn can
can be
be broken
broken down
down further,
further, and we can associate to rr
n a set of
resentation
irreducible constituents. Sometimes these irreducible constituents
constituents are
are deterdetermined
mined up
up to
to equivalence,
equivalence,and
and sometimes
sometimesnot.
not. Even
Even when
when the
the irreducible
irreducibleconconstituentsare
are determined,
determined,they usually
usually do not determine
n . Thus
Thus we will also be
stituents
determine jr.
concerned
concerned with
with the
the extension
extension problem: Given a set SS of irreducible
irreducible representations,
tations, which
which representations
representationshave
have SSas
astheir
their set
setof
ofirreducible
irreducibleconstituents?
constituents?
There
Thereis
is also
alsothe
the problem
problemof
of determining
determiningthe
theirreducible
irreducibleand
andindecomposable
indecomposable
representations
representationsof
of the
the group.
group.
ItIt isis possible
Expossibleto
to give
giveaacategorical
categoricaldefinition
definition of
of indecomposability
indecomposability (cf.
(cf. Exercise
ercise 1.5).
1.5).There
There is
is also
also aauniform
uniformdefinition
definition of
of irreducibility
irreducibilityfor
for the
the classes
classes
of
of representations
representationsconsidered
considered most
most frequently
frequently(cf.
(cf.Exercise
Exercise1.6).
1.6).IIhave
havechochosen
sen however
however to relegate
relegate these
these definitions
definitions to the
the exercises
exercises and
and to
to make
make the
the
appropriate definitions of indecomposability
indecomposability and irreducibility
irreducibility for
for each
each catcatappropriate
egory in the
the chapter
chapter discussing
discussing the
the elementary
elementary representation theory
egory
theory of the
category.
category. This process
process begins
begins in
in the
the next
next chapter,
chapter, which discusses
discussespermutation
permutation
representations.
representations.
However
particular interest.
interest. A representation
representation of a group G on
However one case is of particular
itself via
via conjugation
conjugation (in
(in the
the category
category of
of groups
groups and
and group
grouphomomorphisms)
homomorphisms)
is
is irreducible
irreducible if G
G possesses
possessesno
no nonidentity
nonidentity proper
proper normal
normal subgroups.
subgroups. In
In this
this
case
case G
G isis said
saidto
to be
be simple.
simple.To
To my mind the simple
simple groups
groups and
and their
their irreducible
irreducible
linear and
and permutation
permutation representations
representationsare
are the
the center
center of
of interest
interestin
in finite
finite group
group
linear
theory.
theory.
Exercises
Exercises for
for chapter
chapter11
1.
-subgroups
1.Let
LetGGbe
beaafinite
finitegroup,
group,jrn aaset
setofofprimes,
primes,052the
theset
setofofnormal
normal7rn-subgroups
of
of G,
G, and rrthe
theset
setof
of normal
normal subgroups
subgroupsX
X of G with G/X aa rr-group.
n-group. Prove
(1)
E 0.
Hence
(S2)
isisthe
thenHK
HK
E a.
Hence
(a)
theunique
uniquemaximal
maximalmember
member
(1) IfIfH,
H,KKEE0 then
of0.
of a.
(2)
(2) IfIf H,
H,KK EE Pr then
then HHflnKK EEP.r.Hence
Hence IHEP H
Hisisthe
theunique
uniqueminimal
minimal
member
member of P.
r.
2.
bebethethecategory
2. Let
Let e,8,
categoryofofsets
setsand
andfunctions,
functions, 82
the category
category of
of vecvec'2 the
tor
ti?, the
thecategory
category of
of groups
groupsand
and
tor spaces
spaces and
and linear
linear transformations,
transformations, and 03
homomorphisms.
LetFF =
= (A,:
< i <_( n)
homomorphisms. Let
(Ai: 1 _(
n) be
be aa family
family of
of objects
objects in 1k.
gk.
Prove
Prove
(1)
Thenthe
thecoproduct
coproduct C
C of
of FFisisthe
thedisjoint
disjointunion
unionof
ofthe
the
(1) Let
Let kk ==1.1.Then
sets A,
Aiwith
with c1:
ci:AA,i+
inclusion map.
sets
-). C the inclusion
map. The
The product
productPP of
of FF isis
I
nH,,
.
representations
Abstract representations
11
11
product Al
. .-xxAn
A, with
with pi:
pi:PP-+
+AiAithe
theprojection
projectionmap
map
the set product
Al x ...
pi :(al, .. .,a,) H ai.
Pi:(a1,...,an)Hai.
Let k =
=2.
2. Then
Then C =
=PP ==i @:='=,Ai
is the
direct
sum
thesubspaces
subspaces
(2) Let
1 Ai is the
direct
sum
ofofthe
,
0)
and
pi: P
Ai,
Ai +
-). C defined by aici
ai ci =
= (0,
Ai, with
with ci
ci::Ai
(0,.. .. ..,,ai,
ai,...... , 0 )
P+
Ai the
the projection
projection map.
map.
Ai
Letkk==3.3.Then
Thenthe
theproduct
productPPofofFFisisthe
thedirect
directproduct
productAAl1 xx ...... xx An.
A,.
(3) Let
Ai the
the projection
projection map.
map. (The
(The coproduct
coproduct turns out
out to
to be
be
with pi: P + Ai
so-called free
free product
product of
of the
the family.)
family.)
the so-called
3. Let
G,
I
G,and
andfor
forg gEEGGdefine
defineg7r:
gn: H -).
+ H by
by
Let G
G be
be aagroup,
group,HH<I
x(gn)==xg,
xg,xxEEH.H.Let
Let-'&
categoryofofgroups
groupsand
andhomomorphisms.
homomorphisms.
x(gir)
bebe
thethecategory
Prove
CG(H). itn isis the
the reprerepreProve nit is
is aa -'-representation
&-representation of
of G
G with
with kernel
kernel CG(H).
sentation by
by conjugation
conjugationof
of GG on
on H.
H. If
If H ==G,
the
image
of
G
under
G, the image of G under itn
the inner
inner automorphism
automorphism group of G
G and
and isisdenoted
denoted by
by Inn(G).
Inn(G). Prove
Prove
is the
Inn(G) <
<Aut(G).
I Aut(G). Define
Define Out(G) =
=Aut(G)/Inn(G)
Aut(G)/Inn(G) to be
be the
the outer
outer autoautoG.
morphism group of G.
4. Let
C
Let-'6'bebea acategory,
category,FF==(A1:
(Ai:11<5i i<5n)n)a afamily
familyof
ofobjects
objects in
in -t',
8,
C and
and
P the
the coproduct
coproductand
andproduct
productof
ofthe
thefamily
familywith
withcanonical
canonicalmaps
mapscici:: Ai
Ai + C
C
For ai
ai E
ai to
P + Ai, respectively. For
E Aut(Ai) define Ei
to be
be the
the unique
unique
and pi: P
member of Mor(C,
Mor(C, C) with
with ai
aicici =
= ciEi
ci ai and
andccij =
= ci
ai for
cjEi
for all i # j.
j. Define
Define
ai
Ei EE Mor(P,
Mor(P,P)
P )dually.
dually.Prove
Prove the
the map
n
0: fAut(Ai)
Aut(X)
i=1
(a1,...,CIO
(a1, . . . , a,) H
Ha1
El ...
. . .an
En
is an injective group homomorphism
homomorphismfor
forXX=
= C and P.
P.
Exercise 1.2,
letn7r:
-> Aut(X)
5. Assume
Assumethe
thehypothesis
hypothesisand
and notation
notationof
of Exercise
1.2, and let
: GG+
be a gk-representation,
-'k-representation,where
whereXX==CCififkk=
= 11 or
or2,2,and
andXX== P
P if kk =
= 3.
3.
Prove
Prove the following
following are
are equivalent:
equivalent:
Thereexist
exist-1'k-representations
&k-representationsr1:
ni: G
G + Aut(Ai
Aut(Ai),), 115< i <5 n,
n, such
such that
(a) There
n = $4, where 4
is the injection
injection of
of Exercise
Exercise 1.4
1.4 and
0 is
7r
n
n
definedby
byg$
g* =
= (girl
$ : G + F1 Aut(Ai) is defined
(gnl,, ...
. . ., ,gJrn).
gn,).
i-1
decomposable.
(b) jrnisisdecomposable.
(If k ==1,1,transitivity
transitivityisisthe
thesame
sameasasindecomposability.
indecomposability. See
See chapter
chapter 2 for
the definition
See chapters
chapters 55 and 4 for
definition of transitivity.
transitivity. See
for the
the definitions
definitions of
of
decomposability when
when kk =
= 2 and 3.)
6. Assume
Assume the
the hypothesis
hypothesis and
and notation of Exercise 1.2,
1.2, let X be an
an object in
Aut(X)
a
-'k-representation.
A
-ik-equivalence
-'k,
-ek,and
and p:
p: G
G+
&k-representation. A &k-equivalencerelation
relation
that - isispreserved
on X isis an
an equivalence
equivalence relation
relation on X such that
preserved by
by the
the
operationson
onXX ifif kk =
= 22or
or 33(i.e.
(i.e. ifif yyisisan
ann-ary
n-aryoperation
operation on
on X
X and
and
operations
-
-
12
12
Preliminary results
results
Preliminary
-
xi ^-- yi
yi then
then y(xl,
y(x1,..... ., ,xn)
x,) ---y(yl,...
y(y1,., .y,,)).
. ,y,)).Define
DefineGrr
G ntotopreserve
preserve ^ifif xx^- yyimplies
impliesxgTr
xgn -- yglr
ygn for
for each
each g cE G.
G.Prove
Prove that
that (a)
(a) and
and (b)
(b) are
are
equivalent:
equivalent:
(a)
relation on
on X.
X.
(a) Gzr
G npreserves
preservesno
nonontrivial
nontrivial gk-equivalence
k-equivalence relation
(b)
is
an
irreducible
-'k-representation.
(b) is an irreducible gk-representation.
(See chapters 5 and
and 44 for
for the
the definition
definition of
of an
an irreducible
irreducible -'k-repregk-representation when
when kk ==22and
and3.3.AAwl&l-representation
irreducibleififititisis
sentation
1-representation isisirreducible
primitive,
primitive, and
and primitivity
primitivity isisdefined
definedin
in chapter
chapter2.)
2.)
7.
or:
7. Let
Letn,n,
0 :G
G -*
+Aut(X)
Aut(X) be
befaithful
faithful-'-representations.
&-representations.Prove
Prove rrn isisquasiquasiequivalent to or
o ifif and
n isis conjugate
in Aut(X).
Aut(X).
equivalent
and only ifif G
G7r
conjugate to
to G
Gao in
8.
(Gi
: i iEEI)I)
a family
8. Let
Let GGbe
be aa group
group and 99==
(Gi:
a familyofofsubgroups
subgroupsof
ofG.
G.Prove
Prove
(1)
is connected
(J9,_).
(1) The
Thegeometry
geometry F
r ==1'(G,
r(G,O5F)
is connectedif ifand
andonly
onlyififGG==
(F).
(2)
- Frby
(2) For
For g cE G,
G, define
define ggir:
n : rF +
by(Gix)g'r
(Gix)gn==Gixg.
Gixg.Prove
Proveitnisisaarerepresentation
r.
presentation of
of G
Gas
asaagroup
groupof
of automorphisms
automorphismsof
of F.
-
Permutation representations
develops the elementary theory
theory of
of permutation
permutation representations.
representations. The
The
Section 5 develops
foundation for this theory is the notion of the transitive permutation represenrepresentation. The
The transitive
transitive representations
representations play the role of
of the
the indecomposables
indecomposables in
in
the theory.
theory. It will develop that
that every
every transitive
transitive permutation
permutation representation
representation of
of a
group G is
is equivalent
equivalent to a representation by right multiplication on the set of
of G. Hence the study
study of
of permutation
permutation representations
representations
cosets of some subgroup of
of G is equivalent to the study of the subgroup structure of G.
Section 6 is devoted
devoted to aa proof of
of Sylow's
Sylow's Theorem.
Theorem. The
The proof
proof supplies
supplies aa
nice application of the techniques developed in section 5. Sylow's
Sylow's Theorem is
one of the most important results in finite group theory. It is the first
first theorem
in the local theory of finite groups. The local theory studies
studies a finite group from
the point of view of its
its p-subgroups
p-subgroups and the normalizers of these
these p-subgroups.
5 Permutation
Permutationrepresentations
representations
In this section X is a set, G a group, and n:
7r:GG +
-* Sym(X)
Sym(X)isis aa permutation
permutation reprepresentation of G. Recall Sym(X) is the symmetric group on X; that is Sym(X)
Sym(X)
is the group of all permutations
permutations of X. Thus Sym(X)
Sym(X) is the automorphism group
of X in the
in that
that
of
the category
category of sets
sets and
and functions,
functions, and 7r
n is a representation
representation in
category.
For
X and
and aaEESym(X)
Sym(X) write
write xa
x a for
for the
the image
image of x under
under aa.. Notice
Notice
For xx cE X
that, by definition of multiplication in Sym(X):
= (xa)p
x(ap) =
x E X,
aa,, Pp EE Sym(X).
Sym(X).
I'll often
and write
writexg
xg for
forx(gn),
x(g7r),xx E
E X, g E G.
often suppress
suppress the representation 7r
n and
G.
One feature of this notation is that:
x(gh) = (xg)h
x E X,
g, h E G.
The relation
relation -- on
by xx -- yyififand
on X
X defined
defined by
andonly
only ifif there
there exists g eE G
G with
with
xg ==yyisisan
anequivalence
equivalencerelation
relation on
onX.
X.The
Theequivalence
equivalenceclass
class of
of xx under
under this
this
relation is
xG=fxg:geG}
and is called the orbit of xx under
under G. As
As the
the equivalence
equivalence classes
classes of an
anequivalence
equivalence
relation partition a set, X is partitioned by
by the
the orbits
orbits of
of G on
on X.
Permutation representations
14
be a subset of
of X. G is said to act on Y
Let Y
Y be
Y if Y
Y is a union of orbits of G.
Notice G acts
when yg
yg E Y
acts on Y precisely when
Y for each y E Y, and each gg E G.
G.
Further
IY is
is a permutation
permutation of Y
Y for each g EE G,
Further if G
G acts
actson
onYY then
thenggly
G, and
andthe
the
restriction map
map
G -f
+Sym(Y)
Sym(Y)
gHgly
g I+ glr
is aa permutation representation
representationwith
with kernel
kernel
Gy={g(=-G:yg=y
G y = { g € G : y g = y forall yEY}.
~ E Y ) .
Hence G
Gyy <
L] G when
when G
G acts
acts on
on Y.
Y. Even
Even when
when G does
does not
not act
act on
on Y,
Y, we
we can
can
consider
consider
G(Y) =
= {g
(g EE G:
G: Yg
Yg =
=Y},
Y),
where
yg: yy EE Y}.
subgroups of G called
pointwise
where Yg =={(yg:
Y). GY
Gy and G(Y)
G(Y) are
are subgroups
called the
thepointwise
stabilizer of
of Y
Y in
in G,
G,respectively.
respectively. G(Y)
G(Y) isis
of Y in
stabilizer of
in G and the global stabilizer
the largest subgroup
G~ for the image
image of G(Y)
G(Y) under
under
subgroup of G
G acting
acting on
on Y.
Y. Write
Write GY
the restriction
restriction map
map on
on Y.
Y. We
We have
have seen
seenthat:
that:
Therestriction
restrictionmap
mapof
of G(Y)
G(Y)on
onYY is
is aapermutationrepresentation
of G(Y)
G(Y)
(5.1) The
permutation representation of
with kernel
kernel Gy
Gy and image GY2
- G(Y)/Gy,
G(Y)/Gy,for
foreach
eachsubset
subsetYY of
of X.
X.
x EE X write Gx
G, for G{x}.
G{,). Next
For X
Nextfor
forSSGc G define
Fix(S) =
= xx
={x
{xEE X:
X: xs =
ffor
o r all
all
s EE S}.
S).
Fix(S)
fixed points of S. Notice Fix(S) = Fix((S)).
Fix((S)). Also
Fix(S) is
is the
the set
set of
offuedpoints
(5.2) If
thenGGacts
actson
onFix(H).
Fix(H).More
Moregenerally
generallyGGpermutes
permutesthe
the orbits
orbits
If H
H a9GGthen
H of
of cardinality
cardinality c,
c, for
for each
each c.
c.
of H
X, I'll
I'll sometimes
sometimeswrite
writeCG(Y)
CG(Y)and
and NG(Y)
NG(Y)for Gy
G y and
and G(Y),
G(Y),respecrespecFor Y C X,
Cx(G)
for Fix(G).
Fix(G). Usually
Usually this
this notation
notationwill
will
tively, and
and I'll
I'll sometimes write Cx(G)
for
possesses aa group
group structure
structure preserved by G.
be used only when X possesses
(5.3)
Let P be the set of all subsets
of X.
X. Then
Thenaa:: GG+
- Sym(P)
(5.3) Let
subsets of
Sym(P) isisaaperperP where
where ga:
ga:YY + Yg
Yg for each g E G and
and
mutation representation
mutation
representation of
of G
G on P
YcX.
Y E X.
IT is aa transitive
transitive permutation
permutation representation
representation ifif G
orbit on
on X;
X;
7r
G has just
just one orbit
=y.
y. GGwill
willalso
also be
be
equivalently for
for each
eachx,
x, yy E X there exists
exists gg E
E G with xg
xg =
said to
to be
be transitive
transitive on
on X.
X.
said
Permutation representations
15
Here's one
one way
way to generate
generate transitive representations
representations of G:
G:
(5.4) Let H <5 G.
Sym(G/H)
is is
a transitive
G.Then
Then a:
a :G
G -+
Sym(G/H)
a transitivepermutation
permutationreprerepresentation of
of G on the coset
coset space
space G/H,
G/H, where
ga:
ga:Hx
Hx i-+
I+ Hxg.
Hxg.
multiplication. H is
is the
the representation
representationof
of G
Gon
on the
the cosets
cosets of H by right multiplication.
is the
the
a is
stabilizer of the coset H in
in this
this representation.
representation.
We'll
of G
G is equivalent
equivalent to a
We'll soon
soon see
see that
that every
every transitive
transitive representation
representation of
of G by right multiplication on the cosets of some
representation of
some subgroup.
subgroup.
permutation representations
representations of
of G.
But first here is another way to generate permutation
Sym(G)
isisa apermutation
Sym(G)
permutationrepresentation
representationof
of G
G on itself,
(5.5) The map aa:: G -+
where
ga:xHxg
ga: x I+ xg x,gEG.
x , g E G.
a isis the
by conjugation.
conjugation.For
ForSS c G,
G ,the
theglobal
global
the representation
representation of
of G on itself by
stabilizer of S in G
G is
is NG(S),
NG(S),while CG(S)
CG(S)is the pointwise stabilizer of S.
of Exercise
Exercise 1.3.
1.3.Recall
RecallNG(S)
NG(S)=
=
Notice that 5.5 is essentially a consequence of
(g
CG(S)=
= (g
sg =
{g E
E G:
G:Sg
Sg =
=S},
S ) ,Sg
Sg =
=(sg:
{sg:s EE S},
S), and CG(S)
{g E
E G:
G:sg
=ss for
for all
all s EE
S}. By
By 5.3,
5.3, G is also
S].
also represented
represented on the power set of G,
G , and
and evidently,
evidently, for
S C G,
G ,the
the set
set SG
sG==(S9:
{Sg:ggEEG}
G )of
of conjugates
conjugatesof
of SS under
under G
G is
is the
the orbit
orbit of S
under G with respect to this representation.
representation.
(5.6) Let Y
Y CX
X and
and gg EE G.
G.Then
Then G(Yg)
G(Yg)==G(Y)9
G(Y)gand
and Gyg
Gya ==(GY)g.
(Gy)g.
transitiverepresentation
representationand
andlet
letxx EE X
X and
and H
H=
= Gx
(5.7) Assume
Assume 7r
n is aa transitive
G,.. Then
ker(7r) = n Hg = kerH(G)
gEG
normal subgroup
subgroup of
of G
G contained
contained in
in H.
H.
is the largest normal
it is aatransitive
transitivepermutation
permutationrepresentation,
representation,
E X,
G,
(5.8) Assume n
letlet
x Ex X,
HH
==G,,
of H by right multiplication.
and let aa be the representation of G on the cosets of
Define
Define
8:G/H
G/H+
-X
B:
X
Hg i-+
I-+ xg.
Then P
B isis an
an equivalence
equivalence of the permutation
permutation representations
representations aa and
and 7r.
n.
representations
Permutation representations
16
Proof. We
We must
must show
show ,B
/3 is
with X
X and
and that,
that,
Proof.
is a well-defined
well-defined bijection
bijection of
of G/H
G/H with
for each aa,, g E G,
G, (Ha)O(ga)
(Ha)B(ga) ==(Ha)gnf.
(Ha)gnp.
Both
computationsare
arestraightstraightfor
Both
computations
forward.
forward.
(5.9) (1)
(1)Every
Every transitive
transitive permutation representation of
is equivalent
equivalent to a
(5.9)
of G is
representation
representation of
of GGby
byright
rightmultiplication
multiplicationon
on the
thecosets
cosets of
of some
somesubgroup.
subgroup.
n': G
G ->
+=Sym(X')
Sym(X1)and
andrrnare
aretransitive
transitiverepresentations,
representations, xx E
E X, and
and
(2) IfIf jr':
x' EE X',
X', then
thenn7tisisequivalent
equivalenttotojr'
n'ififand
andonly
onlyififGx
G, isisconjugate
conjugate to
to G,,,
G,t in G.
G.
Proof.
If,B:
8: X
X +=
-> X'
Proof.Part
Part(1)
(1)follows
followsfrom
from 5.8.
5.8. Assume
Assume the
the hypothesis of (2). If
is an equivalence of n
jr and n'
and,
is
n' then
thenGx
G, ==GxA
GXB
and,by
by 5.6,
5.6, Gxp
GXB
is conjugate
conjugate to
Gx,
Gx is
G,? in G. Conversely if G,I
is conjugate
conjugate to
to Gx
G, in G,
G, then by 5.6 there is y EE X
X
with
with Gy
G, ==Gx'
G,Iand
andby
by5.8
5.8both
bothjrnand
andjr'n'are
areequivalent
equivalentto
tothe
therepresentation
representation
of
G,. and
andhence
henceequivalent
equivalentto
to each
each other.
other.
of G
G on
onthe
thecosets
cosetsof
of Gx'
Let
of n to
I ) be
be the
the orbits
orbits of G on
on X
X and
and ni
ni the
the restriction
restriction of
to Xi.
Xi. By
By
Let (Xi: i E I)
q isisaapermutation
permutation representation
representation of
Xiis an
an orbit
orbit of
5.1, ni
5.1,
of G
G on
on Xi
Xi and, as Xi
G,
G, ni
ni is
iseven
evenaatransitive
transitiverepresentation.
representation. (ni:
(ni:ii c=
E II)
) is the family of transitive
transitive
q.
Evidently:
constituents of
constituents
of nn and we
we write
writenn =
= Eic] ni.
Evidently:
xi,,
(5.10)
(xi:: i E I)
I)of
of nnare
areuniquely
uniquelydetermined
determinedby
by
(5.10) The
Thetransitive
transitiveconstituents
constituents (ni
n,
n ,and
andififn'n'isisa apermutation
permutationrepresentation
representationofofGGwith
withtransitive
transitiveconstituents
constituents
(nj:j jE EJ),J),
thenn n
equivalenttoton'n'if ifand
andonly
onlyififthere
thereisisaabijection
bijection aa! of
of II
(jr:
then
is is
equivalent
with
with JJsuch
suchthat
that nta
ni6(isisequivalent
equivalent to ni for each i E I.
I.
So
So the
the study
study of
of permutation
permutation representations
representationsis
is effectively
effectivelyreduced
reduced to
to the
the study
study
of
of transitive
transitiverepresentations,
representations,and
and5.9
5.9says
saysininturn
turnthat
thatthe
thetransitive
transitivepermutation
permutation
representations
representationsof
of aa group
group are
are determined
determined by
by its
its subgroup
subgroup structure.
structure.
The transitive
transitiverepresentations
representationsplay
play the
the role
role of
of the
the indecomposable
indecomposable permutapermutaThe
tion
tion representations.
representations.For
Forexample
examplesee
seeExercise
Exercise1.5.
1.5.
transitiveon
onXXthen
thenXXhas
hascardinality
cardinalityIG
IG: :Gx
G, IJ for each x EE X.
X.
(5.11) IfIfGGisistransitive
(5.11)
Proof.
Proof.This
Thisisisaaconsequence
consequenceof
of 5.8.
5.8.
(5.12)
G.Then
ThenSShas
hasexactly
exactlyIG
J GNG(S)I
:NG(S)Iconjugates
conjugates in
in G.
G.
(5.12) Let
Let SS CGG.
Proof.
Proof.We
Weobserved
observedearlier
earlierthat
thatGGisistransitively
transitivelyrepresented
representedon
onthe
theset
setSG
SGof
of
viaconjugation,
conjugation,while,
while,by
by5.5,
5.5,NG(S)
NG(S)isisthe
thestabilizer
stabilizerof
ofSSwith
with
conjugatesof
ofSSvia
conjugates
respect
respectto
tothis
thisrepresentation,
representation,so
sothe
thelemma
lemmafollows
followsfrom
from5.11.
5.11.
Permutation representations
representations
17
Let pp be
be aaprime
primeand
and recall
recall that
that aa p-group
p-group is
is aa group
group whose
whose order
order is
is aa power
power
of p.
p.
(5.13) If G is aa p-group
power of
of p.
p.
(5.13)
p-group then
then all
all orbits of G on X have order a power
Proof. This
This follows
follows from
from 5.11
5.1 1 and
and the
the fact
fact that
that the
the index
index of any subgroup of
G divides the order of G.
G.
-
(5.14) Let
X is
is finite.
finite.Then
Then1IXI
x1 - (Fix(G)I
IFix(G)I mod p.
(5.14)
Let G
G be a p-group and assume X
Proof. As
5.14 follows
Proof.
Asthe
thefixed
fixedpoints
pointsof
ofGGare
areits
itsorbits
orbitsof
of length
length1,1,5.14
followsfrom
from5.13.
5.13.
Here are a couple
couple applications of 5.14:
(5.15) Let
Let G
G and
and H be p-groups
(5.15)
p-groups with
with H
group homomorphism.Then
ThenCH
CH(G)
(G) #; 1.1.
and let
let a:
a: G -+ Aut(H)
Aut(H) be a
# 1 and
Proof. aa isisalso
of G on H.
Proof.
alsoaapermutation
permutation representation
representation of
H .By
By5.14,
5.14,CHI
(HI -.
JFix(G)I
mod
p.
But
Fix(G)
=
CH(G)
in
this
representation,
while
IHI
IFix(G)I mod p. But Fix(G) =
in this representation, while IHI =-- 00
claimed.
mod p as H is
is aa p-group with H # 1. So CH(G)
Cn(G) # 1, as claimed.
(5.16) If G is a p-group with G # 1, then Z(G) # 1.1.
(5.16)
Proof. Apply
Proof.
Apply5.15
5.15totothe
therepresentation
representationof
of GGon
onitself
itself by conjugation
conjugation and recall
Z(G) =
=CG(G).
CG(G).
The
Sylow's
The following
followingtechnical
technicallemma
lemmawill
will be
beused
used in
in the
the next section
section to
to prove Sylow's
Theorem:
(5.17) Let
for each
each xx E
E X that there exists aa p-subgroup
p-subgroup
(5.17)
Let X
X be
be finite
finite and assume for
P(x)
= Fix(P(x)).
Fix(P(x)). Then
P(x) of
of G
G such
such that {x}
{x) =
(1) GGisistransitive
transitive on
on X,
X, and
and
XIzz- 11 modp.
mod p.
(2) 1/XI
Proof. Let X ==YY++ZZbebea apartition
Proof.
partitionofofXXwith
with GGacting
acting on
on Y
Y and
and Z.
Z. Let
Let
0 and
V=
=Y
Y#0
and pick
pick y EE Y.
Y. For V
Y or Z and H <_(GGdenote
denoteby
by Fixv(H)
Fixv(H)the
the
fixed points
points of
of H
H on
on V.
V. By
By hypothesis
hypothesis(y)
(y}== Fix(P(y)),
Fix(P(y)), so 1 =
= (Fixy(P(y))I
IFixy(P(y))l
0modp.But
and0=
and
0 =IFix,(P(y))I.Hence,
IFix,(P(y))l. Hence,by5.14,IY)
by 5.14, IYI r 1modpandIZI
1 mod p and I ZI -. 0
mod p. But
mod p, aa contradiction.
if ZZ isisnonempty,
I Y[YI
I r 1 mod
nonempty,then,
then,bybysymmetry,
symmetry,
contradiction. Thus
Thus
18
Permutation representations
Y=
=X
Y
X and,
and,as
asIYI
IY 1 =E 1 mod p, (2)
(2) holds. Since
Since we could have chosen Y
Y to be
an orbit of G on X,
(1)
holds.
X , (1) holds.
be aa partition
partition of X.
X . QQ isisG-invariant
G-invariantifif GGpermutes
permutes the
the members
members of
Let Q be
Q.
regard QQ as
as aa subset
subset of
of the
the power
powerset
set PP of
of X
X and represent
Q . Equivalently
Equivalently regard
G on PP as
subset Q
Q of
of P with
asin
in 5.3;
5.3; then
then Q
Q isis G-invariant
G-invariant if G acts on the subset
with
that ifif Q
Q is G-invariant then
respect to this representation. In particular notice that
permutationrepresentation
representationofofGGon
onQQ.. QQ is nontrivial
nontrivial ifif
there is a natural
natural permutation
Q0{{x}:xEX)andQ0{X}.
Q
# { I x ) : x E XI and Q # 1x1.
X.. G is imprimitive on X
X if there exists a nontrivial
nontrivial GLet G be transitive on X
invariant
Q of
of X.
X .In
Inthis
thisevent
eventQQisissaid
saidtotobe
beaasystem
systemof
of imprimitivity
imprimitivity
invariant partition Q
for G on X.
X.G
G isis primitive on
on X
X ifif itit isis transitive
transitive and
and not
not imprimitive.
imprimitive.
(5.18)
X.
(5.18) Let G be transitive on XX and yy E X.
If Q
of imprimitivity
imprimitivityfor
for G
G on
on XX and
and yy EE Y
E Q,
(1) If
Q is a system of
Y E
Q , then G
G is
is
on Q
Q,, the stabilizer H of
of Y
Y in G
G is
is aa proper subgroup
subgroup of G
G properly
properly
transitive on
containing G,,
Gy,YYisis an
an orbit
orbit of
of H
H on
= IG
HI, and
containing
on X,
X , IXI
1x1 ==IYIIQI,
IYllQl, IQI
lQl =
I G : HI,
IYl
=
I
H
:
Gyl.
IYI = IH:GyI.
If G
Gy
{Yg:
(2) If
y<
<H
H <<GGthen
thenQQ= =
{Yg:ggEEG)
G)isisaasystem
systemof
ofimprimitivity
imprimitivity for
X,, where YY =
= yH and
G on X
andHH isisthe
thestabilizer
stabilizerof
of Y
Y in G.
The proof is left as an
an exercise.
exercise. As
As aa direct
direct consequence
consequenceof
of 5.18
5.18we
wehave:
have:
(5.19)
(5.19) Let G be transitive on X
X and xX EE XX.. Then
Then G is primitive
primitive on
on XX if and
only if Gx
G, is a maximal subgroup
subgroup of G.
on XX,, let I1xX
andlet
letxx EE XX.. Then
Then there is
is
Let G be finite and transitive on
1 I>>1,1,and
a sequence
Gx==Ho
Ho(<HI
Hl 5< .....<(HH,,= =
GG
with
withH,Himaximal
maximalininHi+i.
Hi+l.This
This
sequence G,
gives rise to a family of primitive permutation representations:
representations: the representations of Hi+I
Hi+l on the cosets of Hi.
Hi. This
This family of primitive representations can
be used to investigate the
the representation
representation n
it of G on
on X.
X.
play the
the role
role of
of irreirreFrom this point of view the primitive representations play
ducible
ducible permutation representations.
representations.See
See also
also Exercise
Exercise 1.6.
1.6.
I close this section with two useful
useful lemmas.
lemmas. The proofs are
are left
left as
as exercises.
exercises.
(5.20) Let G be transitive on
, and
is transitive
transitive on
(5.20)
on XX,, xXEEXX,
andHH 5
< G. Then H is
X
and only
onlyififGG =
= G,
GxH.
X if and
H.
(5.21)
transitiveon
onXX,
=G
Gx, and
and K
K5
< H.
(5.21) Let G be transitive
, xXEEXX,
, HH=
H.Then
Then NG(K)
NG(K)is
transitiveon
onFix(K)
Fix(K)ififand
andonly
onlyififlCG
KGnflHH== KK".
transitive
H.
.
Sylow's Theorem
19
Sylow's Theorem
6 Sylow's
group. IfIf nn is
is aa positive
positiveinteger
integer and
andpp aa prime,
prime, write
write
In this section G is a finite group.
np for the
the highest power
power of
of pp dividing n.
n. n,
np is
is the
thep-part
p-part of
of n.
n,
A Sylow
p-subgroup of
I GI GI I,.p. Write
Write Sy1p(G)
Syl, (G)
Sylow p-subgroup
of GGisisaasubgroup
subgroupofofGGofoforder
order
for the set of Sylow p-subgroups of G.
G.
section we
we prove:
prove:
In this section
Theorem. Let
Then
Sylow's Theorem.
LetGGbe
be aa finite
finite group and p a prime. Then
(1) Sylp(G)
Syl,(G) isisnonempty.
nonempty.
(2) GGacts
actstransitively
transitively on
on Sylp(G)
Syl,(G) via
via conjugation.
conjugation.
G : NG(P)I
NG(P)Ir
- 11 mod
(3) ISylp(G)I
ISyl,(G)I =
=IIG
mod p for
for PP EE Sylp(G).
Syl,(G).
(4) Every
Every p-subgroup
p-subgroupof
of GGisiscontained
contained in
in some
some Sylow
Sylow p-subgroup of G.
G.
Let P
of G
G and
and Q
0 the
r be
be the
the set
set of all
all p-subgroups
p-subgroups of
the set
set of
of all
all maximal
maximal p-
subgroups
partially order
order rr by
0 be
subgroups of G; that is partially
by inclusion
inclusion and let Q
be the
the maximal
maximal
members
of this partially
partially ordered
ordered set.
set. ItIt follows
follows from
from 5.3
5.3 and
and 5.5
5.5 that
that G
G is
members of
represented as
as a permutation group via conjugation on the power set of G, and
represented
it is evident that G acts on r,
F, Q,
2, and
representation.
and Sy1p(G)
Sylp(G)with respect to this representation.
Let R EE 0.
Q.Claim
ClaimRRisisthe
theunique
unique point
point of 0Qfixed
fixed by
by the subgroup R of G.
For ifif R fixes
fixes Q
Q E 0Qthen,
(Q), so, by
by 1.7.2,
1.7.2, RQ
RQ 5
< G and,
then, by
by 5.5,
5.5, R
R <5NG
NG(Q),
and,
by
IRQI== IRIIQIIIR
IRIIQI/IRnnQl.
Q.Thus
ThusRQ
RQEEr,
F,so,
so,asasRR5< RQ
RQ 2
? QQ and
by 1.7.3,
1.7.3, [RQl
and
we conclude
concludeRR =
=R
Q=
= Q.
R, Q EE S2,
C2, we
RQ
Q.So
Sothe
theclaim
claimisis established.
established.
that for
for each
each RR EE Q
0 there is a p-subgroup P(R)
I've shown that
P(R) of
of G
G such
such that
unique point
pointof
of Q
0 fixed
by P(R); namely
R=
= P(R).
fixed by
namely R
P(R). So
Soititfollows
follows
R is the unique
from 5.17 that:
(i) G
0, and
G isis transitive
transitive on Q,
and
(ii)
p.
(ii) IQ
1Q1 I-1mod
E 1 modp.
ThenPP EE Syl,(G),
Sylp (G), so, as G is transitive
Let P EE 0Qand
andsuppose
supposeI I PP I=
I =I IGGII,. p.Then
(G), we
(G). On
on 0Qand
andGGacts
actsononSylp
Syl,(G),
wehave
haveS2C2CESylp
Syl,(G).
Onthe
theother
otherhand
handasasI R
I RII
divides
GJfor
I foreach
eachR RE EQ0ititisisclear
clearSyl,(G)
Sylp(G)Gc 0,
= Sylp(G).
divides JI G
Q,so
soSZ
Q=
Syl,(G). Thus
Thus
(i) implies parts (1) and (2)
(2) of
of Sylow's
Sylow's Theorem,
Theorem, while
while (ii)
(ii) and
and 5.12
5.12 imply
imply
part (3). Evidently
Evidently each
eachmember
memberofofrr is contained in
in aa member
member of
of Q,
0, so, as
part
SZ== Syl,(G),
Sylp(G), the fourth part of Sylow's Theorem
Q
Theorem holds.
holds.
Theorem it remains to show IIPI
PI =
= IGI
So to complete the proof of Sylow's Theorem
I G 1,p
for P EE 0.
and let
let M
M ==NA(P).
for
Q.Assume
Assume otherwise
otherwise and
NG(P).By
By(i),
(i),(ii),
(ii),and
and 5.12,
5.12,
I G:MI
:MG
I- 11mod
M I p=
G I p,and
andhence
hencepp divides I[MIPI.
M/P I. Therefore,
Therefore,
IG
modp,p,sosoI /MIp
=I IGl,,
(Exercise2.3),
2.3),there
thereexists
existsaasubgroup
subgroupRIP
R/PofofMIP
M/P of
of
by Cauchy's Theorem (Exercise
I
P
I
is
a
power
of
p,
so
P
<
R
E
I',
contradicting
order
p.
But
now
I
R
I
=
I
RIP
I
I
P
I
is
a
power
of
p,
so
P
<
R
E
r,
contradicting
order p. But now I R I = I R/P I
PES2.
P E Q.
This completes the proof of Sylow's Theorem.
Theorem.
20
Permutation representations
Next a few consequences
consequencesof Sylow's
Sylow's Theorem.
Theorem.
(6.1)
Let P EE Sylp(G).
Then P 5
aG
SylP(G). Then
G ifif and
and only
only if
if PPisisthe
theunique
uniqueSylow
Sylow
(6.1) Let
p-subgroup
p-subgroup of
of G.
G.
Proof. This
This isis because
because GGacts
actstransitively
transitively on
on Sylp(G)
Sylp(G)via
via conjugation
conjugation with
with
NG(P)
NG(P)the
the stabilizer
stabilizer of
of PP ininthis
thisrepresentation.
representation.
Lemma
Lemma 6.1
6.1 and
and the
the numerical
numerical restrictions
restrictions in
in part
part (3)
(3) of
of Sylow's
Sylow's Theorem
Theoremcan
can
be
be used
used to
to show
show groups
groups of
of certain
certainorders
ordershave
have normal
normal Sylow
Sylow groups.
groups. See
Seefor
for
example
example Exercises
Exercises 2.5
2.5 and
and 2.6.
2.6.
(6.2)
Let H
H5
a G and P
(6.2) (Frattini
(Frattini Argument) Let
P EESylp(H).
Syl,(H). Then
Then G
G ==HNG(P).
HNG(P).
Proof. Apply
Apply5.20
5.20to
to the
the representation
representation of G on
on Sylp(H),
SyI,(H), using
using Sylow's
Sylow's TheTheProof.
orem
orem to get H transitive
transitiveon
on Sylp(H).
Syl,(H).
Actually
Actually Lemma 6.2 is aa special
special case
case of
of the
the following
following lemma,
lemma, which has aa
similar
similar proof,
proof, and
and which
which II also
alsorefer
refer to
to as
asaaFrattini
FrattiniArgument:
Argument:
(6.3)
Argument)Let
LetKKbe
beaagroup,
group,HH5aK,
K, and
andXX E
C H. Then K =
=
(6.3) (Frattini
(Frattini Argument)
HNK
orbits of equal
HNK(X) if and only
only ifif XK
x ==XXH.H .Indeed
IndeedHHhas
hasI KI K: HNK
:HNK(X)
(X)I1 orbits
equal
x K ,with
withrepresentatives
representatives (X-Y:
(XY: yyEc Y),
set of
of coset
coset
length on
on XK,
length
Y), where
where YY is a set
representatives
HNK(X)in
in K.
K.
representatives for
for HNK(X)
(6.4)
(6.4) Let H a5G
G and
and PP EESylp(G).
Syl,(G). Then P n
nH
H EE Sylp(H).
Syl,(H).
Exercises
Exercises for
for chapter
chapter22
1.
1. Prove
ProveLemma
Lemma5.18.
5.18.
2.
2. Prove
ProveLemmas
Lemmas5.20
5.20and
and5.21.
5.21.
3.
3. Prove
ProveCauchy's
Cauchy'sTheorem:
Theorem:Let
LetGGbe
beaafinite
finitegroup
groupand
and ppaaprime
primedivisor
divisorof
of
G 1. Then
Then G contains
contains an
(x)
(GI.
an element
elementof
of order
order p.
p.(Hint:
(Hint:Prove
Proveppdivides
dividesI CG
1 CG(x)l
for
for some
somexx EEG#.
G'. Then
Thenproceed
proceedby
byinduction
inductionon
on !GI.)
4.
4. Let
LetGGbe
beaafinite
finitegroup
groupand
andppaaprime.
prime.Prove:
Prove:
(1)
G/Z(G)isiscyclic,
cyclic,then
thenGGisisabelian.
abelian.
(1) IfIfG/Z(G)
(2)
A=
(2) IfIfGJGI
=p2
P2then
thenGG=2Zp2
ZpzororEp2.
Ep2.
5.
5. (1)
(1) Let
LetIGI
/GI==pem,
pem,pp>>m,
m,ppprime,
prime,(p,
(p,m)
m) ==1.1.Prove
ProveGGhas
hasaanormal
normal
Sylow
Sylow p-subgroup.
p-subgroup.
(2)
p and
q prime.
Prove
(2) Let
LetIGI
I G I ==pq,
pq,
p and
q prime.
ProveGGhas
hasa anormal
normalSylow
Sylowp-subgroup
p-subgroup
or
oraanormal
normalSylow
Sylowq-subgroup.
q-subgroup.
Sylow's Theorem
21
6.
6. Let
LetI IG
G II ==pq2,
pq2,where
whereppand
andqqare
aredistinct
distinctprimes.
primes.Prove
Proveone
oneofofthe
thefollowing
following
holds:
holds:
(1)
and
(1) qq>>p p
andGGhas
hasa anormal
normalSylow
Sylowq-group.
q-group.
(2)
(2) pp>>q qand
andGGhas
hasa anormal
normalSylow
Sylowp-group.
p-group.
GI I=
= 12 and G has a normal
(3) IIG
normal Sylow
Sylow 2-group.
2-group.
(3)
7.
7. Let
LetGGact
acttransitively
transitively on
on aa set
set X,
X, x EE X,
X, and
and PPEESylp(G,,).
Syl,(G,). Prove
Prove NG(P)
Nc(P)
transitive on
on Fix(P).
Fix(P).
isis transitive
8. Prove
ProveLemmas
Lemmas6.3
6.3and
and6.4.
6.4.
8.
9.
9. Prove
Provethat
thatififGGhas
hasjust
just one
oneSylow
Sylow p-subgroup
p-subgroup for
for each
each ppEE7r(G),
n(G), then
then G
G
thedirect
directproduct
product of
of its
its Sylow
Sylow p-subgroups.
p-subgroups.
isis the
3
Representations of groups on groups
Chapter 3 investigates
investigates representations
representations in the category of groups
groups and homomorhomomorChapter
of groups.
phisms, with emphasis on the normal and subnormal subgroups of
In section 7 the concept
concept of
of an
an irreducible
irreduciblerepresentation
representationis
is defined,
defined, and
and the
the
Jordan-Holder Theorem is established. As a consequence,
Jordan-Holder
consequence, the composition
factors of
of a finite group
group are
are seen to be an invariant
of the group, and
and these
factors
invariant of
composition
composition factors
factors are
are simple.
simple.
The question arises as to how much the structure of a group
group is controlled
controlled
by its composition
composition factors.
factors. Certainly
Certainlymany nonisomorphic
nonisomorphic groups
groupscan
can have
have the
the
composition factors,
same set of composition
factors, so
so control
control is far from
from complete.
complete.To
To investigate
investigate
consider extensions
extensions of a group G by aa group
group A.
A.
this question further we must consider
Section 10 studies split extensions
extensions and introduces
introduces semidirect
semidirectproducts.
products.
Section 9 investigates
investigates solvable and nilpotent groups. For finite groups this
groups all of whose composition
composition factors
amounts to the study of groups
factorsare,
are, in
in the
the first
first
case, of prime order and, in the second,
second, of order p for
for some
some fixed
fixed prime p.
p.
Commutators, characteristic
characteristic subgroups,
subgroups,minimal
minimal normal
normal subgroups,
subgroups,central
central
products are
are also
also studied.
studied.
products, and wreath products
7 Normal
Normal series
In this section G and
and A
A are
are groups,
groups, and
and n:
n :A
A -p
+Aut(G)
Aut(G) isisaarepresentation
representation
of A
I'll also
A in
in the
the category
category of
of groups
groups and
and homomorphisms.
homomorphisms. I'll
also say
say that
that AA
acts as a group
Observe
that
7r
is
also
a
permutagroup of
of automorphisms
automorphisms on G.
G. Observe that n
also a permutation representation,
representation, so we can use the terminology, notation, and results from
chapter 2.
2.
A normal series of
of length
length n for G is
is aa series
series
1=Go<G1a...4G.=G.
l = G o i G 1 g . - .a G , = G .
A-invariant ifif A acts on each Gi
Gi..
The series is A-invariant
(7.1) Let
: 00 j
<i j
< n)
Let (Gi
(Gi:
n) be
be an A-invariant normal series and H an A-invariant
subgroup of G. Then
I H:AA+
-aAut(H)
Aut(H) isis aarepresentation
representationof
of A
A on
on H.
H.
(1) The
Therestriction
restriction7r
n lH:
(Giflf lH:
H :00<jii<jn)n)isisan
anA-invariant
A-invariant normal
normal series for H.
(2) (Gi
(3) IfIfHH< GGthenTrG/H:
then T G I HA-4
A: +Aut(G/H)
Aut(G/H)isisa arepresentation,
representation,where
wherea7rG/H:
anc/ff:
Hg t+HH(ga)
H(ga)for
foraaEEA,
A,ggEE G.
G.
Normal series
23
then(GiH/H:
(GiH/H:00<5i i<_<n)n)isisananA-invariant
A-invariant normal
normal series
series for
(4) IfIfHH<9GGthen
GIH.
G/H.
anA-invariant
A-invariant subgroup
subgroup ofofGG and
f l H is
is an
an
(5) IfIf XXisisan
and HH 9< G,
G, then
then X
x fl
XHIH is
is an
an A-invariant
A-invariant subgroup of
A-invariant normal
normal subgroup of
of X and XH/H
GIHwhich
which is
is A-isomorphic
A-isomorphic to
/ ( X fl
n H).
H).
G/H
to XX/(X
subgroupH
H of
of GGisis subnormal
subnormalin
inGGifif there
thereexists
exists a series
Go<9GG1
A subgroup
series H ==Go
1 5 . ..
aG
G,==G.G.Write
WriteHH<9<9GGtotoindicate
indicateHHisissubnormal
subnormal in
in G.
<
-
(7.2)
X=
=Go
G,G,
then
eitherX ==GGoror(XG)
(7.2) If
IfX
Go<9G1
G1 49...
. .<9G.G,= =
theneitherx
(XG)<_< G,_1
Gn-1 # G,
G,
(xG)
# G.
G.
so (Xo)
(7.3)
Let X
X and
and H
H be
beA-invariant
A-invariantsubgroups
subgroups of G with X <
9<<IG.
G.Then
Then
(7.3) Let
(1) There
Thereexists
existsan
anA-invariant
A-invariantseries
series
(1)
X=Go<G1<...
X fln HH<S<HH..
(2) X
(3) IfIf H
<<
H <9GGthen
then XH/H
XHIH 9
9 G/H.
GIH.
Proof. Part
Part(1)
(1)follows
followsfrom
from7.2
7.2and
andinduction
inductionon
onthe
thelength
lengthnnof
ofaasubnormal
subnormal
series X =
= Go
< -. <
G,G,since
isisA-invariant.
GO< G1
G1 9
9G
G,==
since(XG)
(XG)
A-invariant.Parts
Parts (2)
(2) and
are straightforward.
straightforward.
(3) are
a
-
(Gi: 0 (
(n)
n) isis the
the family
family of
of factor
factor
The family of factors
factors of a normal series (Gi:
<i<
groups (Gi+1/Gi:
(Gi+1/Gi: 00 <5 ii <n).
5 n).Partially
Partiallyorder
orderthe
theset
setof
of normal
normal series
series for
for G
G via
via
(Hi:0<i <m)<(Gi:O<i <n) if Hi=GI(i)
each 00 <5ii <_<m,
m,and
andsome
somej j(i).
for each
(i ).
The representation
representation n
it is said to be irreducible
irreducible if G
G and
and 11are
are the
the only
only AAinvariant
normal subgroups
subgroupsofof G.
G. We
We also
also say
say that
that G is
An Ainvariant normal
is A-simple.
A-simple. An
composition series for G is a normal series (Gi:
(Gi: 00 5
<i (
< n) maximal subject to
being A-invariant and
and to
to Gi # Gi+1
<i (
< n. Of
Gi+I for 0 (
Of particular importance is
the case
to
be
simple
if
it
is
1-simple.
case A
A ==1.1.GGisissaid
said to be simple
1-simple.Similarly
Similarlythe
the composition
composition
series for
for G
G are
areits
its 1-composition
1-compositionseries.
series.
series
(7.4) If
If G
G isisfinite
finite then
then G
G possesses
possesses an
an A-composition
A-composition series.
(7.5)
series ifif and
and only
only ifif
(7.5) An A-invariant
A-invariant normal series is an A-composition
A-composition series
its factors
factors is
is aa nontrivial
nontrivial A-simple
A-simple group.
group.
each of its
24
Representations of
of groups on groups
< ii 5<n)
n) and
and (Hi: 00 5< ii5<m)
m) be
be AJordan-Holder Theorem.
Theorem.Let
Let (Gi
(Gi:: 0 5
compositionseries
seriesfor
forG.
G.Then
Thennn =
=m
composition
m and
and there
there exists
exists a permutation
permutation aa of
of
{i:0
< i <n}
of A
A on
on GiU+l/
GQ+1/GQ
{i:0 i
c n) such
such that
that the
the representations
representations of
Giu and Hi+1/Hi
Hi+I/Hi
equivalent for
for 00 5
<i c
< n.
are equivalent
H). I'll show:
Proof.
= HmW1
and kk ==min{i
Proof. Let H =
min{i:: Gi
Gi 5 H).
show:
(a) G/H
G I HisisA-isomorphic
A-isomorphicto
to Gk/Gk_1,
Gk/Gk-l,and
(a)
/ Xi
(b) (Xi
(Xi==Gia
Gi, nf H:
l H:00<5i i<cn)n)isisan
anA-composition
A-compositionseries
seriesfor
forHHwith
withXi+1
Xi+1/Xi
A-isomorphic
to Giatl/Gia
Gi,,+1 / Gi,,forfor
0<
i<
wherei ia a==i iififi ic<kk-- 11
A-isomorphic to
05
i(
n n--1,1,where
and
a n d i aia=i+1for
= i + l f o r ii>k-1.
z k - 1.
Suppose (a)
(a) and
and (b)
(b) hold.
hold.By
Byinduction
inductionon
onn,n,nn-- 1 =
=m - 1,1,and
andthere
there isis aa
permutation 3j,8 of
of {i:
{i:005< ii c
<n - 11
Hi+1/Hi.
1) with Xip+1/Xifl
XiS+l/Xia A-isomorphic to Hi+l/Hi.
Hence n =
= m and
and the permutation
permutation aa of ti:
{i:00 <
5 i <<n)
n}defined
defined below does the
Theorem:
trick in the Jordan-Holder
Jordan-Holder Theorem:
is=i,Ba ifi<n-1,
l)a =
= kk -- 1,
1,
(n
( n -- 1)a
remains to
to establish
establish (a)
(a)and
and(b).
(b).First,
First,asasGi
Gi5<HH for
for ii c
< k, Gi =
= Gia
Gia =
So it remains
Xi/XiPl, is A-isomorphic to Gi,,
Gia//Gia_1.
Gia-1.
Xi, so certainly Xi/X1_1,
If Gk fnl H -$ Gk-1,
Gk_1, then,
then, as
as (Gk
(Gk n
n H)Gk-l/Gk-l
H)Gk_1/Gk_1 is an A-invariant normal
of the
the A-simple
A-simplegroup
group Gk/Gk-1,
Gk/Gk-1, we have
have Gk
Gk=
= (Gk
(Gkf
n H)Gk-1
H)Gk_1
< H,
subgroup of
l
i
of k.
k. So
So Gk
Gkfln H
H=
= Gk_1.
hand, ifif jj >2k,
contrary to the definition of
Gk-1. On the other hand,
k,
then Gj
H,, so
=HG1,
then
G j -$ H
so aa similar
similar argument
argument using
using 7.3 shows
shows G =
HGj, and hence
hence
G/H
A-isomorphic
/(G1 nn H).
Gj /(Gj n H)
G I H=HGj/H
=HGj/His is
A-isomorphictotoGj
Gj/(Gj
H).In
In particular Gj/(Gj
H)
A-simple,so,
so,ififGGjnnHH 5
<G
Gj_1,
then G
Gj_i
l(Gj nnH)
is A-simple,
j-1, then
j-1 /(G
H )isisaa proper A-invariant
A-invariant
j/(G
normal subgroup of
of the
the A-simple
A-simplegroup
groupGGj
/(Gj fl
n H), and hence Gj-1
j_1=
=G
Gjj fl
n
H<
G/H
5 H.
H. But
But then
then jj==kkby
bydefinition
definition of
of k. Moreover G
/ H isisA-isomorphic
A-isomorphic to
Gk/(Gk n H)
Gk/(Gk
H)=
=Gk/Gk-1,
Gk/Gkel,so
SO(a)
(a) holds.
holds.
By thelastparagraph,
the last paragraph,G
Gjj nnH
above,GG3
H))
H Gj_1
Gjel for jj >>k.
k. So, as above,
j =_((Gj
G j nn H
Gj_1, and
and hence
hence Gj/GjP1
Gj/G1_1 is A-isomorphic
A-isomorphictoto(Gj
(Gjnn H)/(Gj-1
H)/(G1_1 nn H), comGj-1,
pleting the proof of (b).
(b).
The Jordan-Holder
Jordan-Holder Theorem says
says that the factors of an A-composition series
of G are (up
equivalence and order) independent
independent of the series, and hence are
(up to equivalence
an invariant of
of the representation
representation n.
n. These factors are the compositionfactors
factors
of
n. IfIf AA==1,1,these
thesefactors
factorsare
arethe
thecomposition
composition factors
of the representation
representation n.
of G.
G.
Characteristic
Characteristicsubgroups
subgroups and
and commutators
commutators
25
(7.6) Let X
of the finite
X be
be an
anA-invariant
A-invariant subnormal
subnormal subgroup
subgroup of
finite group G.
G.
Then
Then
(1) The
areaasubfamily
subfamilyof
ofthe
theA-composition
A-composition
TheA-composition
A-compositionfactors
factorsof
ofXXare
factors of G.
G.
(2) IfIf XX<9GGthen
thenthe
theA-composition
A-composition factors
factors of G
G are
are the
the union
union of
of the
the
A-composition factors of
of X
X and
and G/X.
G/X.
Proof.
Proof. There
Thereisisan
anA-invariant
A-invariantnormal
normal series
series containing
containing X by 7.3, and as G is
finite this series
series is contained
contained in
in aa maximal
maximal A-invariant
A-invariant series.
series. Thus
Thus there
thereisis an
an
A-composition
A-composition series
series through
through X,
X,so
sothat
thatthe
theresult
resultisisclear.
clear.
8 Characteristic
Characteristicsubgroups
subgroupsand
and commutators
commutators
A subgroup H of
of aa group
group G
G isis characteristic
characteristicin
inGGififHHisisAut(G)-invariant.
Aut(G)-invariant.
Write H char
char G
G to
to indicate
indicate that
that H
H isisaacharacteristic
characteristic subgroup of G.
(8.1)
(8.1) (1)
(1) If H char
char K
K and
and K
K char
char G,
G, then
then H
H char
char G.
G.
K<
a G,
G, then H <9G.
G.
(2) If
If H
H char K and K
(3) If H char
char G
G and
and K
K char
char G,
G, then
then HK char G and ((H
H fl
f l K)
K) char
char G.
G.
A group G
G is
is characteristically
characteristicallysimple
simple if G
G and
and 11are
arethe
the only
onlycharacteristic
characteristic
subgroups
subgroups of G.
G. A
A minimal normal subgroup of G is a minimal member of the
set of nonidentity normal
normal subgroups
subgroups of G,
G, partially
partially ordered
ordered by inclusion.
inclusion.
characteristically simple finite group, then G is
is the
the direct
direct
(8.2) If
If 11 # G is aa characteristically
(8.2)
product of isomorphic
isomorphic simple
simple subgroups.
Proof. Let
Let H
H be
beaaminimal
minimalnormal
normal subgroup
subgroup of G
G and
and M
M maximal
maximal subject
subject
to M
<
G
and
M
the
direct
product
of
images
of
H
under
Aut(G).
M 9 G and M the direct product of images
H under Aut(G). Now
Now
X=
=(Ha:
( H a a:aEEAut(G))
Aut(G))isischaracteristic
characteristicin
in G,
G,so
soby
by hypothesis
hypothesis X =
=G.
G. Hence,
Hence,
G, there is
a M.M.As
AsHa
H aisisa aminimal
minimalnormal
normal subsubif M #
0 G,
is aa Ec Aut(G)
Aut(G) with
withHHa
group of
of G and H
Haa fl
Haa flnM
< M xx
nM
M <9G,
G, we
we conclude H
M ==1.
1. But
But then M <
Ha
H a <9G,
G,contradicting
contradictingthe
the maximality
maximality of
of M.
M.
So
= M.
So G =
M. Thus
Thus G
G=
=H
H xxKKfor
forsome
someKK <5G,
G,sosoevery
everynormal
normal subgroup
subgroup of
H is
is also
also normal
normal in G.
G. Thus
Thus H isissimple
simple by
by minimality
minimality of H,
H , and
and the
the lemma
lemma
is established.
established.
(8.3) Minimal
Minimalnormal
normalsubgroups
subgroupsare
arecharacteristically
characteristicallysimple.
simple.
Proof. This
Thisfollows
followsfrom
from 8.1.2.
8.1.2.
26
ups on groups
Representations of groups
(8.4) (1)
(1) The
The simple
simple abelian
abelian groups
groups are
are the
the groups
groups of prime
prime order.
(2)
is characteristically
characteristically simple, finite,
(2) If G is
finite, and
and abelian,
abelian, then
then G
G2
= Epll
EP" for
and some
some integer
integer n.
some prime p and
For x,
x , y cE G,
G ,write
write [x,
[ x ,y]
y]for
forthe
thegroup
group element
element x-1
X - ' ~y-1xy.
- ' X ~ . [x,
[ x y]
,y]isisthe
thecomcommutator
mutator of x and
and y.
y. For
For X,
X , YY <5G,
G ,define
define
[X,
[ X ,Y]
Y ]==([x,
( [ xy]:
,y]:xx EE X,
X, yy EE Y).
Y).
For z cE Z <5G
G write
write [x,
[ x ,y,
y,z]
z ]for
for [[x,
[ [ xy],
,y ] ,z]
z ]and
and [X,
[ X ,Y,
Y , Z]
Z ] for [[X,
[ [ X Y],
,Y ] ,Z].
Z].
(8.5) Let G
G be
be aa group,
group, a,
a ,b,
b,cc EE G,
G ,and
and X,
X, Y
Y<
5 G.
G.Then
Then
(1)
[a,
b]
=1
if
and
only
if
ab
=
ba.
(1) [a,b]= 1 if andonly ab=ba.
(2)
(2)[X,
[ XY]
, Y ]==11ififand
andonly
onlyifif xy
xy ==yx
yx for
forall
allxxEEXX and
andyy cEY.
Y.
(3)
H His aisgroup
homomorphism
b]a
(3) IfIf a:a :GG-+
a group
homomorphismthen
then[a,[a,
b]=
a=[act,
[aa,bet]
ba]and
and
[X,
[ X ,Y]a
Y ] a=
=[Xa,
[ X a Yet].
,Y a ] .
(4) [ab,
[ab,c]
c] =
[a,c]b[b,
clb[b,Cc]] and
= [a,
and [a,
[a, bc]
be] =
= [a,
[a, c][a,
c][a, bIC.
b]'.
(4)
(5)
( 5 ) XX<5NG(Y)
N c ( Y )ififand
andonly
onlyifif [X,
[ X Y]
,Y ]<5Y.
Y.
(6)
(6) [X,
[ X Y]
,Y l==[Y,
[Y,X]
XI :a9 (X,
( X ,Y).
Y).
Proof. II prove
=
prove (6)
(6)and
and leave
leave the
the other
other parts
parts as
as exercises.
exercises. Notice [a,
[a,b]-1
b]-' =
[b,
[b,aa],] ,so
so [X,
[ X ,Y]
Y ]==[Y,
[Y,X].
XI. Further,
Further,to
toprove
prove [X,
[ X ,Y]
Y ]:a
9 (X,
( X ,Y),
Y ) ,itit will
will suffice
suffice to
[ x ,y]Z
ylZ EE [[X,
X , YY]] foreachx
y EE Y
x ,y]-'
show [x,
for each x E X, Y
Y,, andz
and Z EE XU
X U YY.. As
As[[x,
y]-1 =
=
[y,
may assume zz EE Y
Y.. But,
But, by
by (4),
(4), [[x,
y]Z== [[x,
z]-1[x, yz] E [X,
[ y ,xx],
] , we may
x ,ylZ
x ,z]-'[x,
[ X ,Y],
Y],
so the proof is complete.
complete.
so
Let G
G be
be aagroup,
group,x,
x ,yyEE G,
G ,and
and assume
assume zz ==[x,
[ x ,y]
y]centralizes
centralizesxx and
and y.
y.
(8.6) Let
Then
Then
(1) [xn
[ x nym]
,ym]=Znm
= znmfor
forall
alln,n m
, mE7L.
E Z.
(2)
(yx)'
=
Zn(n-1)12ynxn
for
all
(2) (yx)" = ~ " ( ~ - ' ) / ~ yfor
" xall
" 00<5nnEE71.
Z.
Proof.
Proof. Without
Without loss
loss G
G ==(x,
( x ,y),
y ) ,so
sozz EE Z(G).
Z(G).zz ==[x,
[ x ,y]
y]so
soxy
xY==xz.
xz. Then,
Then,for
for
nn EE 1,
(xn)y
Z(G). Thus
Z,
(xn)Y==(xy)n
(xY)"==(xZ)n
(xz)" ==xnZn
xnzn as z E Z(G).
Thus [x",
[ x ny]
,y]==Zn.
zn. Similarly
Similarly
[ x ,ym]
ym]==zm,
[ x ,ym]n
ymIn=
=zmn,
[x,
Zm, so
So [xn,
[xn, ym]=
ym] _ [x,
Zmn, and
and (1)
(1) holds.
Part (2)
==(yx)nyx
(2)isis established
established by
by induction
induction on
onn.n.Namely
Namely(yx)n+1
(YX)"+'
(yx)"yx==
Zn(n-1)/2 y n x n yx, while
xny ==yxfZn,
yxnzn,so
so that
that the
the result
result holds.
holds.
Zn(n-1)12ynxnyx,
while by
by (1)
(1) xny
(Three-Subgroup Lemma) Let X, Y,
Y , Z be subgroups
subgroups of a group
group G with
with
(8.7) (Three-Subgroup
[X,Y,Z]=[Y,Z,X]=1.Then
[ X , Y, Z ] = [Y, Z , XI = 1. Then [Z,X,Y]=1.
[ Z , X , Y ]= 1.
Proof.
Proof.Let
Letxx EEX,
X, yy cEY,
Y ,and
and zz cEZ.
Z.AAstraightforward
straightforwardcalculation
calculationshows:
shows:
z)-1a(y,
[ x ,y-1,
y-', zly
zlY = x=x-1y-1xz-1x-1yxy-1zy
- ' ~ - ~ x z - ~ x - ' ~ x =a(x,
~=-a'(zx ~,y,
y, z ) - ' ~ (z,
z~, x),
x, ) ,
(*)
(*) [x,
Solvable and nilpotent groups
27
where a (u, vv,, w)=uwu-'vu.~pplyingthepermutations(x,
w) = uw u-1 vu. Applying the permutations (x, y, z)and(x,
z) and (x, z, y)
wherea(u,
to (*) and taking the product of (*) with these two images, we conclude:
[x, Y-'z]y[Y, z-1, x]Z[z, x-1, Y]X =1.
Y, Z]
Z]=
_ [Y,
[Y,Z,
Z, XI
X]=
=1,
we
As [X, Y,
1, also [x, y-1,
y-', z]
z] _=[y,
[y, z-1,
z-', x]
x] ==1,1,so
soby
by (**)
(**) we
get [z, x-',
x-1, y]
x-1],
y] ==1.1.Finally
Finallyas
as[Z,
[Z,X]
XIisisgenerated
generatedby
by the
the commutators
commutators [z, x-'1,
X,ititfollows
followsfrom
from8.5.1
8.5.1that
thatyycentralizes
centralizes[Z,
[Z, XI.
X]. But then, by
by 8.5.2,
8.5.2,
z EE Z, xX EEX,
[Z, X,
X, Y]
Y]=1.
= 1.
The commutator
commutator group or derived group of a group G is the subgroup
subgroup GM
G(') =
=
[G, G].
GI. Extend
Extend the
the notation
notationrecursively
recursivelyand
anddefine
defineG(")
G(")==[G(n-1),
[G("-'), G(n-1)]
G("-')I for
DefineG(O)
G(0)==GGand
andGm
G-=
nn > 1. Define
= f°_ 1 G(O
G(').
nzl
(8.8) Let G be a group and
and H
H5
< G. Then
(8.8)
(n).
(1)
H(")5
G(").
(1) H(n)
<G
(2) IfIf a:
Xx(").
(n).
a:G
G + X is a surjective
surjective group
group homomorphism
homomorphismthen
thenG(n)a
~ ( " =)=a
(3) G(n)
G(") char
char G.
G.
(4) GM
G(')<5HHififand
andonly
only if H
H a9GGand
andG/H
G/Hisisabelian.
abelian.
A group G is
is perfect
pe$ect ifif G
G ==G(1).
G(').
(8.9) Let
group G
G with
with LL perfect
perfect and
and [X,
[X, L,
L, L] =
= 1.
(8.9)
Let X
X and
and L be subgroups
subgroups of a group
1.
Then [X, L] =
=1.
1.
Proof. [L,
[L,X,
X,L]
L]==[X,
[X,L,
L ,L]
L]and
andby
byhypothesis
hypothesisboth
both are
are 1.
1. So
So by the ThreeSubgroup Lemma,
Lemma,[L,
[L, L,
L, XI
X] =
=1.1.Butby
= [L,
[L,, X]
= 1.
But by hypothesis
hypothesis L =
[L ,L],
L], so [L
XI =
1.
9 Solvable
Solvable and
and nilpotent
nilpotent groups
groups
A group G is
is solvable
solvable ifif itit possesses
possessesaa normal
normal series
serieswhose
whose factors
factorsare
are abelian.
abelian.
(9.1) A
Agroup
groupGGisissolvable
solvableifif and
andonly
onlyifif Gin)
G(") =
=11for
for some
some positive integer n.
Proof. IfIfG(n)
< i <5 n)
Proof.
G(") ==1,1,then
then (G(n-`):
(G("-'): 0 5
n) isis aa normal
normal series
serieswith
with abelian
abelian facfactors by 8.8.4.
< n) is such a series
8.8.4. Conversely if (G1:
(Gi: 005< ii 5
series then, by 8.8.4
8.8.4 and
and
induction on
< Gn_i,
so
= 1.
induction
on i,i, G(`)
G(') 5
GnWi,
so G(n)
G(") =
1.
(9.2) A
A finite
finitegroup
group is
is solvable
solvable ifif and
and only
only ifif all
all its
its composition
composition factors
factors are
are of
prime order.
order.
28
Representations of groups on groups
Proof. If all
all composition
composition factors are of prime order then a composition series for G
G isis aanormal
normalseries
seriesall
allof
of whose
whosefactors
factorsare
areabelian.
abelian.Conversely
Conversely if
(Gi
such a series then the
(Gi::00 <i i5<n)n) isis such
the composition
composition factors of the
the abelian
abelian
group Gi+i
/ Gi are also abelian,
Gi+1/Gi
abelian, and hence, by 8.4.1, of prime order. Then, by
7.6, the composition
composition factors
factors of G
G are
areof
of prime
prime order.
order.
(9.3)
(1) Subgroups
Subgroups and
and homomorphic
images of
of solvable
(9.3) (1)
homomorphic images
solvable groups
groups are
are
solvable.
solvable.
(2) IfIf H
H<9GGwith
with HHand
andG/H
G/Hsolvable,
solvable,then
thenGGisissolvable.
solvable.
(9.4)
(9.4) Solvable minimal normal
normal subgroups
subgroups of finite
finite groups
groups are
are elementary
elementary
abelian p-groups.
Proof. Let
LetGGbe
befinite
finiteand
andM
Maasolvable
solvableminimal
minimal normal
normalsubgroup
subgroupof
of G.
G. By
By 9.1
Proof.
and solvability
solvability of M,
M,M(1)
M(') # M. Next, by 8.3,
8.3, M
M isis characteristically
characteristicallysimple.
simple.
MM char M, we
= 1. Thus M is abelian
we conclude
conclude M(l)
M(') =
abelian by 8.8. Then
So, as M(')
8.4.2 completes
completes the proof.
L 1(G)==G,
G,and,
and,proceeding
proceedingrecursively,
recursively,define
defineL,(G)
Ln(G)==[L,-l
[L,-1(G),
G]
(G), GI
Define L1(G)
<nE
c Z. G
(G) =
= 1 for some 11 5
< n E Z.
Z. The
for 1 <
G isis said
said to
to be
be nilpotent
nilpotentifif Ln
L,(G)
nilpotent group
groupisismm- 1,
=min{i:
class of a nilpotent
1, where m =
min{i :Li(G)
Li(G)==11.
1).
(9.5) (1)
(1) Ln(G)
L,(G) char
charGGfor
foreach
each1 1<n
5 n(=-E Z.
Ln(G)
(2) Ln+1(G)
Ln+l(G) iLn(G).
(3)
Ln(G)/Ln+l (GI i
(GI).
(3) Ln(G)/Ln+1(G)
< Z(G/Ln+l
Z(G/Ln+1(G))
Proof. Part
Proof.
Part(1)
(1)follows
followsfrom
from8.5.3
8.5.3by
by induction
induction on
on n.
n. Then
Then (1)
(1) and 8.5.5 imply
(2) while 8.5.1 and 8.5.3 imply (3).
Define Zo(G)
Z0(G) =
=11and
andproceeding
proceedingrecursively
recursivelydefine
defineZn
Z, (G) to be the preimage
Zn(G) char G.
in G of Z(G/Zn-1(G))
Z(G/Z,-l (G))for
for11<,<nn EE Z.
Z.Evidently Z,(G)
(9.6) G is nilpotent
nilpotent if and only if G
G=
= Zn(G)
Z,(G) for some 0 <nn E
E Z.
G isis
(9.6)
Z. If G
= min{n:
= Zn(G)}.
min(n: G =
Z,(G)}.
nilpotent then the class of G is k =
Proof. II first
(G) <
(G)
first show
show that
that ifif GGisisnilpotent
nilpotentof
of class
class m
m then
then L,n+i-i
Lm+l-,(G)
5 Zi
Zi(G)
>0
for 0 < i <
5 m.
m. For i ==00this
this follows
follows directly
directly from the definitions, while ifif ii >
and Lm+2-i(G)
L.+2-i(G) <iZi-i(G)
Zc-l(G) then
then[L,n+i_j(G),
[Lm+l-i(G), G]
GI =Lm+2-i(G) <Zi_i(G),
5 Zi-l(G), so,
so,
by 8.5,
8.5,Lm+i-,(G)Zc-i(G)/Zl-~(G)
Lm+i-i(G)Zj-i(G)/Zi-i(G) <iZ(G/Zi-1(G))
by
Z(G/ZL-i(G))==Z1(G)/Zi-1(G),
Zz(G)/Zi-l(G), and
and
(G) 5
< Z,
Zi (G).
(G). So the claim follows by induction on i, and we see
hence L,n+i_i
Lm+l-,(G)
see
Z,n(G)=L1(G)=G,
sok<m.
Zm(G)= L1(G)=G, so
ksm.
Semidirect products
products
29
Next let's see that ifif Z,(G)
Zn (G)==GGfor
forsome
some005<n
then Li+l
L;+1(G)
< Z,+(G)
Zn-i (G)
n Ec Z, then
(G) 5
for each 0 <
(G) <5
5 i <5 n. Again
Again the case i ==00 isis trivial,
trivial, while
while ifif ii >>00and
and Li
Li(G)
Zn_i+l (G) then Li+l(G)
Li+1(G) =
= {Li
(G), GI
G] 5
< {Zn-i+l
(G), G]
(G) by 8.5.5,
zn-i+,(G)
[Li(G),
[zn-i+l(G),
GI <
I Zn_i
Zn-i(G)
8.5.5,
establishing
Ln+1(G)
<) Zo(G)
= 1,
establishing the claim.
claim. In particular L
n + ~ ( G5
Zo(G) =
1, so
so G
G is
is nilpotent of
class m <
5 n,
n, so
so that
that m <_( k.
k.
m ifif and
and only
onlyifif G/Z(G)
G/Z(G) is
(9.7) 11 # G is nilpotent of class m
(9.7)
is nilpotent
nilpotent of class
m-1.
m - 1.
Proof This
Proof.
Thisisisaadirect
directconsequence
consequenceof
of 9.6.
(9.8) p-groups are
(9.8)
are nilpotent.
nilpotent.
so, by
by 5.16,
5.16,
Proof. Let
LetGGbe
beaaminimal
minimal counter
counterexample.
example. Then certainly G # 1, so,
by minimality
minimalityof
of G,
G, G/Z(G)
G/Z(G) is
Z(G) # 1. Hence, by
is nilpotent,
nilpotent, so, by 9.7, G is
nilpotent, contrary
contrary to the choice
choice of G.
G.
nilpotent,
(9.9) Let
homomorphic images
images
Let G
G be
be nilpotent
nilpotent of class m. Then subgroups and homomorphic
(9.9)
of G
G are
are nilpotent
nilpotent of
of class
class at
at most
most m.
m.
(9.10) If G
(9.10)
G is
is nilpotent and H is
is aa proper subgroup
subgroup of G, then H isis proper
proper in
in
NG (H).
NG(H>.
NC(H) =
=H <
< G. Then Z(G) _(
< NG(H)
NC(H) =
=H
H,, so H*
=
Proof. Assume NG(H)
H* <
< G*
G* =
G/Z(G). By
G/Z(G).
By9.7
9.7and
andinduction
induction on
on the
the nilpotence
nilpotence class of G,
G, H*
H* < NG*(H*).
But, as Z(G)
Z(G) <5 H,
H, NG.(H*)
NG*(H*)==NG(H)*,
NG(H)*,so H <<NC(H),
NG(H),aacontradiction.
contradiction.
(9.11) A finite group is
is nilpotent
nilpotent ifif and
and only
only ifif it is the direct product
product of
of its
(9.11)
Sylow groups.
direct
Proof. The
Thedirect
directproduct
product of
of nilpotent
nilpotent groups
groups is nilpotent, so by 9.8 the direct
product
product of
of p-groups
p-groupsisisnilpotent.
nilpotent.Conversely
Converselylet
letGGbe
benilpotent;
nilpotent;we
wewish
wishto
to show
show
G is the direct product of its
its Sylow
Sylow groups.
groups. Let
Let P
PE
Syl,(G). By
E Syl,(G).
By Exercise
Exercise 2.9
it suffices to
to show
show PP < G.
(P) <
G. If
If not, M
M ==NG
NG(P)
< G,
G, so,
so, by 9.10, M <<NG(M).
Nc(M).
But, as P
= M,
P < M,
M, {P)
{P}==SylP(M),
Syl,(M), so
so PP char
charM.
M.Hence
HenceNG(M)
NG(M)<
5 NG(P)
NG(P)=
M,
contradiction.
a contradiction.
a
10 Semidirect
Semidirect products
products
of
In this section
section A
A and
and G
G are
aregroups
groupsand
andrr:
n:A
A + Aut(G) is a representation of
A as a group of group automorphisms
automorphisms of G.
A
30
Representations of
groups on groups
ofgroups
H a9G.
to H in G is a subgroup
K of G with
Let H
G. A
A complement
complement to
subgroup K
with G =
=HK
HK
and H f1
K=
=1.1.GGisissaid
nK
saidtotobe
bean
anextension
extension of
of aa group
group X by aa group
group Y
Y if there
exists HH 9
a G with H Z
- XXand
Y.Y.The
exists
and G/H
G / H- Z
Theextension
extensionisissaid
said to
to split
split if H
H
in G. The following construction can
can be
be used
used to describe
has a complement in
split extensions.
Let S be the set product A x GGand
anddefine
define aa binary
binary operation
operation on S by
(a,g)(b,h)=(ab,gb"h)
, b E A , g, h E G
(a, g)(b, h) = (ab, gbrh) aa,bEA,g,hEG
gb" denotes the image of G
where gbk
G under
under the
the automorphism
automorphismb7r
bn of G.
G. We call
S the semidirectproduct
semidirect product of G
Denote S by S(A,
G by
by A
A with
with respect
respect to
to 7r.
n . Denote
S(A,G,
G,7r).
n).
(10.1)
(10.1) (1)
(1) SS==S(A,
S(A,G,
G,rr)
n )isisaagroup.
group.
A
-+
S
and
G: +
-+ SS are injective
(2) The
maps
The maps u ~ :+ and o-G:
u ~G
injective group
group homomorhomomorphisms, where
i-+ (a,
(a, 1)
phisms,
whereorA:
o;l:aa H
1) and aG:
u ~g i-+
:H (1, g).
g).
Goo in
(3) GaG
GuGa9SSand
andAorA
A- is a complement to GuG
in S.
S.
(4) (1,g)fa'1?=(1,ga")forgEG,aEA.
(4)
(1, g)(atl)= (1, gar) for g E G, a E A.
Observe that ifif nit is
is the
the trivial
trivial homomorphism
homomorphism then the
the semidirect
semidirectproduct
product is
is
just the direct product of A
A and G.
G.
(10.2)
<H
H,, and B a complement to
toGGin
in H. Let aa:: B -+
+
(10.2) Let H be a group, G 9
Aut(G)
b EE B,
Aut(G) be
be the
the conjugation
conjugationmap
map (i.e., ba:
ba: gg -+
+gb
gbfor
forb
B, gg EE G;
G;see
see Exercise
Exercise
Define B:
,B:S(B,
S(B,G,
G,aa)) +
-+ H by
1.3). Define
by (b,
(b,g),B
g)B =
=bg.
bg. Then
Then,B
B is an isomorphism.
isomorphism.
We see from 10.1 and
and 10.2 that the semidirect products
products of
of G by A are preextensions of G by A. Moreover
cisely the split extensions
Moreover the representation
representationdefining
defining the
the
semidirect product is a conjugation
conjugation map.
semidirect
Under the hypotheses of
of 10.2, I'll
I'll say that H is
is aa semidirect
semidirect product of G
an isomorphism.
isomorphism.
by B. Formally this means
means the
the map
map ,B
B ofof 10.2 isis an
), i =
=1,1,2,
2, be
Si ==S(A1,
S(Ai, GI
Gi,, ,ri
xi),
besemidirect
semidirectproducts.
products.Then
Then there
there exex(10.3)
Let Si
(10.3) Let
ists an
-+ S2
@: Sl
S1+
S2with
withA1orA,0=A2orA2
Aiu~,@
= A z ~ Aand
, G1orG,0=G2o'o2
Gluc1@= G z ~ G ,
an isomorphism
isomorphism0:
7r1isisquasiequivalent
quasiequivalentto
ton2
n2in
inthe
thecategory
category of
of groups
groups and
and homohomoif and only if nl
morphisms.
morphisms.
It is
= S(A,
S(A,G,
G,7r1)
nl) and
and S2
S2 =
=
is not
not difficult
difficult to
to see
seethat
thatsemidirect
semidirectproducts
productsSiS1=
ir2) can
can be isomorphic without Jr1
being quasiequivalent
to n2.
72. To
S(G, A, n2)
nl being
quasiequivalent to
investigate
isomorphic we need to know more about
about how
S1and
and S2
S2are isomorphic
investigatejust when
when Sl
Aut(SS)
actson
onits
itsnormal
normal subgroups
subgroups isomorphic
isomorphicto
to G,
G, and how
how the stabilizer
Aut(Si) acts
in Aut(Si) of such a subgroup
subgroup acts
acts on
on its
its complements. This
This latter
latter question
question is
is
considered in chapter
chapter 6.
6.
Semidirect products
31
of interest
interest to
to generate
generate
It's also
also easy to cook up nonsplit extensions, and it is of
conditions which insure that an extension
extension splits.
splits. The following is perhaps the
most important
important such
such condition:
condition:
(10.4)
abelian normal
normalp-subgroup
p-subgroup
(10.4) (Gaschutz'
(Gaschiitz'Theorem)
Theorem)Let
Letppbe
beaa prime, V
V an abelian
Sylp(G). Then
Then G splits over
over V if
if and
and only
only ifif P
of a finite group G, and P E
E Syl,(G).
splits over V.
V.
Proof.
< P.
P. Hence
Hence if H
H isis aacomplement
complement to V
V in
in G
G then
then by
by the
the
Proof. Notice V 5
Modular Property
PropertyofofGroups,
Groups,1.14,
1.14,PP==PP n
fl G
G=
=P
P fl VH=
Modular
VH =V(P
V(P fl H) and
and
P
P.
P flnH
Hisisaacomplement
complement to V
V in P.
suppose Q is aa complement
complement to
to V
Vin
inP.
P. Let
Let C?G =
= G/ V
observe
Conversely suppose
V and observe
= Q.
P Z Qr
Q. Let
Let X
X be
be aa set
set of
of coset
coset representatives for V in G. Then the map
H xf isisaabijection
of this
this map
mapby
byaaH
H xa.
xH
bijectionof
of X
X with
with G
C? and
and II denote
denote the inverse of
xu.
Then
Then
(i)
(i)
=xaby(a,
xaxb =
xaby(a, b)
b) for
for a,
a , bb E G,
and for some y(a,
y (a,b)
b)EE V.
V.
Next using
c)c)= =
xaxbCY
using associativity
associativity in
in GGand
andGGwe
wehave
havexabcy(a,
xabcy (a,bc)y(b,
bc)y (b,
xaxb, y
xaU(xbxc)
=C
(xaxb)xc
b)xc
=~xabc
c)y
(b, c)
C)= X
(X~X
=
( )~ a ~ b==)xab
x~a cby~(a,
( a~
, ) X=CxabxcY
x ~ ~ x(a,~bb)x`
)X
YC (=xabcy(ab,
, y (ab,c)y
(a, b)x
b)q,from
fromwhich
whichwe
weconclude:
conclude:
y(ab, c)y(a,
c)y(a, b)x1=
b)XC F y(a,
y(a, bc)y(b,,c)
bc)y(b, c) for all a,
a , b,
b, ccEE G.
G.
(ii)
Now choose X =
Y, where Y
Y is
is a set of
of coset
coset representatives
representativesfor
for PP in G.
=QQY,
= gx,,
gxa, so:
Then, for g E Q and a E G, xga
xsa =
(iii)
(iii)
x s = gg
xg
and y(g,
y ( g ,a)
a )==1l
for
f o ralla lglEg Q
~ Q and
a EEG.
U
G.
Now (ii) and (iii)
(iii) imply:
imply:
y(gb, c) = y(b, c) ffor
y(gb,c)=y(b,c)
o rall
a l lb,b c, cE~G
G and
and
(iv)
(iv)
g EEQ.Q .
~
Next for C
G define
define p(c)
,B(c)==n,,y
I1yEpy(y,
c). By
By (iv),
(iv), p(c)
,B(c)isisindependent
independentof
of the
the
c EE G
y(7, c).
representatives. But
choice of the set
set Y
? of coset representatives.
But if b E G then Yb
?b is another
another set
set
representatives for
for Q in G, so:
of coset representatives
fl
NO=
p(c)
= fl y(yb,
y(gb, c)
c) for all
all b,
b, cc EE G.
G.
(v)
(v)
yFk€ 9
~
As V is abelian
abelian we conclude
conclude from (ii) that
(flv(bc)) (flv(b))
yEY
yEY
_
fj Y(y, bc)
yEY
(flYbc))
yEY
Representations of groups on groups
Representations
32
32
and then appealing to (v) we obtain
(vi)
(vi)
j(c)j(b)
=,(bc)y(b,
= jG
IG :: P1.
PI.
j ? ( ~ ) j ? (',b=
) ~j?(bc)y(b,
~
cc)') ~for all b, c E G, where m =
As PP EE Sylp(G),
Hence we
we can
Syl,(G), (m,
(m, p)
p)==1.1.Thus
Thusmmisisinvertible
invertiblemod
modI V
I V 1.
I. Hence
? ( ~ ) - ~ ,-for
'for
, cc EEG.
power of
of (vi)
(vi) we
we
define
define a(c)
a(c) =
= j,6(c)-'"-'
G.Then
Thentaking
takingthe
the-m-'
-m-1 power
obtain:
(vii)
a(c)a(b)
c)-1
a ( ~ ) a ( ,b=a(bc)y(b,
=a(bc)y(b,
)~~
c)-'
for all
all b,
b, cc EE G.
G.
Finally
define y,
y,,=
=x,a(a)
xaa(a) for aii EE G and
E G).
G}. H will be
Finally define
and set
set H
H=={y,,:
{y,: a E
be
shown to be a complement to V in G.
G. This
This will complete
complete the proof.
ybcfor
forall
allb,b,ccEEG.G.But
Butyby,
ybyC
=xba(b)x0a(c) =
_
yby, ==yb,
=xba(b)xca(c)
It suffices to show
show YbYc
c)a(b)x`u(c)=ybca(bc)-ly(b,
c)a(b)X°a(c).
Then,
xbx,a(b)XCa(c) =xbcy(b, c)a(b)xca(c) = Ybca(b~)-l
y (b, c)a(b)xca(c).
Then,
V isisabelian,
abelian,(vii)
(vii)implies
impliesybYC
yby, ==Ybc,
yb,, as desired.
as V
desired.
xbx`a(b)X`a(c)=xbcY(b,
Theorem in
in section
section 18
18 is another
The Schur-Zassenhaus
Schur-Zassenhaus Theorem
another useful
useful result on
splitting.
splitting.
11 Central
Central products
products and
and wreath
wreath products
products
(11.1)
< n)
n} be a set of subgroups of G.
(11.1) Let
Let {G1:
{Gi: 11_(< ii _(
G. Then
Then the
the following
following are
are
equivalent:
equivalent:
(1) G ==( (Gi:
(1)
G i : l1( i<i
( n<n)
) a n and
d [ G[Gi,
i , G jG]
] = =1
l f o for
r i # ij .0 j.
asurjectivehomomorphismofG1
(2) Themap7r:
Themapn: (x1,
(XI,...
. . .,,x")
x,) H
H x1
xl ..... .x"
x, is
isasurjectivehomomorphismof
G1x
=D
D onto G with
=G
=1,
. . x G"
G, =
with Diir
Din =
Giand
and Di
Diflf
lker(7r)
ker(n) =
1, where
where Di
consists of those elements of D with
with 11 in all
all but the ith
ith component.
component.
If either
conditionsofof 11.1
11.1holds,
holds,then
thenGG isis said
said to
to be
be a
either of the equivalent conditions
product of
of the subgroups Gi, 1 (<i i < n. Notice
Notice that
that the kernel of
of the
central product
homomorphism nit of 11.1
of G1 x . . . x
x G,.
G.
11.1 is contained in the centre of
(11.2) Let (Gi
< n) be a family of groups such that Z(G1)
Z(G 1)S
- Z(G1)
(11.2)
(Gi:: 11 _(< ii _(
Z(Gi) and
and
Aut(Z(Gi))=AutAut(G,)(Z(Gi))
for
1
<i
<n.
Then,
up
to
an
isomorphism
Aut(Z(Gi)) = A U ~ ~ " ~ ( ~ , ) ( Zfor
( G1~() i) (n. Then, up to an isomorphism
mapping
Gi to Gi
mapping Gi
Gi for
for each
each i,i,there
thereexists
exists aaunique
uniquecentral
central product
product of
of the
the
groups Gi in which Z(G1)
Z(G1)==Z(G1)
Z(Gi) for
for each
each i.i.
Proof. Adopt the notation of 11.1
11.1 and identify
identify Gi with
with Di.
Di.By
By hypothesis
hypothesis
there are isomorphisms ai:
ai: Z(D1)
<ii <n.
Z(D1) + Z(Di),
Z(Di), 11 5
5 n.Let
LetEEbebethe
thesubgroup
subgroup
of D generated by
by z(z-'ai),
z(z-lai), z EE Z(D1)1
< ii 5
<n.
Z(D1)l <
n. Observe
Observe E isis aa complement
complement
to Z(Di)
): 1 5< ii 5
< n)
n) =
= Z(D) for each i.i. Thus
Thus D
DIE
Z(Di) in Z =
=(Z(Di
(Z(Di):
I E is
is aa central
central
groups Gi with Z(G1)
= Z(G1)
Z(Gi) =
Z(G1) for each i,i, by 11.1.
11.1.
product of the groups
and wreath
wreath products
Central products and
33
33
Z(G1)
each
Next assume G is a central product of the G,
Gi with Z(G
1)=
= Z(Gi)
Z(G,) for each
and let
let 7r:
n : D + G be
be the
the surjective
surjectivehomomorphism
homomorphism supplied
supplied by
by 11.1.
11.1.Let
Let
i,i , and
PI:Z(D1)
Z(Dl) + Z(Di)
Z(Dl)be
be the
theisomorphism
isomorphismwhich
which isis the
the composition
compositionof
of irIz(o,):
nlziD,):
Pi:
Z(D1)
- Z(G1)
Z(D1) +
Z(G1)and
and(7r(z(o;))-1:
(n[Z(D,))-l:Z(G,)
Z(G,) + Z(Di).
Z(D,).Observe
Observe
ker(7r)
= (z(z-1pi):
Z(D1), 1 5
<i<
ker(n) =
(z(z-'~i):zz E
E Z(D1),
5 n)
n )= A
A
is a complement to
to Z(Di)
Z(Di) in Z for
for each
each i,i, and
and of
of course
course G
G=
2D/A.
DIA.To
Tocomcomplete the proof
Aut(D) with
with Diy
Di y=
= Di
Di and
and Ey =
A. Notice
proof I exhibit y E Aut(D)
=A.
Notice yy
induces an isomorphism of
of D
DIE
I E with D/A
DIAmapping
mappingGi
Gito
toGi,
Gi,demonstrating
demonstrating
uniqueness.
uniqueness.
= (ai)-lBi: Z(Di)
Z(Di),so
SOthat
that8iaiEE Aut(Z(Di)).
Aut(Z(Di)).By
Byhypothesis
hypothesis
Let
Z(D,) + Z(D,),
Let 8i =(a1)`1,Bi:
there
= ai.
8i. Define y:
y: D
yi EE Aut(Di)
Aut(Di)with
withyiyiII z (z(oi)
D,) =
D +D
D by
by
there is
is yi
(x1,...,xn)H(x1,x2Y2,...,xnYn)
and observe y E
so that Ey
A. Thus
E Aut(D) with (z(z-1ai))y
( ~ ( z - l a ~==
) z(z-1,Bi),
)z(z-'pi),
~
E y==A.
Thus
the proof is
is complete.
complete.
central product
product of
of the
the groups
groups
Under the hypotheses of 11.2, we say G is
is the
the central
G,
1 **G2
Gi with
with identified
identijied centers,
centers,and
andwrite
writeGG==GG1
G2**. . .. .. ** Gn.
G,.
be aa group
groupand
and 7r:
n: G + Sym(X)
Sym(X) aa permutation
permutation representation
representation of
of G
G on
on
Let L be
direct product
product D
D of n copies of L. G acts as a group
={1,
(1, .... ..,,n}.
n ) . Form the direct
X=
of automorphisms
automorphisms of D via
via the
the representation
representation aadefined
definedby
by
-
xng-in).
ga: (xi,
(XI,...
. . .,,xn)
x,) H (xig-'n,
(xlg-I,, ...
. . .,, xng-I,).
The wreath
of LL by
by G
G(with
(withrespect
respectto
to7r)
n ) is defined
defined to be the semidirect
wreath product of
semidirect
product S(G, D, a). The wreath
wreath product
productisisdenoted
denotedby
byLwr
LwrGGororLwr,G
Lwr,G or
product
Lwr,G.
Lwrr
G.
(11.3)
G be
(11.3) Let W
W=
=Lwr,
Lwr,G
be the
the wreath
wreath product
product of
of LL by
by GGwith
withrespect
respect to
to Jr.
n.
Then
(1) W
product of
of D
D by
= L1 x . . . x Ln
W is
is aa semidirect
semidirect product
by G where D =
L, is a
copies of L.
L.
direct product of n copies
(2) GGpermutes
: 11 < i <5nn }) via conjugation and the
permutesAA=={Li
{Li:
the permutation
permutation
representation
of G
G on A is
is equivalent
equivalent to r.
n.That
Thatisis(L1)g
(Li)g==Lign
Lignfor
for each
each
representation of
g E G and 1 ( i c n .
gEGandI<i<n.
(3) The
Thestabilizer
stabilizer Gi
Gi of
of ii in
in G
G centralizes
centralizes Li.
Li.
Exercises for chapter
chapter 33
A1) 1)
assumeAAisis represented
represented
1. Let
Let GGand
andAAbebefinite
finitegroups
groupswith
with(I (IGG1,I , ((A
==1,1,assume
on G
<i<
n)
(Gi:: 0 (
I
n ) is
is an
an A-invariant
A-invariant
G as
asaagroup
groupofofautomorphisms,
automorphisms,and
and(Gi
normal series for G such
/Gi for 00 <
such that
that A
A centralizes
centralizes Gi+1
Gi+1/Gi
(i <
<n.
n. Prove
Prove
34
34
Representations
Representations of
of groups
groups on
ongroups
groups
A
when (1
(IAA1, G
j) #
01.1.(Hint:
A centralizes
centralizes G. Produce
Produce a counter
counter example
example when
(GI)
(Hint:
Reduceto
tothe
the case
casewhere
whereAA isisaap-group
p-groupand
anduse
use5.14.)
5.14.)
Reduce
2. Let
LetGGbe
beaafinite
finitegroup,
group,ppaaprime,
prime,and
and XXaap-subgroup
p-subgroupof
of G.
G.Prove
Prove
2.
(1)
(1) Either
Either X
X EE Sylp(G)
Sylp(G) or
or X
X isis properly
properly contained
contained in
in aa Sylow
Sylow p-subgroup
p-subgroup
of NG(X).
NG(X).
of
G and
and
(2) If G
G is
is aa p-group
p-group and
and X
X isis aamaximal
maximal subgroup
subgroup of G, then X 4 G
(2)
( G : X I=p.
=p.
JG:X1
3.
3. Prove
Provelemmas
lemmas10.1
10.1and
and10.3.
10.3.Exhibit
Exhibitaanonsplit
nonsplitextension.
extension.
4.
4. Let
Letppand
andqqbe
beprimes
primeswith
withpp>>q.q.Prove
Proveevery
everygroup
groupof
of order
orderpq
pqisisaasplit
split
extension
by Z,.
7Lq.Up
Uptotoisomorphism,
isomorphism,how
how many
many groups
groups of order pq
extensionofof7LP
Zp by
pq
exist?
exist? (Hint:
(Hint: Use
Use10.3
10.3and
andExercise
Exercise1.7.
1.7.Prove
Prove Aut(Zp)
Aut(Z,) isiscyclic
cyclicof
oforder
order
pp --1.1.You
Youmay
mayuse
usethe
thefact
factthat
thatthe
themultiplicative
multiplicativegroup
groupofofaafinite
finitefield
fieldisis
cyclic.)
cyclic.)
5.
< i <5 n,
5. Let
LetGGbe
beaacentral
centralproduct
productofofnncopies
copiesG1
Gi,, 1 5
n, of
of aa perfect
perfect group
group L and
let
bean
anautomorphism
automorphismofofGGofoforder
ordern npermuting
permuting{G1
{Gi::11<5ii<5n}
n )transitransilet aa!be
tively.
Prove
CG(a!)
=
KZ
where
K
=
~
~
(
a
!
)
(
Z
'
)
L/
u
for
some
u
5
Z(L)
tively.
CG (a) = KZ where K = CG (aP) - L/ U for some U < Z(L)
and Z=CZ(G)(a).
Z = CZ(G)(~).
FurtherNA,t(G)(GI)
NAut(~)(G1)
nn C(K) <C(G1).
iC(G1).
and
Further
6. IfIf AA acts
actson
onaagroup
groupGGand
andcentralizes
centralizes aa normal
normal subgroup H of G
G then
then
6.
[G,A]
A1<,
6 CG(H).
CG(H).
[G,
a
4
Linear representations
Chapter 4 develops
develops the
the elementary
elementary theory
theory of
of linear
linearrepresentations.
representations. Linear
Linear
representations
modules over the group
representationsare
are discussed
discussed from
fromthe
the point
point of view of modules
ring. Irreducibility and indecomposability are
are defined,
defined, and
and we
we find
find that
that the
Jordan-Holder
Jordan-Holder Theorem
Theorem holds
holds for
for finite
finite dimensional
dimensionallinear
linear representations.
representations.
Maschke's Theorem is established in section
section 12.
12. Maschke's
Maschke's Theorem
Theorem says
says
that, if G is
a
finite
group
and
F
a
field
whose
characteristic
does
not
divide
is a finite group
F a field whose characteristic
the order of G,
representationsofofGG over
over FF are
G , then
then the
the indecomposable
indecomposable representations
are
irreducible.
Section
Section 13
13 explores the connection
connection between finite dimensional
dimensional linear
linear reprerepresentations
sentations and matrices. There is also a discussion
discussion of the special linear group,
the general
general linear
corresponding projective groups.
particular
linear group,
group, and the corresponding
groups. In particular
we find that the special linear group is generated
generated by its transvections
transvections and is
almost
almost always perfect.
Section 14
14 contains a discussion
discussion of the dual
dual representation
representation which will be
needed in section
section 17.
17.
12 Modules
Modules over the group ring
Section 12
over aa field
field FF using the group
group ring
ring of
of
12 studies linear representations over
G over F.
F .This
Thisrequires
requiresan
an elementary
elementaryknowledge
knowledge of
of modules
modules over rings. One
reference for this material is chapter 3 of Lang [La].
space over
over F.
F. The group of autoThroughout section 12, V will be a vector space
morphisms of V in the category of
of vector
vector spaces
spaces and
and F-linear
F-linear transformations
transformations
general linear group GL(V). Assume n:
n: G +
-* GL(V)
is the general
GL(V) is a representation
this category.
category. Such representations will be called FG-representations
of G in this
FG-representations
and V will be called the representation
representation module for
for n.
n. Representation
Representation modules
FG-representations will
for FG-representations
will be
be termed
termed FG-modules.
FG-modules.
= F[G]
Let R =
F [GI be
be the
the vector
vector space
space over F with
with basis
basis G
G and
and define
define multiplication on R to
of G. Hence a
to be
be the
the linear
linear extension of the multiplication of
agg,
F and
typical member
agg,where
whereag
a, cEF
and at
at most
most aa finite
finite
member of R is of the form
ag are nonzero.
nonzero. Multiplication
Multiplicationbecomes
becomes
number of the coefficients a,
xgEG
(ag)(bhh)=
a
h
agbhgh.
a ,hEG
Linear representations
representations
3366
As is well known (and easy to check), this multiplication makes R into a ring
with identity and, as
as the
the multiplication
multiplication on.
on R commutes with scalar multiplimultiplication from
from FF,, R is
=F[G]
F[G]isisthe
thegroup
groupring
ring or
or group
group
is even
even an F-algebra. R
R=
algebra of G over F.
F.
Observe that V becomes a (right) R-module
R-module under
under the
the scalar
scalar multiplication:
multiplication:
v
(Eagg) = Eag(v(gn)) vEV, Eagg c R
R-module
representation a:
a: G -+ GL(U)
an R-module then we have a representation
Conversely if U is an
defined by
by u(ga)
u(ga) =
product of
of uu Ec U by
by g Ec R.
=ug,
ug, where
where ug
ug is the module product
GL(V1), ii =
= 1,
2, are
V1-+
- V2
ni: G -+ GL(Vi),
1,2,
are FG-representations
FG-representationsthen
then,B:
B: V1
V2
Further
Further ifif 7r,:
is an equivalence
equivalence of the representations
isomorphism of
representationsprecisely
preciselywhen
when,B
B isis an isomorphism
anFG-homomorphism
FG-homomorphism if
the corresponding
Vl -+ V2
Vz isis an
corresponding R-modules.
R-modules.Indeed
Indeedy:y:V1
and only if y is an R-module homomorphism
homomorphism of the corresponding R-modules.
R-modules.
Here y:
V2isisdefined
definedtotobe
bean
FG-homomorphismififyy is
is an
anF-linear
F-linear map
map
Vl -+ V2
an FG-homomorphism
Here
y:V1
commuting
of G in
in the
the sense
sensethat
thatv(gnl)y
v(gn1)y =
commuting with the actions of
=vvyy (gn2)
(gn2)for each
vE
V1,and
andggEcG.
G. In
In the
the terminology of
of section 4, the FG-homomorphisms
E Vl,
are the G-morphisms.
G-morphisms.
The upshot of these observations is that the study of FG-representations is
equivalent to
to the
the study
study of
of modules
modules for
for the
the group
group ring
ring FIG]
F[G] =
= R. I will
will take
take
equivalent
both points of view and appeal to various standard theorems on modules over
rings. Lang [La] is a reference for
for such
such results.
results.
Observe
also that
that V
V is an abelian group
group under
under addition
additionand
andnn is a repreObserve also
sentation of G on V in the category of groups and homomorphisms. Indeed n
representation
induces a representation
n': F#
F' x G
G -+ Aut(V)
that category
category defined
defined by
by(a,
(a, g)n': vv i-->
av(g7r), for
for aa E
E F'
F#,
Here F
F#
in that
H av(gn),
, g cE G..
G. Here
' is
group of
of F
F.. Two FG-representations
FG-representations nn and
the multiplicative group
andor
a are equivalent
equivalent
if n'
jr' and
and a'
a'are
areequivalent,
equivalent,so
sowe
wecan
canuse
usethe
theresults
results of
of chapter
chapter 33
if and only if
to study FG-representations. In the case where F
F isis aa field
field of
of prime
prime order
order we
we
can say
say even more.
more.
(12.1) Let F
(12.1)
F be
be the
the field
field of
of integers
integers modulo p for
for some
some prime p and
and assume
assume
V is of finite dimension. Then
(1) VV isis an
and n is
of G in
an elementary
elementary abelian p-group and
is aa representation
representation of
the category
category of groups and
and homomorphisms.
homomorphisms.
(2) IfIfUUisisan
an elementary
elementaryabelian
abelianp-group
p-groupwritten
writtenadditively,
additively,then
then U
U is
is a vector space FFUUover
overFF,, where
wherescalar
scalarmultiplication
multiplicationisisdefined
definedbyby((p)
((p) +n)u
n)u=
= nu,
nu,
nEl,uEU.
~ E ~ , u E U .
(3)
U) is equal to the group
(3) GL(F
GL(FU)
group Aut(U)
Aut(U) of
of group
group automorphisms
automorphisms of U.
U.
Indeed if W is an elementary abelian p-group then the group homomorphisms
homomorphisms
transformations
from
F UF into
FW
W.
U into
intoW
Ware
areprecisely
preciselythe
theF-linear
F-linear
transformations
from
U into
F W.
from U
+
Modules over the group ring
37
(4)
V defined using the construction in part (2) is pre(4) The
Thevector
vectorspace
spaceFFV
cisely the vector space V.
V.
F is
the FG-representations
FG-representations
As a consequence of 12.1,
12.1, if F
is a field of prime order, the
are the same as the representations of G on elementary
elementary abelian p-groups.
p-groups.
subspace U of V
V is
is an
an FG-submodule
FG-submodule of V
V if
if U
U isisG-invariant.
G-invariant.
A vector subspace
U is an
an FG-submodule if and only if U is an
an R-submodule
R-submodule of the R-module
V.. From 7.1 there
there are
are group
group representations
representationsof
ofF#
F# xx G on U
V
U and
and V/U.
V/U.These
These
representations are
are also
also FG-representations
FG-representations and
and they
they correspond
correspond to
to the
the Rrepresentations
and V/
V /U.
modules U and
V is irreducible or simple ifif 0 and V
V are the only FG-submodules. A comV
V is
is aa series
series
position series
series for V
O=Va<Vi<...<V,,=V
of FG-submodules such that each
/ Viisis a simple FGof
each factor
factor module
module V,+1
Vi+I/V;
in section
section 7.
7. The
The
module. This corresponds to the notion of composition series in
family (Vi+l/
(Vi+1/ Vi:
Vi:005<ii < n)
is
the
family
of
composition
factors
of
the
series.
n) is the family of composition factors
If V
that V
V possesses
possesses a composition
composition
V is of finite
finite dimension
dimension then it is easy to see that
series. Appealing to the Jordan-Holder Theorem,
Theorem,established
establishedin
in section
section7,
7, and
and
to remarks above, we get:
(12.2) (Jordan-Holder
(Jordan-Holder Theorem
Theoremfor
for FG-modules)
FG-modules) Let V be
be aa finite
finite dimen(12.2)
V possesses
possesses aa composition
composition series
series and
and the
the composicomposisional FG-module. Then V
tion factors are independent (up to order and
tion
and equivalence)
equivalence) of the choice of
composition series.
series.
The
restrictions ni
composition series
Therestrictions
ni==7rnIIv,v,/v,-,,
/ v , _ , , 0 << ii <5n,n,ofof7rn toto the
thecomposition
series(Vi:
(Vi:0 <5
i <5n)n)ofofaafinite
finitedimensional
dimensionalFG-representation
FG-representation7r
n are
are called
called the
the irreducible
irreducible
n.They
They are
are defined
defined only
only up
up to order
order and
and equivalence
equivalencebut, subsubconstituents of jr.
ject to
to this
this constraint,
constraint, they
they are
are well
well defined
defined and unique by the Jordan-Holder
Jordan-Holder
Theorem.
Theorem.
V
FG-submodules U
U and W of
V is decomposable
decomposable if there exist proper FG-submodules
of VV
=U
n2
with V =
U®
@ W.
W .Otherwise
OtherwiseVV isisindecomposable.
indecomposable.I'll
I'llwrite
write7rn==7r1
nl + n
2 if
V=
= Vl
I v; is equivalent
Vl ®
@ V2
V2 with Vl
Vl and
andV2
V2FG-submodules
FG-submodulesofofVVand
and7rnlv,
equivalent to
ni.
ni.Observe
Observe that if a ==al
a1+ a2
a2isisan
anFG-representation
FG-representation with
with ai
ai equivalent
equivalent to
ni
to aa..
ni for
for ii =1
= 1and
and2,2,then
then7r
n is
is equivalent to
As in section
section 4,
4, an
an FG-module
FG-moduleVV isissaid
saidto
tobe
be the
the extension
extensionof
of aa module
module X
X by
by
module YY ififthere
thereexists
existsaasubmodule
submoduleUUofofVVwith
withUUZ= XX and
and VV// U
UZ
= Y.
Y. A
a module
complement to U in V is an FG-submodule
FG-submodule W with V =
=U
U®
@ W.
W . The extension
is said to split if U possesses
possesses a complement in V.
V . As in chapter 3, we wish to
investigate
investigate when extensions
extensions split.
split.
+
+
Linear representations
38
An R-module
xr: rrEe R]
R }for
forsome
somexx E V. EquivaR-module VVisiscyclic
cyclicififVV==xxRR==f {xr:
= (xG)
lently V =
(xG) is
is generated
generated as aa vector
vector space by the images of x under G.
The element xx is said to be a generator for the cyclic module V. Notice that
irreducible modules are cyclic.
irreducible
cyclic.
(12.3)
If V =
=xR
(12.3) (1) If
x Risiscyclic
cyclicthen
thenthe
themap
maprri-+
H xr
x r isissurjective
surjectiveR-module
R-module
R: xxrr =
= 01.
homomorphism from R onto V
V with kernel A(x) =
=f(rr E R:
0).
images of
of cyclic
cyclic modules
modulesare
are cyclic,
cyclic, so
so the
the cyclic
cyclic RR(2) Homomorphic
Homomorphic images
images of
of R.
modules are precisely the homomorphic images
(3) V is irreducible if and only if A(x) is a maximal right ideal of R.
Given R-modules
R-modules U
U and
and V, HomR(U,
V) denotes
denotes the
the set
set of
of all R-module
Given
HomR(U, V)
R-module
homomorphisms
of U
U into V.
V) is an
homomorphisms of
V. HomR(U,
HomR(U, V)
an abelian
abelian group
group under the
the
definition of addition:
addition:
following definition
+
+
u(a
E U,
U, a,
a B/? EE HomR(U, V).
u(a +,8)
/?)==ua
ua+ u,8
u/? u E
HomR(U, V) is even an R-module when scalar multipliIf R is commutative, HomR(U,
cation
cation is defined
defined by
u(ar) = (ur)a u E U, r E R, a E HomR(U, V).
(V,V)
V)=
= EndR(V)
EndR (V)isisaaring,
ring,where
wheremultiplication
multiplication isis defined
defined to
to
Finally HomR
HomR(V,
composition. That is
be composition.
u (a
,8) _ (ua),8
u E V, a, ,8 E EndR (V ).
In the language
languageof
of section
section4,4,HomR
H o m (U,
(U,
~ V)
V) ==MorG
Morc (U, V)V).
V be
be R-modules
R-modules and aa EE HomR(U,
HomR(U,V).
V).
(12.4) (Schur's Lemma) Let U and V
(12.4)
Then
(1) If
(1)
If U
U isis simple
simple either aa = 00 or
or aaisisan
an injection.
injection.
(2) If
If V
V is
is simple
simple either a =
=00 or
or aa isis aa surjection.
surjection.
(3) IfIf UU and
and VVare
aresimple
simplethen
then either
either aa ==00or
or aaisisan
anisomorphism.
isomorphism.
(4) If
If VVisissimple
simplethen
then EndR(V)
EndR(V)is a division ring.
The module V is
is aa semisimple
semisimple R-module
R-module ifif V is the
the direct
direct sum
sum of
of simple
simple
submodules.The
The socle
socle of
of V is the submodule
by all the
submodules.
submodule Soc(V) generated
generated by
simple submodules of V.
V.
(12.5) Assume
Assume Q is a set of simple
of V and A C_ Q such that
that
(12.5)
simple submodules
submodules of
V=
= (0)
A.Then
Thenthere
thereexists
existsr rGc SZ
0 with
(Q)and
and (A)
(A)=
=®AEA A.
with A
AC
E Fr such
such that
that
V=
= ®bEr B.
B.
a,,
eAEA
39
Modules over the group ring
Proof. Let SS be
be the
the set of rrCEQ
C2 with
with AA C_
E Fr and
and (F)
(r)==®BErB.
@,,,B. Partially
Partially
that ifif C is a chain in S then UrEC
order S by inclusion. Check that
U,, Fr is an
an upper
bound for C in S.
Zorn's Lemma there
there isis aa maximal
maximalmember
memberrr of
S. Hence by Zorn's
S. Finally prove
prove V =
= (r).
(F).
(12.6) The
(12.6)
The following are
are equivalent:
(1) VVisis semisimple.
semisimple.
(2) V
=Soc(V).
(2)
v=
Soc(V).
(3) VVsplits
splitsover
over every
every submodule
submodule of V.
V.
Proof. The
Proof.
Theequivalence
equivalenceof
of (1)
(1)and
and(2)
(2)follows
followsfrom
from12.5.
12.5.
complement U to Soc(V)
Soc(V)
Assume
Assume(3)
(3) holds
holds but
but V
V # Soc(V).
Soc(V). By
By (3)
(3) there is a complement
in V. Let xX EE u#,
U#, II a maximal
maximal right ideal of R containing
containing A
(x)
_
(r
E
R:
x
= 01,
A(x) = (r ER: x rr =
and W the image of II in
in xx R under the homomorphism of 12.3.1.
12.3.1. By (3) there
Z to W
is a complement
complement Z
W in
in V.
V.By
Bythe
theModular
Modular Property
Property of
of Groups,
Groups, 1.14,
1.14,
M isacomplementto
is a complement toWinxR.ThenM
W in x R. Then MZxR/W,soMissimple
x R/ W, so M is simple
Z nflxxRR==M
by 12.3.3. Hence
Hence M
M5
< Soc(V), so 0 $ M
M<5Soc(V)
Soc(V)flnUU==0,0,aacontradiction.
contradiction.
Thus
Thus (3)
(3) implies
implies (2).
(2).
Finally
U isis aa submodule
submoduleof V
V with
with no
no complecompleFinally assume
assume V
V is
is semisimple
sernisimpleand
and U
Soc(U) =
= ®AEoA
@,,,A for
forsome
someset
setof
of simple
simplesubmodules,
submodules,so
so by
by
ment in V. Now Soc(U)
12.4 there
there isisaaset
setrr of
of simple
simplesubmodules
submodulesofofVVwith
withAAECrFand
andVV== ®BEr B
B..
Then W =
= (F
(r--A)A)isisaacomplement
complementto
to Soc(U)
Soc(U) in
in V.
V. Hence U # Soc(U).
Soc(U).By
By
Property of
of Groups,
Groups, 1.14,
1.14, U
U ='SOC(U)
= Soc(U) CB
® (U fl W). Thus U fl W
the Modular Property
W
has no simple
simple submodules.
submodules.
Choose
(U nfl w)#
W )Ososothat
thatx xhas
hasnonzero
nonzeroprojection
projectionon
onaaminimal
minimal number
number
x EE (U
ChooseX
members of
of rr,, and
E rF such
such that
thatxxa
nn of members
and let A E
a # 0,
0,where
where u:
a:xR
x R -+
+A
A is
is the
the
projection of
of x RR onto
onto A.
A.For
For00 0 yy EE Xx R,
R, the
the set
set supp(y)
supp(y) of
of members
membersof
ofrr
upon which y projects
projects nontrivially is a subset of supp(x). Thus by minimality
-+ A is an injection, and hence, by
of n, supp(x) =
=supp(y).
supp(y). Therefore
Therefore a:
a:xx R +
simple, whereas
whereas it has already
isomorphism. But
xR
12.4.2, a is
is an
an isomorphism.
Butthen
thenx
R Z A is simple,
been observed
observed that U
U fln W
Whas
has no
no simple
simplesubmodules.
submodules.
+
+
+
+
(12.7)
Submodules and
and homomorphic
homomorphic images
images of
of semisimple modules
(12.7) Submodules
modules are
semisimple.
semisimple.
(12.8)
let U be an FG-submodule
FG-submoduleof
of V,
V, and
andififchar(F)
char(F) =
=
(12.8) Assume G is finite, let
Sylp(G).
p >>00 assume
assume there
there is an FP-complement W to U in V for some P EE Syl,(G).
Then V splits over U.
Proof.
of V with
with V
V=
=U
U@
® W,
W, and
and ifif char(F)
char(F) =
=
Proof. Let
Let W
W be
be aa vector
vector subspace of
p >>00choose
If char (F)
(F) =
= 0 let
chooseW
Wto
tobe
be P-invariant
P-invariant for
for some
some PP EE Sylp(G).
Syl,(G). If
Linear representations
representations
40
P=
=1.
1. Let
Let X
X be
be aaset
setof
ofcoset
cosetrepresentatives
representatives for PP in
inGGand
andlet
let7r:
n: V
V -+
-+ U
U
be the projection
projection of
of V
V on U with
respect
to
the
decomposition
V
=
U
®
W.
with respect
decomposition = @ W.
= jG
P) and define
define8:
9: V
V-+
-+ V
Vby
by89=
=(Ex,,
(r_xexx-'nx)/n,
x-17rx)/n, where the sum
Let n =
IG : PI
takes place
placeininEndF(V).
EndF(V).As
As(p,
(p,n)n)=1,
x-1, and
= 1, l1ln
/ n exists in F.
F . Also
Also x, x-',
and n
takes
are in EndF(V),
so 80 is
memberofofEEndF(V).
Wisis PPEndF(V), so
is aa well-defined
well-defined member
n d ~ ( v ) As
. AsW
invariant,hh7r
forallallh hE EP,P,sosoififx xH
H h,
hx is a map
map from
from X
X into
into P then
invariant,
n ==n7rh
h for
= (EXEX
x-1(hx)-17rhxx)/n. That
60' =
(Ex,, x-'(h,)-'nh,x)/n.
Thatisis96' isis independent
independentof the choice of coset
representatives X
X of
of P
P in G.
representatives
Claim 06 E EndR(V).
EndR (V).As
Asthe
themultiplication
multiplication in
in R
R is aa linear
linear extension
extension of that in
Claim
G and 80 E
E EndF(V),
EndF(V), it suffices to
to show
showg6
g0=
= Bg
Ogfor
forall
allggEEG.
G.But
Butgo
gO==(Ex,,
(rxex
((xg-1)-17rxg-1)g/n
~ ~ - ' ) - ' n x ~ - ' )=
~=
/Og
nBgasasXg-1
xg-'isisaaset
setof
ofcoset
cosetrepresentatives
representatives for P in
in G
G
and 80 is independent
independent of the choice of X.
It remains
remains to observe
is the
the identity
identity on U, as W =
= ker(7r),
n is
ker(n), and
and as
as U
U is
is
observethat,
that,as
as 7r
G-invariant,
identity on U and U =
O.Hence
Hence B2
02=
= 6.
0. ThereThere=VVB.
G-invariant,we
we also
also have
have 08 the identity
= VO
® ker(6).
ker(0). As
As 80 EE EndR(G),
EndR(G),ker(6)
ker(0) is
is an FG-submodule. Hence,
fore V
V=
VB @
= V0,
V8, ker(0)
ker(8) is
is aa complement
complement to U
U in V.
V. That
That is
is V
V splits
splitsover
over U.
U.
as U =
(12.9) (Maschke's
(Maschke's Theorem)
Theorem) Assume
AssumeGG is
is a finite group and
and char(F) does
(12.9)
does
not divide the order of G.
G. Then
Thenevery
every FG-module
FG-module is
is semisimple
semisimple and
and every
every
FG-extension
FG-extension splits.
splits.
Proof. This
Thisisis aa direct
direct consequence
consequenceof 12.6
12.6 and 12.8.
12.8.
Using 12.9
12.9 and notation and terminology introduced earlier in this section we
have:
(12.10)
(12.10) Assume G is aa finite
finite group and char(F) does not divide the order of
G -+
-->GL(V)
GL(V) be
be aa finite
finite dimensional
dimensional FG-representation.
FG-representation. Then
G. Let 7r:
n: G
Then
(1) 7r
(7ri:11(<ii 5
< r).
n=
=yi_1
E;=,7ri
niisisthe
thesum
sumof
of its
its irreducible
irreducible constituents (xi:
(2) IfIf aa==Ei_1 aiaiis is
another
finite
another
finitedimensional
dimensionalFG-representation
FG-representation with
with
constituents(ai:
(a1:11(< ii <
to aa if and only
(s)
s)then
then 7r
n is equivalent
equivalent to
irreducible constituents
=s and
and there is a permutation aa of {1,
{1,2,
r) with ni
equivalent to aiv
ai,
if r =
2, ..... .,,r}
it equivalent
for each ii..
xi=,
So, in this special case, the study of
of FG-representations
FG-representations is essentially reduced
irreducible FG-representations.
to the study of irreducible
FG-representations.
Let V
V be
be aa semisimple
semisimpleR-module
R-moduleand
and SSaasimple
simpleR-module.
R-module.The
ThehomogehomogecomponentofofVVdetermined
determinedbybyS Sisis(U:
(U:UU(<V,
V,UU Z= S).
S).V
V is
is homogehomogeneous component
neous if it is generated
generated by
by isomorphic
isomorphicsimple
simple submodules.
submodules.
(12.11) Let V be a semisimple
(12.11)
semisimple R-module.
R-module. Then
Modules over the group ring
41
41
homogeneous then
then every
every pair of
of simple
simple submodules
submodules of
(1) IfIf VVisishomogeneous
of V is
isomorphic.
isomorphic.
thedirect
directsum
sumof
ofits
itshomogeneous
homogeneous components.
components.
(2) VVisisthe
eAGQ
Proof.
®A,12 AAfor
simple submodules
submodules
Proof.As
AsVVisissemisimple,
semisimple,VV==
forsome
someset
setS2
S2 of simple
of V.
supp(T)the
theset
setof
ofsubmodules
submodulesin
in
V. Let
Let TT be
be aa simple
simplesubmodule
submoduleof
of VVand
and supp(T)
SZ
uponwhich
which T
T projects nontrivially. If A E
E supp(T)
supp(T) then the
S2 upon
the projection
projection map
map
a:
by Schur's Lemma. But if V
a:TT -->
-+ AA is an isomorphism by
V is
is homogeneous
homogeneous
then, by 12.5,
= SS for
AZ
for some
some simple
simple R-module
R-module S and
and all
all
12.5, we may choose A
E S2.
Hence(1)
(1)holds.
holds.
AE
0. Hence
Similarly if SS is
is aasimple
simpleR-module,
R-module, HHthe
thehomogeneous
homogeneouscomponent
component of
of
V determined by
by S, and K the
the submodule
submodule of V
V generated
generated by the
the remaining
remaining
homogeneous components, then,
then, as
as V is
is semisimple, V
V=
= H + K.
K. Further
Further ifif
<H
12.7. But now, by (1), S =TT Z= Q
HnK#O,wemaychooseT
H f1
n KKbby
y 12.7.Butnow,by(l),SZ
Q
H
f1 K 0, we may choose T 5
for some
some simple
simple R-module
R-moduleQQ determining
determiningaahomogeneous
homogeneouscomponent
componentdistinct
distinct
S, aa contradiction.
contradiction.
from that
that of
of S,
from
+
(12.12) Let
andUUaasimple
simpleFH-submodule
FH-submoduleof
of V.
V. Then
Then
Let H
H <9GGand
(1) Ug
G.
Ugisisaasimple
simpleFH-submodule
FH-submoduleof
of VVfor
foreach
eachgg EE G.
CG(H)then
then U
U isis FH-isomorphic
FH-isomorphic to
to Ug.
Ug.
(2) IfIf ggEE CG(H)
(3) IfIf XXand
andYYare
areisomorphic
isomorphicFH-submodules
FH-submodulesof
of VVthen
thenXg
Xgand
andYg
Yg are
are
FH-isomorphic
FH-isomorphic submodules for each g Ec G!'
(12.13) (Clifford's
(Clifford's Theorem) Let
LetVVbebea afinite
finitedimensional
dimensionalirreducible
irreducibleFGFG9G.
G. Then
Then
module and H <
semisimpleFH-module.
FH-module.
(1) VVisisaasemisimple
(2) GGacts
actstransitively
transitivelyon
onthe
theFH-homogeneous
FH-homogeneouscomponents
componentsof
of V.
V.
(3) Let
bean
anFH-homogeneous
FH-homogeneouscomponent
component of
of V.
V. Then
Then NG(U)
NG(U)is
is irreirreLetUUbe
NG(U).
ducible on U and HCG(H)
HCG(H) <5 NG(U).
Proof.
Proof.By
By12.12.1,
12.12.1,GGacts
actson
onthe
thesocle
socleof
of V,
V, regarded
regardedas
asan
an FH-module.
FH-module. Thus
Thus
(1) holds by 12.6
12.6 and the irreducible
irreducible action of G. By
By 12.12.3,
12.12.3, G
G permutes
permutes
the homogeneous
homogeneous FH-components
FH-componentsof
of V.
V.Then,
Then,by
by 12.11.2
12.11.2and
andthe
theirreducible
irreducible
action of G,
G, G
G isis transitive
transitive on
on those
those homogeneous
homogeneouscomponents.
components. By
By 12.12.2,
12.12.2,
HCG(H) acts
HCG(H)
actson
on each
eachhomogeneous
homogeneous component.
component. Then 12.11.2
12.11.2 and the irreducible action of G completes
completes the proof of (3).
(3).
Observe that
that n
7r can
can be
be extended
extended to
to aa representation
representation of
of R
R on V (that is to an
an
Observe
F-algebra homomorphism
homomorphismofofRRinto
intoEndF(V))
EndF(V))via
vian 7r:
aggH
HC
> ag(g7r).
: C>agg
ag(gn).
Indeed
Indeed for r EE R and v E V,
V, v(r7r)
v(rn) ==yr
vrisisjust
just the
theimage
imageof
of vv under the module
Linear representations
representations
42
product of
V. Further
of vv by
by r in the R-module V.
ker(n)
vr ==O0 for all vv EE V1.
V).
ker(ir) =
= ((rr E R: yr
V is said to be afaithful
faithful R-module
R-module ifif nit is an injection on R. As G generates
generates R
R
as an F-algebra, R,r
Rn isisthe
thesubalgebra
subalgebraof
of EndF(V)
EndF(V)generated
generatedby
by G7r.
G n . We
We call
call
Rzr
enveloping algebra
algebra of the representation nn..
Rn the enveloping
(12.14) EndF(V) = CEfldF(v)(R7r) = CEndF(v)(GJr).
(12.15) If G is finite and n
it is irreducible then
then Z(Gn)
Z(Gir) is cyclic.
(12.15)
Proof. Let
Let E =EndF(V).
=EndF(V)
Proof.
=EndF(V). As
As 7r
n isis irreducible,
irreducible, D =
EndR(V) is
is aa division
division
ring by Schur's Lemma.
= Z(G7r)
CE(G7r)==DDby
by12.14.
12.14.Also
AlsoDD 5
<
Lemma. Z =
Z(Gn) 5< CE(Gn)
CE(GTr)(
< CE(Z),
CE(Z), so
so Z <
CE(Gn)
5 Z(D).
Z(D).Thus
Thusthe
the sub-division-ring
sub-division-ring K of D generagenerated by Z isis aa field.
field. Now
Now ZZ isisaafinite
finitesubgroup
subgroup of
of the
the multiplicative
multiplicative group
group of
the field K, and hence K is
is cyclic.
cyclic.
I conclude
conclude this section by recording two results whose proofs can be found in
section 3 of chapter
chapter 17
17 of Lang
Lang [La].
[La].
(12.16) Let
-* GL(V)
irreducible finite dimensional
dimensional FG-representaFG-representaLet7r:
n: G +
GL(V)be
be an
an irreducible
tion. Then Rn
Rir is isomorphic
as
an
F-algebra
to
the
ring
of
all
m
by m
m matrices
isomorphic
F-algebra to the ring
EndFG (V)=
= D, where m =
= dimD
(V). Further
Further F
F is
over the division ring EndFG(V)
dimD(V).
is in
in the
the
centre of D.
D.
(12.17)
-* GL(V)
(12.17) (Burnside) Assume F
F isis algebraically
algebraically closed
closed and
and 7r:
n: G +
GL(V) isis
an irreducible finite
finitedimensional
dimensionalFG-representation.
FG-representation.Then
ThenEndF(V)
EndF(V)=
=R
Rir
n and
F=EndFG(V).
F = EndFG(V).
13 The
The general
general linear
lineargroup
groupand
andspecial
special linear
linear group
group
F is
In this section F
is aa field,
field,nn isis aapositive
positive integer,
integer, and V
V is
is an
an n-dimensional
n-dimensional
over F. Recall the group of vector space automorphisms of
of V is
vector space over
GL(V). As the isomorphism
isomorphism type of V depends only on
the general linear group GL(V).
n and
GL"(F) for GL(V).
and F,
F ,the
thesame
same is
is true
true for
for GL(V),
GL(V), so
so we can also write GLn(F)
GL(V).
(13.1)
Let Fnxn
F""" denote
the F-algebra
F-algebra of
of all
all nn by
over FF,, let
(13.1) Let
denote the
by n matrices
matrices over
let
= (xi
basis for
for V, and
and for
for gg EEEndF(V)
EndF(V)let
letMx(g)
Mx(g)=
_
X=
(xl .....
. . ,x")
xn) be an ordered basis
(gig)be
bethe
thematrix
matrixdefined
definedby
byxig
xi g==Cj
>j gig
xj, gij
gij E
E F.
F. Then
(gij)
gijxj,
(1) The map Mx: gg H
HMx(g)
Mx(g)isisananF-algebra
F-algebraisomorphism
isomorphism of
of EndF(V)
EndF(V)
with FnXn.
F"'Hence
thethe
map
restricts
to to
a group
Hence
map
restricts
a groupisomorphism
isomorphismofofGL(V)
GL(V)with
withthe
the
of all
all nonsingular
nonsingular nn by
by nn matrices
matrices over
over F.
F.
group of
The
The general linear
linear group
group and
and special
special linear
lineargroup
group
43
43
(2) Let
LetY
Y=
=(yl,
(yl, ...
. . ., , yn)be
beaasecond
secondordered
orderedbasis
basisof
of V,
V,let
let hh be
be the
the unique
unique
(2)
element
element of GL(V)
GL(V) with
with xih
xih==yi,
yi ,11 <
ii <
5 n, and
and BB==My(h).
My(h). Then
Then Mx
Mx ==
h*My
MyB*
h*My= M
y B*isis the
the composition
composition of
of h*
h* with My
My and
and of
of My
My with
with B*,
B*,where
where
h*
h* and
and B*
B* are
are the conjugation automorphisms
automorphismsinduced
induced by h and
and B
B on
on EndF(V)
EndF(V)
and F'1
F n x, nrespectively.
,respectively.
and
Because of
of 13.1,
13.1,we
we can
can think
think of
of subgroups
subgroupsof
of GL(V)
GL(V) as
asgroups
groupsof
of matrices
matrices
Because
FGifif we
we choose.
choose.II take
take this
this point
point of
of view
view when
when itit isis profitable.
profitable. Similarly
Similarly an
an FGrepresentation
representation itn on
onVVcan
canbe
bethought
thoughtof
of as
as aa homomorphism
homomorphism from
from G
G into
into the
the
group
groupof
of all
all nn by
by nn nonsingular
nonsingular matrices
matricesover
over G,
G,by
by composing
composingitn with
withthe
theisoisomorphism
Mx. itn isis equivalent
equivalent to ir':
n': G
G -*
+GL(V)
GL(V) ifif and
and only
only if 7r'=
n' =irh*
nh*for
for
morphism Mx.
some
this
when
Jr'Mx = Tr
Mx B *
GL(V),and
andbyby13.1.2
13.1.2
thishappens
happensprecisely
precisely
whennfMx
= nMxB*
somehh EEGL(V),
for
for some
somenonsingular
nonsingularmatrix
matrixB.
B.This
Thisgives
givesaanotion
notionof
ofequivalence
equivalencefor
for`matrix
'matrix
representations'.
homomorphismsaa and
and a' of
representations'. Namely two homomorphisms
of G
G into
into the
the group
group
of
all
n
by
n
nonsingular
matrices
over
F
are
equivalent
if
there
exists
a
of all n by n nonsingular matrices over F are equivalent if there exists anonnonsingular matrix
matrix B
B with
with a'a'==aaB*.
B*.
singular
Let's
Let's see
seenext
nextwhat
whatthe
thenotions
notionsof
ofreducibility
reducibility and
anddecomposability
decomposabilitycorrecorrespond
spond to
to from
fromthe
thepoint
pointof
ofview
viewof
ofmatrices.
matrices.
(13.2)
n : GG +GL(V)
GL(V) be
be an
an FG-representation,
FG-representation, U
U an
anFG-submodule
FG-submoduleof
of
(13.2) Let
Let7r:
m)
V,V
=
V/
U,
and
X
=
(xi:
1
<
i
<
n)
a
basis
for
V
with
Y
=
(xi:
1
<
i
<
v , ~ V / U ,andX=(xi: 1 5 5 n ) a basis
V
Y =(xi: 1 5 i 5 m)
aa basis
basis for
for U.
U.Then,
Then,for
forggEEG,
G,
Mx(gn) =
[My(g7r l u)
0
A(g)
Mx(g7r(g7r v)
L
for some
some n --mmby
bymmmatrix
matrixA(g).
A(g).
for
Of
Of course
coursethere
thereisisaasuitable
suitableconverse
converseto
to13.2.
13.2.
(13.3) Let
Let ir:
n :G
G -*
+GL(V)
GL(V)be
bean
anFG-representation,
FG-representation,UUand
and W
W FG-submodules
FG-submodules
of
(xi
: 11<5 i < n) aa basis
of V
V with
with V=U
V = U®@IW,W,and
andXX==
(xi:
basis for V
V such
such that Y=
Y=
(xi: 1 5< ii <
5 m)
m) and
and Z
Z ==(xi:
(xi: m <<ii<5n)n)are
arebasis
basisfor
forUUand
andW,
W,respectively.
respectively.
Then
G,
Then for
for gg EE G,
Mx(gn)
[
My(gnly)
0
fi
Mz(gn)w)
Again
Again there
thereisisaasuitable
suitableconverse
converseto
to13.3.
13.3.
Recall
Recallthe
thenotion
notionof
ofgeometry
geometrydefined
definedin
insection
section3.3.We
We associate
associateaageometry
geometry
PG(V)
PG(V) to
to V,
V,called
calledthe
theprojective
projective geometry
geometry of
of'V.
V. The
The objects
objects of
of PG(V)
PG(V)are
are
the proper nonzero subspaces
subspaces of V,
V, with
with incidence
incidencedefined
defined by
by inclusion.
inclusion.If
If UU
44
Linear representations
of V,
V, the
theprojective
projectivedimension
dimensionofofUUisisPdim(U)
Pdim(U)==dimF(U)
dimF(U)- 1.
is a subspace of
1.
The type function
function for PG(V) is the
the projective dimension
dimension function
function
1,....,
- 1).
Pdim:
Pdim: PG(V) + I= {0,
{O,1,
. . ,n 1).
PG(V) is said to be of dimension
1. The
The subspaces
subspaces of projective dimension
PG(V)
dimension nn - 1.
0, 1,
arereferred
referredtotoasaspoints,
points, lines,
lines,and
andhyperplanes,
hyperplanes, respectively.
respectively.
1, and n --22are
Forg EEGL(V)define
gP:PG(V)Ug,
GL(V) definegP:
PG(V) +PG(V)bygP:
PG(V) by gP: U
UH
t-,
Ug,for
forU
U EPG(V).
E PG(V).
of GL(V) in the catEvidently P:
P: GL(V)
GL(V) + Aut(PG(V)) is a representation
representation of
egory of geometries. (See the discussion in section 3.) Denote the image of
GL(V) under P by
by PGL(V).
PGL(V).PGL(V)
PGL(V)isisthe
theprojective
projective general
general linear
lineargroup.
group.
The notation PGL,(F)
PGLn(F) is
is also
also used
used for
for PGL(V).
PGL(V).
Visa
A scalar transformation of V
is amember
membergg of
of EndF(V)
EndF(V) such
such that
that vg
vg =
=av
av
for all vv in V and some a in F
F independent
independent of v. A scalar matrix is a matrix
of the form aaI,l , aa EE F,
F,where
where IIisisthe
theidentity
identitymatrix.
matrix.
(13.4)
(1) Z(EndF(V))
(13.4) (1)
Z(EndF(V)) is
is the
the set
setof
ofscalar
scalartransformations.
transformations. The image of
Z(EndF(V)) under Mx is
Z(EndF(V))
is the
the set
set of scalar
scalar matrices.
(2) Z(GL(V))
is
the
set
of
nonzero
Z(GL(V)) is the set of nonzeroscalar
scalar transformations.
transformations.
(3)
Z(GL(V)) = ker(P).
(3) Z(GL(V))=ker(P).
projective general
general linear
linear group PGL(V)
PGL(V) is
By 13.4,
13.4, the projective
is isomorphic
isomorphicto
to the
the group
group
bynnnonsingular
nonsingular matrices
matrices modulo
modulo the
the subgroup
subgroup of
of scalar
scalarmatrices.
matrices. Often
Often
of all nn by
it will be convenient to regard these groups as the same.
same.
Given any
can be
be composed
composed with P
any FG-representation nn:: G + GL(V), n can
homomorphismnnP:
P : G + PGL(V). Observe that nnPP isis aa represenrepresento obtain aa homomorphism
tation of G on the projective
projective geometry PG(V).
EndF (V)define
definethe
thedeterminant
determinant of
of yy to
tobe
bedet(x)
det(x) =
= det(Mx(y)).
det(Mx (y)). That
For y EE EndF(V)
is the determinant of y is
is the
the determinant
determinant of
of its
its associated
associated matrix.
matrix.Similarly
Similarly
define the
the trace
trace of
of y to be
(y)). So
So the
the trace
trace of
of y is the trace
define
be Tr(y)
Tr(y) =
=Tr(Mx
Tr(Mx(y)).
If A is a matrix and B
B isis aanonsingular
nonsingular matrix
matrix then
then
of its associated matrix. If
=~ det(A)
and Tr(AB)
=~ Tr(A),
independent of
of
ddet(AB)
e t ( ~=
det(A)
)
T ~ ( A=
)
so det(y) and Tr(y) are independent
the choice
13.1.2.
choice of basis X by 13.1.2.
the special
special linear group SL(V)
Define the
SL(V) to be the set of elements of GL(V) of
determinant 1. The
The determinant map
map is
is a homomorphism
homomorphism of
of GL(V) onto the
determinant
multiplicativegroup
group of
of F with
multiplicative
with SL(V)
SL(V) the
the kernel
kernel of
of this
this homomorphism,
homomorphism,so
so
normal subgroup
subgroupof
ofGL(V)
GL(V)and
andGL(V)/SL(V)
GL(V)/SL(V) Z
= F#.
SL(V) is a normal
F'. Also
Also write
write
SLn(F) for SL(V). The image
image of
of SL(V)
SL(V) under
under P is
SL,(F)
is denoted
denoted by PSL(V) or
PSLn
(F). The group PSL(V) is the projective
projective special linear
linear group. Sometimes
PSL,(F).
Sometimes
PSLn(F)
Ln(F).
PSL,(F) is denoted by L,(F).
,
The general linear group and
and special
special linear
linear group
group
45
45
Prove
Prove the
the next
next lemma
lemma for
for GL(V)
GL(V)and
andthen
thenuse
use5.20
5.20to
toshow
showthe
theresult
resultholds
holds
SL(V).See
Seesection
section15
15for
forthe
thedefinition
definitionof
of2-transitivity.
2-transitivity.
for SL(V).
for
(13.5)
(13.5) SL(V)
SL(V)isis2-transitive
2-transitiveon
on the
the points
points of
of PGL(V).
PGL(V).
For
For v in V and a in
in EndF(V), [v, aa]] =
=va
va --v visisthe
thecommutator
commutatorof
of vvwith
with a.
a.
This
This corresponds
corresponds with the notion of commutator
commutator in section 8. Indeed we can
form the semidirect
semidirect product of V by GL(V) with respect to the natural representation, and in this group the
the two
two notions
notions agree.
agree. Similarly,
Similarly,for
forGG 5< GL(V),
[V, G]
= ([v,
GI =
([v, g]:
gl: v EE V,
V, gg EE G)
G)
and, for g E
g] =
_ [V,
E G, [V, gl
[V, (g)].
(g)l.
A
A transvection
transvection is an element
element tt of GL(V)
GL(V) such that
that [V, t] is aa point
point of
of
PG(V), Cv(t)
is
a
hyperplane
of
PG(V),
and
[V,
t]
<Cv(t).
[V,
t]
and
Cv(t)
Cv(t) is a hyperplane of PG(V), and [V, t] I
Cv(t). [V, t] and Cv(t)
are called the center and axis of t,t, respectively.
Let x,xnEEVV-- Cv(t). Then
respectively. Let
Then
[x,,, t] ==Xx1
t] and we choose xi
xi EECv(t),
Cv(t), 1 <
<i <
< n, so that X =
I generates [V, t]
=
(xi: 1 5
< ii <5n)n)isisaabasis
basisof
of V.
V. Then
Then
MX(t)=
1
0
0
0
1
0
1
0
1
so evidently
evidently t is
is of
of determinant
determinant 11 and
and 13.1.2
13.1.2implies
implies GL(V)
GL(V)isistransitive
transitiveon
onits
its
transvections.
Write diag(al,
diag(al, ....,
. . ,an)
a,) for
forthe
thediagonal
diagonal matrix
matrix whose
whose (i,
(i, i)-th
transvections. Write
entry isisaai.
> 22let
i . IfIfnn >
let
A = {diag(1, a, 1, ... , 1): a E F#}
n = 2let
let
and ififn=2
and
A=
= (diag(a,
{diag(a, a): a EE F#).
F') .
Then A <5CGL(v)(t)
and either
CGL(~)(t)
either det: A +F#
F' is
is aa surjection
surjection or
or nn ==22 and
and some
some
element of
of F is
is not
not aa square
square in F.
F . In
Inthe
thefirst
first case
case GL(V)
GL(V) =
=ASL(V),
ASL(V), so, as
GL(V) is
is SL(V)
SL(V) by
by 5.20. Further
Further ifif nn >
> 22
is transitive
transitive on its transvections, so is
and s is the transvection
with Cv(s)
Cv(s) =
= Cv(t)
Cv(t) and [xn,
=x2, then st is also
transvection with
[x,, s] =x2,
also
a transvection. So,
So, as
as SL(V)
SL(V)isistransitive
transitiveon
onits
itstransvections,
transvections,t =
t =s-1(st)
s-'(st) is a
commutator
b EE F#
(b) be the
SL(V).On
Onthe
theother
otherhand,
hand,ifif nn ==2,2,then
thenfor
forb
F' let
lettt(b)
commutatorof
of SL(V).
transvection with
with x2t(b)
x2t(b)=
= x2
x2 +bxl
bxl and
andgg=
= diag(a,
diag(a, a-I).
a-1). Then t(b)g
t(b)9 =
= t(a2b).
t(a2b).
Thus, setting
setting bb =
= (a2(a2_1)-l'
thenaacanbe
can bechosen
chosenwith
witha2
a 2 # 1. Thus,
- I)-',
Further if
if IF
F >I >3 3then
we have [t(b),
[t(b), g]
g] =
=t,t,and
and again
again tt isisaacommutator
commutatorof
of SL(V).
SL(V).
+
Linear representations
representations
46
We have
have shown:
shown:
(13.6) (1)
(13.6)
(1)Transvections
Transvectionsare
areof
ofdeterminant
determinant 1.
1.
(2) The
The transvections
transvections form a conjugacy class of GL(V).
of SL(V) or n =
= 2 and
(3) Either
Either the transvections
transvections form a conjugacy class of
F contains
F
contains nonsquares.
(4) IfIf IF
IFII >>33or
ornn>>2,2,then
theneach
eachtransvection
transvectionisisin
inthe
thecommutator
commutatorgroup
group
of SL(V).
SL(V).
(13.7)
(13.7) SL(V)
SYV)isis generated
generatedby
by its
its transvections.
transvections.
P r o o fLet
. Let
theset
setofofn-tuples
n-tuplesow==(XI,
(x1,. ..... ,, x,-l,
x,,-,, (xn))
Proof.
i-2 0bebethe
(x,)) such
suchthat
that(x
(xll , ...
. . .,,
xn) is
is a basis for V. Let TT be
x,)
be the
the subgroup
subgroup of G
G ==SL(V)
SL(V) generated
generated by the
transvections
of G.
G. I'll
I'll show T
on a.
0. Then,
transvections of
T is transitive on
Then, by
by 5.20,
5.20, G
G ==TG,,,.
TG,.
1, so
so the
the lemma
lemma holds.
But GU,
G , == 1,
remains to
to show
show T
T is
is transitive
transitiveon
ona.
0. Pick
Pick aa =
= (yi,
It remains
(yl, ..... ., ,Yn-1,
y,-1, (Yn))
(y,)) E i-2
0
such that
that aa 04 wT,
m,m,and,
oT,yi
yi =xi for
fori i<i
and,subject
subjecttotothese
theseconstraints,
constraints, with m
maximal. Let
Let U ==((xi:
x =xn+1> Yy=y,+l,
= Y,n+i,and
and W=
W = (U,x,
(U, x, y).
y). Then
x i : ii <im),
rn),x=x,,+l,
or 2.
dim(W/ U) =
=k=
= 11 or
= 22 and
and let
let H
H be a hyperplane
hyperplaneof
ofVVcontaining
containingUUand
andxx-- y but
Suppose kk=
not x.
x. Let
Let tt be
bethe
thetransvection
transvectionwith
withaxis
axisHHand
and[y,
[y,t]t]==xx-- y.
y. Then
Thenyt
yt =
= x and
xi
t
=
xi
for
i
<
m,
so
at
E
wT
by
maximality
of
m.
Then
a
E
wT,
contrary
xit =xi
( m, so a t E oT by maximality of m. Then a E oT,contraryto
to
the choice of a.
a.
Soo k ==l 1.
for some a E F#.
As
S
. SSuppose
u p p o s em
m = nn --l 1.
. AAs
s k k= l=, a1,xax
-y- y E U
Uforsomea
F#.As
m=
= nn - 11and
0. 0.
SoSo
there
rn
and aa ##w,
o,ax
ax --yy# #
thereisisa atransvection
transvection tt with
with axis
axis U
U
and
[y,
t]
=
ax
y.
Now
at
=
w,
contradicting
a
0
off.
So
m
<
n
1,
and
and [y, t] = a x - y.
a t = o,contradicting a 4 oT.So m < - 1, and
hence there is z E
E V --W.
W.An
Anargument
argumentininthe
thelast
last paragraph
paragraph shows
shows there are
transvections
andt twith
withUU (
< Ca(t)
Cv(t) fl
r l Cv(s),
Cv(s), ys =
=z, and zt =x. But
But now
now
transvections ssand
=xi for
forii<5mmand
andyst
yst==
contradictingthe
thechoice
choice of aa..
xist =xi
x,x,contradicting
(13.8) If n >222then
then SL,(F)
SL,(F)isisperfect
perfectunless
unlessnn==22and
andIFIFII ==22or
or 3.
3.
(13.8)
Proof. Let
contained
Prooj
LetGG==SLn(F).
SL,(F).By
By13.7
13.7ititsuffices
sufficesto
to show
show transvections
transvectionsare contained
in GM,
G('), and this follows
follows from
from 13.6.4.
13.6.4.
14 The
The dual representation
representation
In section 14, V continues
continues to be
be an
an n-dimensional
n-dimensional vector space over F
F and
and
7r:
G+
-+ GL(V) is an FG-representation
FG-representation of a group G.
n: G
Let (K:
(Vi:-oo
-oo <<i i<<oo)
oo)bebea asequence
sequenceofofFG-modules
FG-modules and
and
...-+ V_1
VoC"') V1-+ ...
47
dual representation
The dual
of FG-homomorphisms.
FG-homomorphisms. The
The latter
latter sequence
sequenceisis said
said to
to be
be exact
exact ifif
a sequence of
ker(ai+1)
=K
Viai
ker(ai+l) =
a i for each i. A short
short exact
exact sequence
sequence is an
an exact
exact sequence
sequence of
the form
form 00+
-+UU3
-V
-+ 0. The maps 00 -+
are forced
forced to
V5 W+
+UU and
andW
W-->
-t O
0 are
be trivial. Observe that the hypothesis
hypothesis that the sequence
sequence is exact
exact is
is equivalent
equivalent
to requiring
that aa be an injection,
and Ua
Ua =
= ker(p).
requiring that
injection, /?$ a surjection, and
ker(/?).Hence
Hence
- V/
essentially00+
-+Ua
Ua +
-+ V
V+
-+ V/
V/ Ua +
-+ 0 with
with
WE
V/Ua
Uaand
andthe
thesequence
sequence is essentially
-+ V/Ua
Ua -+
+VV inclusion
inclusion and V +
V/Uathe
thenatural
naturalmap.
map.The
Thesequence
sequence is said
said to
split if V splits
splits over Ua.
Ua.As
Asisiswell
wellknown,
known,the
the sequence
sequencesplits
splits ifif and
and only
only ifif
there is y EE HomFG(W,
V) with
with y/?
yp =1,
HomFG(W, V)
= 1,and
andthis
thiscondition
conditionisis equivalent
equivalent in
turn to the existence
HomFG(V, U) with a6
aS =
= 1.
existence of S
6 E HomFG(V,
1.
= HomF(V,
HomF (V,FF)
and recall
recall from
from section
section 13 that
that V* is
is aa vector
vector space
space
Let V
V**=
) and
dimF
(V * ).
over F.
F.We
Wecall
callV*
V*the
thedual
dualspace
spaceofofV.V.ItItisiswell
wellknown
knownthat
thatn n= =
dimF(V*).
If
If U
U isisananF-space
F-spaceand
anda EHomF(U,
a €HomF(U,V)
V)define
definea*a*EHomF(V*,
€HomF(V*, U*) by
by
xa* ==ax,
XU*
a x , xx EE V*.
V*.
(14.1) Let
dimensional F-spaces,
F-spaces, a E
HomF(U, V),
U, V,
V,and
and W
W be finite dimensional
E HomF(U,
V), and
Let U,
,B
HomF(V, W). Then
/? EEHomF(V,
H y*
y*isisan
anF-space
F-spaceisomorphism
isomorphismof
of HomF(U,
HomF(U, V) with
(1) The map
map y i-+
HomF(V*, U*).
HomF(V*,
(2) (afi)*
(a/?)* = /?*a*.
(2)
(3) If U4
WW
is exact
then
U 3VV- 5
is exact
thensosoisisW*
W*
%U*.
U*.
V*
V*-*
n: G
an FG-representation
FG-representationthen,
then, from
from14.1,
14.1,7r*:
n *: G + GL(V*)
If 7r:
G+
-+ GL(V) is an
GL(V*)
H (g-17r)*.
is also an FG-representation,
FG-representation, where
where 7r*:
n*: gg H
(g-ln)*. The
Therepresentation
representation7r*
n*
is called the dual
dual of
of 7r
n and the representation
representation module V*
V* of 7r*
n * is called the
dual of the representation
representation module V
V of
of 7r.
n.
Given aa basis
basisXX=
= (xi:
(xi :11_(<i i5< n)
n) for
for V,
V,the
thedual
dualbasis
basis2X==(ai
(zi::115< ii 5
< n)
: xj
Notice T i aixi
definedby
byxiii:
x ji-+
H Sid
Jij. .Notice
a i f i isis the
the unique
unique member
member of
of V*
V*
of X isis defined
mapping xi to ai for each i.i .
xi
-+ GL(V) be an FG-representation
(14.2) Let
Let 7r:
n: G +
FG-representation and X a basis for V. Then
MX(g7r*)
= ((MX(g7r)-1)T
where BT
M2 (gn *) =
M (gn)-l)T,
~
B~ denotes the transpose of aa matrix
matrix B.
B.
representation, then
just the composicomposiBy 14.2,
14.2, if 7r
n is viewed as a matrix representation,
then 7r*
n * is just
tion of n7r with
with the
the transpose-inverse
transpose-inverse map
map on
on GL,(F).
GLn(F). As the transpose-inverse
map is of order
order 2, we conclude
conclude
(14.3) (7r
*)*isisequivalent
equivalentto
tonrrfor
foreach
eachfinite
finitedimensional
dimensionalFG-representation
FG-representation 7r.
(n*)*
n.
There
ere is a more concrete way to see this.
Linear representations
representations
48
(14.4) Let U and V
(14.4)
V be finite dimensional F-spaces. Then
with xu0
xv9 =
= vx
thereexists
existsaaunique
unique element
element vB
v0 E (V*)*
(V*)*with
(1) For
Foreach
eachvv EEVVthere
for all x E V*.
V*.
(2) The
Themap
map 0:
0: vv H
HvB
v0 isis an
an F-isomorphism
F-isomorphism of
of VVwith
with (V*)*.
(V*)*.
(3) For
For each
each aaEEHomF(U,
HomF(U, V),
V), aB
a 0 ==B(a*)*.
@(a*)*.
(4) 00defines
definesan
anequivalence
equivalenceofof7rn and
and(7r*)*.
(n *)*.
Proof. Toprove
XX
= (xi
: 1 51i <i
5 n)
be a basis for V,V,8X=(zi:
= (ai: 11 5
5 n)
prove(1)
(1)letlet
= (xi:
<n)beabasisfor
< ii <
its dual basis,
basis, and
and 2X==(Xi
(fi ::115< i <5n)n)the
thedual
dualbasis
basisofofX8inin(V*)*.
(V*)*.Let
Let
xv ==vx
x EE V*
vv==Y'i aixi
a.x.
, ,EEVVand
andv.fr==_Yi biii
biziEE(V*)*.
(V*)*.Then
Thenx8
vx for
forall
allx
V* ifif and
only if zi
= vv.xi
for all
all i.
i. Further ixii Dv=
= bi
bi and
and vii ==ai,
Xi is
ii8v =
i i for
ai,so
sov0
v0==_Yi aiaiTi
uniquely determined.
determined.
aixiH
H 1: aiXi,
As 0: _Yaixi
ai.fi,(2)
(2) holds.
holds. Part
Part (4)
(4) follows
follows directly from (2) and
(3). To prove
prove (3)
(3) we must show (u9)(a*)*
= (ua)0
(ua)B for each u E U. But, for
(&)(a*)* =
=xa*(u9) =
= ax(u0)
ax(u9) =
x EE V*,
V*, x(u9)(a*)*
x(u0)(a*)*=xa*(u0)
=u(ax)
u(ax) =
=(ua)x
(ua)x ==x((ua)B),
x((ua)0), comcompleting the proof.
xi
xi
x
xi
x
constructive proof of 14.3.
be useful
useful in
in the
the proof
proof
Notice 14.4
14.4 gives a constructive
14.3. It will also be
of the next lemma.
U, V,
V, and
and W
Wfinite
finitedimensional
dimensional FG-modules.
FG-modules.
(14.5)
group and
and U,
(14.5) Let G be aa group
Then
Then
(1) HomFG(U*,
HOmFG(U*,V*)
V*) =
=(a*:
{a*:aaEEHomFG(V,
HomFc(V, U)}.
U)}.
B
(2) 0 + U
sequence of FG-modules
FG-modules if
(2)
V+
W + 00 is an exact sequence
if and
and only
only
U-%4 V
-W
B*
if 00 + W
W** -+
U**-+
the
-+ VV*
* %U
-+00 is.
is. The
The first
first sequence
sequence splits
splits ifif and
and only if the
second
second splits.
splits.
(3) VVisisirreducible,
irreducible,indecomposable,
indecomposable,semisimple,
semisimple,and
and homogeneous if and
only if V*
V* has the
the respective
respectiveproperty.
property.
Proof. Part
Part (1)
(1) follows
follows from
from 14.1.1
14.1.1and
and 14.1.2.
14.1.2. The
The first
first part
part of
of (2)
(2)follows
follows
from 14.1.3
14.1.3 and 14.4. The second part follows from the remark
remark about splitting
14.1.2. Part (3) follows
follows from (2), since
since the
at the beginning of this section and 14.1.2.
properties in (3) can be described in terms of
of exact
exact sequences and the
the splitting
of such sequences.
sequences.
(14.6)
dimensional mod(14.6) (1)
(1) Let
Leta:a:VV--*
+UUbe
bean
anFG-homomorphism
FG-homomorphismof
of finite dimensional
ules.
[G, U] _(
Vaa if and
and only
only ififker(a*)
ker(a*)<(CU.
CU*(G),
while [G,
[G, V*]
V*] _(
ules. Then [G,
<V
(G), while
<
U*a*
and only
onlyififker(a)
ker(a) _(< Cv(G).
U*a* ifif and
CV(G).
(2) If
If U
U isis aa finite
finite dimensional FG-module then
then U =
= [U, GI
G] if and only ifif
CU.(G)
=
0,
while
U*
=
[U*,
G]
if
and
only
if
CU(G)
=
0.
Cu*(G)=
U * = [U*, GI
CU(G)=
The dual representation
49
Proof.We
Wehave
havethe
theexact
exactsequence
sequence
Proof.
V-0-'> U - U/Va --> 0
so, by
by 14.1.3,
14.1.3,
so,
0--> (U/Va)*-->
is
be the
the representation
representationof
ofGG on
on U/
U/ Va.
Va. Then [G, U] <
5 Va
Va
is also
also exact.
exact. Let 7r
n be
ifif and
and only
only if Gn
G n==1.1.This
Thisisisequivalent
equivalentto
to Gn*
G n *==11which
whichininturn
turnholds
holdsifif and
and
only
only if G
G centralizes
centralizes (U/Va)*.
(U/ Va)*.As
As(U/Va)*
(U/ Va)*isisFG-isomorphic
FG-isomorphictotoker(a*)
ker(a*)by
by
the
the exactness
exactnessof
of the
thesecond
secondseries
seriesabove,
above,the
the first
firstpart
part of
of (1)
(1)holds,
holds,while
whilethe
the
secondfollows
followsfrom
fromthe
thefirst
firstand
and14.4.3.
14.4.3.
second
Let
U
#
[U,
GI
=
V
anda:
V
-+
UUtheinclusion.
theinclusion. By
(l), ker (a*) i:CU*(G).
[U,
G]
=
V
and
a:
-->
By(1),ker(a*)<CU.(G).
Let U
(G). Hence
As
As aaisisnot
notaasurjection,
surjection,ker(a*)
ker(a*) # 00 by
by 14.5.2.
14.5.2.Thus
Thus 00 # Cu.
CU*(G).
Hence by
by
14.4.3, ifif U* #
[U*,G]
GI then
then 00 # Cu(G).
CU(G).
14.4.3,
0 [U*,
-> U/CU(G)
Similarly,ifif 00 # Cu(G),
CU(G),let
let,B:
8: U -+
U/CU(G)be
be the
thenatural
natural map.
map. By
By (1),
(I),
Similarly,
[G,
U*]
<
(U/CU(G))*,8*.
As
,B
is
not
an
injection,
,B*
is
not
a
surjection
[G, U*] 5 (U/CU(G))*B*.As B is not an injection, B*
a surjectionby
by
14.5.2. So
So U*
U* # [G,
[G, U*]. Applying 14.4.3
14.4.3 we see that if 00 # CU.(G)
CU*(G)then
then
14.5.2.
U # [G,
[G,U],
U],completing
completingthe
theproof
proof of
of (2).
(2).
U
The
character of
ofan
anFG-representation
FG-representation7r
n is
is the
the map
map X:
X:G
G -+
-+FFdefined
definedby
by
The character
X
=Tr(g7r).
RememberTr(gn)
Tr(g r)isis the
the trace
trace of
of the matrix
matrix Mx(gn)
Mx(gn) and
~ (g)
( g=Tr(gn).
)
Remember
and isis
independent
'the representation
for'ihe
representationmodule
moduleof
of7r.
n.
independentof
of the
thechoice
choiceof
of the
thebasis
basisXXfor
(14.7)
thedual
dual of
of nr,, and
(14.7) Let
Let 7r
n be
be an
an FG-representation,
FG-representation, n7r*
* the
and Xx and
and X*
x * the
the
characters
= X(g-1)
charactersof
of7r
n and
and 7r*,
n*, respectively. Then X*(g)
x*(g) =
x ( ~ - ' )for
foreach
eachggEEG.
G.
Proof.
= Mx(g-17r
lemma
follows as Tr(A) =
_
Proof. By
By14.2,
14.2,MX(g7r*)
M2(gn*) =
~ ~ ( ~ -)T,' so
nso the
)the
~ lemma
,
T ~ ( Afor
~ )each n by nn matrix
matrix A.
A.
Tr(AT)
Since
Sincecharacters
charactershave
havenow
now been
been introduced
introducedII should
shouldprobably
probablyrecord
recordtwo
twomore
more
properties
propertiesof
of characters
characterswhich
which are
are immediate
immediatefrom
from 13.1
13.1and
and the
the fact
factthat
that conconjugate
jugate matrices
matriceshave
havethe
thesame
sametrace.
trace.
(14.8)
(14.8) (1)
(1) Equivalent
EquivalentFG-representations
FG-representationshave
havethe
thesame
samecharacter.
character.
(2) IfIfXxisisthe
thecharacter
characterofofan
anFG-representation
FG-representationthen
thenX (gh)
(gh)==Xx(g)
(g) for
foreach
each
G.
g, hh EE G.
g,
Remarks.
Remarks.The
Thestuff
stuffininsections
sections12
12and
and13
13isispretty
prettybasic
basicbut
but section
section14
14is
is more
more
specialized.
prepare the way for
for thel-cohomology
the l-cohomology
specialized.Section
Section14
14is
is included
includedhere
here to prepare
in
in section
section 17.
17.That
That section
section is
is also
also specialized.
specialized. Both
Both can
can be
be safely
safely skipped
skipped or
or
Linear representations
50
casual reader. If so, lemma 17.10
proving
postponed by the casual
17.10must be assumed in proving
the Schur-Zassenhaus
Schur-Zassenhaus Theorem in section 18. But
But that's
that's no problem.
of modules over
over rings
rings might
might
The reader who is not familiar with the theory of
want to bone up on modules before beginning
beginning section
section 12.
12.
for chapter
chapter 4
Exercises for
1. Let
Let G
G be
be aafinite
finitesubgroup
subgroup of
of GL(V),
GL(V), where
where VVisisaafinite
finitedimensional
dimensional
field F
F with (char (F),
(F), IGI)
\GI)=
=1.
1.Prove
Prove
vector space over aa field
(1) VV ==[G,
[G, V]
Vl ®
@Cv(G).
Cv(G).
(2) If G
then V =
= (CV(D):
(Cv(D):DD EE A), where
A is the
set of
of
G is
is abelian
abelian then
where A
the set
subgroups D
D of
of G
G with
with G/D
G/D cyclic.
cyclic.
subgroups
E,., nn >>0,0,and
andVV==[V,
[V, G],
GI, then V =
@
,,,
CV(H),where
where
If G =Z Ep",
(3) If
= ®HEr
Cv(H),
rr isisthe
theset
set of
of subgroups
subgroups of G of
of index p.
2. Let
field F, g EE EndF(V),
Let VVbe
beaafinite
finitedimensional
dimensional vector space over a field
EndF(V),
and U a g-invariant subspace
subspace of V.
V. Prove
< U.
V/ Uififand
and only
only if [V,
[V, g] 5
U.
(1) gg centralizes
centralizes V/U
of V onto
(2) The
The map
map vv i-+
H [v,
[v,g]
g]isisaasurjective
surjective linear
linear transformation
transformation of
Cv(g).
[V, g] with kernel CV(g).
=) dimF(CV(g))
g])
(3) dimF(V)
d i m ~ ( V=
dim~(Cv(g)>+dimF([V,
dim~([V,gl).
3. Let
Let G
G be
be aafinite
finitegroup,
group, FFa afield
fieldofofprime
primecharacteristic
characteristicp,
p,and
and.7r
n an
irreducible FG-representation.
FG-representation.Prove
Prove Op(G7r)
O,(Gn) ==1.1.
irreducible
irreducibly
4. Let
LetFFbe
beaafield,
field,rrand
andqqbe
beprimes,
primes, X
X aa group
group of order r acting irreducibly
elementary abelian
abelianq-group
q-group Q,
Q, and
and V =
= [V, Q] a faithful
on a noncyclic elementary
faithful
irreducible
Then dimF(V)
dimF(V) =
= rk where k =
= dimF(Cv(X))
dimF(CV(X)) =
irreducible FXQ-module. Then
dimF(CV(H)) for
for some
some hyperplane
hyperplaneHH of
of Q.
dimF(Cv(H))
5. Let
permutation representation
Letaa:: G
G -->
+ Sym
Sym(I)
(I)be
beaapermutation
representationof
ofaafinite
finitegroup
groupG
G on
on
a finiteset
setII,, and
andlet
let F
F be
be aafield
fieldand
andVVan
anF-space
F-spacewith
withbasis
basisXX=
= (xi
(xi::i iEEI).
I).
afinite
The FG-representation n
7r induced
induced by
byaa is the representation on
on V with
g7r:
xi H
H xiga
I. V
permutation module
g n : xi
xigafor
for each
each g E G and i E I.
Vis
is called
called the
thepemzutation
module
of a.
a.Let
Let Xx be
be the
the character
characterof
of7r.
n. Prove
Prove
(1) X(g)
~ ( gisis)the
thenumber
numberof
offixed
fixedpoints
points of
of got
g a on IIfor
foreach
eachggEE G.
G.
(2) (>gEG
of G on I.
I.
(CgsGX(g))/IGI
x(g))/lG I isisthe
thenumber
number of
of fixed
fixed points of
(g)2)/ I GI Iisisthe
(3) If
onII then
then(C,,,
(E gEG Xx (~)')/IG
thepermutation
permutation rank
rank
If G
G is transitive on
permutation rank.)
rank.)
of G on I.
I.(See
(Seesection
section15
15for
for the definition of permutation
on I.
I. Let
6. Assume
Assumethe
the hypothesis
hypothesisof
of the
the previous exercise with G transitive on
zZ =
Cis[
xi,
Z
=
(z),
and
= G-+iEI Xi, Z = (z),
+
U=
{(xi:
IEI
=O,a1 EF
iEI
U is the core of the permutation
permutation module V.
V. Prove
(1) ZZ ==CV
(G) and
Cv(G)
and U
U ==[V,
[V,G].
GI.
.
The dual representation
representation
5511
(2) If
If W
W is
is an
an FG-module,
FG-module, i E I,
=Gi
Gi isis the
the stabilizer
stabilizer in G of
of i,i , w
wE
I, H =
Cw(H),
Cw (H),and
and W
W ==(wG),
(wG),then
thenthere
thereisisaasurjective
surjectivehomomorphism
homomorphism of
V onto W.
(3) Assume
Assume pp ==char(F)
char(F)isisa aprime
primedivisor
divisorofof111.
I I I. Then V does not split
over U, V does not
notsplit
splitover
over2,
Z,and
andififOP(G)
O' (G)==GGthen
thenHH'(G,
1(G,U/Z)
U/Z) #
#
of the
the 1-cohomology group
group H';
H1; in
0. (See section 17 for a discussion of
particular use 17.11.)
particular
17.1 1.)
7. Let
LetFFbe
beaafield,
field,UUaa2-dimensional
2-dimensionalvector
vectorspace
spaceover
over FFwith
withbasis
basis {x,
{x,y},
y),
G=
= GL(U),
and
V
=
F[x,
y]
the
polynomial
ring
in
x
and
y
over
F.
Prove
GL(U), and V = F [x, y] the polynomial ring in
F.
(1) Irn isisan
by f (x, y)g7r
=
anFG-representation
FG-representation on V
V where Ir
n is defined by
y)gn =
ff(xg,
(xg, yg)
for f EE VVand
g EE G.
yg)for
andg
G.
(2) GGacts
actson
onthe
the(n
(n+ 1)-dimensional
1)-dimensionalsubspace
subspaceVn
Vn of homogeneous
homogeneous polynomials of degree
nomials
degree n.
n. Let
Let nn
nnbe
bethe
therestriction
restrictionofof7rn totoVn.
Vn.
(3) IfIf char(F)
for pp 5
< n $ -1
char(F)==pp>>0,0,prove
prove7rn
n,,is not irreducible for
-1mod
mod p,
7rnisisirreducible
irreduciblefor
for005
<nn <
< p.
but nn
(4) ker(7rn)
the group
groupof
ofscalar
scalartransformations
transformationsaaIIofofUUwith
withaaEE F
F and
ker(nn)is the
+
an=1.
(Hint: In
transvections in G with center (x)
In(3)
(3) let
let TT be
be the group
group of transvections
(x) and for
of M
M=
=V
Vn
generatedbybyyjxn-1,
y1xn-j,0 05<jj 5
< i.i .
i <5nnlet
letMi
Mi be
be the
the subspace of
n generated
[yixn-1 +Mi-2,
+Mi_2, T]
< ii <5n.n.Conclude
Prove [ylxn-l
TI =
=Mi_1/Mi_2
Mi-l/Mi-z for all 1 i
ConcludeMo
Mois
is
contained
in
any
nonzero
FG-submodule
of
M
and
then,
as
Mo
is
conjugate
is conjugate
contained
FG-submodule
and then,
to (yn)
(yn)under
under G,
G, conclude
conclude Mi
MI is
is contained
contqined in any
any such
such submodule
submodule for
all i.)
8. Let
<
=V
LeV
t Vbe
b eaavector
v e c t ospace
r s p aover
c e o av field
e r a f Fi eand
l d F0a=n Vo
d O <= V1
Vo~V5
l ~Vn
V
~n~ =
Va
sequence of
sequence
of subspaces.
subspaces.Let
LetG
G be
beaa subgroup
subgroupof
of GL(V)
GL(V)centralizing
centralizingVi+1
Vi+' / Vi
Vl
each i,i, 00 5
< i << n.
for each
n. Prove
Prove
at most
mostnn - 1.
(1) G
G isis nilpotent
nilpotent of class at
1.
of V
Vthen
thenCU
Cu(G)
(G)## 0 and
and [U,
[U, GI
G] <
<
(2) IfIf 00 ##UUisisaa G-invariant
G-invariant subspace of
U.
9. Let
vector space
spaceover
overaafield
fieldFFwith
withnn2> 2, G
G=
Let VVbe
be an
an n-dimensional
n-dimensional vector
=
GL(V),
:1 j< (ji <i),T={Vi:
G
L ( V ) ,X ==((xi:
x i : 115 <i
i ~<nn)
) a ab abasis
s i s f for
o r VV,
, VVi
l ==( x(xj
j : l5
),T={V,:
1<
n},
1 5i<
i <
n),and
andY=
Y ={(xi):
{(xl):1<
1 5i<
i 5n}.
n).Prove
Prove
of type
type II =
= {0,
(1) TT isis aa flag
flag of PG(V) of
{O, ..... .,,n --1}.GT
lj.GT==BBisisthe
the group
group
of lower triangular matrices and B is
is the
the semidirect
semidirect product
product of
of the
the
subgroups U and H where
where U
U consists
consistsof
of the
the matrices
matrices in B with
with 11 on
the main diagonal
diagonal and H isisthe
the group
groupof
of diagonal
diagonalmatrices.
matrices.
of F'
P.. UUisis nilpotent
Z(U)
(2) HH==Gy
G isisthe
thedirect
directproduct
productof
of n copies of
nilpotent and Z(U)
is the root group
transvection. (The root group of a transvection
transvection t
group of a transvection.
consists
GL(V) with
with Cv(t)
Cv (t) 5
< Cv(G)
Cv (G) and
and [V,
[V,g]
g] 5
< [V, t].)
consists of those g EE GL(V)
F II >>22then
(Y) =
(3) NA(Y)
NG(Y)isisthe
thesemidirect
semidirectproduct
productofofHHby
bySn.
Sn.IfIfI IF
thenNG
NG(Y)
NG(H).
Linear representations
52
(4) If
If FFisisfinite
finiteof
of characteristic
characteristicp then
then U
U EE Sylp(G).
Syl,(G).
(5) BB==NG(U).
NG(U).Indeed
IndeedVV
Vi is the unique object of type i fixed by U.
(6) The
Theresidue
residue Fs
rsofofaaflag
flagSSof
of corank
corank 11isis isomorphic
isomorphicto the projective
line over FF,, and (G5)rs
( G s ) ~ s2 PGL2(F).
PGL2(F).(See
(See section
section 33 for
for the
the definition
definition
of residue.)
residue.)
10. Let FFbe
projectiveline
lineover
overF.
F.Let
LetGG =
=
be aafield
field and
and FI-==FFUU{oo}
{co}the projective
GL2(F)
by 22 matrices
matrices over
over F,
F, and for
G L ( F ) be the group of invertible 22 by
A = (aij) =
a1,1
a1,2
a2,1
a2,2
(=- GL2(F)
define @(A):
O(A):I-F +
-+ F
I- by
O(A): z H
a1,1z + a2,1
a1,2z + a2,2
by convention
conventionalto
a/oo =
= 00 for
i /a1,2. Pick
for aa EE F#
F#and
andooo(A)
co@(A)==ai,al,l/al,2.
Pick aa
where by
2-dimensional vector
vectorspace
space V
Vover
overFF,, and
and identify
identify
basis B =={x1,
{xl,x2}
x2}for aa 2-dimensional
GL(V) with G via the isomorphism
isomorphism MB::GL(V)
GL(V) +
-+ G. Let Q
GL(V)
!2 be the points
of the projective geometry of V.
V. Prove
Prove
(1) For
For A
A EE G,
G,O(A)
@(A)isisaapermutation
permutation of
of 1,.
r.
0: G +
-+ G* is a
(2) G*
G* =={O(A):
{@(A):A EE G}
G}is
is aa subgroup
subgroup of Sym(F),
Sym(r), and @:
surjective
group homomorphism
homomorphismwith
withkernel
kernelZ(G),
Z(G), so
so @0 induces an
surjective group
+ G*,
G*, where G =
PGL(V).
isomorphism 6:
: G -->.
= PGL(V).
a: Q
-+ Fr by
oo and
anda:
a: F(hxl
F(a.x1+x2)
+x2)HH h? for ha. EE F.
F.
(3) Define
Define a:
!2 +
by a:
a:Fx1
Fxl i-±
H co
bijection such
suchthat
that(wg)a
(tog)a =
= (oa)$(g) for each
Then aa is a bijection
each Co
o EE!20 and
g EE G. Hence
Henceaaisisan
anequivalence
equivalenceof
of the
thepermutation
permutation representations
representations
C on Q
!2 and G* on r.
of O
F.
c
c.
5
Permutation groups
This chapter derives a number of properties of the alternating and symmetric
groups
S ofoffinite
oups An
A, and S,
finitedegree
degreen.
n. For
Forexample
examplethe
the conjugacy
conjugacyof
of elements
elementsin
in
An and S,
S,, is determined, and
and it is shown that A,
An is simple if
if n >2 5.
5.Section
Section
15 also contains a brief discussion of multiply
multiply transitive
transitive permutation groups.
groups.
Section 16
16 studies rank 3 permutation
permutation groups.
groups.
alternating groups
15 The
The symmetric and alternating
Let X be aa set
set and
and SS the
the symmetric
symmetric group on X. A
A permutation group on X
X
of S. Let G be a permutation group on
on X.
X. In this section
section X
X is
is a subgroup of
G are
are finite.
finite.
assumed to be of finite order n. Thus S is of order n!, so S and G
let H
H=
ordermmand
andHH=
= {gl:
<
Suppose gg E S and let
=(g).
(g).Then
Thengg is
is of finite order
{g':00 5
}. Further H has a finite number
numberof
oforbits
orbits(x,
(xiH:
H:115< ii 5
< k), and the orbit
i <<mm}.
xi
H is of finite order
order I,.
li. Let
Let H,
Hl =
= Hx;
in H of
x, H
H,, be
be the
the stabilizer in
of xi.
x,. By
By 5.11,
5.1 1,
} is a set
H : Hi 1, so, as H== (g)iscyclic,H,
(g) is cyclic, Hi=
= (gli)and{g-':Os
(gli) and {gJ : 0 <j j < liI,}isaset
=IIH:H,I,so,asH
I,li =
representatives for H,
Hi in H. Hence,
5.8,x,xiHH=
= {xi
g1:
of coset representatives
Hence, ,by
by 5.8,
{x,g-'
:005< jj <<liI,}.
}.
Therefore g acts
Therefore
acts on
on xi
x, H as
as the
the following
followingcycle:
cycle:
-1).
gIXix
= (xl,
xig, xlg2,
xig2, ...- ., xig1
g
l x , ~=
(XZ,
xzg,
Xzg 1,-1).
9
This
notation indicates
indicates that g:
g:x,
xig'
xi gj+l
gJ+1for
for005<jj <
< 1,
li - 11and
This notation
g-' i-+
H x,
andg:
g: xi
x, gli -1HH
gl
fixes
of H partition X, we
xi.
x, .The
The last fact holds as gb fixes xi. Further, as the orbits of
can describe
describe the action of g on
on X
X with
with the
the following
following notation:
notation:
glg-l
(x1, xlg, ... , xlg1 -1) (X2, X29, .. -X29 12-1) ... (xk, xkg, ..
,
xkgit-1).
This is the cycle
g. It describes g, and the decycle notation for the permutation
permutation g.
of representative
representative xi
xi for
for the
the ith
ith orbit and the
scription is unique up to a choice of
ordering
1 , 2, ..... ., ,n},
theset
setofofintegers
integers{{1,2,
n},the
the
orderingof
of the
the orbits.
orbits.For
Forexample,
example,ififXXisisthe
representative
th orbit, and
xi could
couldbe
bechosen
chosento
to be
be the
the minimal
minimal member
memberof
of the
the iith
representativexi
< xk.
the orbits ordered so that xl
x1 < x2 <
< ... <
xk. If so, g can be written uniquely
each partition
partition of
of X and each ordering of the
in cycle notation, and conversely each
partition and the
the members
members of the partition,
partition, subject
subject to these constraints,
constraints, defines
some member of S in the cycle
cycle notation.
notation.
By convention the terms
terms (xi) corresponding
corresponding to
to orbits
orbits of
of H
H of length 1 are
omitted.
Thus for
for example
example ifif n =
= 55 we
2)(3, 4)(5) as
we would
would write
write gg ==(1,
(1,2)(3,4)(5)
as
omitted. Thus
Permutation groups
54
g ==(1,
2)(3, 4). Notice g isis still
(1,2)(3,4).
stilluniquely
uniquely described
described in this
this modified
modified cycle
notation.
notation.
gt;-l) is
is a member
member of S.
Subject to this
this convention,
convention,gig1== (xi,
(xi, xig,
xt g,....,
. . ,x!
xig''-')
S. The
The
elements gl, ...
. . ., ,gk
gk are called the cycles of g. Also g is said to be a cycle ifif H
has at most one orbit of length greater than 1.
1. Notice the two uses of the term
`cycle' are compatible.
'cycle'
compatible.
subset A of X let Mov(A) be the set of points
points of X moved by A. Here
Here
Given a subset
Mov(A) =
= Mov((A))
x in X is
is moved by A if ax # x for some aa E A. Notice Mov(A)
Mov((A))
and X is the disjoint
disjoint union of Mov(A)
Mov(A) and Fix(A).
Fix(A). Cycles
Cycles cc and
and d in
in SS are
are said
said
fl Mov(d)
Mov(d) isis empty.
empty.
to be disjoint if Mov(c) fl
(15.1)
Let A, B
Bg
C S with
fl Mov(B)
Mov(B)empty.
empty.Then
Thenaab
= ba
ba for all
with Mov(A)
Mov(A) n
b=
(15.1) Let
aaEAandbEB.
~ A a n d B.
b ~
cyclesofofgg EE s'
S.. Then
(15.2)
gi,, ...,
(15.2) Let gl
. . .,grgr be
be the nontrivial cycles
Then
gigj =gjgi
= gigs fori
for i## j.
(1) gjgi
(1)
=gi
gl .. .. .. gr
nontrivial cycles.
cycles.
g, is the product in S of its nontrivial
(2) g =
(3)
(3) If g ==clcl.... .cs
.cswith
with{cl,
{cl,.... ., .cs)
, c,)a asetsetofofnontrivial
nontrivialdisjoint
disjointcycles
cyclesthen
then
{cl,...,C5)
(gl,...,g,).
Iclt . . - , c s )=_ Igl,
- .. , g r ) .
(4) The order
common multiple
least common
multipleof the
the lengths
lengths of
of its
its cycles.
cycles.
order of g is the least
(4)
written uniquely
uniquely as
as the
the product
product of
of nonBy 15.2, each member of S# can be written
trivial
trivial disjoint
disjoint cycles,
cycles, and these cycles
cycles commute, so the order of the product is
immaterial.
For g cESSdefine
defineCycg
Cyc,totobebethe
thefunction
functionfrom
from7L+
Z+into
into1L
Zsuch that Cycg(i)
Cyc,(i)
of cycles
cycles of
of gg of
of length i. Permutations g and h are said to have
is the number of
structure if Cyc,
Cycg =
= Cych.
the same cycle structure
Cyc,.
(15.3) Let g, h E S with
(15.3)
g =(al,...,aa)(bl,...,bp)...
Then
(1) gh
g' ==(alh,...
bah)....
(alh, . . ,. ,aah)(blh,
a,h)(blh, ..... .,,bah).
. ..
(1)
ifssand
(2) s EE SS isis conjugate
conjugate to g in SS if
if and
and only if
andgghave
havethe
the same
same cycle
cycle
structure.
structure.
A transposition is an element of S moving exactly
exactly two
two points
points of
of X.
X. That is a
transposition is a cycle
cycle of length
length 2.
2.
(15.4) SS is
(15.4)
is generated by its transpositions.
transpositions.
55
The symmetricand
and alternating
alternating groups
Proof. By 15.2 ititsuffices
sufficesto
to show
show each
each cycle
cycle isis aaproduct
productof
oftranspositions.
transpositions.
But (1,2,
(1, 2, ....,
2)(1, 3) .... .(l,
. . ,m)
m) ==(1,
(1,2)(1,3).
(1,m).
m).
A permutation is said to be an
an even
even permutation if it can
can be
be written
written as
as the
the
product of an even number of transpositions, and to be an odd permutation
permutation if
it can be written
written as the product of an odd number of transpositions.
transpositions.Denote
Denote by
by
Alt(X) the
the set
set of
of all
all even
even permutations
permutations of X.
X.
(15.5) (1) Alt(X)
Alt(X) isis aa normal
normal subgroup
subgroup of
of Sym(X)
Sym(X) of
of index 2.
(15.5)
permutationisiseven
evenifif and
and only
only ifif itit has
has an
an even
even number of cycles
cycles of
(2) AApermutation
length. A permutation is
even length.
is odd if and
and only
only if it has an odd number of cycles
cycles
of even length.
Proof. Without
loss X
X=
= (1,
ring R
R=
=
Proof.
Without loss
{I, 2, ..... .,,n).
nj. Consider
Consider the polynomial
polynomial ring
7L[xl,
Z[xl, ..... .,,xn]
x,] in nn variables
variables xi
xi over
overthe
thering
ring1L
Z of integers.
integers. For
For Ss EE SS define
define
sa:R
R by
f(x1,...,xn)sa
= f(x1s,...,xn5).Check
that a:S-+
sa: R +
R by
f(x1,. . . ,x,)sa
= f(xls,. . .,x,,). Checkthat
a : S + Aut(R)
Aut(R)
homomorphisms.
is a representation
representation of S in the category of rings and ring homomorphisms.
Consider the
the polynomial
polynomialP(xl,
P(x1,....
xn) =
= P EE RRdefined
.. ,, x,)
defined by
n
P ==P(x1,
P ( x l , ....
. . , x,)
= = H (X
x'n)
- xi).
(xjj -xi).
l1<i<j<n
'i<jin
For sS EE S,
S, ssa
the factors
factorsxxj j--xixi of
of PP up to aa change
For
a permutes
permutes the
change of sign, so
Psa
(1,2),2),then
thenP tPta
P
s a ==PPoror-P.
-P.Moreover
Moreoverifift tisisthe
the transposition
tran6position (1,
a ==- -P.
P.
Thus A
A=
Thus
={{ P,
P , - PP)) isisthe
the orbit
orbit of P under
unde; S.
S. Let A =
=Sp
Spbe
be the
the stabilizer
stabilizer
is also the kernel of a so A a S and, by2.11,
by 2.11, (IS:
Al( = JAI
s .A
AisalsothekernelofasoA<ISand,
S:A
( A /==22..
of P iinn S.
form aa conjugacy
conjugacyclass
classofofS,
S,so,
so,asast t EE SS -A
A and
and
By 15.3, the transpositions form
A <IS,
4 S, each transposition
- A.
= 2,
transposition isisinin SSA. But, as IS:
IS: Al
A( =
2, the
the product
product of m
of SS - AA isis in
m isis even.
even.So
SoAA== Alt(X)
Alt(X)and
andSS-- A
elements of
in A
A if and only ifif m
(1), and, since we saw during the
is the set of odd permutations. This proves (I),
proof of 15.4 that a cycle of
of length
length m
m is
is the
the product
product of
of m
m- 11transpositions,
proof
transpositions,
(2) also holds.
The group Alt(X) is the alternating
alternating group
group on
on X.
X. Evidently
Evidently the isomorphism
isomorphism
type of Sym(X)
Sym(X) and Alt(X)
Alt(X) depends
depends only
only on
on the
the cardinality
cardinality of X,
X, so
so we
we may
may
= n.
n. The
write Sn
S, and An
A, for Sym(X)
Sym(X) and
and Alt(X),
Alt(X), respectively,
respectively, when IX
(XII=
groups Sn
S, and An
A, are the symmetric and alternating groups
groups of
of degree
degree n.
n.
by Xm
X1 the set product of m copies of X.
If m is aa positive
positive integer, denote by
X.
If G isis aapermutation
permutation group on X, then
then G
G isis also
also aapermutation
permutation group
group on
X'
H (x1g,
Xmvia
viag:g :(x1,
(xl ,...
. . ., ,x,n)
x,) H
(xlg, ...
.. .,,xng)
xmg)for
for gg cEGGand
and xi
xi cEX.
X.Assume
Assume G
G
on X. Then the orbits of G on X2
is transitive on
x2are
are called
called the orbitals of G.
G.
the diagonal
diagonal orbital
orbital {(x, x):
x):xx Ec X}.
permutation rank of
of a
One orbital is the
X). The permutation
transitive permutation group G is defined to be the number of orbitals of G.
56
Permutation groups
(15.6) Let
transitivepermutation
permutationgroup
groupon
onX,
X,xXEEGG,, and
and(xi
(xi: 1: 15<i i5< r)
r)
(15.6)
Let G
G be a transitive
representatives
forthe
theaction
actionofofHH=
= G,
Gx on
on X.
X. Then
Then {(x,
{(x,xi):
xi):11_(<ii 5
< r}
r } is a
representatives for
set of representatives
representatives for
for the
the orbitals
orbitals of
of G
G and
and ((x,
x , y) E (x,
(x, xi)G ifif and
and only
only ifif
y EE xi H. In particular r is
is the
the permutation rank of G.
regular permutation representation
The regularpermutation
representationof
of aa group
group H
H is
is the
the representation
representation of H
H
byright
rightmultiplication.
multiplication.Apermutation
A permutationrepresentation
representationn:
n :H
H+
-)- Sym(Y)
on itself by
Sym(Y)
semiregular if and only if the identity element is the only element
element of H fixing
is semiregular
fixing
Equivalently H,
Hy =
= 11for
points of Y. Equivalently
for all
all yy in
in Y.
Y.
(15.8)
representation nn of
(15.8) A permutation representation
of finite
finite degree
degree is semiregular
semiregular if and
constituents of
of n
n are regular.
only if the transitive constituents
A regular normal subgroup of G is a normal subgroup of G which is regular
on X.
(15.9) Let G be transitive
transitiveon
onX,
X,xx EEX,
X, and
andH
H I_(
< G.
(15.9)
G. Then H
H is
is regular
regular on X
X
if and only if G,
Gx is a complement to
to H in G.
(15.10) Let H
K by a subgroup A
A
H be
be the
the split
split extension of a normal subgroup K
(15.10)
of H
of H
H on
on the
the cosets
cosetsofofA.
A.Then
ThenKKE
=K
Kitn
H and
and let jr
n be the representation of
and Kit
of HHn.
K n isis aa regular normal subgroup of
n.
(15.11) Let H be
of G and X
E X.
X. Then the
the map
map
(15.11)
be aa regular
regular normal subgroup of
xE
a:
by aa:: hh H
a : H + X defined
defined by
Hxh
xhisisananequivalence
equivalenceof
of the
the representation
representation of
Gx on
on H
H via conjugation
Gx on X.
G,
conjugation with the representation of G,
Given
a positive integer
integer m, G is said to
to be
bem-transitive
m-transitive on
on X
X ifif G
G acts
actstransitively
transitively
Given apositive
consistingofofthe
them-tuples
m-tuplesall
allof
of whose
whoseentries
entriesare
are distinct.
distinct.
on the subset of
of Xm
X' consisting
permutation rank 2. Also
Notice G is 2-transitive
2-transitiveif and only if it is transitive of permutation
for kk 5
< m.
m-transitivity implies k-transitivity for
m.
(15.12)
(1) Let m >
(15.12) (1)
2 22 and
and xx EE X.
X.Then
ThenGGisism-transitive
m-transitive on X if and
and only if
G is transitive and
and G,
Gx is
is (m
(m - 1)-transitive
1)-transitive on X --{x}.
{x}.
(2) Sym(X)
Sym(X) is
is n-transitive
n-transitive on X.
(3) Alt(X)
Alt(X) is (n --2)-transitive
2)-transitiveon
on X.
X.
(4) If
2)-transitiveon
on X
X then
then G
G ==Sym(X)
Sym(X)or
orAlt(X).
Alt(X).
If G
G is
is (n
(n --2)-transitive
Proof. The
Thefirst
first three
three statements
statements are
are straightforward.
straightforward. Prove the fourth by induction on n using (1) and the following observation:
If G is transitive on X
observation: If
and Gx
)x or Alt(X),
Alt(X )xthen
thenGG =
= Sym(X)
G, ==Sym(X
Sym(X),
Sym(X) or Alt(X), respectively. The
observation
followsfrom
from5.20
5.20plus
plusthe
thefact
factthat,
that,ififG,GX5<Alt(X)
Alt(X)and
andnn 2> 4,
observation follows
,
The symmetric
symmetric and
and alternating
alternating groups
57
then G =
=(Gy:
(G,: Yy EE X)
X) <5Alt(X),
Alt(X),since
sinceAlt(X)
Alt(X)a1Sym(X)
Sym(X)and
andGGisis2-transitive
2-transitive
on X.
(15.13)
of G.
(15.13) Let G be
be m-transitive
m-transitive on X and H aa regular
regular normal
normal subgroup of
Then
(1) If
If m
m ==22then
thennnisisaapower
powerof
ofsome
some prime
prime p and
and H
H isisan
anelementary
elementary
abelian p-group.
(2) If
of 22 or
or nn =
= 3 and G =
If m
m ==33then
theneither
either nn is
is aa power of
=Sym(X).
Sym(X).
(3) If
m >2 44t then
m == 44==nna and
Sym(X).
Ifm
henm
n d GG =
=Sym(X).
Proof.
m2
> 2. Let xx E X and K =
By 15.12,
15.12, KK isis (m
(m-- 1)Pro05 We
We may take m
=Gx
G,.. By
1)transitive on
on X
X- {x},
by 15.11, K
K acts (m
(m - 1)-transitively
transitive
{x},and then, by
1)-transitively on H#
H'
conjugation. In particular K is
via conjugation.
is transitive on H#.
H'.
Cauchy's
Let p be
be aa prime divisor
divisor of n. As H
H is
is regular
regular on
on X,
X, nn ==I H
IH1.I . So by Cauchy's
Theorem there
thereisis hh EE H
H of order p. Thus,
Thus, as K
K isistransitive
transitive on H#,
H', every
every
Theorem
element of
of H
H#
Theorem that
that HH
element
' is of order
order p.
p. We
We conclude
conclude from Cauchy's
Cauchy's Theorem
is a p-group.
p-group. So n =
=JHJ
IHI is aa power
power of p. By
By 9.8,
9.8, H
H isissolvable
solvable and
and as K
K
on H',
H#, H is
of G. So, by 9.4, H isis
is transitive on
is aa minimal
minimal normal subgroup of
elementary
ntary abelian.
abelian.
Thiss completes
completesthe
theproof
proofofof(I),
(1),so
sowe
wemay
maytake
takemm2>3.3.Let
Letyy=
= xh.
xh. By
By
15.2,
transitiveon
onXX - {x,
= CK(h)
15.2, KY
K , isistransitive
{x,y}
y } and
and so,
so, by 15.11,
15.11, KY
K, =
CK(h)isis transitive
transitive
CK((h)), so either (h)
on H --{1,
(1,h}
h}via
via conjugation.
conjugation. But CK(h)
CK(h) CK((h)),
(h) ==11,
(1,h}
h}
=3.
3. In
In the
the first
first case p ==22and
andin
in the
the second
second G ==Sym(X)
Sym(X)by
by 15.12.4.
15.12.4.
or n =
This completes the
the proof
proofofof(2),
(2),so
sowe
wemay
maytake
takemm2>4.4.Let
LetggEEHH - (h).
(h).
By 15.11 and
and 15.12,
15.12,CK(g)
CK(g)f?nC,y(h)
CK(h)== JJ is
on HH --{1,
is transitive
transitive on
(1, g,
g, h)
h) via
via
But JJ centralizes gh,
gh, so n =
= 4.
conjugation. But
4. Hence
Hence G ==Sym(X)
Sym(X)by
by 15.12.4.
15.12.4.
Recall
Recall the
the definition
definition of aa primitive
primitive representation
representationfrom
from section
section5.
5.
(15.14) 2-transitive representations
(15.14)
representations are
are primitive.
H a GG then
then H
H isistransitive
transitive on X and
(15.15) If G
(15.15)
G isis primitive
primitive on
on X
X and
and 11 # H
H for
each x E X.
=Gx
G,H
foreachx
G=
Proof. Let
of G, while
while as
Pro05
Let X
xE
E X. By 5.19, M =
=GX
G, is a maximal
maximal subgroup of
M by
by 1.7. Thus
ThusMH
MH =
=M
M or G.
H a1G,
G, MH
MH isis aasubgroup
subgroup of G containing M
In the latter case
case H
H is transitive
transitiveon
onXXby
by5.20.
5.20.InInthe
theformer
formerHH5< M
M,, so
so HH 5
<
ker(r)
ker(n)by
by5.7,
5.7,where
where rnisisthe
therepresentation
representationof
of G
G on
on X.
X. But,
But, as G <5S,S,jrn isisthe
the
identity map on G and
and in
in particular
particular is
is faithful.
faithful. This
This is
is impossible
impossible as
as H
H # 1.
1.
(15.16)
groupA,
A is simple ifif nn 2
> 5.
5.
(15.16) The alternating group
5588
Permutation groups
Proof.
H ==G.
Proof. Let nn >
2 5,
5, G
G ==Alt(X),
Alt(X), and
and 11 ## HHa_aG.
G.We
Wemust
must show
show H
G.By
By
15.12, G is (n
(n - 2)-transitive
2)-transitiveon
on X,
X, so,
so, by 15.14
15.14 and 15.15, H is transitive on
H, where K ==G,andx
Gx and X E
= H nflKKt hthen,
X aand
n d G == K
KH,whereK
c XX.. IIff 11 =
e n , bby
y 115.9,
5 . 9 ,H iiss
regular on X. But this contradicts
contradicts 15.13 and the
the fact
fact that
that G
G is
is (n
(n - 2)-transitive
2)-transitive
with n >
>_ 5.
5.
K.But
ButKK==Alt(Y),
Alt(Y),where
where Y
Y=
=X
X--(x),
(x),so,
so,by
byinduction
induction on
on
So 110#HHflnK.
n, either K
K is
is simple
simpleorornn=
= 5.
5. In
Inthe
theformer
formercase,
case,asas1 1# ¢HHf fll KK_a4K,
K,KK =
=
K4
< H,
H.H.Thuswemaytaken
Thus we may take n ==5.5.HereatleastHflK
Here at least H fl K
H nfl K
H , so
s o G ==KH
K H= =
is
transitive on Y by 15.12,
and 15.15.
So 4 ==I lYl
Y (==(H
fI K)y I
istransitiveony
15.12, 15.14,
15.14,and
15.15.So4
l HfInKK: (H
:(HnK),I
for y EE YY by
( ==( (HH::H
(, so
by 5.11.
5.11. Similarly
Similarly 55 ==(X
(XI
H flnKKI,
so 20
20 divides
divides the order
order
of H.But
H. But (S(
= 5! =
while IS:
so IGI
IS1 =
=120
120while
IS: G(=
GI =2,2,so
IGI==60.
60.Thus,
Thus,asasIHI
[HIdivides
divides
(G(, I1H(
= 20 or 60. In the latter case H ==G,
IGI,
HI =
G,so
sowe
wemay
mayassume
assumethe
theformer.
former.
Exercise 2.6,
2.6, H
H has a unique
unique Sylow
Sylow5-group
5-groupP.
P. Hence
Hence PP char H _a
4 G,
By Exercise
G,
so, by 8.1, P
P 4 G.
G.This
Thisisisimpossible
impossibleasaswe
wehave
havejust
just shown
shownthat
that44 divides
dividesthe
the
nontrivial normal subgroup
order of any nontrivial
subgroup of G.
G.
a
(15.17) (Jordan)
(Jordan)Let G
G be
be aa primitive
primitive permutation group
group on
on aa finite set X and
suppose
nonempty subset of X such that IX
( X-- YYI (>> 1 and
and Gy
Gy is
is transitive
transitive
suppose Y is a nonempty
onX
Y. Then
Then
on
X --Y.
(1) G
G isis2-transitive
Z-transitiveon
on X, and
and
(2) ifif Gy
- YY then
primitiveon
onXX- (x}
Gyisis primitive on X then Gx
G, is primitive
(x) for xx cE X.
X.
Proof. Let
YYand
Let Fr ==XX-andinduct
inducton
on(Y(.
I Y I.IfIf(YI
1 Y I ==11the
theresult
result isis trivial.
trivial. So
So
assume IIYYI (>> 11and
andlet
letxx and
and yy be
be distinct
distinct points
points of
of Y.
Y. By
By Exercise
Exercise 5.5, there is
assume
ggEGwith
E G with xxEYgandyVYg.LetH=(Gy,Gyg)andS2=FUFg.Then
E Yg and y $ Yg. Let H = (Gy, GYg)and C2 = r U r g . Then
H <_(Gx
H acts
on 0. Suppose I Irl
r I >>(Y
1. Then r fl rg is nonempty so H
G, and
andH
actsonC2.Suppose
IYI.Thenrnrgisnonemptyso
H
is transitive
transitiveon
onC2.0. As
AsHH 5
< G,
of G,.
G. Since
G,,r rU U(y}
(y}isiscontained
contained in
in an
an orbit of
Since
holds for
for each
each yy EE Y
- (x},
this holds
Y(x},GGisis2-transitive
Ztransitive on X by
by 15.12.1.
15.12.1. Further ifif
primitive on
on rF and
and Q is a G,-invariant
Gx-invariantpartition
partitionofofX'X'=
=X - {x},
Gy is primitive
(x), then
then
ffor
o rZZEEQQeither
eitherIZ(Zf?flrr[(4<11ororrr _
c Z. As
followsthat
thatIZ(
(Z (=
G
As 1171
Irl 2> ((YY ((ititfollows
=11
Gx is primitive on XI.
X'.
IX'I. Hence G,
or IX'(.
assume Irl
Ir( <
So assume
_( (Y(
lYI and let a,
a,yy be
be distinct
distinct points
points of F.
r.By
By Exercise
Exercise 5.5
5.5
thereish
= (Gy,
(Gy,Gyh),F'
thereis h EEGGwithy
with y Ecr rhbut
h but aa $Vr hrh.LetK
. Let K =
GYh),r' ==rrUrh,
Urh,
and
Y' =
= X --F'.
Fh,
and Y'
r'. Then
Then K
K <_(Gy,,
Gy/,and
and as
as y cE rrflf?
rhK
, Kisistransitive
transitive on F'
r'
and
( F>( >
(rI 4
< (Y(
andy
and Irl
Ir'IF'-r(.As
- rl. As Irl
IYI and
y EcrrflFh,wehave
n r h , we have Y'
Y' # 0.IfGy
0. If G y is
is
primitive on
on rF then, as (I rF1 (>> I (r'r' - Fr 1,
primitive on
on r'
F' by
1, K is primitive
by an
an argument
argument
in the
induction
onon
I YIY1,I, the
the last
last paragraph.
paragraph.So,
So,replacing
replacingYYby
byY'Y'and
andapplying
applying
induction
the
result
result holds.
Jordan's Theorem
investigating finite alternating groups and
Theorem is
is aa useful
useful tool for investigating
symmetric
5.7, and
symmetric groups. See
See for
for example
example Exercises
Exercises 5.6,
5.6,5.7,
and 16.2.
16.2.
.
59
59
Rank 3 permutation groups
Rank 33 permutation
permutationgroups
groups
16 Rank
In this section
section G
G isis aa transitive
transitive permutation
permutation group
group on
on aa finite
finite set X of order
order n.
n.
Recall
Recall the
the definition
definition of an orbital in the preceding section. Given an orbital
AP of A
A is
A of G, the paired
paired orbital AP
AP = {(y, x): (x, y) E A).
Evidently AP
of G
G with
with (AP)P
(Ap)p =
= A.
Ap is an orbital of
A. The
The orbital
orbital A
A is
is said
said to
to be
be
AP.
self paired
paired ifif A
A=
= AP.
(16.1)
(16.1) (1)
( 1 ) AAnondiagonal
nondiagonalorbital
orbitalA
A of
of GGisisself
self paired
paired ifif and
and only if (x,
( x ,y)
y) is a
cycle in some gg E G, for (x,
( x ,y) E
E A.
A.
(2)
possessesaanondiagonal
nondiagonalself
selfpaired
pairedorbital
orbitalififand
and only
only ifif G
G isis of
of even
even
(2)GGpossesses
order.
order.
(3)
( 3 )IfIf GGisisofofeven
evenorder
orderand
and(permutation)
(permutation)rank
rank33then
thenboth
bothnondiagonal
nondiagonal
orbitals of G
G are
areself
self paired.
paired.
orbitals
Recall
3.
Recall the
the definition
definition of
of aa graph
graph from
from section
section3.
(16.2) Let
Let AA be
beaaself
self paired
paired orbital
orbital of
of G.
G. Then
Then A
A isis aasymmetric
symmetricrelation
relation on
X, so g
-9 =
=(X,
( X A)
, A )isisaagraph
graphand
andGGisisaagroup
groupof
of automorphisms
automorphismsof
of -9
g transitive
transitive
g.
on the
the edges
edges of
of -9.
on
,<
61
In the remainder
remainder of this section
section assume
assume G
G is
is of
of even
even order
order and of permutation
rank
3.
Hence
G
has
two
nondiagonal
orbitals
A
and
F.
nondiagonal orbitals A and r.By
By 16.1.3
16.1.3both
both A
A and
andFr
rank 3. Hence G has
are
are self paired. For xx cE X,
X ,Gx
G, has
has two
two orbits
orbits A(x)
A ( x )and
and F(x)
r ( x )on
onXX --{x},
{ x )where
,where
A(x)
= {y
this
A(x) =
( y cE X:
X: (x,
( x , y) cE A}
A ) and
and rr(x)
( x )=={y
{ ycEX:
X:(x,
( x ,y)
y) cEr);
r);
thisholds
holdsby
by
15.6. By
By 16.2,
16.2, g
' ==
(X,
a graph
( XA)
, Ais) is
a graphand
andGGisisa agroup
groupofofautomorphisms
automorphismsof
of-9
g
transitive on the edges of -9.
Notice
A(x)
is
the
set
of
vertices
adjacent
to
x
in
g.
A ( x ) is
set
adjacent
this gaph.
graph.Defined
Define x1=={{x}UA(x),k
IA(x)l,l =
= IF(x)I,,1
= IA(x)nA(y)I
x ) U A ( x )k, == lA(x)l,l
Ir(x)l,h =
IA(x)nA(y)l
for
for y EE A(x),
A ( x ) ,and
and µp==I A(x)n
l A ( x ) A(z)I
nA(z)lfor
forZZEEF(x).
r ( x )As
.AsGx
G, isistransitive
transitiveon
on A(x)
A(x)
and F(x),A
r ( x ) ,h and
and µpare
arewell
welldefined.
defined.As
As GGisistransitive
transitive on
on X
X these
thesedefinitions
definitions
independent of the choice of xx..
are independent
(16.3) (1) n=1+k+l.
(2) µl = k(k - A - 1).
Proof.
Proof.Part
Part(1)
( 1 )isistrivial.
trivial.To
Toprove
prove (2)
( 2 )count
countIQI
IC2I in
in two
two different
differentways,
ways, where
where
0S2 =={(a,
c A(a),
{(a,b)
b )::bb,, xx E
A(a),b cE F(x)}
r(x))
X.
for fixed xx E
c X.
Permutation groups
60
60
(16.4)
(16.4) If kk <511then
then the
the following
following are
are equivalent:
equivalent:
imprimitive.
(1) GG isisimprimitive.
(2)
(2) ha,=k-1.
=k-1.
(3)
(3) a
p=
=0.
0.
(4)
c G} =
=Q
system
imprimitivity
(4) x1
xL =
=y1
y' for
for y cE A(x),
A(x), {(x1)g:
{(x')~: gg E
C2 is aasystem
of of
imprimitivity
for G, and
G
is
2-transitive
on
Q.
and is 2-transitive C2.
Proof.
for G
G and x cE 90 EES,
Proof: IfIf SS isis aasystem
system of
of imprimitivity
imprimitivity for
S,then,
then, by
by 5.18,
5.18,
Gx
acts on
on 8.
9. So,
So, as
as G,
Gxisis transitive
transitiveononA(x)
A(x)and
andr(x),
F(x),either
either0 9== x'x1 or
G, acts
or
dividesn,n,so
soasask
< 1,
= x1.
09 =
=(x)
{x)UU F(x).
r(x).By
By 5.18,
5.18,101
101 divides
k5
1,e9 =
x'. Now
NOW (4)
(4) holds. If
1f
x1
x' ==y1'y then
then(2)
(2)and
and(3)
(3)hold.
hold.Also
Also(2)
(2)isisequivalent
equivalentto
to(3).
(3). Finally
Finally ifif (2)
(2) holds
holds
then x'x1 ==y1,
y', sosoQC2isisaasystem
systemof
ofimprimitivity
imprimitivity for
for G
G and
and hence
hence (1) holds.
(16.5)
µ 0#00ororkkthen
then GGisisprimitive.
primitive.
(16.5) If p
Proof.
wemay
maytake
takek k>>1.1.Let
Let,2 µ==1l-- (k
(k -,X
- 1).
µ=
Proof: By
By 16.4 we
h1). If ,2
=00then,
then, by
= µ,
p , contrary
contraryto
to hypothesis.
hypothesis. Hence,
Hence, by 16.4
16.4 and symmetry between
16.3.2, kk =
A and F,
r ,G
G isis primitive.
primitive.
connected.
(16.6) If
If G is primitive
primitivethen
then ' isisconnected.
(16.6)
Proof.
_ (Gx,
Proof.By
By5.19,
5.19,Gx
G, isismaximal
maximalininG,
G,sosoGG==
(G,, Gy)
G,) for
for x
5.2 completes the proof.
Exercise
# y. Now Exercise
(16.7) Assume G is primitive. Then either
(16.7)
(1)kk=land
or
= l a n d pµ=,k+l=k/2,
= h + l =k/2,or
(1)
(2) d ==(X
(2)
(A --µ)2
p)2+ 4(k
4(k--µ)pis
) isa asquare
squareand
and setting
setting
D ==2k
2k + (A
(A --µ)(k
p)(k+ 1)1)we
wehave:
have:
(a) d1/2
d112divides D
D but
but 2d1j2
2d1I2does not if n is even, while
(b) 2d1/2
2d1f2divides D
D if
if n is
is odd.
odd.
+
+
+
Proof. Let
That is
is A
A=
= (axy)
Proof:
Let A
A be
be the
the incidence matrix for s-9.
?. That
(a,,) is the n by n
rows and
and columns
columns are
are indexed
indexedby
byX,
X, and
and with
witha,,
axy=
= 1 ifif (x, y)
matrix whose rows
is an edge of -9' while
while a,,
axy =
= 00 otherwise.
otherwise. Let B be
be the
the incidence
incidence matrix
matrix for
and II the
thenn by
(X, rF),
) , JJ then
the nby
bynnmatrix
matrixall
allof
of whose
whose entries are 1, and
by n identity
matrix. Observe:
Observe:
(i) AA isis symmetric.
symmetric.
A + BB =
(ii)
(ii) I + A
= JJ..
+ +
.
Rank 3 permutation groups
61
(iii)
AJJ =
(A-- kkI)J
(iii) A
= kJ,
k J , so
so (A
I ) J ==0.
0.
A2 =
k I ++ ,kA
A A + pAB.
B.
(iv) A2
= kI
definitions. As
As lA(x)l
(A(x)e=
= k,
The first two statements are immediate from the definitions.
each row
row of A has k entries
equal
to
1.
Thus
(iii)
holds.
By
(i)
the
(x,
y)-th
entries
1.
(iii)
(i) the (x, y)-th
product of
of the
the xth
xth and yth rows of
of A. But this inner
entry of A2 is the inner product
product just counts IA(x)
lA(x) fl
nA(y)e,
A(y)l, so
so (iv)
(iv) holds.
Next (ii), (iii), and (iv)
(iv) imply:
(v)(A-kI)(A2-(,l-A)A-(k-ll)I)=0,
(v) (A - ~ I ) ( A ' - (A - p)A - (k - p)I) = 0,
so the minimal
minimal polynomial
polynomial of
of A
A divides
divides
p(x) _ (x -k)(x2 -()-A)x -(k-bl)).
+
The roots
roots of
of p(x) are k, s,
=
The
s, and
and t,t ,where
where ss ==((A
((A --A)
p ) + d1/2)/2
d1I2)/2 and
and t =
((),
- A)
Let
((A p ) --d1/2)/2.
d1I2)/2.
Letme
mebebethe
themultiplicity
multiplicityofofthe
theeigenvalue
eigenvalue e.
e.
= kc
if ci
ci =
= cl
Claim mk
Mk== 1.1.Indeed
Indeedfor
forc c==(el,
(cl,.....,
Claim
. , c,), cA =
kc if
if and
and only if
cl
each i.i. As
As cA
cA =
= kc,
=
c1 >
for all i. To
To prove this we can take IIcll
>_ Ici
lcil for each
kc, kc1
kcl =
Eiciail, so,
= ci
Ziciail,
so, as
as exactly
exactly k of the ail are
are 11 and
and the
the rest
rest are 0, c1
cl =
ci for
for each
each
i cE A(1).
that clcl =
= ci
A(1). But
But now
now as
as -9 is
is connected
connected it follows
follows that
ci for
for all
all ii EE X,
X,
completing the proof of the
completing
the claim.
claim.
Mk =
= 11 itit follows
follows that:
that:
As mk
(vi)m,.+mt =n - 1 =k+1.
0:
Also, as A is of trace 0:
(vii) kk+
+ m,s
m,s +
+ mtt =
= 0.
0.
Now (vi) and (vii)
(vii) imply:
imply:
+
(viii)m,
m,.==((k
((k++l)t
l)t +k)/(t
k)/(t - s).
(viii)
Of course
coursett - s ==-d1/2,
((A
A)p -) d1/2)/2,
and
-dl/', t t==
((A- - d1I2)/2,
andms
m,isisananinteger,
integer. Thus
Of
(ix) (Dd-'I2
(Dd-1/2 - (k
(k++l))/2
1))/2isisan
aninteger,
integer,where
where
+
+
D=
=2k+(A-A)(k+1).
2k (A - p)(k I).
D
either dd isisaasquare
squareororDD=
= 0. If
If D
D=
In particular either
=00then
then A
p ==Ah +
+11and 1 == k,k,
so 16.7.1 holds by 16.3.2. If d is a square
then
16.7.2
holds
by
(ix).
square
16.7.2
Remarks. Wielandt
Wielandt [Wi
[Wi 2]
21 is
is aa good
goodplace
place to
tofind
findmore
moreinformation
information about
about
permutation groups. The material in section 16 comes from Higman [Hi].
Section 16 is somewhat technical and can be safely omitted by the novice.
On the other hand the results in section 15 are reasonably basic.
Permutation groups
62
Exercises for chapter
chapter 5
1. (1)
A5 has no
no faithful
faithful permutation
permutation representation
representation of degree
degree less
less
(1) Prove A5
than 5.
5.
(2) Prove
Prove that,
that, up
up to
to equivalence,
equivalence, A5
A5 has unique transitive representations
representations
of degree 5 and 6. Prove both are 2-transitive.
2-transitive.
= S5.
S5.
(3) Prove
Prove Aut(A5) =
(4) Prove
Prove there
there are
are exactly
exactly two
two conjugacy
conjugacy classes
classes of
of subgroups
subgroups of
of S6
S6 isomorphic
morphic to
to A5.
A5. Prove
Prove the
the same
samefor
forA6Ag.
group on
on X,
X, A a self paired
orbital of
of
2. Let
Let G
G be
be aatransitive
transitive permutation
permutation group
paired orbital
(x, y)
y) EE A,
A, #
-9=
= (X,
(X, A)
E) the graph
on X determined
by A,
A, and
and H =
G, (x,
graph on
determined by
=
(G, Gy).
(G,,
G,). Prove
Prove
(1) xH
XHUUyH
yHisisthe
theconnected
connectedcomponent
componentof
of -9
9 containing
containing x,
x, and
and
(2)
-9
is
connected
precisely
when
one
of
the
following
holds:
is
connected
precisely
when
one
of
the
following
holds:
(2)
(i) G
=H,
H, or
or
G=
H 4== 22 aand
partition. {xH, yH}
(ii) IIGG ::HI
n d'gisisbipartite
bipartitewith
withpartition.{xH,
yH)(i.e.
(i.e.{xH,
{xH,yH}
yH]
partitionofofXXand
andAA2c_(xH
(xHxxyyH)
is aa partition
H) UU(y(yH
H xx xxH)).
H)).
3. Let
Let SS and
and A
A be
be the
the symmetric
symmetricand
and alternating
alternating groups of degree n, respectively,
andlet
letaa EE A. Prove
Prove aaA
tively, and
A #0 as
as precisely
precisely when (*) holds
(*) Cyc,(2m)
Cyca(2m) =
= 0 and Cyc,(2m
Cyca(2m - 1)
1)<511for
foreach
eachpositive
positive integer m.
in which case as==aaA
Awith(a) =
= (b)andlaAI
inwhichcaseas
AUUbAforsomeb
bAforsomeb EEAwith(a)
(b) andlaAI_=(bAl.
IbAI.
4. Let
A
be
the
alternating
group
on
a
set
X
of
finite
order
n
>
3
and
let
V
Let A be the alternating group
order n > 3 and
V be
be
the set of subsets of X of order 2. Prove
(1) AAis
is aa rank
rank 33 permutation
permutation group
group on V.
V.
(2) A
A,isisa amaximal
maximalsubgroup
subgroupof
ofAAfor
for V
v E V.
rank 3 group
group on
onaaset
setQ2 of
of order
order 10
10then
thenGG%
- A5
(3) If
If G
G is
is aa primitive rank
A5 or
and the
the representation
representationof
of G
G on
on S2
0 is
S5 and
is equivalent
equivalent to its
its representation
representation
on V.
V.
5. Let
Let G
G be
beaaprimitive
primitive permutation
permutation group
group on a finite set X and let x and
and yy
be distinct points of X. Let Y be a nonempty proper subset of X and define
= {Yg:
and T(x)
T(x) =
= nzEs(x) Z.
S(x) =
{Yg:gg E G, x E
E Yg}
Yg} and
2.Prove
(1) T(x) =={x),
{XI,and
and
but yy 60 Yg.
(2) there
there exists gg E G with x E Yg but
6. Let
Let GGbe
beaaprimitive
primitive permutation
permutation group
group on a set X of finite order n. Prove
(1) If Y
C X such
such that
thatGGyyisisprimitive
primitive
andIYI
(Y==m
with 11 5<
(1)
YE
onon
XX
- Y-Yand
m with
m<
5 nn --2,2,then
thenGGisis(m
(m+ 1)-transitive
1)-transitive on X.
(2) IfIf GGcontains
containsaatransposition
transpositionor
or aa cycle
cycle of length 3 then G contains the
alternating group on X.
(3)
Mov(a)nflMov(b)J
Mov(b)l
then[a,
[a,b]b]isis aa cycle of
(3) If a,
a , bb E
E G with
with IIMov(a)
= =1 1then
of
length 3.
(4) Let
YC
2X
X be
be of
of minimal
minimal order
order subject
subject to Gy
Gy ==1.1.Prove
Prove either
either G
G
Let Y
n/2.
contains the alternating group on X or IIYYI (<
5 n/2.
1.
n,,,,,,
+
Rank 33permutation
permutation groups
groups
Rank
63
63
(5)
(5)IfIfGGdoes
doesnot
notcontain
containthe
thealternating
alternatinggroup
groupononX,X,prove
provej Sym(X)
iSym(X): :GGII >3
[(n
[(n+ 1)/2]!
1>/21!
7.
7.Let
LetXXbebeaaset
setofoffinite
finiteorder
ordernn<55,5,AAthe
thealternating
alternatinggroup
groupon
onX,
X,and
andGGaa
proper
propersubgroup
subgroupofofA.
A.Prove
Proveone
oneofofthe
thefollowing
followingholds:
holds:
(1) JA:
IA :G)
GI >>n.n.
(1)
(2) (A:GI
1A:GI =nand
= n a n d G=Ax
G =A, forsome
for somex
E X.
(2)
xEX.
(3) (A:GI
1A:Gl ==n=6andG-A5.
n=6andGSA5.
(3)
Provethat
thateither
either
8.8.Prove
(1)
(1)An
A, has
hasa aunique
uniqueconjugacy
conjugacyclass
classofofsubgroups
subgroupsisomorphic
isomorphictoto A,-1, or
or
(2) nn==
andAn
A,has
hasexactly
exactlytwo
twosuch
suchclasses.
classes.
(2)
6 6and
+
6
Extensions of groups and
and modules
modules
Chapter
considers various questions
Chapter 6 considers
questionsabout
about extensions
extensionsof groups
groups and modules,
modules,
particularly the conjugacy of complements
complements to
subgroup
most particularly
to some fixed normal subgroup
in a split
split group extension. Suppose G is represented
represented on an abelian group or
F-space V
V and
and form
form the
the semidirect
semidirectproduct
product GV.
GV. Section 17
17 shows there is a
between the
the set of conjugacy
of complements
complements to
to V in GV
bijection between
conjugacy classes of
GV
and the 1-cohomology
group H1(G,
H1(G, V).
V). IfIf V is an F-space so is H1(G,
1-cohomology group
H'(G, V).
V).
Cv(G) ==00there
thereisisaalargest
largestmember
member of
of the
the class of FG-modules
Moreover ifif C"(G)
that CU(G)
Cu(G) ==00and
U such that
andUUisisthe
theextension
extension of
of VVby
by aa module
module centralized
U isis the
thelargest
largest member
member of this
this class
class then
then
by G. Indeed it turns out that if U
U/V = H1(G,
V). Further
Further the
the dual
dual of
of the
the statement
is also
also true:
true: that
that is
is if
U/VE
H1(G, V).
statement is
= [V,
that U*
U* =
= [U*,
V=
[V,G]
GIthen
thenthere
thereis
is aa largest FG-module U* such that
[U*,G]
GI and
and
U* is the extension
extension of
of an FG-module Z by V
V with
with G
G centralizing
centralizing Z.
Z. In
In this
this
ZE
= H1(G,
case Z
H1(G,V*).
V*).
Maschke's Theorem are then used to prove the
These results together with Maschke's
Schur-Zassenhaus Theorem, which gives
Schur-Zassenhaus
gives reasonably
reasonably complete
complete information
information
extensions of
of aa finite group
group B by a finite group A when the orders of
about extensions
A and B are
A
are relatively
relatively prime.
prime. The
The Schur-Zassenhaus
Schur-Zassenhaus Theorem
Theorem is
is then
then used
used
to prove Phillip Hall's extended
Hall's
extended Sylow Theorem for solvable groups. Hall's
Theorem supplies a good illustration
illustration of how restrictions on the composition
composition
factors
group can
can be
be used
used to
toderive
derivestrong
stronginformation
information about
about the
the group.
group.
factorsof a finite group
I have chosen to discuss 1-cohomology
1-cohomology from a group theoretical point of
view. Homological algebra
arguments in section
algebra is kept to a minimum. Still the arguments
section
17 have a different flavor than most in this book.
17 1-cohomology
1-cohomology
In this section p is a prime, F
characteristic p,
F is
is aa field of characteristic
p, V
V is
is an
an abelian
abelian group
group
written additively, G is a finite group, and ir:
n : G -->
+ Aut(V)
Aut(V) is
is aa representation
representation
of G on V.
V.
product S(G,
S(G, V,
V, n7r)
Vby
by G
G with
with respect
respect to
to nit and
Form the semidirect product
) ofof V
identify G and V
identify
V with subgroups
subgroupsof
of S(G,
S(G,V,
V,7r)
n ) via
via the
the injections
injectionsof
of 10.1.
10.1.Then
Then
S(G, V,
= GV
V, 7r)
n) =
GVwith
with VV49GV
GVand
andGGisisaacomplement
complementto
toVVtotoGV.
GV.
fromGG into
into V
V is
is aa function
function y:
y:GG +
- VVsatisfying
A cocycle
cocycle from
satisfyingthe
thecocycle
cocycle
condition
E G.
(gh)y ==(gy)p`
(gylh + by
hy g, h E
+
.
1-cohomology
65
Notice the cocycle condition forces each cocycle
cocycle to
to map
map the
the identity
identity of
of G
G to
the identity
identity of V.
V.
r(G, V)
G to
to V and
and make
makeF(G,
r(G, V)
Let F(G,
V)denote
denotethe
theset
set of
of cocycles from G
V) into
a group
group (again
(again written
written additively)
additively) via:
y, 6S EEUI'(G,V),g
EG.
g(y + 6) = gy + g6 y,
G , V), g E
G.
g(Y+S)=gy+gS
A=
automorphisms of
ofGV,
GV,and
andlet
letU(G,
U(G,V)
V)=
_
Let A
=Aut(G
Aut(GV)
V) be
be the group of automorphisms
CA(GV/
v)
n
CA(V).
CA(GV/V) n CA(V).
over F
F and n
7r isis an
an FG-representation
FG-representation then
then F(G,
r(G, V)
If V is a vector space over
V)
F via:
is also a vector space over F
via:
= a(gy) Yy EE r(G,
dg(ay)
ay) =
U G , V),
V), g E
E G, a E
E F.
(17.1)
For y EE r(G,
(17.1) For
F(G,V)
V)define
defineS(y)
S(y)==fggy:
{ggyg: gE EG}.
G}.Then
Thenthe
themap
mapS:
S:yyr-*
I-+
S(y)
S(y) is
is aa bijection
bijection of r(G,
F(G,V)
V)with
withthe
theset
setof
ofcomplements
complementsto
to VVin
in GV.
GV.
Proof. The
Thecocycle
cocyclecondition
condition says
says that
that S(y)
S(y)isisaasubgroup
subgroupof
ofGV.
GV. Evidently
Evidently
S(y) is
is aa complement
complement to V in
in GV
GV and
and SSisisinjective.
injective. Conversely
Conversely ifif C is
then, for
for gg EE G,
G, gV
gV n
nC
c contains
contains a unique
unique element gv and
and
a complement
complement then,
y: g r-*
cocycle with
withS(y)
S(y) =
= C.
I-+ v is a cocycle
C.
(17.2)
(G, V) define
defineyy0:
--i GV
E I'
F(G,
4: GGV
V -+
G V by
by
(17.2) For y E
YO: gv H g((gY) + v).
the map
map4:
0:yy I-+
H yO
isomorphismofofF(G,
r(G, V) with U(G, V).
V
Then the
y 4 isis aa group
group isomorphism
For
uO-1: g I-+
r-* g-'gU.
g-lg"
Foruu E U(G,
U(G, V)
V ) and
a n d g E G,
G,u&':g
Proof. The
implies yyO
homomorphism.(-y)4
(-y)O is an
The cocycle
cocycle condition
condition implies
4 isis aa homomorphism.
for yyO,
so yyO
Aut(GV).
centralizes V
V and
andGGV/
inverse for
4, so
4 EE Aut(G
V). By definition yyO
4 centralizes
V/ V,
r-* g-'gU.
g-lg". An
so yyO
4 E U(G,
U(G,V).
V). For
For U
uE
EU
U define
define ul/r:
u+: gg I-+
Aneasy
easycheck
check shows
shows
ui/r is
is a cocycle
cocycle and
and i1r==&',
0-1, so 4
0 is an isomorphism.
u+
isomorphism.
+
(17.3) U(G,
V. Indeed
(17.3)
U(G,V)
V)acts
actsregularly
regularly on
on the
the set
setof
of complements
complementsto
to V
V in
in G
GV.
G"
GU=
=S(uo-1)
S(u4-') for
foreach
each u EE U(G,
U(G, V).
V).
Proof. As
As UU==U(G,
U(G,V)V)acts
actsononVVititpermutes
permutesthe
thecomplements
complementsto V
V in GV.
GV.
By definition
of the
the maps
maps40 and
and S,
S, GU
G" =
= S(uO-1)
dkfinition of
S(u4-') for
for uu EE U.
U. Hence,
Hence, as
as SS and
and
40 are
are bijections,
bijections, the
the action
action of U
U isis regular.
regular.
Lemmas
and 17.3
of the
to V in
in
Lemmas 17.1
17.1 and
17.3 give
give descriptions
descriptions of
the complements
complements to
GV in terms
terms of r(G,
F(G,V)
V)and
andU(G,
U(G,V),
V),while
while17.2
17.2and
and17.3
17.3give
givethe
the correscorrespondence between the two
two descriptions.
descriptions. The
The next
next few
few lemmas
lemmas describe
describethe
the
66
66
Extensions of groups and modules
representations
representations of G on rF(G,
( G , V)
V) and
and U(G,
U(G, V),
V),and
andshow
show these
these representations
representations
are equivalent.
equivalent.
are
(17.4) For gg EE G and
byh(yg)
h(y8) ==[(hg-'
[(hg-i)y]g,
(17.4)
and yy EE r(G,
F(G, V) define
defineyg:
yg:GG+
- VVby
)y]g,
G. Then
Then Fr
it is
yg.
for h EE G.
is aa representation
representationofofGGon
onr(G,
F(G, V),
V), where
wheregit:
gfr:yy H
H yg.
If tr
n is an
an FG-representation
FG-representation so is ft.
5.
(17.5)
For vv EE V
V define
definevva:
-+ V
V by
byg(va)
g(va) =
= [g,
(17.5) For
a : GG+
[g, v] =
=vv --vg.
vg.Then
Thenthe
the
map a:
a:vv H va
va isis aa G-homomorphism
G-homomorphismof
of V
V into
into F(G,
r ( G ,V)
V)with
withkernel
kernel Cv(G).
CV(G).
map
If
FG-representation then a
a isis an
If 7r
n is an FG-representation
an FG-homomorphism.
FG-homomorphism.
Proof. Lemma
Lemma8.5.4
8.5.4says
saysvu
vaisisaacocycle.
cocycle.The
Therest
rest isisstraightforward.
straightforward.
Proof.
If V
V is
is an
an FG-module
FG-module then so is F(G,
r ( G , V)
V) and
and 0# induces
induces an
an F-space
F-space structure
structure
If
U(G, V)
V)which
which makes
makes 0#into
intoan
anF-space
F-spaceisomorphism.
isomorphism. That
That is, for
for uu EE
on U(G,
U(G, V)
g" =
= gv then
V) and
and a EE F,
F, au
a u ==(a(uo-1))O.
(a(u4-I))#. Equivalently
Equivalently ifif gu
then g°"
gaU=
=
g(av). This
This isis the
the F-space
F-spacestructure
structureon
on U(G,
U(G,V)
V)implicit
implicitininthe
theremainder
remainder of
of
g(av).
the section.
section.
the
(17.6) Let c:
Aut(GV)
V) and d: Aut(G
Aut(U(G, V))
V)) be
(17.6)
c: G
GVV+
- Aut(G
Aut(GV)
V)nf1N(V)
N(V)+
- Aut(U(G,
the conjugation
conjugation maps. Then
a
V- F(G, V)
U(G, V)
and 0# are
is a commutative diagram, the maps c, aa,, and
areG-homomorphisms,
G-homomorphisms, and
and
V is
is an
an FG-module
FG-module the
the maps
maps are
are FG-homomorphisms.
FG-homomorphisms. Here
it
if V
Here#,
0, aa,, and fr
17.2, 17.5,
17.5, and
and 17.4,
17.4,respectively,
respectively, and Jr,
n,it,
the
are defined in 17.2,
Fr, and
and cd are the
r ( G , V),
V),and
and U(G,
U(G,V),
V),respectively.
respectively.
representations of G on V, F(G,
Proof. Let
Let v,
v, w
w EE V,
V, U
u E U(G, V),
V), and g, h E G.
G. Then
Proof.
vc: gw
g°w = g(v - vg + w) = (gw)(va)O,
so c =
=ao
a#and
andthe
thediagram
diagramcommutes.
commutes.
FG-homomorphism if V is
By 17.5, a is
is aa G-homomorphism,
G-homomorphism,and
and even
even an FG-homomorphism
an FG-module. By 17.2,
0 is
17.2,#
is aa homomorphism.
homomorphism.
Next
agcd:hvbyH
H ((hv)g-'"g
= hg-'ugv.
hg-'"gv.Hence
Hence(hv)(yg#)
(hv)(ygo)== h(hy8
h(hyg + v) _=
Next ugcd:
h ~ ) ~ - " g=
h(((hg-i)y)g
v)
=
(hg-l(hg-')y)gv
=
((hg-l)(y#))gv
=
(
h
~
) ( ( ~ # ) gso
SO
~~),
h(((ha ')y)g + v) = (hg '(ha ')y)gv = ((hg ')(yo))gv =
ygo
=
0 is
therefore #
yg# = (yO)gcd,
(y#)gcd,and therefore
is aa G-homomorphism.
G-homomorphism.
+
+
(hv)((yO)gcd)
1-cohomology
67
Suppose V isis an
=
Suppose
anFG-module.
FG-module. For
ForaaEEF,F,(au)g`d
(au)gcd==((a(uO-1))c)g`d
((a(u@-'))@)gcd=
((a(uO-1))9)0 ==(a((ug`d)0-1))O
==a(ugcd),
((a(u@-'))g)@
(a((ugcd)@-I))@
a(ugcd),so
socd
cdisisan
an FG-representation.
FG-representation.
a@
andaaand
and0@are
areG-homomorphisms
G-homomorphisms(or
(orFG-homomorFG-homomorFinally, as
as cc =
= ao
and
phisms),
c. Therefore
Therefore the
the proof is
is complete.
complete.
phisms), so
so is
is c.
By 17.6,
r(G, V)/
V)/ Va
Va =SU(G,
U(G,V)/
V)/Vc.
Vc.The
Thefirst
jirstcohomology
cohomologygroup
group of
of the
the
17.6, F(G,
representation
H 1(G,V)
V)2= 1'(G,
V a =SUU(G,
(G, V)/
V c. This is an
representation n7risis H'(G,
r ( G , V)/
V)/Va
V)/Vc.
an
additive group
group and, ifif V is an
(G, V) is even a vector space
additive
an FG-module, H1
H'(G,
F.
over F.
(G, V)
V) is in one to one correspondence with
with
The next lemma says that H1
H1(G,
the set
set of conjugacy
conjugacy classes
classesof
of complements
complementsto
toVVininGV.
GV.
1 (G, V) acts
of conjugacy
conjugacy classes
classes of complements
complements
(17.7) HH1(G,
acts regularly on the set of
to V in GV via
(Vc)u: (G")'
(Gw)v H
H (Gwu)v
(G"')'
U,
U(G, V).
u, wW EE U(G,
V).
In particular the number of conjugacy classes of
of complements
complementsto
to V
V in
in GV
GV is
IH1(G, V)I.
IH1(G,V)I
Proof. This
that U(G, V) acts on V.
Proof.
Thisisisaaconsequence
consequenceof
of 17.3
17.3 and the fact that
ii
(17.8) [Gcd,
(17.8)
[Gcd,U(G,
U(G,V)]
V)]<5Vc.
Vc.
Proof. For g EE G,
u] =
_ [gc,
G, U
uE
E U(G, V), [gcd, u]
[gc,u]
u] EE U(G,
U(G,V)
V)f1
n GVc, as
U(G, V)
Aut(GV).But
ButU(G,
U(G, V)
V) n
fl GVc
GVc =
= Vc.
V) and
and GVc
GVc are
are normal
normal in Aut(GV).
Vc.
(17.9) Assume
Assumeeither
eitherCv(G)
Cv(G)=
= 0 or G ==OP(G).
Op(G).Then
Then CU(G,v)(Gcd)
Cu(~,v)(Gcd)== 0
(17.9)
and ifif CV
Cv(G)
= 0 then
then c:
c: V
V+
- U(G,
and
(G) =
U(G,V)
V)isisan
aninjection.
injection.
G] <
= [u, Gcd] =
= [u,
Proof. If UuEECU(G,v)(Gcd)
then
CU(G,v)(G~d)
then 1 =
[u, Gc], so [u, GI
ker(c) =
= Z(GV)
X
G] <
ker(c)
Z(GV) ==CZ(G)(V)
Cz(~)(V)
x Cv(G). I'll show
show [u, GI
(G,
G,so
sothat,
that, by
by
= 1,1,and
= 0.
and hence
hence CU(G,v)(Gcd)
CU(~,v)(Gcd)
=
0.
17.3, uu =
If Cv(G)
G. IfIf G =
If
CV(G)=
=00 then
then [u,
[u,G]
GI <5CZ(G)(V)
CZ(G)(V)5<G.
= OP(G)
OP(G) then
then G
G=
=
OP(GZ(GV)) char GZ(GV),
Op(GZ(GV))
GZ(GV), so
so [u,
[u,G]
GI <5G.
G.So
Sothe
theclaim
claimisis established.
established.
Cv(G) is the kernel of
of c,
c, cc is
is an
an injection
injectionififCv(G)
Cv(G) =
= 0.
As CV(G)
(17.10)
If G is aa finite
then H1(G,
H '(G, V) =
= 0.
(17.10) If
finite p'-group and V an
an FG-module, then
0.
Hence G is
is transitive on the complements
complements to V
V in GV
GV in this case.
68
Extensions of groups and modules
Proof. By
c.
By Maschke's
Maschke's Theorem,
Theorem,the
the FG-module
FG-module U ==U(G,
U(G,V)V)splits
splitsover
overVVc.
Let W be an FG-complement
FG-complement to
to Vc
V cininU.
U.By
By17.8,
17.8,[Gcd,
[Gcd,W]
W] 5< W n
fl Vc
v c=
= 0.
0.
by 17.9,
17.9, W
W=
= 0. Thus H'(G,
H 1(G, V)
V) G
-W
So, by
W==0,0,and
and17.7
17.7completes
completesthe
the proof.
proof.
Lemma 17.10
17.10 will be used to prove
Theorem in the next
Lemma
prove the
the Schur-Zassenhaus
Schur-Zassenhaus Theorem
section.
section.
(17.11) Let V be
be an
anFG-module,
FG-module,B:0: V
V+
- WWananinjective
injectiveFG-homomorphism,
FG-homomorphism,
(17.11)
VB and
=0.
0.Then
Then there
there exists
exists an injective FGand assume [G, W] <5 V18
and Cw(G)
Cw(G) =
homomorphism
y: W
W+
- U(G,
homomorphism y:
U(G,V)V)making
makingthe
thefollowing
followingdiagram
diagramcommute:
commute:
V
W--> U(G, V)
Y
dimF(W/ VB)
V,B)5< ddimF(H1(G,
In particular dimF(W/
i m F ( ~(G,
' V)).
Proof. Let
Let r'n'bebethe
therepresentation
representationof
ofGGon
onWWand
andconsider
considerthe
thesemidirect
semidirect
products H ==S(F#
G,G,
V,V,
7r).
AsAs,BBisisan
S(F#xxG,G,W,
W,7r')
n')and
andSS==S(F#
S(F' x x
IT).
an injecinjecFG-homomorphism it induces by 10.3
homomorphism
tive FG-homomorphism
10.3 an injective group homomorphism
,J: S + H
H which is the identity on
on F'
F# xx G. As [G, W] <
B:
5 V,8,
VB, (GV),B
(GV)B 4 H, so
so
conjugation map
map eeof
of H
H on
on (G
(GV)B
V),Bcomposed
composedwith
with(,!-I)*:
(0-1)*:Aut((G
Aut((GV)B)
V),B)+
the conjugation
Aut(GV) =
=A
A maps
maps H
Hinto
intoA.
A. As
As Cw(G)
Cw(G)==0,0,the
therestriction
restrictionyyofofe(,B-1)*
e(Bw')*
to W
of W into U ==U(G,
W is
is an
an injection
injection of
U(G, V).
V). As
As yy isisthe
thecomposition
composition of
of
(F# x G)-homomorphisms,
is aa
(F'
G)-homomorphisms, y is an (F#
(F' xx G)-homomorphism,
G)-homomorphism,and hence is
preserving the multiplication by F.
F. In
G-homomorphism preserving
In particular
particular if the multiplication u H
i-+ ua,
ua, aa eE F#,
F', Uu EE U
U of
of F#
F' on
onUUisisthat
that of
of the
the F-space structure
on U defined earlier, then y is
FG-homomorphism. For
is an FG-homomorphism.
Forthis
this we
we need
need to show
show
gUa = ga-'ua = ga-'ua =
gU' =
= gau
for gg Ee G.
= gv then gaU
gau =
= gav, while
gaUfor
G. But
But if
if gU
gU=
while
gua
= (gv)a
gun=
(guy =
gav as
as [a,
[a,g]
g] ==11and
and Va
va =
= gav
= av.
By =
=c.
c. Keeping
Keeping in
in mind
mind that
that,B:
B: S + H isis an
an injecinjecIt remains to show
show ,By
tive homomorphism
homomorphismtrivial
trivialon
onF#
F#xx G,
G, we
we see
see that
that for
for v,
v, w
w EE V and
and g E G
we have
( g v wP)fi-1
B w ~ ) ~=
~ l w ~ ) B -=
=
'
we
have(gw)vB)'
(gw)°Py==(((gw)~)VB)~-'
(((gw)P)' )P-1=
= (gv
=- l (g[g, vvP]wP)f-1
g[g, v]w
= g"w
v]w =
gVw==(gw)v
(gw)' ==(gw)"`.
( g ~ ) ~So
SO
' .,By
By ==ccas
asdesired.
desired.
Lemma 17.11
says that
thatififCV(G)
Cv(G) =
= 00 then
17.1 1 says
then U(G,
U(G,V)
V)isis the
the largest
largest extension W
0.
of V such that
that [W,
[W,GI
G] 5< V and Cw(G) =
= 0.
of the dual V* of a finite dimensional FG-module V
Recall the definition of
given in section 14, and define U*(G,
U*(G, V)
V) to
to be
be (U(G,
(U(G,V*))*.
V*))*.
1-cohomology
69
(17.12) Let V be a finite
finitedimensional
dimensionalFG-module,
FG-module,B:0:W
W+
- VVaasurjective
(17.12)
surjectiveFGFGhomomorphism, and assume ker(,B)
CW(G) and
and W
W=
= [W,
ker(B) 5< Cw(G)
[W, G].
GI. Then
Then there
there
exists a surjective
surjectiveFG-homomorphism
FG-homomorphismy:y:U*(G,
U*(G,V)
V)+
-W
Wmaking
making the
the following diagram
diagram commute:
commute:
Here e* is
is the
the dual
dualof
ofthe
theconjugation
conjugationmap
mape:e:V*
V*+
- U(G,
U(G,V*)
V*)and
and e*
e* is a surjective FG-homomorphism with
with H'(G, V*)
V*)=Eker(e*)
ker(e*)<5CU.(G,v)(G).
Cv*(c,v)(G).Thus
dimF(ker(,B))
dimF(ker(B)) 5< ddimF(H'(G,
i m F ( H 1 ( ~V*)).
V*)).
,
==
0 0byby14.6.
(G),
Proof. As
AsW
W==[W,
[W,G],
GI,CW.(G)
Cw*(G)
14.6.Similarly,
Similarly,asasker(,B)
ker(B) <5 Cw
Cw(G),
*] 5
<V
*,B*byby14.6.
14.6.Let
Lete:e:V*
V*+- U(G,
[G, W
W*]
V*B*
U(G,V*)
V*)be
be the
the conjugation
conjugationmap.
map. By
By
there is
is an injective
injective FG-homomorphism
FG-homomorphismS:
S:W*
W*+
-+ U(G,
U(G, V*) such that
17.11 there
the diagram
diagram commutes:
commutes:
W*
W*
6
U(G, V
V*)
U(G,
*)
applying 14.1.2
conclude the
thefollowing
followingdiagram
diagramcommutes,
commutes,
Then applying
14.1.2and 14.4.3
14.4.3we conclude
e*
Y
= y,8
where y ==S*.
S*.As
As S6 is injective, y is
is surjective,
surjective, so e* =
yB is
is surjective
surjective since
fi
by hypothesis.
hypothesis.H'(G,
H1 (G,V*)
V*)==U(G,
U(G, V*)/V*e,
V*)/ V*e,sosoH'(G,
H'(G, V*) E
=
j
3 is surjective by
H'
(G, V*)*
[U(G, V*),
H'(G,
V*)*=Eker(e*)
ker(e*)by 14.5.
14.5.As [U(G,
V*), G]
GI <5V*e,
V*e,GGcentralizes
centralizesker(e*)
ker(e*)
by 14.6.
14.6.
that ifif V =
= [V,
Lemma 17.12 says that
[V,G]
GIthen
then U*(G,
U*(G,V)
V)isisthe
thelargest
largestextension
extension
of aa module
module ZZ by
by V
V with
with ZZ 5
< CW(G)
and W
W=
= [W,
W of
Cw(G) and
[W, G].
GI.
70
Extensions of groups and modules
modules
Extensions
18 Coprime
Coprime action
(18.1)
and
(18.1) (Schur-Zassenhaus
(Schur-ZassenhausTheorem)
Theorem)Let
LetGGbe
be aa finite
finite group,
group, H
H9
< G, and
assume
(i)
(9 (IHI,
(IHl, IG/HI)
IGIHI) ==1,1,and
and
(ii) either
either H or
or G/H
G / Hisissolvable.
solvable.
Then
Then
(1)
(1)
(2)
(2)
G splits
splits over H,
H , and
and
G is transitive on the complements to
to H
H in G.
Proof. Let
LetGGbe
beaaminimal
minimalcounterexample.
counterexample.Assume
Assume first
first that H
H isissolvable
solvable
Proof.
and let M be a minimal normal subgroup of G contained in H
H.. By 9.4, M is an
G/M. By
Byminimality
minimality
elementary abelian
abelian p-group
p-group for
for some
someprime
primep.
p.Let
LetGG =
= G/M.
of G,
complement X to H
G, there
there is a complement
H in
in G
G and
and G
G isis transitive
transitiveon
on the
the complements
complements
H. Also
Also if Y is a complement
to H.
complement to
to H
H in G then Y
F is a complement to
to H
H in G,
so it suffices to show
show X
X splits over M
M and X is transitive on its complements
complements to
M. Hence, by
M.Now
Now Gaschutz'
Gaschiitz' Theorem,
Theorem,
M.
by minimality
minimalityofofG,
G,GG=
= X and H
H ==M.
10.4,
to H in G.
10.4, says that G splits over H. Let Y be a complement to
By
By 12.1
12.1the
the conjugation
conjugationmap
map c:
c: Y
Y -+ Aut(M)
Aut(M) is
is aa FY-representation, where
the field
field of
of order
order p.
p. Hence
Henceby
by 17.10,
17.10,GGisistransitive
transitiveon
on the
the complements
complements
FF isisthe
t oH
H= M
nG
to
M iin
G..
Assume next that
H isis solvable.
solvable. Let G* =
G I Hand
andK*
K *aaminimal
minimalnormal
normal
Assume
that GI
G/H
= G/H
subgroup of G*.
G*. By 9.4, K*
K* is an elementary
elementary abelian p-group for
for some
some prime
prime
subgroup
p. Let
Let PPEESylp(K)
Sylp(K)and
and observe
observe that
P is
is aa complement
complement to
K. By
By aa
p.
that P
to H
H in K.
HNG(P).
Frattini Argument,
Argument,6.2,
6.2,GG=
= KNG(P), so
soas
asKK =
= HP, also G =
= HNG(P).
IfYisacomplementtoHinGthenK
Y=
H ( Kfln Y)
Y)
If
Y is a complement to H in G then K==KKnflGG== K
K nflHHY
= H(K
K fln YY isis aa complement
complement
by the Modular Property
Property of
of Groups, 1.14.
1.14. Then
Then RR =
=K
H in
in K,
K,so
soRREESylp(K).
Sylp(K).Hence,
Hence,by
bySylow's
Sylow'sTheorem,
Theorem,there
there is k EE K
K with
with
to H
R~ = P . As K 9 G, R = K n Y 9 Y, so
y k I NG(P).
Thus,
if NG(P)
splits
Rk=P.AsK<G,R=KfY<Y,soYk
< NG(P).
Thus,
if NG(P)
splits
NH(P)and
and isis transitive
transitive on
on its
its complements
complements to NH(P),
NH(P),then
then the
thetheorem
theorem
over NH(P)
9 G.
G.
holds. So by minimality
minimality of
of G,
G, P <
Finally
let G =
= G/P.
G/P.IfIfYYisisa acomplement
complementto
to HHininG,
G,then
thenYYcontains
containsaa
Finally let
complement to
Sylow p-group
p-group of
of G
G and
and hence
hencePP I
< Y. Also
Also Pf isis aa complement
to H
H in G.
is
Moreover, by
by minimality
minimalityof
of G,
G, there
there isis aa complement
complement Cto
to H
H in G and G is
complement to
G and
and
transitive on
on its
its complements
complementstoto H.
H. Now
Now C
C is a complement
to H in G
E G with
with Cg ==Y,
P,sosoCg
Cg ==Y,Y,completing
completingthe
theproof.
proof.
there is g E
c
cg
The Odd
Odd Order
Order Theorem
Theorem of
of Feit
Feitand
andThompson
Thompson [FT]
[FT]says
saysthat
that
Remarks. The
of odd
oddorder
orderare
aresolvable.
solvable.Notice
Noticethat
thatif if
(IAl,
either
groups of
(I A
1, IIGJ)
G 1)== 1 then either
A I or
G I is odd,
IAI
orI IGI
odd, so A
A or
or G
G isissolvable.
solvable. Thus
Thus the
the Odd
Odd Order
Order Theorem
Theorem says
says
hypothesis
hypothesis (ii) of the Schur-Zassenhaus
Schur-Zassenhaus Theorem
Theorem can
can be removed.
Coprime action
7
711
n,,,
be aa set of primes.
of aa positive integer
=r[pen pep,
pep,
Let n be
primes. The
The n-part
n part of
integernnisisn,n, =
where
the prime
prime factorization
factorizationof
of n.
n. Given
Given a finite group G, n ((G)
G)
= nnisisthe
wherenppep
rl p pep =
denotes the set of prime factors
factors of II G
G ) is
GI. 1.GGisisaan-group
n -groupififnr((G)
is aa subset of
of
n . A Hall
n-subgroup
G is
is aa subgroup
subgroupof
oforder
orderI /GI,.
7r.
Hall 7r
-subgroup of G
G Jn .nn'' denotes the set of
primes not in nn..
n-subgroups.
The following lemma gives a useful characterization of Hall n-subgroups.
Hall n-subgroup
n-subgroup of the finite
finite group
(18.2)
(18.2) H
H is aa Hall
group GG ifif and
and only
only ifif H
H is a
n-subgroup
of G
G and
and IG:
(G:HI,
HI, =
= 1.
n-subgroup of
1.
(18.3) Let
-subgroup of
of G. Then
G be
be aafinite
finite group
group and
and H
H aa7r
n-subgroup
(18.3)
Let G
(1) If a:
then H
Haa isis aa rr-subgroup
of Ga. If
(1)
a : G -+ Ga
G a isis aa homomorphism
homomorphism then
n-subgroup of
a is
Haa isisaa Hall
H is
is a Hall n-subgroup of G and a
is surjective
surjective then H
Hall n-subgroup
n-subgroup
of Ga.
(2) If
If H isis aa Hall
of GG and
andHH (
<K
K(
< G then
H is aa Hall
(2)
Hall n-subgroup
n-subgroup of
then H
Hall
n-subgroup of K.
K.
7r-subgroup
(18.4) IfIf pp and
Hall p'p'- and
and 9'-subgroups
q'-subgroups
andqq are
are distinct
distinct primes and H and K are Hall
of a finite group G,
G, respectively,
respectively, then
(1)
(1) G=HK,and
G =HK,and
(2) H
H flnKKisisaaHall
Hall (p,
(p,q}'-subgroup
9)'-subgroup of
of G,
G, aa Hall p'-subgroup of K, and a
Hall 9'-subgroup
q'-subgroup of
of H
H..
Proof. This
Thisfollows
followsfrom
from 1.7.3.
1.7.3.
(18.5)
beaafinite
finitesolvable
solvable group
group and n aa set
set
(18.5) (Phillip
(Phillip Hall's Theorem) Let
LetGGbe
of primes. Then
Then
(1) GGpossesses
possessesaaHall
Hall n-subgroup.
n-subgroup.
(2) GGacts
actstransitively
transitivelyon
onits
its Hall
Hall n-subgroups
n-subgroupsvia
via conjugation.
conjugation.
(3) Any
Anyn-subgroup
n-subgroupofofGGisiscontained
containedininsome
someHall
Hall7r-subgroup
n-subgroup of
of G.
G.
Proof.
Proof. Let
LetGGbe
beaaminimal
minimalcounterexample
counterexampleand
and M
M aa minimal
minimal normal subgroup
subgroup
of G. By 9.4, M
M is
is aap-group
p-group for
forsome
someprime
primep.p.Let
LetG*
G*=
= GIM.
G/M. By
By minimality
minimality
of G,
G, G*
G* satisfies
satisfies the theorem. In particular
particular G*
G* possesses
possesses aa Hall
Hall n-subgroup
n-subgroup
H*.
H*. Also
Also ifif X
X isis aa n-subgroup
n-subgroupof
ofGGthen
thenso
soisisX*,
X*,so
soX*
X*isiscontained
containedin
in some
some
conjugate H
H*g
of
H*,
and
hence
X
<
Hg.
*g of H *, and hence X 5 Hg .
Suppose pP E n.
n . Then
Then HHisisaaHall
Halln-subgroup
n-subgroupof
of G.
G.Further
FurtherX
X isis contained
contained
in the Hall n-subgroup Hg of G. Indeed if X is aa Hall n-subgroup of G, then
X=
= Hg
1IXJ
x1=_ IH91
lHgl so X
Hg isis aa conjugate
conjugate of H.
H.
Thus
we
may
assume
p
n.
By
the
Schur-Zassenhaus
Thus
assume 6 n . By the Schur-Zassenhaus Theorem,
Theorem, 18.1,
18.1,
there is a complement K to M in H,
H , and
and H
H isistransitive
transitive on
on such complements.
complements.
72
of groups
groups and modules
Extensions of
= IIGlnt
G I,' these complements are precisely
But as (H*
I H*JI =
precisely the
the Hall
Hall7r-subgroups
n-subgroups
of G
contained
in
H.
Therefore
G
possesses
Hall
7r-subgroups
and H
H is
G contained
H . Therefore G possesses Hall n-subgroups and
is
transitive on
on the
the Hall
Hall n-subgroups
7r-subgroupsofofGGcontained
containedininH.
H. So,
So, as
as Xg
Xg 5
< H,
transitive
H, X
X
to K if X
of G. This shows G is transiis conjugate
conjugate to
X isisaaHall
Hall7r-subgroup
n-subgroup of
transiof G, X
tive on its Hall 7r-subgroups.
n-subgroups. Finally
minimality of
Finally ifif H # G then, by minimality
is contained
of Hg, which
n-subgroup of
which is
is also
also aaHall
Hall7r-subgroup
n-subgroup
contained in a Hall
Hall 7r-subgroup
of G.
SoassumeH== G.ThenG
G.ThenG =
=KMsoXM=XMf1G
SoassumeH
K M s o X M = X M n G =XMf1KM=
=XMnKM=
(XMf K)M
(XMn
K)Mby
bythe
theModular
ModularProperty
Propertyof
of Groups,
Groups, 1.14.
1.14.Now X and XMn K
K are
are
Hall
7r-subgroups
of
XM,
so,
by
(2),
there
is
y
E
XM
with
XY
==
XMl
K
<
K.
Hall n-subgroups of XM, so, by (2), there is y E XM with XY = XM n K 5 K.
Thus
Thus the proof is complete.
complete.
A converse of Phillip Hall's Theorem also holds, as we will see soon. In particular the hypothesis of solvability
solvability is necessary to insure the
the validity
validity of the
the
theorem. Hall's Theorem
Theorem is
is aa good
good example
example of how restrictions
restrictions on the compocomposition factors of
of a finite group can lead to significant restrictions
restrictions of
of the
the global
global
structure of the group.
structure
Now
Now to a proof
proof of aa converse
converse to Phillip
Phillip Hall's Theorem.
Theorem. The proof
proof will
to Bumside's
Burnside's paqb-~heorem
pagb-Theorem that finite groups G with 17r(G)I
= 2 are
appeal to
Jn(G)I =
solvable. The
The proof
proof of
of Bumside's
Burnside's Theorem will be
be postponed
postponed until
until the
the chapter
chapter
on character
charactertheory.
theory.
(18.6) Let
finite group possessing
possessing aHall
a Hallpl-subgroup
p'-subgroup for each p EE 7r(G).
LetGGbe
be aafinite
x(G).
Then G is
solvable.
is solvable.
By 9.8,
9.8, G
G is not
not a p-group. By
Proof. Let
Let G
G be
be aa minimal
minimal counterexample.
counterexample. By
> 2.
Burnside's pogo
paqaTheorem,
Theorem,35.13,
35.13,I7r(G))
In(G)J # 2. Thus j7r(G)I
In(G)I >
2.
(G)and
andHH aa Hall
Hall pl-subgroup
p'-subgroup of
of G.
G. By
By 18.4,
18.4, H
H nfl K is a Hall qlq'Let p EE 7r
n(G)
subgroup of
of H for each prime q distinct from p and each Hall ql-subgroup
q'-subgroup of
K of
of G.
G. Therefore
Therefore H
H satisfies
satisfiesthe
the hypothesis
hypothesis of the lemma, and hence H is
is
solvable by minimality of G. Let M be aa minimal
minimalnormal
normal subgroup
subgroup of
of H.
H. By 9.4,
9.4,
solvable
M is an r-group
r-group for
for some
some prime
primer.
thereisisqqEEn(G)
7r(G)-- (p,
r. As 17r(G)I
[n(G)I >> 22 there
(p, r}.
r }.
q'-subgroup of G. As q 54
Let K be a Hall qf-subgroup
# r, K contains a Sylow r-subgroup
of G. Hence, by Sylow's Theorem, M is contained
contained in some conjugate of K,
K,
by 18.4. So,
So,as
asM
M5a H,
H, X
X=
=
we may take to be K. As q # p, G =
which wemay
=HK
HKby
=) (MHK)
=~(MK)
((MG)
M ~=
( M ~=
( )M K )<5K.K.Hence,
Hence,asassubgroups
subgroupsof
ofsolvable
solvable groups
groups are
course X
X <I
< G and
conclude from 18.3
G// X
solvable, X is solvable. Of course
and we conclude
18.3 that G
X
hypothesis of the lemma,
lemma, So, by
by minimality
minimality of
of G,
G, G/X
G/ X is
satisfies the hypothesis
is solvable.
solvable.
Therefore
Therefore G
G is
is solvable
solvable by 9.3.
Coprime action
73
= 1.
(18.7) (Coprime Action) Let
LetAAand
andHHbebefinite
finitegroups
groupswith
with(IA
(I A1,
1, I1HI)
HI) =
1.
Assume A is represented
of automorphisms of
of H
H and either A or
represented as a group of
H is
is solvable.
solvable. Let p be
be aa prime.
prime. Then
(1)
A-invariant
Sylowp-subgroup
p-subgroup of
of H
H..
There
exists
an
A-invariant
Sylow
(1)
Sylow p-subgroups
p-subgroups of G.
CH(A) is transitive on the A-invariant Sylow
(2)
(2) CH(A)
Every A-invariant
p-subgroup of
of H
H is
(3)
(3) Every
A-invariant p-subgroup
is contained
contained in
in an
an A-invariant
A-invariant
Sylow p-subgroup of H.
H.
bean
normalsubgroup
subgroupofofHHand
andH*
H*== HIK.
H/K. Then
an A-invariant normal
(4)
(4) Let K be
CH*(A)=
NH*(A)
=
CH(A)*.
= NH*(A) = CH(A)*.
Proof. Form
Formthe
thesemidirect
semidirectproduct
product G
G of
of H
H by
byAAwith
with respect
respect to
to the
the represenrepresentation of A on H, and identify A and H with subgroups
subgroups of G via the injections
of 10.1. Then H <g G,
G, A
A is
is aa complement
complement to H in
in G,
G, and
and the action of A on H
is via conjugation.
conjugation.
NG(A)
= NG(A)
f1AH=ANH(A)
NG(A) =
NG(A)n
AH =ANH(A) by
by the
the Modular
Modular Property of Groups,
Groups, 1.14.
<An
n H ==1,1,so
= A x CH(A).
NH(A)] 5
so NH(A)
NH(A)=
=CH(A)
C,y(A) and
and NG(A)
NG(A)=
C,y(A).
Also [A, NH(A)]
Let PP EE Sylp(H).
6.3, G =
= HNG(P).
Let
Sylp(H). By aa Frattini
Frattini Argument,
Argument, 6.3,
HNG(P). Then
Then by
Schur-Zassenhaus
there is
is aa complement
complementBBtotoNH(P)
NH(P) in
in NG(P)
NG(P) and
and Bg
Bg =
=A
Schur-Zassenhaus there
of H.
=QQisisan
anA-invariant
A-invariant Sylow
Sylow p-subgroup
p-subgroup of
for some g E G. Hence Pg =
NG(Q)
f1 NG(Q)
NG(Q) by
by Schur-Zassenhaus,
Schur-Zassenhaus, so,
NG(Q)is
is transitive
transitive on
on AG n
so,by
by 5.21,
5.21, NG(A)
NG(A)
transitive on the A-invariant Sylow p-subgsoups
p-subgroups of
(A) ==AAC,y(A)
CH (A)
is transitive
of H.
H.As
AsNG
NG(A)
and A <5NG(Q),
NG(Q),CH(A)
CH(A)isisalso
alsotransitive
transitive on
on the
the A-invariant
A-invariant Sylow p-subgroups of H
H by
by 5.20.
5.20.
A-invariant p-subgroup
p-subgroup of
of H
H.. If R 4 Sylp(H)
Let R be a maximal A-invariant
Sylp(H)then,
then, by
by
Exercise 3.2, R 4 Sylp(NH(R)).
Syl,(NH(R)). But NH(R)
NH(R) is
is A-invariant
A-invariant so by (1)
(1) there
there is
is
an A-invariant Sylow p-subgroup of NH(R),
NH(R), contradicting
contradicting the maximality of
establishes (3).
R. This establishes
(3).
Assume the hypothesis
Assume
hypothesisof (4) and let X* ==CH.
CH*(A).
(A). We
We have already shown
X**=
5 AX and, by Schur-Zassenhaus,
Schur-Zassenhaus, AAX =
so
X
= NH.(A).
NH. (A).Now
NowAK
AK<AX
= AAK , so
6.3, AX
AX =
= KNAX(A).
HenceXX =
= KNx(A)
by a Frattini
Frattini Argument,
Argument, 6.3,
KNAx(A).Hence
KNx(A) by
by the
the
Modular Property
Property of
of Groups,
Groups, 1.14.
1.14. So
So as
as CH(A)* 5
< CH*(A),we conclude
Modular
CH.(A) =
= X*
(A)* ==CH(A)*,
CH*(A)
X* ==Cx
Cx(A)*
CH(A)*,completing
completing the proof of (4).
Again the hypothesis
hypothesis that A or G is solvable can be
be removed
removed from
from the
the statement
statement
of 18.7,
18.7, modulo
modulo the
the Odd
Odd Order
Order Theorem.
Theorem.
Remarks.
Remarks. The
Thematerial
materialininsection
section18
18isisbasic,
basic,while
while that
that in
in section
section 17
17 is more
specialized. Thus the reader
reader may
may wish to skip or postpone section 17. If so,
17.10 must be assumed in proving
proving the
the Schur-Zassenhaus
Schur-Zassenhaus Theorem.
Extensions of
of groups aand
n d modules
74
Exercises for chapter
chapter 66
1. Let A
A be
be aasolvable
solvablegroup
groupacting
acting on
onGG==XY
XYwith
withYYa9G,
G,XXand
andYAYA1GI)== 1.
1.Then
Then CG(A)
CG(A) =
= CX(A)Cy(A).
invariant, and (IAA,
(I A(, IGI)
Cx(A)Cy(A).
2. Prove
jr, under the assumption that
that
Prove18.7
18.7with
with pp replaced
replacedby
by aa set
set of primes n,
H is
is solvable.
solvable.
group on
onaaset
setIIofoffinite
finiteorder
ordern n>>2,2,let
letFF=
= GF(2),
3. Let
Let G
G be the alternating group
of the
the representation
representationon
onI,
I, and define
define Z
let V be the permutation module of
V as
as in
in Exercise
Exercise 4.6. Prove
and the core U of V
of V.
V. In
Inparticular
particularUU =
=
(1) 0,
0, Z,
Z, U,
U, and
and V
V are
are the
the only
only FG-submodules
FG-submodules of
(U + Z)/Z
Z)/Zisisan
anirreducible
irreducibleFG-module.
FG-module.
prove H'(G,
H1(G, U)
U) =
= 0.
(2) If
If n is odd prove
indecomposable FG-module,
FG-module, H'(G,
H 1 (G,U)
U)2= F
F,,
(3) IfIf nn is
is even prove V is an indecomposable
and V =
= U(G,
U(G, U).
D).
In (2)
(2) and
and (3)
(3) let
let H be the stabilizer of
of a point
point xx of II and
(Hint: In
and proceed
proceed
by induction
inductionon
onn.n.IfIfnnisisodd
oddprove
proveHHcentralizes
centralizesw wE cU(G,
U(G,0U)
) -- U,
0 , and
centralizes aa complement
complement W
W
appeal to Exercise 4.6. If
If n is even prove H centralizes
to U in U(G,
U)
and
W
is
a
hyperplane
of
CU(G,
U)(Hy)
for
x
0
y
E
I.
Use
U(G, 0 ) and W is a hyperplane of CLI(G,~)(Hy)
for x # y E I. Use
this to conclude dim(
dim(W)
< 1.)
W) 5
1.)
4. (Alperin-Gorenstein)
(Alperin-Gorenstein) Let
Let FFbe
beaafield
field of
of characteristic
characteristic p,
p, G
G aa finite
finite group,
V an FG-module,
FG-module, and A a G-invariant collection
collection of pf-subgroups
p'-subgroups such
suchthat:
that:
(1) V
V=[V,X]foreach
XEA,and
(1)
= [V, XI for each X
E A, and
joining X
X to
to Y
Y ifif [X,
[X, Y]
Y] =
= 11 is connected.
(2) the
the graph
graph on A obtained by joining
Prove H'(G,
H1(G, V)
V) =
= 0.
0.
r-group acting on
on an
anrf-group
r'-group G.
G. Then
Then G
G=
= (CG(B):
5. Let
Let A
A be an abelian r-group
(CG(B):
cyclic).
B<
(A,
A, A/B
AIBcyclic).
+
7
Spaces with forms
forms
Chapter 7 considers pairs (V, ff)) where
whereVVisisaafinite
finitedimensional
dimensionalvector
vector space
space
and ffisisaanontrivial
nontrivialsesquilinear,
sesquilinear,bilinear,
bilinear, or quadratic
quadratic form on
over a field F
F and
V. We'll be primarily interested in
in the
the situation
situation where
whereAut(V,
Aut(V,ff)) is large; in
satisfiesone
oneofofseveral
severalsymmetry
symmetryconditions
conditions(cf.
(cf.Exercises
Exercises 7.9,
7.9,
that event ff satisfies
7.10,
9.1, and 9.9). Under suitable restrictions on F,
7.10,9.1,
F , such
such pairs
pairs are
are determined
determined
in sections
sections 19,
19, 20,
20, and
and 21.
21. For
For example
exampleifif FF is finite the
the
up to isomorphism in
isomorphism types are listed explicitly in section 21.
isomorphism
It turns out that such spaces
spaces satisfy
satisfy the Witt property: that is, if X and Y are
subobjects of
of (V,
(V, ff)) and
then aa extends to an
and a:
a:X
X ->
-+Y
Y is
is an
an isomorphism,
isomorphism, then
automorphism of (V, ff ).
the representation
representation of
ofAut(V,
Aut(V, ff)) on
automorphism
). As a result the
on (V,
(V, ff )
investigating Aut(V,
Aut(V, ff).).
is particularly useful in investigating
The groups Aut(V, ff),),certain
certainnormal
normalsubgroups
subgroupsof
of these
these groups,
groups, and
and their
their
images
images under
under the
the projective map of section
section 13
13 are called the classical
classical getups.
groups.
Section 22 derives various properties of the classical
classical groups. For example
example for
by their transvections or reflecsuitable fields they are essentially generated by
much later
later in
in section
section 41
41 that
that
tions, and are essentially perfect. It will develop much
PG is simple.
if G is a perfect finite
finite classical group then the projective group PG
Theorem for finite simple groups says that,
Conversely the Classification Theorem
that, by
by
measure, most of the finite
some measure,
finite simple
simple groups
groups are
are classical
classicalgroups.
groups.
This chapter
chapter is one of the longest
longest and most complicated
complicated in the book. Moreover the material covered here is in some
some sense
sense specialized
specialized and
and tangential
tangential to
to
much of the
the other
other material
material in this book.
book. Still,
Still, as
as I've
I've indicated,
indicated, the
the classiclassical groups and their representations
representations on the associated spaces (V, Q)
Q )are
are very
very
important,
important, so
so the
the effort
effortseems
seems warranted.
warranted.
19 Bilinear,
and quadratic
quadratic forms
Bilinear, sesquilinear, and
In this section
vector space
space over
over aa field
field F
F and 08 isis
section V
V is
is an
an n-dimensional
n-dimensional vector
of F. A sesquilinear form
form on V with respect to 8
9 is a map
an automorphism of
forallx,y,zeV
and
F::
ff:: V
V x V->Fsuch
V -+ F suchthat,
that, for
all x, y, z E V
andall
allaa EE F
+
.f
(x, Yy +
+ z)
z) =
= .ff (x,
+ ff(x,
f (x9
(x, y)
Y)+
(x, z)
.f
(y, z)
f ((x
x ++Yy,, zz)) =
=.ff (x,
( x ,z)
Z)+ .ff(y,z)
.f (ax, y)=af(x,
y) = of (x, y)
f(ax,
.f
= ae f.f(x,
(x,y).
y)
f (x, ay) =
0=
= 1.
Usually I'll
I'll write (x,
(x, y)
y) for
for ff (x, y).
The form f isis said
said to
to be bilinear if 8
1. Usually
Spaces
forms
Spaces with
withforms
76
I'll always
always assume
assume 08 is
is of
of order
order at
at most
most 2.
2. There
There is
is little
little loss
loss of
of generality
generality
in this assumption
assumption since we are interested in forms with big symmetry groups.
Exercises 7.10 and 9.9 make
make this
this comment
comment more
more precise.
precise.
and ff (x,
(x, y)
y)==ff (y,
(y,x)
x)for
forall
allx,x,yyininV.
V.ff
f isissymmetric
symmetric ifif ff is
is bilinear
bilinear and
is skew
symmetricif iff f isis bilinear
bilinearand
andff(x,
(x,y)y)=_--ff (y, x) for all x, y in
skew symmetric
in V.
V.
Finally
and ff (x, y) =
= ff (y,
Finally f isishermitian
hermitiansymmetric
symmetric if 08 is an involution
involution and
(y, x)8
x)'
for all xx,, y in V.
V. I'll always
always assume
assume that ff has
hasone
oneof
ofthese
thesethree
three symmetry
symmetry
conditions. One consequence of this assumption is that
(*)
(*
For
allx,x,yyininV,
V,f(x,
f(x,y)=0
onlyififf(y,x)=O.
f(y,x)=0.
For all
y)=Oifif and
andonly
On the other hand
hand if
if (*) holds
holds then
then Exercise
Exercise 7.10
7.10 shows
showsthat
thatff (essentially)
(essentially)
satisfies one
one of
of the three symmetry conditions. Further
Further ifif our form has a big
automorphisms then Exercises 7.9 and 9.1 show we may as well take
group of automorphisms
satisfyone
oneof
of the
the conditions.
conditions.
ff totosatisfy
If
(x, y)
y)==00I'll
I'll write
writexx1Iyyand
andsay
saythat
thatxx and
andyy are
are orthogonal.
orthogonal.For
For X C
5V
V
Ifff (x,
define
X'={vEV:x±v forallxEX}
subspace of V and that
= (x)'.
(X )1. Indeed
Indeed
X' 1-isisaa subspace
that X
X' 1-=
and observe
observe that
that X
(19.1) For
For X
V,x'x1-==ker(a),
ker(a),where
whereaaEEHomF(V,
HomF(V,F F)
definedby
by yya
=
) isis defined
a=
(19.1)
x EEV,
dim(x1-)>_> nn - 11with
(y, x). Hence dim(xL)
withequality
equality precisely
precisely when x 0$ V1-.
v'.
V
calledthe
theradical
radicalof
ofV.
V.Write
WriteRad(V)
Rad(V) for
for v'
V. J-.We
Wesay
sayffisisnondegenerate
nondegenerate
VL-Lisiscalled
if Rad(V) =
=0.
0.
The form f will
willbe
be said
said to
to be
be orthogonal
orthogonal if
if ffisisnondegenerate
nondegenerateand
and symsymmetric,
and ifif in addition,
addition, when
when char(F)
char(F) =
= 2, ff (x, x) =
= 0 for all x in
metric, and
in V.
V. The
The
saidto
to be
be symplectic
symplectic iiff f isisnondegenerate
nondegenerateand
and skew
skew symmetric,
symmetric, and
form ff isissaid
in addition when
whenchar(F)
char(F)=
= 2,
2, ff (x,
(x, x)
x) =
= 00 for
for all
allxx in
in V.
V. Finally
Finally ff is
is said
said to
i f f isisnondegenerate
nondegenerateand
andhermitian
hermitian symmetric.
symmetric.
be unitary if
A few words to motivate these definitions. I've
I've already indicated why the
symmetry
assumptions are
are appropriate.
appropriate.For
For any
any space
space V,
V, V
V=
= Rad(V)
Rad(V) @
®U
symmetry assumptions
subspace UU such
such that
that the
the restriction
restrictionofof ff to
for some
some subspace
to UU isisnondegenerate.
nondegenerate.
Thus there
there is little loss in assuming
Thus
assuming ff totobebenondegenerate.
nondegenerate.Besides,
Besides, from
from
Exercise
9.1, if (V, f)
Exercise 9.1,
f ) admits
admitsan
anirreducible
irreduciblegroup
group of
of automorphisms
automorphisms then
that ififchar(F)
char(F) =
= 2 then
(essentially) f isis forced
forced to
to be nondegenerate. Observe that
symmetry and skew symmetry are the same.
same. Also,
Also, ififchar(F)
char(F) # 2 and
and ff is
is skew
symmetry
symmetric, notice ff (x,
(x,x)
x)==00for
forall
allxx EE V.
V.This
Thismotivates
motivates the
the requirement
requirementthat
that
symmetric,
ff (x,
char(F) =
(x, x)
x) ==00 for
for all
all x E V when char(F)
=22 and
and ff isisorthogonal
orthogonalor
orsymplectic.
symplectic.
Indeed
generality
shows that
that this
this assumption
assumptionleads
leads to little loss of generality
Indeed Exercise
Exercise7.9.3 shows
satisfiedifif(V,
(V, ff)) admits
and is always satisfied
admitsan
an irreducible
irreduciblegroup
group of automorphisms.
automorphisms.
.
Bilinear, sesquilinear, and quadratic forms
77
(19.2)
(19.2) Let ff be
benondegenerate
nondegenerate and U <5 V.
V. Then
Then dim(U1)
dim(^') ==codim(U).
codim(U).
Proof.
Proof.The
Theproof
proofisisby
byinduction
inductionon
on m ==dim(U).
dim(U).The
Thelemma
lemmaisistrivial
trivialifif m
m ==0,
0,
so take m
m>
> 0.
V- U1.
0. Then as ff isisnondegenerate
nondegenerate there exists xx EE V
u'. By
By 19.1,
19.1,
=Un
fl x'x1 is a hyperplane
hyperplane of
of U.
U. By
Byinduction
inductionon
onm,
m,dim(wL)
dim(W1)==nn-- m
m + 1.
W=
Let
W; then
W' flnu1.
u'. As
As x E W'
u', U1
U' isisa ahyperplane
hyperplane
Let uUEEUU-- W;
then U'
U1==W1
W1 - u1,
W1 by
by 19.1. So
So dim(^')
dim(U1)=dim(W1)
m,m,completing
= dim(wL) --11==nn-completingthe
theproof.
proof.
of W'
+
vector xx EE V
V isis isotropic
isotropicififff (x,
(x, x)
x)=
= 0.
0. I've
I've already
alreadyobserved
observedthat,
that,ififff is
A vector
symmetric and char(F)
char(F) isis not
isotropic. Recall that
not 2,
2, then
then every
every vector is isotropic.
skew symmetric
orthogonal form
this is part of the defining hypothesis of a symplectic
symplectic or orthogonal
form when
when
= 2,
char(F) =
of V is totally
totally isotropic
isotropic ififthe
therestriction
restrictionofoff f to U is trivial,
A subspace U of
equivalentlyififUU 5< U1.
therestriction
restrictionofofff to
to U
U is
u'. U
U is
is nondegenerate
nondegenerate ifif the
or equivalently
nondegenerate,or
orequivalently
equivalentlyUunflU'U1=
= 0.
=Rad(U) =
nondegenerate,
(19.3) Let ffbe
(19.3)
benondegenerate
nondegenerateand
and UUaasubspace
subspaceof
of V.
V. Then
Then
=U@
® u'
U1.
(1) U
U isisnondegenerate
nondegenerate if and only ifif V =
.
(2)
(u')'
=U.
U.
(2) (U1)1
=
U1 is nondegen(3) IfIf UUisistotally
totallyisotropic
isotropic then each complement to U in U'
erate.
erate.
then dim(U)
dim(U) < n/2.
U isis totally
totally isotropic then
(4) If
If U
Proof. Parts
Parts (1),
(I), (2),
(2), and
and (4)
(4) are
are easy
easy consequences
consequences of 19.2,
19.2, while (2) imimplies
plies (3).
(3).
Assume ff isis symmetric.
form on
on V
V associated
associatedtotoff is aa map
symmetric. A quadratic
quadratic form
map
Assume
-* F
E V and
and aa EE F,
F, Q(ax) =
= a2Q(x)
F such
such that,
that, for all x, y E
a 2 e ( x )and
and
Q: V +
Q(x + Y) = Q(x) + Q(Y) + f (x, Y).
forces Q(x)
Q(x)=
= ff (x, x)/2,
x)/2, so the
that ifif char(F) # 2 this definition
definition forces
Observe that
quadratic form
form is
is uniquely
uniquely determined
determinedby
byff,, and
quadratic
and hence
hence adds
adds no
no new
new inforinformation. On the other
other hand,
hand, ifif char(F)
char(F)==22there
thereare
aremany
manyquadratic
quadraticforms
forms
associated to
to ff.. Observe
that f isisuniquely
uniquely determined
determined by Q, since
associated
Observe also that
f(x, Y) = Q(x +Y) - Q(x) - Q(y)
A symplectic
space (V,
(V,ff)) is
symplectic space
is aa pair
pair consisting
consisting of a vector space V and a
symplecticform
formff on
on V.
V. A
A unitary
unitary space
spaceisisaapair
pair(V,
(V,f f)
with ff aa unitary
) with
unitary
symplectic
form. An orthogonal space is a pair (V, Q)
Q) where
where Q
Q is
is aa quadratic
quadratic form on V
with associated bilinear form
form ff..
78
78
Spaces with
with forms
In the remainder
of this section
assume (V,
(V, ff)) isis aa symplectic
remainder of
section assume
symplectic or
or unitary
unitary
space,
or Q isis aaquadratic
form ff
space, or
quadratic form
form on
on VVwith
with associated
associated orthogonal
orthogonal form
and (V, Q) is
is an
an orthogonal
orthogonal space.
space. The type
type of V
V isis symplectic,
symplectic, unitary,
unitary, or
orthogonal, respectively.
orthogonal,
respectively.
A vector
= 0 when V is
vector v in V is singular if vv is
is isotropic
isotropic and also Q(v) =
orthogonal space.
totally singular
is totally
totally
an orthogonal
space.AA subspace
subspaceUU of
of V is totally
singular ifif U is
isotropic
and also
also each
each vector
vector of
of U is
of V is
isotropic and
is singular.
singular. The Witt
Witt index
index of
is the
the
maximum dimension of a totally singular
singular subspace
subspaceof V.
V. Notice
Notice 19.3
19.3says
saysthe
the
of V is at
at most
most n/2.
n/2.
Witt index of
isometry of
of spaces
spaces (V,
(V, ff)) and
An isometry
and (U,
(U, g)
g) isis aanonsingular
nonsingular linear
linear transfortransformation a:
a: V +
-* U such that g(xa,
g(xa, ya)
mation
ya) ==ff(x,
(x,y)
y)for
forall
allx,x,yyEE V.
V. A
A similarsimilarity
is
a
nonsingular
linear
transformation
a:
V
-+
U
such
that
g(xa,
ya) =
=
ity
nonsingular linear transformation a: V + U such that g(xa, ya)
.l(a) ff (x, y) for all x, y E V and some A(a)
independentofofxx and
and y.
y. If
h(a)
h(a) EEFF#
' independent
and (U,
(U, PP)) are
are orthogonal
orthogonalI'll
I'llalso
alsorequire
requireP(xa)
P(xa)==Q(x)
Q(x)or
orP(xa)
P(xa) =
=
(V, Q) and
X(a) Q(x), in
inthe
therespective
respectivecase.
case.Forms
Formsff and gg (or
(or P and Q)
h(a)Q(x),
Q) on V
V are
are said to
equivalentifif(V,
(V,ff)) and (V, g)
g) (or
(or (V,
(V, Q)
Q) and
and(U,
(U, P))
P)) are isometric.
isometric.ff and
be equivalent
O(V, ff))
g (or PP and
and Q)
Q)are
aresimilar
similarififthe
thecorresponding
corresponding spaces are similar.
similar. O(V,
of isometrics
isometriesof
ofthe
thespace,
space,while
whileA(V,
O(V,ff)) (or
(or O(V, Q)) denotes the group of
A(V, Q)) denotes the
thegroup
groupofofsimilarities.
similarities.Evidently
EvidentlyO(V,
O(V,
L(V, ff).).
f )f)9aA(V,
Let
X = (xi : 1p< ii p
< n)
be a basis of V. DefineJ J== J(X,
J(X, f)
LetX=(xi:l
n)beabasisofV.Define
f ) ttoo bbee tthe
h e n bby
yn
matrix JJ ==(Jib)
= ff (xi,
(Jij) with
with Ji1
Jij =
(xi, xj).
xi). Observe
Observe that JJuniquely
uniquely determines
determines the
form
form ff ..
Suppose
(yi 1: 1p<i ip< n) is a second
basisfor
forV,
V,let
letyiyi== >j aijxj,
Suppose YY==(yi:
second basis
aijxj,
aiJ
F,, and
= (aid).
AB=
= ((ar)
transpose of A. Observe
Observe
aij E F
and A
A=
(aij). Set Ae
a t )and
andlet
let AT
AT be the transpose
AOT=ATe
= ATOand
andJ(Y,
J(Y,ff)) = AAJATB.
AeT
J A ~ ' . Further
xi
(19.4)
A form
formgg on
on V is similar
to ff ififand
(19.4) A
similar to
andonly
onlyififthere
thereexists
exists aa basis
basis
Y
n) of
of V
V with
with J(Y,
J(Y, g)
g) ==AJ(X,
h J(X, f)
f )for
forsome
someXh EE F#.
F'. EquivaEquivaY=
= (yi:
(yi :115<ii 5
< n)
lence holds precisely
precisely when
when hAcan
canbe
bechosen
chosentotobe
be1.1.IfIfQQand
and PP are quadratic
forms associated to
to ff and
to P
P precisely when
when
and g,
g, respectively,
respectively, then Q is similar to
Y can
canbe
bechosen
chosensosothat
thatJ(Y,
J(Y,
and P(yi)
P(yi)=).Q(xi)
Y
g)g)=).J(X,
= hJ(X, f f)
) and
= hQ(xi) for
for each i,
= 1 in case of equivalence.
equivalence.
h=
with X
Proof.
a: (V, f)f)+(V,g)beasimilarityandletyi=xiaandY=(yi:
-* (V, g) be a similarity and let yi = xi a and Y = (yi : 1 p
<
Proof. Let
Leta:(V,
_ h,kJ(X,
i <pn).
n).Then
ThenJ(Y,
J(Y,g)
g) =.l(a)J(X,
= h(a) J(X,f).
f ).Conversely
Conversely if
if J(Y,
J(Y, g) =
J(X, ff)) let
a E GL(V) with
similarity with h(a)
h. Of course the
with xia
xia =
= yi. Then aa isis aa similarity
A(a) =
=X.
arguments extend to quadratic
same arguments
quadratic forms.
forms.
,kA
(X, f f)AT
O for
andonly
onlyifif J(X,
J(X,g)g)_=
hAJJ(X,
)ATe
(19.5) A
Aform
formgg on
on V
V is similar
similar to ff ififand
nonsingular matrix
F#, with hX=
= 1 in
incase
case of
ofequivalence.
equivalence.
h EE F',
some nonsingular
matrix A
A and
and some
some A
Bilinear, sesquilineal;
sesquilinear, and quadratic
quadratic forms
forms
Bilineal;
79
79
Proof. This
Thisfollows
followsfrom
from19.4
19.4and
and the
the remark
remark immediately preceding it.
One can see
see from
from the
the preceding
preceding discussion
discussionthat
that equivalence
equivalence of
of forms
formscorrecorresponds
sponds to equivalence
equivalence of
of the
the associated
associateddefining
defining matrices
matricesof
of the
the forms.
forms.
Given aa E
E GL(V)
GL(V) define
defineMx(a)
MX(a)to
to be
be the
the nn by
by n matrix
Given
matrix (aij)
(aij) defined
defined by
by
xia
=
Ej
ajjxj.
xia C aijxj.
(19.6)
Let aa EE GL(V).
GL(V).Then
ThenaaEEA(V,
A(V,f )f)ifif and
and only
only ifif (xis,
(xia, xja)
xja) =
=.l(a)
(19.6) Let
h(a)
(x1,
xj) for
for all ii and j,
j, and
a EE O(V, ff)) precisely
(xi, xi)
and some
some .l(a)
h(a) EE F#, with a
precisely when
when
A(a)
=1.1.IfIf (V,
only ifif aa EE A(V,
A(V, ff))
h(a) =
(V, Q)
Q)isis orthogonal
orthogonal then a E A(V, Q) if and only
and Q(x;a)
Q(xia)==A(a)Q(x;)
h(a)Q(xi)for
foreach
each i,i,with
with A(a)
h(a) =1
= 1for
forequivalence.
equivalence.
(19.7)
Let a E
GL(V), AA==Mx(a),
Mx (a),and
andJJ==J(X,
J(X, f).
f ). Then
Then aa EE O(V,
O(V,f f)
E GL(V),
) if
(19.7) Let
and only if JJ==AJATB.
E A(V,
A(V, ff)) ifif and
AJA~O.U
aE
and only
only ifif AJ
h J==AJATB
A J A ~ ' for
for some
some
A
P.
h EEF#.
(19.8) If
If VV isisnot
notaasymplectic
symplecticspace
spacethen
thenVVcontains
containsaanonsingular
nonsingularvector.
vector.
Proof. Assume
otherwise. Letx E V#
and YEEVV--xx1.
Then
((x,y)-lx,
y)-lx, y),
Proof.
Assumeotherwise.Letx
and
L.~
h e n11==((x,
y),
so without
without loss
loss (x,
(x, y)
y)=
= 1.
1.Now
Now(ax
(ax + by,
by, ax
ax + by) =abs
= abe + bae
bas as
as xx and
and yy
are singular
singular and
and(x,
(x,y)
y)== 1.1.IfIfchar(F)
char(F)## 2 tdke
takeaa =
= bb =
=11 to get ax + by nonsingular. If
If char(F)
char(F) =
= 2 and V is unitary
unitary take
takeaa=
= 1 and
andbb #
# be.
be. Finally if V is
orthogonal and char(F) =
= 22 then as x and y are singular,
singular, Q(x) =
= Q(y)
Q(y) ==0,
0, so
so
Q(x +
+y)y)_=(x,(x,y)y)==1,1,and
andhence
hencexx + yyisisnonsingular.
nonsingular.
+
+
+
+
+
(19.9) Assume
symplectic, andif
and ifV
Visisorthogonal
orthogonalassume
assumechar(F)
char(F)## 2.
AssumeVVis
is not symplectic,
basis X
X=
= (xi
(x1:
of X
Then there exists aa basis
: 115< ii < n)
n) of
of V
V such
such that the members of
are nonsingular
nonsingular and
and distinct
distinctmembers
membersare
areorthogonal.
orthogonal.
is anonisotropicvector x1EV.By
19.3,V=(xl)®(x1)1
Proof. By 19.8there
19.8 thereisanonisotropicvectorxl
E V. By 19.3,
V = (xl)@(xl)'
with (x1)1
By induction on n there is aa corresponding
(xl)l nondegenerate.
nondegenerate. By
corresponding basis
(xi:l
<ii <n)of(xl)1
(xi: 1 <
5 n) of (x1lL.
one of
of 19.9
19.9will
willbe
betermed
termedan
anorthogonal
orthogonalbasis.
basis. An
Anorthonormal
orthonormal
A basis like the one
basis for V
V is
is aa basis
basisXXsuch
suchthat
thatJ(X,
J(X,ff)) =
= I.
I.
(19.10)
of 80 for each
(19.10) If V
V is
is unitary
unitary then
then (x,
(x, x) isis in
in the
thefixed
fixed field
field Fix(O)
Fix(8) of
xEV.
XEV.
Spaces with forms
forms
80
80
isorthogonal
orthogonal assume
assumechar(F)
char(F) #
0 2.
(19.11) Assume
AssumeVVisis not
not symplectic,
symplectic,and if V is
2.
that the
the fixed
fixed field
fieldFix(@)
Fix(9)of
of@0satisfies
satisfiesFix(@)
Fix(O)=_{aas:
{aae:aaEE F].
F).
Assume further that
Then
(1) VVpossesses
possesses an
an orthonormal
orthonormal basis.
basis.
(2) All
All forms
forms on
on VV of
ofeach
eachof
ofthe
theprescribed
prescribedtypes
typesare
areequivalent.
equivalent.
Proof.
basis
Proof. Notice
Notice(1)
(1)and
and19.4
19.4imply
imply (2).
(2).To
To prove (1),
(I), choose an orthogonal basis
in 19.9. Then
Thenby
byhypothesis
hypothesisand
andLemma
Lemma19.10,
19.10,(xi,
(xi,xi)
xi)=a:('
= ai
X as in
+')l+e> for some
a,
ai EE F#. Now replacing xi by aixi,
aixi, we
we obtain
obtainour
our orthonormal
orthonormal basis.
basis.
V is a hyperbolic plane
plane if n =
= 2 and V
V possesses
possesses aa basis X =
=(x1,
(XI,x2) such that
xl and
(xl,,x2)
hyperbolic
and x2 are singular and (xl
x2) =
=1.
1.Such
Such aa basis
basis will be termed a hyperbolic
pair.
(19.12) Let
Letxx EE V#
V be singular
andyyEc V
V- x1.
singular and
x'. Then
Then (x,
( x , y) =
=UU isis aa hyper(19.12)
bolic plane and x is contained in a hyperbolic pair of U.
= 1.
Proof. Let
Let bb ==(x,
(x,y)-B.
y)-s. Then
Then (x,
(x, by)
by) ==1,
1,so
so without
without loss (x, y) =
1. Observe
nondegenerate, so if y is singular
U is nondegenerate,
singular we are done. Thus we may assume
assume each
member of
of U --(x)
(x)isisnonsingular,
nonsingular,so
soininparticular
particularVVisisnot
notsymplectic.
symplectic.Thus,
Thus,
unless V is
isorthogonal
orthogonaland
andchar(F)
char(F)==22,
(ax +y,
(y, Y).
, 00#0 (ax
y, ax
ax + y) =
=aa +a"
+as ++(y,
y).
However
char(F) #
0 2 we may take
take aa =
= -(y, y)/2,
y)/2,and
anduse
use19.10
19.10to
to obtain
obtain
However ififchar(F)
a contradiction.
contradiction.
Thus char(F)
char(F) =
= 2. Suppose V is
is unitary.
unitary.Let
LetddEEFF - Fix(O).
Then ee=
=dd +
Fix(@).Then
dB #
0 0.
= (y, y)/e and
c Fix(O),
soaa + as
ae =
= ce =
=
ds
0. Let c =
and a =
=cd.
cd. By 19.10,
19.10, cc E
Fix(@),so
(y, y), and hence ax
ax + yy isis singular.
singular.
leaves the
the case
case VV orthogonal.
orthogonal.Then
Thenchoosing
choosingaa==Q(y),
Q(y), ax
ax + y is sinThis leaves
gular, completing the proof.
+
+
+
+
+
+
Here's an immediate corollary to 19.12
19.12 and 19.4:
(19.13)
= 2.
(19.13) Let dim(V) =
2. If
If V#
V# possesses a singular
singular vector, then V is a hyperplane. In particular, up to equivalence, there is a unique nondegenerate
bolic plane.
form on V of each type possessing
possessing aa nontrivial
nontrivialsingular
singular vector.
vector.
(19.14) Let
Let U be a totally
totally singular
singular subspace
subspaceofofV,
V,RR==(ri:
(ri 1: 15<i i 5< m) a baba(19.14)
for U, and W a complement
to U in U1.
= (si: 11 5
<
sis for
complement to
u'. Then
Then there
there exists S =
ii <
5 m)
m) : V
V such
such that
that ri,
ri,sisi isisaahyperbolic
hyperbolicpair
pair for
forthe
thehyperbolic
hyperbolic plane
plane
Ui
= (ri,
(ri,si)
si) and
and W'
W1 is the
direct sum
sum of
of the
the planes
planes (Ui:
(Ui:11 5
<
Ui =
the orthogonal
orthogonal direct
i<
5 m).
m).
Witt's
Witt's Lemma
81
B~ 19.3
19.3 we
. ~LeteUoUo
t ==(r,:
Proof. By
we may
maytake
takewW==0.0.Thus
ThusuU==u'U1.
(r,:11<< ii 5
< m).
of ((Uo)'.
complement U1
Ul =
=
~ 0 )Then
Then
~ . there exists a complement
By 19.2, U is a hyperplane of
(r1, sSi)
in U
Uo
is nondegenerate.
nondegenerate. By 19.12
(rl,
l ) to Uo in
: and by 19.3,
19.3, U1 is
19.12 we
we may assume
assume
rl,,s1
rl
sl is
is aa hyperbolic
hyperbolicpair
pairfor
forU1.
U1.By
By 19.3,
19.3,VV==U1
U1®@Ul
u.: L. Finally by induction
induction
(s1:11<< ii <
_( m)
m) in
in (U1)1
( ~ 1 ) ' to satisfy
satisfy the lemma.
on m we may choose (si:
hyperbolic if V is the orthogonal
orthogonal direct
hyperbolic planes.
Define V
V to be hyperbolic
direct sum
sum of hyperbolic
planes.
hyperbolic basis
basis for
foraahyperbolic
hyperbolicspace
spaceVVisisaabasis
basisXX== (xi:
(x1:115<ii 5
< m)
A hyperbolic
m)
such that V
V is
is the
theorthogonal
orthogonalsum
sumof
ofthe
thehyperbolic
hyperbolicplanes
planes(x2i_1,
(x2i-1, x20
x2i) with
with
hyperbolic pair
pair x2i-1,
x2i_1,xi.
xt.We
Wesay
say V
V is
is dejinite
definite ifif V possesses no nontrivial
hyperbolic
singular vectors.
As a consequence of 19.14
19.14 and 19.3
19.3 we
we have:
of V. Then
Then V =
= UU®
(19.15) Let
Let U
U be
be aa maximal
maximal hyperbolic subspace of
@ U1
U' and
and
U1
is
definite.
Moreover
every
totally
singular
subspace
of
V
of
dimension
U' is definite. Moreover
totally singular subspace V of dimensionm
m
is contained in a hyperbolic
hyperbolic subspace
subspace of
of dimension
dimension 2m
2m and
and Witt
Witt index
index m.
m.
(19.16) All
All symplectic
spaces are hyperbolic.
(19.16)
symplectic spaces
hyperbolic. In particular
particular all
all symplectic
sy mplectic
spaces are of even dimension and, up to equivalence, each space of even dimension admits a unique symplectic
symplectic form.
form.
,and the fact that all vectors
Proof. This
This isisimmediate
immediate from
from 19.15,
19.15, 19.4,
19.4,'and
vectors in a
symplectic space are singular.
11
If char(F) ==22and
then (V,
(V, ff)) is symplectic,
whereff is
and(V,
(V, Q)
Q) is
is orthogonal then
symplectic, where
the bilinear form determined
Q. Hence
Hence by
by 19.16:
19.16:
determined by Q.
(19.17) If V
char(F) ==2,2,then
(19.17)
V is orthogonal and char(F)
thenVVisisof
of even
even dimension.
dimension.
20 Witt's
Witt's Lemma
Lemma
This section is devoted to a proof of Witt's Lemma. I feel Witt's Lemma
Lemma is
probably the most important result in the theory of spaces with forms. Here
it is:
is:
Witt's Lemma.
Lemma.Let
LetVVbebeananorthogonal,
orthogonal,symplectic,
symplectic, or
or unitary
unitary space.
space. Let
U and W be subspaces of V and suppose aa:: U + W is an isometry. Then a
extends to an isometry of V.
Before proving Witt's Lemma let me interject
interject an
an aside.
aside. Define an object X in
in
a category i'6?totopossess
property if, whenever Y and Z are subobjects
possessthe
theWitt
Wittproperty
Spaces with forms
forms
82
-+ Z is
of X and a:
a:Y
Y+
is an
an isomorphism,
isomorphism,then aa extends
extends to
to an
an automorphism
automorphism of
Lemma says
says that
that orthogonal
orthogonalspaces,
spaces,symplectic
symplecticspaces,
spaces,and
and unitary
unitary
X. Witt's Lemma
spaces have the Witt property
property in the category of
of spaces with
with forms
forms and isometries. All objects in the category
category of sets
sets and
and functions
functionshave
have the
the Witt
Witt property.
property.
But in most categories few objects have the Witt property; those that do are
are
very well behaved
behaved indeed.
indeed. IfIf X
X is an object with the Witt property and G is
is
then the representation
representation of
of G on X is
its group of automorphisms, then
is usually
usually an
an
excellent tool for studying
studying G.
G.
proof of
of Witt's
Witt's Lemma.
Lemma. Continue the
the hypothesis
hypothesis and
and notation
notation of
of
Now to the proof
number of
of steps.
steps. Assume
Assume the
the lemma
lemma
the previous section. The proof involves a number
counterexample with n minimal.
is false and let V be a counterexample
Let H <5UUand
(20.1) Let
andsuppose
supposea aI /HHextends
extends to an isometry
isometry /3P of V.
V. Then
Then
y ==a$-1:
with yYIH
= 1,
a/3-':UU-++WEB-'
W/3-' is
is an
an isometry
isometry with
J H=
1, and
and a extends
extends to an
of V
V if and only ifif y does.
isometry of
Assume00:#HH<5UUwith
(20.2) Assume
withHHnondegenerate.
nondegenerate. Then
Then
H1
(1) If H
' G ((Ha)1
~ a ) then
' thenaaextends
extendsto
toan
anisometry
isometry of V.
V.
(2) If Ha
(2)
Ha ==HHthen
thena aextends
extendstotoan
anisometry
isometry of V.
Proof. Notice
implies(2).
(2).As
AsHH isis nondegenerate,
nondegenerate,sosoisisHa,
Ha,and
andVV =
=
Proof.
Notice ((1)
1 ) implies
H@
®H1
(Ha)1
isometry.
I H' =
=Ha
Ha®(Ha)1.
@ I ( H ~ ) ' . Let
Let$:/3:H1
H' -++
( ~ abean
) 'be an
isometry.By
Byminimality
minimality
of n, (ajunH±)$-1
extendstotoan
anisometry
isometryy yofofH'.H1.Then
ThenyB:
y$:HH1
-+ ((Ha)1
(alunH~)/3-'
extends
I +
~a)'
is an
I unH± and
and aa IlH~+y/3
y,Bisisan
anisometry
isometryof
of V
V extenan isometry
isometry extending
extendingaalunHl
ding aa..
+
Rad(U) and K aa complement
(20.3) If H isis aa totally
totally singular
singular subspace of Rad(U)
complement to
there exist
existsubspaces
subspacesU'
U'and
andW'
W'ofofVVwith
withKK=
= Rad(U1)
Rad(U')
H in Rad(U),
Rad(U),then there
5 U'
U' such
such that a extends to
: U'U' -+ W'.
and U <
toan
anisometry
isometrya a:
W'.IfIfUU=
='H
H1 then
U'=V.
U'
= v.
Proof. Let
Let (ri:
(r1:115< ii <5 m)
in H
H1
Proof.
m )be
be aa basis
basis for H, X
X aa complement
complement totoHH in
'
containing
K, and
and X'
X' aa complement
complementtotoHa
Ha in
in ((Ha)1
(X ffll U)a.
U)a.
containing K,
~ a ) 'containing
containing (X
thereisis(si:
(s;:115< ii 5
< m)
(s': 1 <
By 19.14
19.14 there
m) and (sj:
5 ii <5m)
m)such
suchthat
that XX' 1 and (X')1
(x')'
are the orthogonal sum of hyperbolic planes (ri,
(ri, si)
s;) and (r;
a, s'),
(ria,
s,'),respectively.
respectively.
Extend a to
U ' = (U,si:1
(U,si:1 i
_( m )
U'=
<ii <m)
by defining
definingsia
s;a =
= si.
s'.
by
Witt's
Witt's Lemma
83
(20.4)
(20.4) V
V isis not
not symplectic.
symplectic.
assume U
U is
is nondegenerate.
nondegenerate.As
AsUUg= W,
W, dim(U)
dim(U) =
=
Proof. By
By 20.3
20.3 we may assume
= W1-.
dim(W), so dim(U1)
dim(^') ==dim(W1-).
dim(wL). Hence,
Hence, by 19.16,
19.16, U1U' E
w'. Then 20.2
contradicts the choice of V as a counterexample.
counterexample.
(20.5) If there
there exists
existsaatotally
totallysingular
singular
subspace
= Ha
(20.5)
subspace
0 #0:HH =
H a of
of Rad(U)
Rad(U) then
aa extends
extends to
to V.
V.
Proof. Let L =
= H1
Proof.
H' and
and LE ==L/H.
LIH.Then
Thenf f(or
(orQ)Q)induces
inducesaaform
form ff of
oftype
type
a: 0
U+W
ff (or
(orQ)
Q)on
onL defined
defined by f ((x,
f , y)
ji) =
=ff(x,
(x,y)
y)and
andthe
the induced
induced map ti:
w
is an isometry, so, by
by minimality
minimality of
of n,
n, tiit extends
extends to
to an
an isometry
isometry fiP of
of EL.. Let
basis of
of L with
with X
x flf l H
H and
and xXflnUUbases
basesfor
forHHand
andU,Urespectively,
, respectively,
X be a basis
and let
let/3P Ee GL(L)
GL(L) be
beaamap
mapwith
with/3lu
8ju==aa and
and3
xj8==ffi
xpfor
forxxEe X
X- U.
and
U. By
By
construction
construction
(x, Y) = (x, Y) = (0, 0) = (xp, YP)
for x, y eEX,X,soso,B/3isisananisometry
isometryofofL.L.Now,
Now,by
by20.3,
20.3,,B,
/3, and
and hence
hence also a,
a,
extends
extends to an isometry of V.
V.
(20.6) Assume H is a hyperplane
hyperplaneof
ofUUwith
withaaIH
= 1. Assume
Assumealso
alsothat
thatHH=
= 00
l=
~
(20.6)
if V is unitary or char(F) ##2.2.Then
Thenaaextends
extendsto
toan
an isometry
isometry of V.
V.
+
Proof. Let u eE U
U --HHand
andset
setKK==UU+ W.W.Assume
Assumeaadoes
doesnot
not extend.
extend.
Suppose U
U=
= W.
W. Then
Then Rad(U)
Rad(U)# 0O 0by
by 20.2,
20.2, and
and as
as aa acts
acts on
on Rad(U),
Rad(U),
Suppose
Rad(U) is
by 20.5.
20.5. Thus
Thuschar(F)
char(F) =
= 22 and
is not totally singular by
and V
V isis orthogonal.
orthogonal.
As
a does not extend, a I u##1,sou
1, so u#0ua.Nowua
ua. Now ua =
= au+h,forsomea
au +h, for some a E F#
Asadoesnotextend,alu
F#
and
As a IH =
= 1,
h). By 20.2, Rad(X) 0. Hence as
andhh E H.
H.AsaIH
l , a aacts
c t son
o nX
X = (u,
(u,h).By20.2,Rad(X)#O.Henceas
each member of V# is isotropic
isotropic (because
(because V
V isisorthogonal
orthogonaland
andchar(F)
char(F) =
= 2),
2), X
X
is totally isotropic. Hence
Henceas
as Q(u)
Q(u)=
= Q(ua),
Q(ua), zz ==uu + ua
uaisissingular.
singular.Therefore
Therefore
either X is
is totally
totally singular
singular or (z)
(z)is
is the
the unique
unique singular
singular point
point in X,
X, and
and hence
hence
is a-invariant. By 20.2,
20.2, H
H contains
containsno
no nondegenerate
nondegenerate subspaces, so ff II HH==0
0
by 19.12.
E Rad(U),
Rad(U), so,
so, by
by 20.5,
20.5, hh is
is nonsingular.
nonsingular.So
So (z)
(z)=_ (za)
(za) is
19.12. Thus h E
= u. Again by 20.5 there is h' EE
singular
singular and zz 4 H. Hence
Hence we may assume z =
H -z'
-z1.
Now
a
acts
on
X'=
(h',
z)
and,
as
X'
is
. Now a acts on X' = (h', z) and, as X' isnondegenerate,
nondegenerate,20.2
20.2supplies
supplies a
contradiction.
contradiction.
So U
U#
: W.
So
W.Let
Let cc ==11ifif(u,
(u,ua)
ua)==0 0and
andcc==(u,
(u,ua)B/(u,
ua)'/(u, ua)
ua)otherwise.
otherwise.
Observe
we can
canextend
extendaatotoan
anisometry
isometrya'a'ofofKKwith
with(ua)al
(ua)a' =
Observe we
=cu,
cu,by
by 19.6.
19.6.
So, by the
first
argument
in
the
previous
paragraph,
char(F)
=
2
and
the first argument
previous paragraph, char(F) = 2 and V is
orthogonal.
Bydefinition
definition
fixes
ua. Now
NowH'
H'=
orthogonal. By
of of
a',a',
a' a'fixes
z z==uu +ua.
= (H, z) is a
+
+
84
84
with forms
Spaces with
forms
hyperplane of
of K with a'(H'
a' (H' =
= 1,
paragraph, a',
a', and hence
1, so,
SO,by the previous paragraph,
also
also a,
a ,extends
extends to
to an
an isometry
isometry of V.
V.
(20.7)
(20.7) V
V is
is orthogonal
orthogonal and char(F) ==2.2.
Proof.
Proof.Assume
Assumenot.
not.By
By20.3,
20.3,we
wemay
maytake
takeUU to
to be
be nondegenerate.
nondegenerate.By
By 19.8
19.8 and
20.4 there is a nonsingular
nonsingular point
point LL in
in U
U.. By 20.6 applied
applied to
to LL in the role
role of
of
U,
extendstotoan
anisometry
isometry of
of V.
V. Then by 20.1 we may take
= 1.
take aaI1 L=
1. But
U ,aaI1 Lextends
now
now 20.2
20.2 supplies
suppliesaa contradiction.
contradiction.
We
position to complete the proof
proof of
of Witt's
Witt's Lemma.
Lemma. Choose
Choose U of
We are now in a position
minimal
a: U
minimal dimension so that an isometry a:
U -*
-+W
W does
doesnot
not extend
extendto
to V.
V.Let
Let
H
to
H be
be aahyperplane
hyperplane of U.
U. By
By minimality
minimality of U,
U , ajH
a l extends
extends
~
to an
an isometry
isometry of
V,
= 1.1.Now
Now 20.6
20.6 supplies
suppliesaacontradiction
contradictionand
and
V, so
so by
by 20.1
20.1 we
we may
may take
takeaaI1 HH =
completes
completesthe
the proof.
proof.
I close
close this
this section
section with some
some corollaries
corollaries to Witt's Lemma.
Lemma.
(20.8)
(20.8) (1)
(1)The
Theisometry
isometrygroup
groupofofVVisistransitive
transitiveon
onthe
themaximal
maximaltotally
totallysingular
singular
subspaces
subspaces of V,
V, and
and on
on the
the maximal
maximal hyperbolic
hyperbolic subspaces
subspaces of V.
V.
(2)
(2) VVisisthe
theorthogonal
orthogonaldirect
directsum
sum of
of aa hyperbolic
hyperbolic space
space H
H and
andaadefinite
definite
space.
space. Moreover H
H isisaamaximal
maximalhyperbolic
hyperbolicspace
spaceand
andthis
thisdecomposition
decompositionisis
unique
unique up to an isometry of V.
V.
(3)
(3) The
Thedimension
dimensionof
ofaamaximal
maximalhyperbolic
hyperbolicsubspace
subspaceof
of VV isistwice
twice the
the Witt
Witt
index
indexof
of V.
V.
Proof. These
Theseremarks
remarksare
areaaconsequence
consequenceof
of Witt's
Witt's Lemma
Lemma and
and 19.15.
19.15.
Proof.
(20.9)
of F
F and NF
K -+
-* F is
(20.9) (1) If
If K
K isisaaquadratic
quadratic Galois
Galois extension
extension of
N;: : K
is the
the
norm
norm of
of K
K over
over F,
F,then
then(K,
(K,NF)
N;) isisa a2-dimensional
Zdimensionaldefinite
definiteorthogonal
orthogonalspace
space
over
over F.
F.
(2) Every
Every 2-dimensional
z-dimensional definite
definite orthogonal space
similar to
to aa
(2)
space over
over F
F isis similar
space
space (K,
(K,NF)
N;) for
forsome
somequadratic
quadraticGalois
Galois extension K of F.
F.
Proof.
Galoisextension
extensionofofFFthen
thenGal(K/F)
Gal(K/F)=
= ((a)
Proof. IfIf K
K isisaa quadratic
quadratic Galois
u )is of
order
NF (a) =
= aa'
It is
order 2 and N;(a)
a a afor
fora ac K.
E K.
It straightforward
is straightforwardtotoprove
prove(K,
(K,NF
N;) )
is
is aa definite
definiteorthogonal
orthogonalspace.
space.
Next
Next aa proof
proof of
of (2).
(2). Let
Let (V,
(V,Q)
Q)be
be aadefinite
definite orthogonal
orthogonal space and {x,
(x, y) aa
basis for
for V.
V. If
If char(F)
char(F)0#2 2then
thenbyby19.9
19.9we
wecan
canchoose
choose(x,
(x,y)y)==0,0,while
whileifif
basis
char(F)
1. 1.
Replacing
char(F)==2 2choose
choose(x,
(x,y)y)==
ReplacingQQbybya ascalar
scalarmultiple
multipleififnecessary,
necessary,
we
assume Q(x)
Q(x)=
= 1.
1.Let
LetQ(y)
Q(y)==bband
andP(t)
p(t)==t2
t2+t(x,
+t(x, y)+b,
y) +b,sosothat
that PP isis
we can assume
Spaces over finite fields
85
over F.
F. As V is definite, P is irreducible. Let K be the
a quadratic polynomial over
for PP over F
F and c cE KKaaroot
splitting field for
rootof
ofP.P .Then
Thenthe
themap
mapxxi-+
H1,
1,yy i-+
Hc
induces an isometry of (V, Q) with (K,
(K, NF
N:).).
(20.10) Assume FFisisalgebraically
(20.10)
algebraicallyclosed.
closed. Then
Then
then,
(1) IfIf char(F)
char(F)0#2 2
then,upuptotoequivalence,
equivalence,VVadmits
admitsaa unique
unique nondegenerate
nondegenerate
quadratic form. Moreover
V has an
an orthonormal
orthonormal basis with respect
respect to that
that
Moreover V
form.
form.
char(F)==22then
thenVVadmits
admitsaanondegenerate
nondegeneratequadratic
quadraticform
form if and only
(2) IfIf char(F)
if n is
is even.
even. The
The form
form is determined
determined up to equivalence and V is a hyperbolic
space with
with respect
respect to
to this
this form.
form.
Proof. Part
Part(1)
(1)follows
followsfrom
from19.11.
19.1 1.To
To prove
prove part (2)
(2) itit suffices
suffices by 19.17
19.17 and
20.8
V an
an orthogonal
orthogonalspace
spaceof
of dimension
dimension22and
andprove
proveVVisisnot
notdefinite.
definite.
20.8 to take V
But as FFisisalgebraically
algebraicallyclosed
closed itit possesses
possesses no
no quadratic
quadratic extensions, so V is
not definite
definite by 20.9.
20.9.
(20.11) If V
V is
is an
an orthogonal
orthogonal space
space of dimension at least 2 then V has a nondegenerate
2-dimensional
subspace.
degenerate Zdimensional subspace.
Proof. If
If char(F) 0# 22this
of 19.9.
19.9.IfIfchar(F)
char(F) =
= 22 then
Proof.
thisisisaaconsequence
consequence of
then
by 19.16
i~ hyperbolic and hence possesses
19.16 the underlying
underlying symplectic
symplectic space is
a hyperbolic
subspace of the
the orthogonal
orthogonal
hyperbolic plane, which is aa nondegenerate
nondegenerate subspace
space V.
V.
21 Spaces
Spacesover
over finite
finite fields
In this section the hypothesis
hypothesis and notation of section 19
19 continue; in particular
V is an orthogonal, symplectic, or
or unitary
unitary space
space over
over FF.. In addition assume F
is a finite
finite field of characteristic
characteristic p.
p.
(21.1) Assume
(21.1)
Assume n ==2.2.Then
Thenup
upto
toequivalence
equivalencethere
there is a unique nondegenerate
definite
quadraticform
formQQ on
on V.
V. Further
Further there
thereisis aa basis
basis XX =
definite quadratic
={x,
(x,y}
y) of
of VV
such
such that:
that:
then (x,
(x, y)
y) =
= 0, Q(x) =
= 1,
F#.
is odd then
1, and -Q(y) isisaagenerator
generator of F'
.
(1) If p is
(2) If p =
y) =
= 1, Q(x)
Q(x) =
= 1, Q(y)
Q(y) =
= b, and
and P(t)
P(t) =
= t2 + t + bb is
(2)
=22 then
then (x, y)
an irreducible polynomial over
over F.
F.
+ +
Proof. By
By20.9
20.9 and
andits
itsproof,
proof, QQisisatatleast
leastsimilar
similarto
tosuch
suchaa form.
form. ItIt is
is then
then an
an
easy exercise to prove
prove forms
forms similar
similarto
to Q
Q are
are even
even equivalent
equivalenttotoQ.
Q.As
As F
F is
finite, it has a unique quadratic extension, so Q is
is unique
unique by 20.9.2.
20.9.2.
forms
Spaces with forms
86
Denote by
by D
D=
= D+
D+and
and QQ==D_
D-the
the(isometry
(isometrytype
typeofofthe)
the)hyperbolic
hyperbolic plane
and the 2-dimensional definite orthogonal space over F,
respectively.
F , respectively. Write
Write
forthe
theorthogonal
orthogonaldirect
directsum
sumof
ofm
mcopies
copiesof
of D
D with kk copies of Q.
Dm Qkfor
Q.
ek
(21.2)
beaafinite
finitefield.
field. Then
Then
(21.2) Let
Let FFbe
(1) Dm
Dmisis aa hyperbolic
hyperbolicspace
spaceof
of Witt
Witt index
index m.
(2) Dii-1
index m
m- 1.
Dm-' Q
Q is
is of
of Witt index
1.
(3) D2m
DZmis
is isometric
isometricto
toQ2m.
(4) Every
Every 2m-dimensional
2m-dimensionalorthogonal
orthogonal space over FFisis isometric
isometricto
to exactly
exactly
Dm-l Q.
one of Dm
Dmor Dm-1
eZm.
Proof. By
By construction
constructionDm
Dmis hyperbolic
hyperbolic and, by 19.3.4,
19.3.4, Dm
Dm is of Witt index
m. By
By construction
constructionQQisisofofWitt
Wittindex
index
Dm-'Q,
Dm-12=UU 5
<
m.
0. 0.
LetLet
VV
2 =Dm-'
Q, Dm-'
V, and Q
=
U1=
W.
Let
X
be
a
maximal
totally
singular
subspace
of
U.
Q 2 U' = W .Let X be a maximal totally singular subspace
As dim(X)
=m - 1,X1=
X x® @
W.W.
ForFor
WwEEW#,
dim(X) =m
1, X' =
W', Q(w)
~ ( w#
#)0,
0, so,
so, for
for xx EE X
As
Q(x
Q(w)# #
0. Thus
X aismaximal
a maximal
totallysingular
singularsubspace
subspace
Q(x + w) ==Q(w)
0. Thus
X is
totally
X1,
of x'
, so
so X
X is
is also
also aa maximal totally singular
singular subspace
subspace of V. Hence, by 20.8,
gDm-1
m-1 Q
Dm,as
as they
they have
have different
different Witt
Witt indices.
indices.
Q is
is not
not isometric
isometric to
to Dm,
+
Let V
spaceover
overF.
F. By
By 20.1
20.11
1 V has aa
V be
be aa 2m-dimensional
2m-dimensional orthogonal
orthogonal space
nondegenerateplane
plane U.
U. By
By 21.1
21.1and
andinduction
inductionon
onm,
m,UU2= D
D or Q and U1
U' 2
nondegenerate
gDm-1
m - 1 or
Dm,Dm-1Q,
Dm-'Q, or
or 0"-'Q2.
complete the
Dm-2 Q2.Thus
Thus to
or Dm-'Q.
Dm-2 Q.So
SoVV 2
= Dm,
to complete
the
D2. This will follow from
from 20.8 if we can
proof of 21.2 it remains
remains to
to show
show Q2
Q2 2 0'.
show Q2 has
UU2=U'U1 2
Q.
hasa2-dimensional
a 2-dimensionaltotally
totallysingular
singularsubspace.
subspace.SoSotake
take
= Q.
Let {x,
y} and
and (u,
{u,v}
v}be
bebases
basesfor
forUUand
andu'U1,
respectively.IfIfchar(F)
char(F) =
= 22
(x, y}
, respectively.
then
by 21.1
21.1we
wemay
maychoose
choose
(u,v)v)==1,1,Q(x)
Q(x)== Q(u)
Q(u) =
= 1,
then by
(x,(x,
y) y)
= =(u,
1, and
and
Q(y) ==Q(v).
Q(v).Then
Then (x
(x+ u,u,yy+ v)v)isisa atotally
totallysingular
singularplane
plane of
of V.
V. So
So take
take
char(F)
char(F) to
to be
be odd.
odd. Then
Then by 21.1
21.1 we may take x, y,
y , u,
u, and
and vv to
to be
be orthogonal
orthogonal
with
Q(x) and
and Q(v)
Q(y). Then again (x + u, y + v)
v) is a totally
with Q(u)
Q(u)=
_- Q(x)
Q(v) =
_ - Q(y).
singular
singular plane.
plane.
+
+
+ +
F isis finite
finite and n ==2m
2misiseven,
even,then
then21.2
21.2says
saysthat,
that, up
up to
to equivalence,
equivalence,there
If F
are exactly two quadratic forms on V, and
and that the
the corresponding
corresponding orthogonal
orthogonal
spaces have
have Witt
Witt index
index m
m and
and m - 1,1,respectively.
respectively. Define
Define the sign of
of these
these
spaces to
to be
be +11 and -1,1,respectively,
respectively,and
and write sgn(Q) or sgn(V) for the sign
of the space. Thus the isometry type
an even dimensional
dimensional orthogonal
orthogonalspace
space
type of an
over a finite field is
is determined by its sign. If V is an orthogonal
orthogonalspace
space of
of odd
odd
dimension over
then, by 19.17,
19.17, char(F)
char(F)isisodd.
odd.Let's.
Let's look
look at
at such
such spaces
spaces
dimension
over FF,, then,
next.
+
(21.3) Let V
V be
be an
an orthogonal
orthogonal space
space of
of odd
odd dimension
dimension over a finite field F.
F.
Then V possesses a hyperplane which is hyperbolic.
hyperbolic.
Spaces over finite fields
87
Proof. The proof is by induction on n. The remark is trivial if n = 1, so take
n > 3. By 19.9, V possesses a nondegenerate subspace U of codimension 2.
By induction on n, U possesses a hyperbolic hyperplane K. If n > 3 then,
by induction on n, Kl possesses a hyperbolic plane W. Then K ® W is a
hyperbolic hyperplane of V.
Son = 3. Choose a basis X = (x;:1 < i < 3) for V as in 19.9. We may
assume V is definite. Thus (x1, x2) is definite, and hence possesses a vector
y such that Q(y) has the same quadratic character as -Q(x3). Thus there is
a E F with a2Q(y) _ -Q(x3). Now ay + x3 is singular and 19.12 completes
the proof.
If V is an odd dimensional orthogonal space over a finite field then, by 21.3,
V possesses a hyperbolic hyperplane H, and, by 20.8, H is determined up to
conjugacy under the isometry group of V. Let x be a generator of Hl and
define the sign of V (or Q) to be +1 if Q(x) is a quadratic residue in F, and
-1 if Q(x) is not a quadratic residue. Then evidently when n is odd there are
orthogonal spaces of sign s = +1 and -1, and, by the uniqueness of H (up to
conjugacy), these spaces are not isometric. On the other hand if c is a generator
for the multiplicative group F# of F, then c Q is similar to Q under the scalar
transformation cl. Moreover (H, Q) is similar to (H, cQ), so (H, cQ) is also
hyperbolic. Hence, as (cQ)(x) = cQ(x) has different quadratic character from
Q(x), sgn(cQ) : sgn(Q). Thus we have shown:
(21.4) Let F be a field of odd order, n an odd integer, and c a generator of the
multiplicative group F# of F. Then
(1) Up to equivalence there are exactly two nondegenerate quadratic forms
Q and c Q on an n-dimensional vector space V over F.
(2) sgn(Q) = +1 and sgn(cQ) _ -1.
(3) Q and cQ are similar via the scalar transformation cl, so O(V, Q) _
O(V, cQ).
(21.5) Let F be a finite field of square order. Then up to equivalence V admits
a unique unitary form f. Further (V, f) possesses an orthonormal basis.
Proof. As F is of square order it possesses a unique automorphism 0 of order
2. Moreover Fix(0) = f aae: a E F}, so 19.11 completes the proof.
The final lemma of this section summarizes some of the previous lemmas in
this chapter, and provides a complete description of forms over finite fields.
88
88
Spaces with forms
forms
over aa finite
finite field
fieldFF of
of order q and
(21.6) Let V be an n-dimensional space over
characteristic p. Then
(1)
which case
case ff
(1) VVadmits
admits aa symplectic
symplectic form ff ifif and
and only
only if nn is
is even, in which
is unique up to equivalence and
and (V,
(V, ff)) is
is hyperbolic.
hyperbolic.
(2) VVadmits
case ff
andonly
only ifif qq is
is aa square,
square, in which case
admits aa unitary form ff ififand
is unique up to equivalence
equivalence and
and (V,
(V, ff)) has
has aa orthonormal
orthonormal basis.
(3) IfIfnnisiseven
even then
then V
V admits
admits exactly
exactly two
two equivalence classes of nondegenerate quadratic
quadratic forms.
forms. Two forms are equivalent precisely when they have the
Dn/2 or 0tnJ2)-l
D(n/2)-1 Q
Q
same sign. IfIf P is
is such
such a form then (V, P)
P) is
is isometric
isometric to 0""
of sign +
+11 and -1,
-1,respectively.
respectively.
(4) IfIfnn isis odd
form precisely
precisely when
when
odd then V
V admits
admits a nondegenerate quadratic form
p is
of forms. All forms
is odd,
odd, in
in which
which case
case there
there are
are two equivalence classes of
are similar.
similar.
22 The
The classical
classical groups
groups
In section 22, continue to assume the hypothesis and notation of section 19.
19.
Section 22 considers
considersthe
theisometry
isometrygroups
groupsO(V,
O(V,ff)) and O(V,
O(V, Q), certain normal
normal
groups, and the images of such groups
subgroups of these groups,
groups under
under the
the projective
projective
P of
of section
section 13.
13. Notice
Notice that one can also regard the general linear group
map P
the isometry
isometrygroup
groupO(V,
O(V,f),
f), where
where ff isis the
as the
the trivial form ff (u, v) =
=00for
for all
all
u, v cE V.
V.The
Thegroups
groups G
G and
and PG,
PG,asasGGranges
rangesover
overcertain
certain normal
normal subgroups
of O(V,
f), are
of
O(V, f),
are called
called the
the classical
classical groups
groups (where
(where ff isistrivial,
trivial,orthogonal,
orthogonal,
symplectic, or unitary). We'll be particularly concerned with classical
classical groups
groups
over finite fields.
fields.
Observe
that if
if two spaces
Observe that
spaces are
are isometric
isometric then their isometry
isometry groups are
isomorphic. This is a special case of an
an observation
observation made in section
section 2. As
As aa
fact the
the isometry
isometry groups are isomorphic if the spaces are only similar,
matter of fact
which is relevant because of 21.6.4. The upshot of these
these observations
observations is that in
in
discussing the classical
classical groups
groups we need only concern ourselves
ourselves with forms
forms up
up
to similarity.
similarity.
Recall,
from 19.16 that
that ifif n is even
Recall, from
even there is, up to
to equivalence,
equivalence, a unique
symplectic form
form f on
). Sp(V)
on V.
V. Write
WriteSp(V)
Sp(V)for
forthe
theisometry
isometry group
group O(V,
O(V, ff).
Sp(V)
is the symplectic
symplecticgroup
groupon
on V.
V.As
As V
V isis determined
determinedby
bynnand
andF,
F, I'll also write
Spn(F)
group over
over F.
F.
Spn(F)for Sp(V).
Sp(V). Spn(F)
Sp,(F) isis the
the n-dimensional
n-dimensional symplectic group
If ff isisunitary
then
O(V,
f)
is
called
a
unitary
group.
Similarly
if
unitary then O(V, f ) is called a unitary group. Similarly if Q is a
quadratic form then O(V, Q)
Q) is
is an orthogonal
orthogonal group.
group.In
In general
general
nondegenerate quadratic
similarity classes of forms on V and hence more than one
there are a number of similarity
group or orthogonal
orthogonal group on V. Lemma 21.6 gives precise
precise information
information
unitary group
moment. In any event I'll write
F is
is finite;
finite; we will consider
consider that case
case in a moment.
write
when F
GU(V) or O(V) for aa unitary
unitary or
or orthogonal
orthogonal group
group on
on V,
V, respectively,
respectively, even
even
The classical groups
89
though
though there may
may be more
more than one
one such
such group.
group. GU(V)
GU(V) is
is (the)
(the) general
general uniunitary
group.
Write
SU(V)
and
SO(V)
for
SL(V)
fl
GU(V)
and
SL(V)
fl
O(V),
tary group. Write SU(V) and SO(V) for SL(V) n GU(V) and SL(V) n O(V),
respectively
respectively (recall
(recall the special
special linear group SL(V)
SL(V) is
is defined
defined and discussed
discussed in
section
SU(V) and
and SO(V)
SO(V) are
are the
the special
special unitary
unitary group
group and
and special
special ororsection 13).
13). SU(V)
thogonal
thogonal group,
group, respectively.
respectively. Write Q(V)
Q(V) for
for the
the commutator
commutator group
group of
of O(V).
O(V).
Suppose
that F
F ==GF(q)
Suppose for the moment that
GF(q)isisthe
thefinite
finitefield
fieldofoforder
orderq.q.Then
Then
write
Spn (F). Also,
Also, from
from 21.6, there is a unitary
unitary form on V precisely
write Spn
Sp, (q) for Sp,(F).
when
square,ininwhich
whichcase
casethe
theform
formisisunique,
unique,and
andIIwrite
writeGUn(r)
GU,(r)
when q ==r2r2isisaasquare,
rather rr ==IF11/2
and SU,(r) for
forGU(V)
GU(V) and
and SU(V).
SU(V). Notice
Notice r # (F1,
IF/, rather
IF^''^ in the
the
and
unitary
case.
If
n
is
odd
there
is
an
orthogonal
form
on
V
only
when
q
is
odd,
unitary case. If n is odd there is an orthogonal form
V only when q is odd,
in
O,(q), SOn(q),
SO,(q), and
and Qn(q)
Q,(q)
in which
which case
case all
all such
such forms
forms are
are similar
similarand
and II write
write On(q),
for
for O(V),
O(V), SO(V),
SO(V),and
and Q(V).
Q(V).Finally
Finallyififnnisiseven
eventhen
thenup
uptotoequivalence
equivalencethere
there
are
Q, on
onV,
V,distinguished
distinguishedby
bythe
thesign
sign
are just two
two nondegenerate
nondegenerate quadratic
quadraticforms
forms Qe
sgn(QE)
= sE ==+1
thetheform.
(q), and
'(q) for
sgn(Q,) =
+1oror-1-1of of
form.Write
Write01(q),
Oi(q),SO'
SOi(q),
andQQi(q)
forthe
the
corresponding
correspondinggroups.
groups.
For each group G we can restrict the
the representation
representationP:
P:GL(V)
GL(V) +
-* PGL(V)
of GL(V) on
PG(V) to
to G
G and
and obtain
obtain the
the image
image PPG
on the projective space PG(V)
G of G
group of
of automorphisms
automorphismsof
of the
the projective
projective space
space PG(V).
PG(V). Thus
Thus for
for
which is aa group
example
PSp,(q), PGUn(r),
PGU,(r), POn(q),
POi(q),PS2n(q),
PQi(q), etc. It will
example we
we obtain
obtain the
the groups
groups PSpn(q),
develop
PSpn(q), PSU,(r),
PSUn(r), and PQi(q)
PQ (q) are
aresimple
simple
develop much later that the groups PSp,(q),
and qq are
aresmall.
small.In
Inthis
thissection
sectionwe
weprove
provethese
thesegroups
groupsare
are(usually)
(usually)
unless n and
perfect
(i.e. each
each group
group is
is its
its own
own commutator
commutator group).
group). This
This fact
fact together
together with
with
perfect (i.e.
Exercise
7.8 is
is used
used in
in 43.11
43.1 1to
to establish
establishthe
thg simplicity
simplicityof
ofthe
thegroups.
groups.
Exercise 7.8
Recall
1, 2, and
and nn - 11are
are
Recall from section 13
13 that subspaces of V of dimension 1,2,
called
called points,
points, lines,
lines, and
and hyperplanes,
hyperplanes,respectively,
respectively, and in general subspaces
subspaces of
V are objects of
of the
the projective
projectivespace
spacePG(V).
PG(V).IfIfVVhas
hasaaform
formffor
or Q,
Q, then
then from
section 19
19 we
we have
have aa notion
notion of
of totally
totally singular
singularand
and nondegenerate
nondegenerate subspace,
subspace,
and hence
hence totally singular
singular and nondegenerate
nondegenerate points, lines, and hyperplanes.
(22.1)
then Cv(g) =
=[V,
[V,g]1.
glL.
(22.1) If g E O(V, ff)) then
Proof.
Proof. Let U =
=Cv(g).
Cv(g).For
For u EE U
U and
and vv EE V,
V, (u, v) =
=(ug,
(ug, vg)
vg) ==(u,
(u,vg).
vg).
Thus as
as
Thus
+
vv + u1
u) =
= (v,
u' =={x
(xEEVV:: (x, U)
(v, u)]
u)}
+
u'. Hence
Hence
we have vg E v + u1.
u' ==U1.
u'.
[v,g]g]EEn u1
[a,
uEU
U
EU
Therefore
[V,g]g] 5
< U1.
g]) =
_
Therefore [V,
u L .But,
But,by
byExercise
Exercise 4.2.3
4.2.3 and
and 19.2,
19.2, dim([V,
dim([V, g])
dim(uL),so
sothe
the proof
proof is
is complete.
complete.
dim(U1),
90
90
Spaces with forms
forms
Recall the definition of
of a transvection in
in section
section 13. I'll
I'll prove the next two
lemmas
lemmas together.
together.
(22.2) O(V, f)
(22.2)
f )(or
(orO(V,
O(V,Q))
Q))contains
containsaatransvection
transvectionifif and
and only
only ifif each
each of
of the
the
following holds:
possesses isotropic
isotropicpoints.
points.
(1) VVpossesses
If (V,
(V, Q)
Q) is
is orthogonal then
2.
(2) If
then char(F)
char(F) =
= 2.
(22.3)
Let G ==O(V,
(22.3) Let
O(V,f)f )(or
(orO(V,
O(V,Q))
Q))and
andassume
assumettisisaatransvection
transvection in
in G.
G.
Then
point and
andCv(t)
Cv(t) =
= U1.
=[V,
[V, t]t] is
is an
an isotropic point
UL.
(1) U =
(2)
Let
AU
be
the
of
transvections
with
center
(u)and
andlet
letRR =
=
Au
be
the
set
of
transvections
with
center
UU==(u)
(2)
RU
= (AU)
of t. Then
Then R#
R# =
= Au,
Ru =
(Au)be the root group oft.
Au, for
for each
each r EERRand
andYy EE V
V
have yr
yr =
= yy + ar(y,
we have
a,(y, u)u
u)ufor
forsome
somear
a, EEF,F and
, andone
oneofofthe
thefollowing
following holds:
+
H ar
a,isisan
anisomorphism
isomorphism of
of R
R with
with
(i) (V,
(V, f)
f )isissymplectic
symplectic and
and the map r i-group of
of FF..
the additive group
(ii) (V,
(V,f)f )isisunitary
unitaryand
andrri--Hare
areisisananisomorphism
isomorphismofofRRwith
withFix(9),
Fix(@),where
where
E F#with
witheeee==-e.
-e.
eE
char(F) =
= 2, R =EZ2,
and a,
at =
=
Z2,U
U isisnonsingular,
nonsingular, and
(iii) (V,
(V, Q)
Q) is orthogonal,
orthogonal, char(F)
Q(u)-l.
Q(u)-'
the center
center of
of
(3) If (V, f)
f )isissymplectic
symplecticor
orunitary
unitarythen
then each
each singular
singular point is the
and G is transitive
transitive on
on the
theroot
rootgroups
groupsof
oftransvections.
transvections.IfIf(V,
(V, Q)
Q)
a transvection and
is orthogonal each
each nonsingular
nonsingular point
point is
is the
the center
center of
of aa unique
unique transvection.
transvection.
(4)
Assume
either
(V,
f)
is
symplectic
and
H
=
G
or
(V,
(4) Assume either (V, f ) is symplectic and H =
ff)) is
is unitary
unitary
SU(V).Then
Thenone
oneof
ofthe
thefollowing
followingholds:
holds:
and H ==SU(V).
(i) R
R 5< H(').
(i)
W)
(ii) nn ==22and
and IFix(9)!
(Fix(0)I <
i3.
3.
(iii)
(V,f)f )isissymplectic,
symplectic,and
and BFI
IF I =
(iii) n ==4,4,(V,
= 2.
Now to the proof of 22.2
Now
22.2 and
and 22.3.
22.3. First
First 22.3.1
22.3.1 follows
follows from 22.1
22.1 and
and the
the
definition of a transvection. In
In particular
particular ifif G possesses a transvection then V
possesses isotropic points,
(u) is an isotropic point of
points, so
so we
wemay
mayassume
assumeUU=
= (u)
By 19.12
19.12 there
thereisisan
anisotropic
isotropicvector
vectorx xEEVV-- U1
U' with
with x,
x, uu aa hyperbolic
hyperbolic
V. By
in the
the hyperbolic
hyperbolichyperplane
hyperplaneWW== (u, x).
x). Let
Let X
X=
= (u,
pair in
{u,x}
x} CEYY be
be aa basis
basis
Y-X
for V with Y
X aa basis for W1.
W L .Let
Let t ==tot, be
bethe
thetransvection
transvectionin
in GL(V)
GL(V) with
with
Cv(t)
of F#.
P. Then,
Cv(t) ==U1
U' and
andxt
xt==xx+ au,
au,where
whereaaisissome
somefixed
fixed member
member of
Then, by
(vlt, vet)
v2t) ==(vl,
(vl,v2)
v2)for
forall
all v1,
vl, v2
v2 E
E X, and if (V, Q)
Q)
19.6, tt E G if and only ifif (vlt,
orthogonalalso
also Q(vt)
Q(vt) =
= Q(v)
to
is orthogonal
Q(v)for
for all
all vv EE X.
X.By
Byconstruction
construction if suffices to
+
The classical groups
91
check these
these equalities
equalitieswhen
whenvvl
= xx ==V2,
check
l =
v2,and,
and,ifif (V,
(V,Q)
Q)isisorthogonal,
orthogonal, for
for
v=
=x.
x . The
Thecheck
checkreduces
reducesto
toaaverification
verification that:
+
(*) a + sae
orthogonalalso
alsoa a==--Q(u)-1
(')
&ae==0,0,and
and ifif (V,
(V, Q) is orthogonal
Q(u)-'
E =
(V,f)f )isissymplectic,
symplectic,and
and sE ==+1
+1otherwise.
otherwise.
&
= -1 ifif(V,
0, where
# 0,
If (V, ff)) isissymplectic
eachaa EE F'
F#,
Au =
symplecticthen(*)
then (*) holds for each
, so A"
={ta:
{t,: a EE F#}.
F').
Also the map a i-group of
of FF..
H to
t, isisan
anisomorphism
isomorphism of R
R with
with the additive group
If (V, f)
(") has
has aa solution
solution if and
and only if there exists e E F#
F#
f ) isisunitary
unitarythen
then(*)
with ee =
= -e,
-e,ininwhich
whichcase
caseaaisisaasolution
solutiontoto(*)
(*) precisely
precisely when a =
=be
be with
with
b E Fix(@).
Fix(B). Observe there is c EE F
F with c # ce
co and,
and,setting
settinge e==c c--ce,
co,eeee==-e.
-e.
Finally if
if (V, Q) is orthogonal
thenaa is
is a solution
solution to
to (*)
(*)ififand
andonly
onlyififaa =
=
Finally
orthogonal then
char(F) =
= 2.
in each
eachcase
caset,:ta:y yHH yy + a(y, u)u for
Q(u) # 0 and
and char(F)
2. Observe
Observe in
Q(u)
each y EE X,
X, and
and hence also for each y E V.
V.
So 22.2 and the first two parts of 22.3 are established.
established. The transitivity statement in 22.3.3
22.3.3 follows
follows from
from Witt's Lemma,
Lemma, so
so itit remains
remains to
to establish
establish22.3.4.
22.3.4.
Assume
be the
the group
groupgenerated
generated by
by the
the transvectransvecAssume the
the hypothesis
hypothesisof
of 22.3.4.
22.3.4. Let
Let LL be
tions
with centers
centersinin W.
W. W'
W1 <
tions with
5 Cv(L),
Cv(L),so
soLLisisfaithful
faithfulon
onW
Wand
andhence
hence
L <5O(W,
Then,
O(W,f).
f).Now,
Now,by
byExercise
Exercise7.1
7.1 and
and 13.7,
13.7, L -ESL2(Fix(0)).
SL2(Fix(@)).
Then,by
by
13.6.4,
either RR 5
< LM
<53,3,and
13.6.4, either
L(') <5HM,
H('),or
orIFix(0)I
/Fix(@)/
andwe
wemay
may assume
assume the latter
with
> 2.
2.
with nn >
vectorinin W'
W1 and
and ZZ =
= (W,
be aa nonsingular
nonsingular vector
(W,v),
v),
If (V, ff)) isisunitary
unitary let
let vv be
while if (V, f)
subspace containing
containing W
W
f )isissymplectic
symplecticlet
letZZbe
beaanondegenerate
nondegenerate subspace
of dimension 44 or
or 6,
6, for
for IIFI
= 3 or 2, respectively.
LetKK =
= CH(Z1),
FI =
resp$ctively. Let
c~(z'), so
SO that
K ==Sp(Z)
Sp(Z)ororSU(Z).
SU(Z).IfIfRR<5KM
K(')then
thenthe
theproof
proofisiscomplete,
complete,so
sowithout
without loss
loss
V=
Z.
But
now
Exercise
7.1
completes
the
proof.
= Z. But now Exercise 7.1 completes the proof.
+
(22.4) Assume
Assumeeither
either(V,
(V,f )f)isissymplectic
symplecticand
and
O(V,f )f) =
= Sp(V),
(22.4)
G G==O(V,
Sp(V), or
(V, ff)) isis unitary
least2,2,Fix(@)
Fix(B)=={aae:
{aae:a aEE F),
F}, and
andGG =
=
unitary of dimension
dimension atatleast
SL(V) n
fl O(V,
O(V, ff)) ==SU(V).
SU(V).Then
Then either
either
(1) GGisisgenerated
in O(V,
O(V, ff),), or
generated by
by the
the transvections in
or
= 4, and
and nn =
= 3.
3.
(2) (V,
(V, f)
f )isisunitary,
unitary,II FII =
Proof. IfIf (V,
(V, f)
f )isissymplectic
symplecticlet
let Fr ==V#
V' and
and0S-2the
theset
setofofhyperbolic
hyperbolicbases
bases of
V. If
If (V, ff)) isisunitary
unitary let
F = {v E V: (v, v) = 1}.
by the
the transvections
transvectionsininO(V,
O(V,ff).
Let T be the group generated by
). I'll show:
show:
(i) TTisistransitive
transitiveon
on Fr unless
unless22.4.2
22.4.2 holds,
holds, and
and
(ii) TT isistransitive
0 ifif (V,
transitive on S-2
(V, f)
f )isissymplectic.
symplectic.
Spaces with forms
forms
92
Observe that the stabilizer in G of w E 7 is trivial,
trivial, so
so (ii)
(ii)implies
impliesTT =
= G.
by Exercise
Exercise 7.1
7.1 the
the lemma
lemma holds
holdsififnn =
= 2,
2, so
so we may assume
assume
Observe also that by
n>2.
n > 2.
Suppose (i)
(i) holds.
holds.IfIf (V,
(V,ff)) is unitary and
and xx E
= SU(xL),
EF
r then
then GX
G, =
su(xL), and,
and,
as n >> 2,2,induction
GXi
< TT unless
unless nn =
= 44 and
= 4,
induction on n implies
implies G,
and IFS
I FI =
4, where
where
Exercise 7.3
7.3 says
saysthe
thesame
samething.
thing.Hence,
Hence,byby(i)(i)and
and5.20,
5.20,GG=
= T.
T. So
So take
take
Exercise
(V, ff)) symplectic
andletletXX== (x,:
(x1:1 15<i i5< n)
n) and
and YY=
= (y,:
(yi: 11 5< i <I
symplectic and
:n)
n) be
be
members
of
R.
We
need
to
prove
Y
=
Xs
for
some
s
e
T.
By
(i)
we
may
members of Q. We need to prove Y = Xs for some s E T. By (i) we may
yl.Claim
ClaimY2
y2 EE x2(TX,).
xz(T,,). As
y2 ==X2
x2 + vv
take xl
xl ==yl.
As(XI,
(xl,xz)
x2)=_ (XI,
(xl, yz)
y2) =
= 1,1,y2
for some v EE (xl)L.
(~1)'. IfIf X2
x2 V
6 vL
vL then there is aa transvection
transvection t with
with center
center (u)
(v)
x2t =
= y2,
and x2t
y2, and
and as
as vv EE (xl)L,
(xl)', ttEETz,.
T,, .IfIfX2
x2 EE vL
V' the
the same
same argument
argument shows
x2 and y2
conjugate to xl + x2
and hence
hence also to each
each other. So the
the
y2 are conjugate
x2 in
in TX,,
T,,, and
x2 =
= y2.
claim holds and hence we may take xz
yz. But now X is conjugate
conjugate to Y
Y in
Sp((x1, x
X2)
inductionon
onn,
n,so
sothe
the lemma
lemma holds.
Sp((xl,
2 ) l-L)
)n flT Tbybyinduction
remains to prove (i). Let x , y E F
So it remains
r and U ==(x,(x ,y).y) If. IfUUisisnondegenerate
nondegenerate
then x E
e y(T fl
O(U,
f
))
since
(i)
holds
when
n
=2.
Thus
we
n
f )) since
when n =2. Thus we may
may take 0 #
Rad(U)and
andititsuffices
sufficestotoshow
showthere
thereexists
existsaa point
point (z) (such
(such that z is
uUEERad(U)
is
nonsingular ifif f isis unitary)
unitary) with
with (x,
(x,z)
z) and
and (y,
(y,z)
z) nondegenerate.
nondegenerate. If n >
> 33just
choose zz E uL with (x,
we can
can assume
assumenn =
= 3, so that
(x, z) nondegenerate. Thus we
(V, f)
is
unitary.
Further
we
may
assume
I
F
I
>
4.
Let
u,
v
be
a
hyperbolic
f ) is unitary. Further we may assume IF I > 4. Let u, v be hyperbolicbasis
f o r yyL.We
L . ~ e mmay
a y t atake
k e=
x x=y
y + u +u.
. ~ h eThen
n n ' =xL=(u,y-v)and
(u,y - v ) a n d i f zif=z=au
a u + y+y-v,
-v,
for
L =
(au --v).
v).ItItsuffices
sufficesto
to choose
choose z so that
are
then (y, z)
z) nfl yyL
= (au
that zz and
andau
au - vv are
nonsingular.
Equivalently,ififTT =
= T ,,(e) is the trace of F over
over Fix(B),
Fix(@),then
nonsingular. Equivalently,
T(a) # 0 or 1. Hence
Hence as
as IIFS
> 4 and T: F
F1 >
F -*
+Fix(B)
Fix(@)isissurjective,
surjective, we can
choose z as desired.
desired.
+
+
TL~(~)
andthe
thep-power
p-powermap
map
A field
field F
F is perjfect
perfectififchar(F)
char(F) =
= 0 or char(F)
char(F) =
= pp >>00and
(i.e. F ==F").
FP).For
Forexample
examplefinite
finitefields
fields and
surjection from
is a surjection
fromFF onto
onto FF (i.e.
algebraically closed fields are perfect.
perfect.
(22.5) Let F
F be
be aaperfect
perfect field
field of
of characteristic
characteristic 2 and (V, Q)
Q) orthogonal
orthogonal of
dimension
dimension at
at least
least 4.
4. Let
Let (u) be a nonsingular point of
of V, tt the transvection
center (u),
(u), and
and G
G=
= O(V,
O(V, Q).
Q). Then
Then
with center
= CG(t)
as Sp(uL/(u))
Sp(u-/(u)) on uL/(u),
(1) G(u)
G(,, =
CG(t)is
is represented
represented as
uL/(u),with
with (t)
(t)the
the
kernel
kernel of this representation.
representation.
from (u).
(u).
(2) CG(t)
CG(t)is
is transitive
transitive on the nonsingular points in uL distinct from
FI >
> 22 then uL =
L, CG(t)].
Ca(t)].
(3) If IIFI
=[u
[uL,
is transitive
transitiveon,
on,its
itstransvections,
transvections,orornn== 4,
(4) Either
EitherG
G isis generated
generated by, and is
= 2,
2, and
and sgn(Q)
sgn(Q)=
= +l.
+1.
IIFS
F1 =
The classical groups
93
Proof.
Gu.
As.Ast ist isthetheunique
Proof. Let
Let U
U ==(u)
(u)and
andHH==
GU
uniquetransvection
transvectionwith
with center
center
U, H ==CG(t)
representation 7t
TrofofHHon
onu'
UL/
= M.
Cc(t)and
andt tisisthe
thekernel
kernelof
of the
therepresentation
/ UU =
M.
Observe that,
that,ifif ff is
form on
on V defined
definedby
by Q,
Q, then
then (M,
(M, 7f))is
is the
the bilinear form
is aa
Observe
symplecticspace,
space,where
wheref(Z,
1(x,5,)y)==ff (x,
(x,y).
y).Let
Let.Z EE M#
symplectic
M# and W =
=(x,
(x,u).
u).As
As
F
with Q(u) ==1,1,and
F is
is perfect
perfect we can choose u with
andW
Wcontains
containsaa unique singular
singular
point (w).
(w ) .For a EE F#,
F#,let
lettat, be
be the
the transvection
transvectionwith
with center
center (aw
(aw+ u).
u) .Then
Thenta7r
tan
transvectionon
onM
Mwith
withcenter
centerw
Wand,
and,for
fory yEEM,
M,?(tan)
y(ta7r)==5,y +a2(y,
is a transvection
a2(ji, w)w
#)a
by 22.3.2.
22.3.2. Thus,
Thus,as
asFF =
= F2,
a aE EF#)
is isthe
F2, Rn
R7t ==(tan:
(tan:
F#)
thefull
fullroot
rootgroup
groupof
of tan
tan in
in
by 22.4,
22.4, H7t
H7r =
= Sp(M).
established. Also
sSp(M)
~ ( M )by 22.3.2. So, by
Sp(M).Therefore
Therefore(1)
(1) is established.
[y, t,]
ta] =
= a(y,
a(y,w)(aw
w)(aw ++u)u)by
by22.3.2,
22.3.2,so,
so,ififIFI
I FI>>2,2,then,
then, choosing
choosing y V
6 wL,
wL,
have uu E
E [y, R], so, as M =
that
we have
=[M,
[M,H7r],
Hn],(3)
(3)holds.
holds.To
To prove (2) observe that
H7r
transitive on M#
NH(W) is transitive on the
H7t is transitive
M# by (1),
(I), so it suffices to show NH(W)
set F
r of
of points
points of
of W
W distinct
distinct from (u)
(u) and
and (w).
(w ). Let
Let w,
ii, ,v6 be
be aa hyperbolic
hyperbolic pair
in M. By
By (I),
(1), for
foraa EE F#
F# there
thereisisg,ga EE HH with
withwg,
wga=
= aw
aw and
and vg,
vga =
= a-1v.
a-'v.
Hence (ga:
(g,: a E F#)
F#)is
is transitive
transitive on F.
r.
It remains to prove (4). Let TT be
be the
thegroup
groupgenerated
generated by
by the
the transvections
transvections
in G. We've
that H
H<
We've seen that
(T,
T,so
soto
to prove
prove TT ==GGititsuffices
sufficesby
by 5.20
5.20 to
to show
show
T is transitive on the set A
A of
of nonsingular
nonsingular points of V.
V. This
This will
wiU also
also show
show
G is transitive on its transvections. Let Z be aa second
second nonsingular
nonsingular point. We
must show
T.This
Thisfollows
followsfrom
from(2)
(2)ifif(Z
(Z+
+ uU)L
show ZZ EE UT
U T.
) contains
~
a nonsingular
point distinct
distinct from
from U
U and
and Z,
Z, so
so assume
assumeotherwise.
otherwise.Then
ThenUU +ZZ =
= (U + Z)L
z)'
and U and Z are the only nonsingular
nonsingularpoint?
point§ininUU++ Z.
Z. This
Thisforces
forcesnn =
= 4 and
and
IF)
pointininA'AL-- (U
(U + Z)
JFI=
=2.
2. Let
Let A =
=Rad(U
Rad(U ++Z).
Z).IfIfBBisisaa nonsingular
nonsingular point
then U +
UTUT.
+BBand
and ZZ++BBare
arenondegenerate,
nondegenerate,so
soU,
U,ZZEEBT
BTand
andhence
henceZ ZE E
T.
Thus no such point
point exists,
exists,which
whichforces
forcessgn(Q)
sgn(Q)== +l.
+1.
+
+
+
+
+
If char(F)
char(F) ;#2 2and
and(V,
(V,Q)Q)isisorthogonal,
orthogonal,then
thena areflection
rejectionon
onVVisisan
anelement
element
r in
O(V, Q)
Q )such
such that
that [V,
[V, r]
r] isis aa point
point of
of V.
V. [V,
[V, r]
r ]isiscalled
called the
the center
center of
of r.r.
in O(V,
(22.6) Let char(F) # 2 and (V, Q)
Q) orthogonal.
orthogonal. Then
(22.6)
(1) IfIf rr isisaareflection
on
V
then
reflection on V thenrrisisananinvolution,
involution,[V,
[V,r]r]isisnonsingular,
nonsingular,
[V, r]-L.
and Cv(r)
Cv(r) = [V,
r]'.
and
(2) IfIf UU==(u)(u)isisa anonsingular
nonsingularpoint
pointofofVVthen
thenthere
thereexists
existsaaunique
uniquereflecreflection r,r,, on
on V
V with
withcenter
centerU.
U.Indeed
Indeedxr,xra==xx-- (x, u)u/Q(u) for
for each x E V.
Let r be
on V.
V. Then
Then [V,
[V,r]r] =
= (v)
Proof. Let
be aa reflection
reflection on
(v)isisaapoint.
point. By
By 22.1,
22.1,
Cv (r) =
Cv(r)
=vL,
v', so
sovL
v' isisaahyperplane
hyperplane by
by 19.2.
19.2.By
By 22.2,
22.2, rr isisnot
not aatransvection,
transvection, so
so
andhence
hencevvisisnonsingular.
nonsingular.
Next
forsome
some110
vV
6 vvL
' and
Next
vr yr==avavfor
# aa E F#
F#and
and
Q(v) =
= Q(vr)
Q(vr) ==Q(av)
Q(av)==a2Q(v),
a2e(v),so
soaa==-1.
-1.Hence
Hencer risisananinvolution.
involution.
Spaces with forms
forms
94
94
let U
U=
= (u)
Conversely let
(u)be
beaanonsingular
nonsingularpoint.
point. By
By (1)
(1) there
there is
is at
at most
most one
one
reflection with
with center U, while
reflection
while aa straightforward
straightforward calculation
calculation shows
shows the
the map
map
(2) is such a reflection.
listed in (2)
on page
page 19 of
of
The proof of the following lemma comes essentially from 1.5.1 on
21.
Chevalley [Ch 2].
(22.7) Let (V,
( V ,Q)
Q )be
be an
an orthogonal
orthogonal space.
space. Then either
(1)
( 1 ) O(V,
O ( V ,Q)
Q )isisgenerated
generatedby
byits
itstransvections
transvections or
or reflections,
reflections, or
(2) IF
1 FI =
I =2,2,n n==4,4,and
andsgn(Q)
sgn(Q)== +1.
(2)
1.
+
result
Proof. If n ==2 2ororIFS
1 FI ==2 the
2 the
resultfollows
followsfrom
fromExercise
Exercise 7.2
7.2 or
or 22.5.4,
22.5.4,
respectively.
= 1,
respectively. IfIfnn =
1, then
then char(F) ##2 2and
andO(V,
O ( VQ)
, Q )isisgenerated
generated by
by the
the
reflection
-I.
Thus
we
may
taken
>
2
and
IF)
>
2.
may take n > 2 and IF I > 2.
reflection - I . Thus
Let T be
or reflections
reflectionsininGG =
=
be the
the group
group generated
generated by all transvections
transvections or
O(V,
EG- T.
O ( V , Q)
Q ) and suppose hh E
T.Pick
Pickhhso
sothat
that dim(Cv(h))
dim(Cv(h))isis maximal
maximal in the
coset hhT.
T.
Suppose yYEEVVwith
withz z== [[y,
nonsingular.yhyh==yy+z
+zand
andQQ(yh)
= Q(y),
y ,hh]] nonsingular.
(yh)=
Q(y),
so Q
Q(z)+(y,
z)=O. In
Inparticulary
andy+z
= yy-(y,
( z ) ( y , z)=O.
particular y $0'z z1and
y +z =
- ( y ,z)z/Q(z)
z ) z / Q ( z )=
=yrz,
yr,,
where
rZ isis the
the transvection
transvectionor
orreflection
reflectionwith
withcenter
center( 2(z).
Thusyh
yh =
= yrz,
where r,
) . Thus
yr,,
so yy EE Cv(hrz).
Cv(h)
= [V,
Cv(hr,). By 22.1,
22.1, C
v(h) =
[ V ,h]L
h]' C
E z1
'z ==Cv(rz),
Cv(r,), so
so Cv(h)
Cv(h) <
5
Cv(hrz).
Hence
dim(Cv(hrr))
>
dim((Cv(h),
y))
>
dim(Cv(h)),
contrary
to
C v (hr,).
dim(Cv (hr,)) > dim((Cv( h ) ,y ) ) > dim(Cv (h)),contrary to
the choice of h.
Therefore [V,
[ V ,hh]] is totally singular. I claim next that T is
is transitive
transitive on
on the
the
maximal
V . Assume not and pick
maximal totally singular
singular subspaces
subspaces of V.
pick two such
such spaces
spaces
M and N such that M
M 0$ NT
NTand,
and,subject
subjectto
tothis
this constraint,
constraint, with dim(M fl
n N)
maximal. Then
ThenM
M#
#N
maximal.
N so
so M
M <<MM+ N,N and
, andhence
henceby
bymaximality
maximalityof
of M
M there
there
nonsingular vector
vectorxx =
= m + nn EEM
is a nonsingular
M + N.
N As
. Asxxisisnonsingular,
nonsingular,(m,
( m n)
, n )54
# 0.0.
Also Q
Q(x)
= (m,
somr,
mrx==-n
-n EE N,
M fln NN 5
< x1
Also
(x)=
( m ,nn),
) , so
N , while
while M
x L ==Cv(rx).
Cv(r,). Thus
Thus
M
nN
N <<(M
N,N and
Mn
nN
N,, Mrx
Mn
( MnnN,
N ,n)n )<5Mrx
Mr,nn
, andthen,
then,by
bymaximality
maximality of M
Mr, E
E
NR. But now M
ME
E NR, contrary to the choice of M and N.
N.
is established.
established. Next there is a maximal
maximal totally
totally singular subspace
subspace
So the claim is
M
with [[V,
M.Then
Then'MM1(< [[V,
h]1=
= Cv(h). Let
LetHH =
= Cc(M1)
M with
V , hh]
] 5< M.
V , hlL
C G ( M L )n
n
Cc(V/M1),
C G ( v / M L )so
,SOthat
thathhEEH.
H .As
AsTTisistransitive
transitiveon
onmaximal
maximaltotally
totally singular
singular
follows that
that G
G=
= NG(M)T
NG(M)Tand
and each
each member
member of G
G has
has aa T-coset
T-coset
subspaces, itit follows
representativein
inHg
H9for
forsome
somegg EE G
G.. Then
Thenas
asHH I<!NG(M),
NG(M), H
HT
T<
a_ G,
G ,so
so each
each
representative
of G
G has a T-coset
T-cosetrepresentative
representativeininHT,
HT, and
andhence
hence GG =
= HT.
member of
By Exercise 4.8,
4.8, H
H is abelian, so
so G
G/T
/ T isis abelian.
abelian. Thus [h,
[ h ,gg]] E TT for
for each
each
g EG
G,, so
so [h,
[h, gg]] 0$ ho.
hG.IfIfhhacts
actsonona aproper
propernondegenerate
nondegeneratesubspace
subspace U of
of
+
+
+
+
The classical groups
95
V then h ==h1h2
hlh2with
withh1
h l EEO(U,
O(U,Q)
Q )and
andh2
h2EEO(U1-,
O ( u L Q).
, Q).By
Byinduction
induction on
on
n,
n , h,
hi EE T,
T ,so
sohhEET.
T. Hence
Hencehhacts
actson
onno
nosuch
suchsubspace.
subspace.InInparticular
particularC1,
C v (h)
(h)
is totally
totally isotropic,
isotropic, so, as [V,
[ V ,h]
h ] isis totally
totally singular,
singular, 22.1
22.1 and
and 19.3.2
19.3.2 imply
imply
Cv(h)
M,Mand
V Vis is
hyperbolic.
Further,
C v ( h )==[V,
[ Vh]
,h ]==
, and
hyperbolic.
Further,for
forVv EEV,V [v,
, [ vh]
,h ]EEv1-,
vL,
since (v,
( v ,[v,
[ v ,h])
h ] )isisnot
not nondegenerate.
nondegenerate.
since
By 19.14
there is
is a totally
totally singular
singularsubspace
subspaceNNofofVVwith
withVV =
=M
19.14 there
M®
@ N.
N.
Let xl
x1 E
As M
M ==[V,
h]h is
E M#.
M'. As
[ V ,h]
h ]_=[N,
[ Nh],
, h ]¢:
,4:nnr+I-+[n,[n,
] isa avector
vectorspace
space
isomorphism of
of N
N and M.
there is
is y2
y2 EE N
N with xl
x1 =
= [y2,
M . Hence there
[y2,h].
h ] .By the
last
remark
of
the
previous
paragraph,
(X1,
y2)
=
0.
As
¢
is
an
isomorphism,
last remark of the previous paragraph, ( x l ,y2) =
@ is an isomorphism,
(N
)¢ ¢gy2y$,
L, so
is yly1 EE N --xix:with
==
X2x20 $y2yj!-.
L. Let
( N --x1
x:)@
SO there
there is
with¢(y1)
@(yl)
Let
X
Then Vl
V1=
= (X)
X ==(x1,
( x lx2,
,x2,Y1,
yl, Y2).
y2). Then
( X )isisnondegenerate
nondegenerate and
and h-invariant,
h-invariant, so
v vl.
V = V1.
For a EE F#
F#define
define ah
a h EEGL(V)
G L ( V )by
by v(ah)
v(ah)==v v++a[v,
a [ vh],h for
] foreach
eachvvEEV.
V.
Notice
Q(v(ah)) =
= Q(v)
Notice Q(v(ah))
Q ( v )as
as [v,
[ v ,h]
h ] EE v1
v' flnM.
M .Therefore
Therefore aahh EE H.
H.Indeed
Indeed
setting
we have
haveM
MX(h)
MXa(ah)and
andJ J(X,
_
setting Xa
X, =
=(axi,
( a x l x2,
,xp, a-1y1,
a - l y l , y2) we
x ( h ) ==Mx,(ah)
( X , f f)) =
J(Xa,
Xg =
J(X,, f).
f ) . So
Sothe
the element
element gg E
E GL(V)
G L ( V ) with
with Xg
=Xa
X, isis in
in GGby
by 19.6
19.6 and
and
hg =
can choose
chooseaawith
witha a-- 1 # 0.
_
=ah.
ah.Now,
Now, as IFS
I FI > 2, we can
0. Then
Then [h,
[ h , g] =
h-'hg
= (a
contrary
to to
anan
earlier
h-lhg ==((-1)h)(ah)
((-l)h)(ah)
= -(a1)h
- lE) hhG,
E hG,
contrary
earlierremark.
remark. The
The
complete.
proof is complete.
Let
construct the
the ClifSord
Cliffordalgebra
algebra CC =
=
( V ,Q)
Q )be
be an
an orthogonal
orthogonalspace.
space. We next construct
Let (V,
C(Q)
C ( Q )of
of (V,
( V ,Q).
Q).The
Thetreatment
treatmenthere
herewill
willbelabbreviated.
be4abbreviated.For
For aamore
morecomplete
complete
discussion
in Chevalley
Chevalley [Ch
[Ch 1].
11. CC isis the
the tensor
tensoralgebra
algebra(cf.
(cf.Lang
Lang
discussionsee
see chapter
chapter22 in
[La], chapter 16, section
section 5)
5)of
ofVV,, modulo
modulothe
therelations
relationsx x@®x x-- Q
Q(x)1
= 0,
(x)l =
0 ,for
for
x EEV.
V .For
Forour
ourpurposes
purposesititwill
willsuffice
sufficeto
toknow
know the
the following:
following:
(22.8) Let
Let (V,
( V ,Q)
Q )be
be an
anorthogonal
orthogonal space
space with ordered basis X. Choose X
X to
be
be orthogonal
orthogonal if char(F)
char(F) ##2 2and
andchoose
chooseXXtotobe
beaahyperbolic
hyperbolicbasis
basisof
of the
theununderlying symplectic space
space ((V,
f)
if
char(F)
=
2.
Let
C
=
C(Q)
be
the
Clifford
V , f ) char(F) = 2. Let C = C ( Q )be the Clifford
algebra
algebra of (V,
( V ,Q).
Q).Then
Then CC isisan
anF-algebra
F-algebrawith
withthe
thefollowing
followingproperties:
properties:
(1)
F-linear map
map p:
p: VV +
-- CCsuch
(1) There
There is an
an injective F-linear
suchthat
that CC isisgenerated
generated as
as
an F-algebra by Vp.
Vp. Write ex for xp if x EE X.
X.
< x, write es ==ex,
{ x l ,...
. . .,,x,"}
x,} E
ex,.... .exm.
. ex,,,.
(2) For S =={x1,
C X with xl
x1 < . - . <
Then (es:
(es: SS E
C X)
basis for
for CC over
over FF.. In
In particular
particulare,eH==11=
= 1, and
X ) is a basis
2".
dimF(C) =
= 2".
and upvp
upvp + vpup =
= (u,
(3) For
For u,
u , v E V,
V , (up)2 =
= Q(u)
Q ( u ) 11 and
( u ,v)
v ) . 1.
1.
Ci be
be the
the subspace
subspace of C spanned
spanned by the vectors
vectors es, S C
E X,
X ,ISI
IS1 =
(4) Let Cl
imod
grading of C
C.. That is C ==Co
imod 2, i ==0,0 1.
, l .Then
Then{C0,
{Co,C1
C 1}}isis aa grading
Co®@C1
C1
Ci+j,for
fori,i,jj EE{0,
{0,11,
11, where
where i + jjisisread
readmod
mod2.2.
and C;Cj
C i C jC Ci+j,
+
+
forms
Spaces with forms
96
(5) If uu isis aanonsingular
of V then
nonsingular vector
vector of
then up is
is aa unit
unit in
in CCwith
withinverse
inverse
Q(u)-lup,
-(vru)p,where
whererur,isisthe
thereflection
reflection
Q(u)-'up, and,
and, for
for vv cEV,V,(vp)uP ==-(vru)p,
in O(V, Q) with center (u).
(u).
or transvection in
(6) The
The Clifford
Clifford group G of C is
is the
the subgroup
subgroup of units
units in
in C
C which
which permute
permute
Vp via conjugation.
conjugation. The representation 7r
n of G on V
V (subject
(subjectto
to the
the idenidentification
of V
V with
with Vp
of G on
tification of
Vp via p) is
is aa representation
representation of
on (V,
(V, Q)
Q) with
with
Gir
=O(V, Q)
Q) if n is
= SO(V,
Q) ifif nn is
is odd.
odd. IfIf u is aa
G
n =O(V,
is even
even and Gir
Gn =
SO(V, Q)
nonsingular
vectorininVVthen
thenupn
up7r==-r,.
-ru.
nonsingular vector
the set
setof
ofunits
unitsininZ(C)
Z(C) and
andZ(C)
Z(C) =
Fex for
(7) ker(,-r)
ker(n) is the
=FF11or
or FF11 + Fex
fornn even
even or
odd, respectively. IfIfnnisisodd,
odd,no
nounit
unitininCCinduces
induces-I-I on C by conjugation.
(8) There is an involutory
algebraantiisomorphism
antiisomorphismt tofofCCsuch
suchthat
thatest
est =
=
(8)
involutory algebra
ex,,
e,,, ....
. .ex, for each S =={x1,
{XI,...,
. . .xm
, x,}} C
G X.
X.
+
multilinear algebracan
algebra can be avoided
avoided
Proof. I'll
I'llsketch
sketchaaproof.
Proof.
proof. If char(F)
char(F) # 2 the multilinear
can be found in
as in chapter
chapter 5, section
section 4 of Artin
Artin [Ar].
[Ar]. A full
full treatment
treatment can
Chevalley [Ch
[Ch 1].
11.
By definition,
definition,CC=
= T/K
By
T/K where
where TT==®°°o Ti(V)
z ( V )isisthe
thetensor
tensoralgebra
algebraand
and
K ==(x(x®x
Q(u)1:
@ x- Q(u)l:XxEEV).
V).InInparticular
particularTo(V)
To(V)==F1
F 1and
andthere
thereisisaanatural
natural
isomorphismpo:
po:V V+-+Tl(V).
T1(V).Then
Then
themap
mapv vH
H vpo
K
p: p:
VV
+-+CCisisthe
vpo + K
isomorphism
induced by
by po,
po, and
and (1) will follow from (2), once that
induced
that part
part is
is established.
established.
ex,n== xl
x1@®.. .®xm
es =
=ex,
ex, ...
. . .ex,,,
- 8 x, + K,K,sosoCiCiisisthe
theimage
image of
of ®j_i Tj(V)
Tj(V)
in C. Hence
of multiplication
in T.
in
Hence (4) follows from (2) and the definition
definition of
multiplication in
The universal property of TT implies
impliesthere
thereisisan
aninvolutory
involutoryantiisomorphism
antiisomorphismtoto
of T
T with
with (xl
(xl8
®... . 8
(9 xm)to
xm)to==x,xm8®. -®. @3
x1.xl.As
of
Astotopreserves
preserves KK ititinduces
induces
t on C.
C. Thus
Thus (8)
(8) holds.
holds. Part
Part (3)
(3) isisaadirect
directconsequence
consequence of
of the
the definition
definition of
of
® xx - Q(u)1
C, since xx 8
Q(u)lEEK.K.An
Aneasy
easyinduction
inductionargument
argumentusing
using (3)
(3) shows
shows
eseT
of the
the elements
elementseeR,
X, for
for each
each S,
S, TT C X,
e s e ~is a linear combination of
~RR
, C X,
X,
spans C.
C. Using
Using the
the universal
universal property of the tensor algebra,
so (eR:RR C X) spans
11 that there is a homomorphic image of
Chevalley shows on
on page
page 39 of [Ch 1]
C of dimension 2", completing
completing the
the proof of
of (2),
(2), and
and hence
hence also
also of
of (1)
(1)and
and (4).
(4).
I omit this demonstration.
demonstration.
Part (5)
(5) is a straightforward
consequenceofof(3).
(3).IfIfchar(F)
char(F)#0 22 then
then (3)
(3)
Part
straightforward consequence
shows
shows e:e' ==(-1)mes
(-l),es for
forxxEEXXand
andSSCGX,X,where
wheremm==BSI
IS]if
if x $ Sand
S and
M=
= IS
-11ifif xx EE S.S.Since
m
IS1ISinceXp
Xpgenerates
generatesCCas
asan
anF-algebra,
F-algebra,(7)
(7)follows
follows in this
char(F) ==22then
case. If char(F)
thenchoose
chooseXXso
sothat
thateach
eachof
of its
its members
members is
is nonsingular.
Then
(3)
shows
[ex,
es]
=
0
if
S
C_
x1,
while
[ex,
es]
=
Q(x)-les+y
Then (3) shows [e,, es] = 0 if E xL, while [ex, es] = Q(x)-'es+, if S
contains the
xL,where
where SS + yy isisthe
the symmetric
symmetric difference of
the unique
uniqueyyininXX- x1,
of
followsthat
that[ex,
[ex,C]C]== (es:
(es:SSG
c x1)
S with {y}.ItItfollows
xL)isis of
of dimension
dimension 2a-1.
2'+'. So, as
2n-1 = dim(C)/2 and ex is an involution, [ex, C] = Cc(ex). Thus (7) holds in
2"-1
= dim(C)/2 and ex is an involution, [ex, C] = Cc(ex). Thus (7) holds in
this case too.
Let G and n
7r be
be as
as inin (6).
(6).For
Forgg EE G
G and
and vVEE V,
V, Q(vgn)l
Q(vgir)1 =
= ((vp)g)2
=
((vp)s)' =
((vp)2)g=
= Q(v)l,
Q(v)1, so
so Q(vgn)
Q(vgir) =
((vp)')g
=Q(v).
Q(v).Hence
Hence Gir
G n <5O(V,
O(V,Q).
Q).Let
LetGo
Go be
be the
$Eo
+
ej-,
+
+
The classical groups
97
subgroup
subgroupof
of GGgenerated
generatedby
by the
theelements
elementsup
upas
asuuvaries
variesover
overthe
thenonsingular
nonsingular
vectors
of V.
V. By
By ((5),
up,-r== -r,,
-ru, so,
G,-r
= Gon ==O(V,
5 ) , upn
so, by 22.7,
22.7, G
n =
O(V, Q)
Q) if
if
vectors of
then -I
-1 isisaaproduct
char(F) ==2.2.Further
Furtherifif char(F)
char(F) # 22 and
and n is even then
product of
of
char(F)
elements
ru cEGo7r
by 22.7,
22.7,GG7r
Goir =
= O(V,
-r,, sosor,
Gonand,
and, again by
n == Gon
O(V, Q).
Q). Finally
Finally
elements-ru,
ifif nn is odd then
then det(-r,)
det(-ru) ==+1,
=
+1,so
soGoir
Gon<5SO(V,
SO(V,Q),
Q),and
andthen,
then, as O(V, Q) =
(-I)
Q),Q),
22.7
says
SO(V,
(-I)x SO(V,
x SO(V,
22.7
saysG07r
Gon==
SO(V,Q).
Q).Then
Thentotocomplete
completethe
theproof
proof
G0ir =
= G7r.
of (6)
(6) itit remains
remains only to observe that,
Gn, so Gon
Gn.
of
that, by
by (7),
(7), -I
-1 $ G7r,
(22.9)
(22.9) Let
Let (V,
(V, Q)
Q) be
be an
an orthogonal
orthogonal space, C =
=C(Q)
C(Q)its
itsClifford
Cliffordalgebra,
algebra, G
G
the
Co
theClifford
Clifford group
groupof
of (V,
(V, Q),
Q),and
andG+
G+==GGfln
Cothe
thespecial
specialClifford
Cliffordgroup.
group.Let
Let
n be the
the representation
representationof
of 22.8.6.
22.8.6. Then
Then
7r
(1) G+
Gf isisaasubgroup
subgroupof
of GGof
ofindex
index2.
2.
char(F) # 22 then
then G+
G+n
SO(V, Q).
Q).
(2) IfIf char(F)
7r =
= SO(V,
char(F)==
thenG+
G+n
ofindex
index 22 in
in O(V,
O(V, Q).
Q).
(3) IfIfchar(F)
2 2then
7r isisof
(4) G+n
Gf ncontains
containsno
notransvections
transvectionsororreflections.
reflections.
Proof.
consequence ofof22.8.4.
Proof.Part
Part(1)
(1)isisaaconsequence
22.8.4.IfIfnnisiseven
eventhen,
then,by
by 22.8.7,
22.8.7, ker(7r)
ker(n) <5
G+
7r is
G+ and,
and, by
by 22.8.6,
22.8.6, G,-r
G n ==O(V,
O(V,Q),
Q),sosoG+
G+n
is of
of index
index 2 in O(V,
O(V, Q).
Q). Also
Also
-1
byby
thethe
proof
-I EEG+n
Gf n
proofofof22.8,
22.8,sosobyby22.8.6
22.8.6each
eachtransvection
transvectionor
orreflection
reflection
ru
r, isis not
notininG+,-r.
G+n.Thus
Thusthe
thelemma
lemmaholds
holds ininthis
thiscase
caseasasreflections
reflections are
arenot
not
in
SO(V, Q).
Q). If
If nn isis odd
odd then
then Gir
G n ==SO(V,
SO(V,Q)
Q)byby22.8.6,
22.8.6,while,
while,by
by22.8.7,
22.8.7,
in SO(V,
G+ker(n).So
Soagain
againthe
the lemma
lemma holds.
holds. 9
G ==G+ker(,-r).
(22.10) Let
Letv1,
vl, ...
. . ,. ,vm
urnbe
be nonsingular
nonsingular vectors
vectors in
in the
the orthogonal
orthogonalspace
space (V,
(V, Q)
Q)
such
such that r,,,
r,, ..... .r,,,
r,,,, ==1.1.Then
Thenthe
theproduct
product Q(vl)
Q(vl)...
. .Q(vm)
. Q(vrn)isisaasquare
square in F.
F.
Proof.
p. As
1, 1,
mm
is iseven,
Proof.Let
Letcc==vlvpl p....vm
. . vmp.
Asr,,,
r,, .... .r,,,,
. r, ==
even,because
becauseby
by22.9
22.9there
there
is
is aa subgroup
subgroupof
of O(V,
O(V,Q)
Q)ofofindex
index22containing
containingno
notransvection
transvectionor
orreflection.
reflection.
). Indeed,
Hence
r,,, ......r,,,
(- l)"r,,
r,,,=1
= 1by
by22.8.6.
22.8.6.SoSoCcEEker(7r
ker(n).
Indeed,as
asm
m is
is even,
even,
Hencec7r
cn ==(-1)m
22.8.7. So cc =
= a . 11 for some aa cE F#.
G+ flnker(7r)
ker(n) =
F# . 1 by 22.8.7.
F#.
c EE G+
= F#
Let
followsthat
thatc(ct)
c(ct) =
= c2
Let tt be
be the
the antiisomorphism
antiisomorphism of 22.8.8. ItIt follows
c2 =
=a2
a 2 . 1.
1.
On the other
p .... .vm
pvm p ...
p as
other hand c(ct)
c(ct)==vivlp
.v,pvrnp
.. .v1
vlp
as t is
is an
an antiisomorphism.
antiisomorphism.
Further
p)2 =
(vi) 1, so
so c(ct)
c(ct) _=(Q(v1)...
1, completing
(vip)'
=QQ(vi)l,
(Q(v1) . Q
. . (vm))
Q(vrn))l,
completingthe
theproof.
proof.
Further(vi
Let
F2=={a2:
{a2:aaE EF#}
F#}bebethe
thesubgroup
subgroupofofsquares
squaresininF#,
F#,and
andconsider
considerthe
the
Let F2
factor group F#/F2.
For
of odd
odd order
orderthen
thenF#/F2
F#/F2
F#/F2.
Forexample
example if
if FFisisaafinite
finite field of
is of
of order
order 2,
2, while
while if FFisisperfect
perfect of
ofcharacteristic
characteristic 22 then
then F#
F#==F2.
F2.Define
Define
.
r,,,,,
.
a map
-+ F#/F2
F'#/F2asasfollows.
follows.For
ForggEE O(V,
O(V, Q),
Q), gg ==rx,
r,, .. . .r,,, for
map 9:
8: O(V, Q) +
suitable transvections or reflections
rX,with
with center
center (xi). (Except in the exreflections r,,
ceptional case of 22.7,
22.7, which
which I'll
I'll ignore.) Define
Define B(g)
9(g) =
= Q(xl)
Q(xl) ..... .Q(xm)F2.
Q(x,)F'.
Observe first that Q(axi)
Q(xi )F2, SO
so the definition
of 89 is
~ ( a x i=a
=a2Q(xi)
) 2 Q(xi) E
E Q(x~)F',
definition of
is inin.
.
ryk
then
dependent of
of the choice of generator xi
xi of (xi).
(xi).Also
Also ifif gg==ry,
r,, .. . .r,,
Spaces with
with forms
98
98
r,, ...
. . r,,,,,
.r,,,,r,rYk. ...
. .ry,,so,by22.10,
1=
= r,,,
ry, , so, by 22. 10,Q(xl).
Q (xi) ..... Q(xm>F2
Q (xm) F2== Q(y1).
Q (yi) .... .Qe(yk)F2.
(Yk)F2
Thus 0 is
is independent
independent of
of the
the choice
choice of
of transvections
transvectionsand
and reflections
reflections too.
too. 00 isis
called the spinor
spinor norm of O(V, Q).
Q). From
From the
the preceding
preceding discussion
discussion it is evident
that.
spinor norm 0 is a group homomorphism
The spinor
(22.11) The
homomorphism of
ofO(V,
O(V, Q)
Q) into
into F#/F2.
F#/F2.
(22.12) If Q
Q isisnot
not definite
definitethen
then the
the spinor norm maps G+n
G+nsurjectively
surjectivelyonto
onto
F#/F2.
F#/F2.
plane U and
F# there exist u,
Proof. V
V contains
contains aa hyperbolic
hyperbolic plane
and for each a E F#
u,
v cE U
U with
with Q(v)
Q(v) =
=11and
and Q(v)
Q(v) ==a.a.Now
Now O(r,ru) = a.
a.
(22.13)
hyperbolic orthogonal
orthogonalspace,
space,let
letrr =
= F(V)
(22.13) Let (V, Q) be a hyperbolic
r ( V ) be
be the
the set
set
of maximal
maximal totally
totallysingular
singularsubspaces
subspacesofofV,V,and
anddefine
definea relation
a relation
onrr by
---on
is isananequivalence
A ---BBififdim(A/(A
dim(A/(AflnB))
B))isiseven.
even.Then
Then--equivalencerelation
relation with
with
exactly two equivalence
equivalence classes.
classes.
Proof. Given
of rr define
Given aa triple A, B, C of
of members of
define
+
S(A, B,
B, C)
C) =
= dim(A/(A
dim(A/(A nfl B))
B)) + dim(A/(A
dim(A/(A nfl C))
C))++ dim(B/(B
dim(B/(B nfl C)).
c)).
6(A,
Observe that the lemma is equivalent to the assertion that S(A,
6(A, B, C)
C) is
is even
even
for each triple A, B, C from F.
r.
Assume the lemma is false and pick aa counterexample
counterexample V with nn minimal.
minimal.
is hyperbolic,
hyperbolic, nn =
= 2m
= 2,
2m is
is even. If m =
=11then
then IF
)rII =
2, so
so the result holds.
As V is
Thus n >
> 1.
1. Let
Let A,
A, B,
B, C
C EE Fr with
with S(A,
6(A, B, C)
C) odd.
odd.
Let
D == A
B nflCCaand
= D1
D. As we
let^
A fln ~
n d suppose
s u p p oD
s e#~00.. ~Let
e tUU=
~ ' aand
n dUu==U/
U/D.Aswe
have seen several times
times already,
already, Q
Q induces a quadratic form
form Q on U.
0 . Further
U is hyperbolic
with A,
A, BB,, C
C EE I'(U),
0
hyperbolic with
r(U), so,
so, by
by minimality
minimality of V, 8(A,
6(A, B,
B , C) is
is
even. As
As 6(A,
S(A, B,
B, C)
C) =
= S(A,
6(A, B,
B, C)
C) we
we have
have aa contradiction
contradiction to the choice of
A, B,
B, C.Hence
C. HenceDD=0.
= 0.
A,
Suppose
nextEE =
=A
let Co
Co =
= (C fl
Suppose next
A fl
n BB # 0 and
and let
n E1)
E') + E.
E.By
By 19.2,
19.2,
Co E
E r.
F. 0 # E ==AAflnBBflnCo,
Co
Co,so,
so,by
bya aprevious
previouscase,
case,S(A,
6(A, B,
B,Co)
Co)is
is even.
even.
But
X nflCo
Co== (X
(X nfl C)
C)++ EE and
and XXnflCCnflEE =
= 0 for
X=
But X
for X
=A
A and
and B,
B, so
so
S(A,
6(A,B,B,Co)
Co)mod
mod2,2,again
againaacontradiction.
contradiction.
6(A, B, C) - S(A,
Thus 6(A,
S(A, B,
B, C)
C) =
= 3m. Hence
Hence m
m is
is odd.
odd. Therefore
ThereforeififT,T,SSEErF with
with TT -ti
Thus
A
fl T # 0, so, by
by the
thelast
lastcase,
case,TT--- S.
- isisan
A -- S then A n
S. Hence -an equivalence
equivalence
Finally let
let R
R E Fr with
relation. Finally
with A
A fln RR aahyperplane
hyperplane of
of A.
A. Then
Then 00 # A n
fl R,
so, by the last case, 8(A,
This shows
shows--- has
6(A, B, R)
R) is
is even,
even, and
and hence
hence B
B -- R. This
has
two classes and completes
completes the proof.
proof.
+
The classical groups
99
(22.14) Let (V, Q) be a hyperbolic
orthogonalspace,
space,let
letGG =
= O(V,
(22.14)
hyperbolic orthogonal
O(V, Q),
Q), and
and
let H be
be the
the subgroup
subgroupof
of G
G preserving
preserving the
the equivalence relation of 22.13. Then
(1) IG:
IG: HI =
=2.
2.
(2) HHisisthe
theimage
imageofofthe
thespecial
specialClifford
Cliffordgroup
groupunder
under the
the map
map of
of 22.8.6.
22.8.6.
(3) Reflections
transvectionsare
areininGG-- H.
Reflections and transvections
H.
Proof. By
set rr of maximal
maximal totally singular
singular
Proof.
ByWitt's
Witt'sLemma,
Lemma,GGisistransitive
transitive on
on the set
subspaces of
of V, so IG: HI is the number of
of equivalence
equivalenceclasses
classesof
ofr.
r. That is
I G:HI
H I== 22 by
by 22.13. It's easy to check that (3)
IG:
(3) holds. Then,
Then, by (1),
(I), (3),
(3), and
and
22.7, H is
is the
the subgroup
subgroupof
of G
G consisting
consistingof
of the
the elements
elements which are the product
of an even number of transvections or reflections, while by 22.9 this subgroup
subgroup
is the image of the special
special Clifford group under the map of 22.8.6.
section with a brief discussion
discussion of some geometries
geometries associated to the
I close this section
geometries are derived in Exercise
classical groups. A few properties of these geometries
7.8, while chapter 14
14 investigates these geometries in great detail.
index m
m of
of (V,
(V, ff)) or
Assume the Witt index
or (V,
(V, Q)
Q)isispositive.
positive. In
In this
this case
case there
there
are some interesting geometries associated to the space and preserved by its
isometry group.
isometry
group. The reader may wish to
to refer to the
the discussion
discussion in
in section
section 33 on
on
geometries.
geometries.
) or (V,
(V, Q)
the geometry
geometry over
The
The polar
polar geometry
geometryrr of
of (V,
(V,ff)
Q) is
is the
overI I =
=
10, 1,
1, ..... .,,m
} whose
(0,
m --1 1)
whose objects
objects of type i are
arq the totally singular
singular subspaces
subspaces of
V of projective dimension i,i, with
V
with incidence
incidence defined
defined by
by inclusion.
inclusion.Evidently
Evidently
O(V, f)
f )isisrepresented
representedasasaagroup
groupof
of automorphisms
automorphismsof F.
r.Indeed
Indeed the
the similarity
similarity
group A(V,
O(V, f)
is
also
so
represented.
f ) is also so represented.
If (V, Q)
Q )is
is aa hyperbolic
hyperbolicorthogonal
orthogonalspace
space there
there is
is another
another geometry
geometry associassociated to (V, Q)
geometry. Assume
Assume the
Q)which
whichisis in
in many ways
ways nicer than the polar geometry.
the
dimension
dimensionof
of V
V is
is at
at least
least 6,
6, so
so that
that the Witt index m of (V, Q)
Q) is
is at
at least
least 3.
3. The
The
oriflammegeometry
geometryrrof
of(V,
(V,Q)
Q)isisthe
thegeometry
geometryover
overI I== (0,
10,1,1,. ..... ,,mm-- l}
orijlamme
1)
whose objects of type ii <<m
m --22are
arethe
thetotally
totallysingular
singularsubspaces
subspacesof
of projective
projective
dimension i, and
and whose
whose objects
objectsofoftypes
typesmm-- 1 and
and m
m- 22 are
are the
the two equivalence classes of maximal
maximal totally singular
singular subspaces
subspaces of (V, Q)
Q) defined
defined by
by the
the
equivalence relation of 22.13. Incidence is inclusion,
except
between
objects
inclusion,
U and W
m- 11and
U
W of type m
and m
m --2,2,which
whichare
areincident
incidentifif UU flflW
W isisaahyperhyperO(V, Q)
plane of U and W.
W. In this case the subgroup of A(V,
Q) of index
index 2 preserving
the equivalence relation of 22.13 is represented as a group of automorphisms
of F.
r.
Remarks. The
Thestandard
standardreference
referencefor
for much
much of
of the
the material
material in
in this
this chapter
chapter isis
Dieudonne [Di].
DieudonnC
[Di]. In particular
particular this is
is aa good
good place
place to
to find
find out
out who
who first
first proved
proved
what in the subject.
subject.
Spaces
forms
Spaces with
with forms
100
100
We
We will encounter
encounter groups generated
generated by reflections
reflections again in sections
sections 29
29
and 30.
30.
Observe that, by 13.8,
13.8, 22.4.4, 22.4, and
and Exercise
Exercise 7.6, almost
almost all
all the
the finite
finite
groups SL,(q),
SLn(q), Sp,(q),
Spn(q), Qi(q),
Q '(q), and SU,(q)
SUn(q) are perfect. This fact is
classical groups
PSL,,(q), PSp,(q),
PSpn(q), PQ:(q),
PQ (q),
used to prove in 43.12 that the projective groups PSL,(q),
are small.
and PSU,(q)
PSUn(q) are simple unless n and q are
Since by some measure most finite simple groups are classical, the study
of the classical
classical groups is certainly important.
important.Moreover
Moreover along
along with
with Lie
Lie theory
theory
(cf. chapter 14 and section 47) the representations of the classical groups on
their
their associated
associated spaces
spacesis
is the
the best
best tool
tool for
for studying
studying the
the classical
classicalgroups.
groups.On
On the
the
other hand the
the study
study of the
the classical
classical groups
groups is
is aa special
special topic
topic and
and the
the material
material
in
in this chapter
chapter is technical.
technical. Thus the casual
casual reader may wish to skip, or at least
postpone,
postpone,this
this chapter.
chapter.
Exercises for chapter
chapter 7
1. Let
LetVVbe
bean
ann-dimensional
n-dimensionalvector
vector space
space over
over F.
F .Prove:
Prove:
(1) If
If n ==22then
then SL(V) =
=Sp(V).
Sp(V).
(2)
order2,2,n n== 2,
2, and
(V, ff)) is aa
(2) If 98 is
is an
an automorphism
automorphism ofofFFofoforder
and (V,
ESL2(Fix(9)).
SLz(Fix(8)).
hyperbolic unitary
unitary space,
space,then
thenSL(V)
SL(V)nnO(V,
O(V,ff)) =
Let IF
I FII =
q2<<oo
ooand
and (V,
(V, f)f )a a3-dimensional
3-dimensionalunitary
unitary space
space over F
(3) Let
= q2
F..
Then there exists a basis X of V such that
0
J(X, f) _
0
0
1
0
1
0
'01
Moreover if P consists of those g E SU(V) with
a
c
Mx(g) = 0
1
b
0
0
1
1
then
q3and
and [P,
[ P ,P]
PIisisa aroot
rootgroup
groupof
ofSU(V).
SU(V).
then PI PI1 ==q3
(4) Let
andand
(V,(V,
f) fa )2m-dimensional
Let IF!
IF] ==q q< oo
< oo
a 2m-dimensionalsymplectic
symplecticspace
space
over F with m
m>
existsaabasis
basisXX== (xi:
(xi: 11 F:
< i <5 2m)
> 1.
1.Then
Then there exists
2m)
such that
J(X, f) _ -IM
0
0
Im
/
I, isis the
the m
mby
by mmidentity
identitymatrix.
matrix. Moreover
Moreover if P consists
consists of
of
where In
The classical groups
101
those gg cE Sp(V)
Sp(V) with
Mx(g) =
1
0
0
a
12._2
0
b
c
I
where a and
=
and cc are
are column
column and
androw
row vectors,
vectors,respectively,
respectively, then
then II P II =
q2'-1.
If
[P, P]
= 22 an
m=
= 33
qZm-'. If q is odd, [P,
PI isis aa root group of Sp(V). IfIf qq =
andd m
then
transvectionand
andP P== [P,
[P, HI
H] where
where HH =
= Sp(U)
contains aa transvection
Sp(U)
then P contains
and U ==(XI,
(xr,x2m)1
~2,)'.
over F.
F. Prove:
2. Let
Let (V,
(V, Q)
Q)be
be aa 2-dimensional
2-dimensional orthogonal space over
subgroupH
H by
by ((r)
= Z2,
(1) O(V,
O(V,Q)
Q)isisthe
thesemidirect
semidirectproduct of aa subgroup
r) E
Z2,where
inverts H.
H.
r inverts
(2) Each
Each element
element of O(V, Q) --HHisisa atransvection
transvectionor
orreflection.
reflection. In particular O(V,
O(V, Q)
Q) isis generated
generated by
by such
such elements.
elements.
(3) If
If O(V,
O(V,Q)
Q)isishyperbolic
hyperbolic then
then HHisisisomorphic
isomorphictotothe
themultiplicative
multiplicative
F# of
of FF..
group F#
(4) IfIf(V,
(V,Q)
Q)isisdefinite
definitethen
then there
there exists
exists a quadratic
quadratic Galois extension K of
F such
that (V,
(V, Q)
Q)isissimilar
similartoto(K,
(K,N;)NF)and
andHHE- {a
{a EE K:
K:aae
aae=
= 11,
I),
F
such that
where (0)
(0)=
= Gal(K/F).
Gal(K/F).
where
3. Let
Let (V,
(V, f)f )bebea a4-dimensional
4-dimensionalunitary
unitary space
space over
over a field F of
of order
order 4, X
X
orthonormalbasis
basisfor
forV,V,AA=_ {(x):
{(x):x
an orthonormal
x cE X},
X I , and G ==SU(V).
SU(V).Prove
Prove
NG(A)°
E27,
and
N
~ ( A -E
)S4,
~S4,Go
GA=E
EZ7,
andNG(A)
NG(A)sisgenerated
generatedby
bytransvections.
transvections. Let
DE
E A,
A, TT the
the subgroup
subgroupof G
G generated
generatedby
b i the
the transvections
transvectionsin G,
G, and FI? the
the
of conjugates
conjugatesofofAAunder
underNG(D).
NG(D).Prove
ProveNNG(D)r
A4and
and)Gr
I Gr1 =
= 54.
~ ( D2
) -~A4
54.
set of
NG(D) 5
< T.
Prove Nc(D)
4. Let
Letqqbe
beaaprime
primepower.
power. Prove
Prove
of order
order qq + 1.
(1) Z(GU,,(q))
Z(GU,(q)) and
and GU,(q)/SU,(q) are cyclic of
1.
(2) Z(SU,(q))
Z(SU,(q)) and
andPGU,(q)/
PGU,(q)/ U,,(q)
U,(q) are
arecyclic
cyclic of order (q + 1,
1, n).
Exercise 4.7
4.7with
withchar(F)
char(F) # 2. Let
5. Assume
Assumethe
the hypothesis
hypothesis and notation of Exercise
= V3,
a ==7r3,
F by
V3, a
n3, and define Q: W -->
+F
W=
+
+
+
+
Q(ax2 + bxy
bxy + cy2) =
= b2 - 4ac.
e(ax2
4ac.
Prove
(1) QQisisaanondegenerate
2+
nondegeneratequadratic
quadraticform
formon
onW
Wwith
withbilinear
bilinearform
form(ax
(ax2
bxy + cy2,rx2
rx2 +sxy
sxy + ty2)
ty2)=
= 2bs
2bs-- 4(at
4(at + cr).
bxy
eachggEEG,G,g agais is
a similarity
of (W,
A(ga)
= det(g)2.
(2) For each
a similarity
of (W,
Q) Q)
withwith
h(ga)
= det(g)2.
thegroup
groupA(W,
A(W,Q)Q)ofofallall
similarities
of (W,
where
(3) (Ga)S
(Ga)S is the
similarities
of (W,
Q),Q),
where
S is S is
the group of scalar
scalar maps on W.
W.
(4) Up
Up to
tosimilarity,
similarity, (W,
(W, Q)
Q) isisthe
theunique
unique3-dimensional
3-dimensional nondefinite
nondefinite ororthogonal space over F
F..
+
+
+
+
+
forms
Spaces with forms
102
102
(5) IfIf FFisisfinite
finiteororalgebraically
algebraicallyclosed
closedevery
every3-dimensional
3-dimensionalorthogonal
orthogonal
over F
F is similar to (W,
(W, Q).
space over
(6) (rrh:
(rrh:rr EE R,
R,hh EEGet)
Ga) ==O(W,
A(W,Q)(l)
Q)") =SL2(F),
L2(F),where
whereRR isisthe
the set
set of
reflections in O(W, Q).
Q).
6. Let
fieldFF with
withnn 2
> 3.
Q )be
bean
ann-dimensional
n-dimensionalorthogonal
orthogonalspace
space over aafield
Let(V,
(V,Q)
(1) Assume
(V,
Q)
is
not
definite
and
if
IF
I
<
3
and
n
<
4
assume
n
=4
Assume (V, Q) is
and if I FI 5 3 and n 5 4 assume n =
and sgn(Q) _=-1.
- 1.Prove
Provethe
thefollowing
followingsubgroups
subgroupsare
areequal.
equal.
(i)
(9 n(V,
Q(V, Q).
Q).
(ii) The
7r of the spinor norm, where G+ is the
Thekernel
kernel in
in G+
G+n
the special
special
Clifford group.
group.
(iii) (rrg:
(rrg:rrreflection
reflectionor
ortransvection,
transvection, g E
E O(V, Q)),
Q)).
(2) If char(F) 0#22prove
= 33 and
proveQ(V,
Q( V, Q)
Q) isis perfect
perfect unless
unless IIFI
FI =
and either
either
(2)
prove O(V, Q)/Q(V,
= 3 orn=4andsgn(Q)
= 4andsgn(Q)_+1.IfFisfinite
= +l.If F is finiteproveO(V,
Q)/ Q(V, Q)
Q)
nn=3orn
and -I
-I EEQ(V,
evenand
andsgn(Q)
sgn(Q)r
- IFI"/z
S E4, and
Q(V,Q)
Q)ififand
andonly
only if
if n is even
1 F In/'
mod 4.
(3) If char(F)
(3)
char(F) ==22and
andFFisisperfect,
perfect,prove
proveeither
either Q(V,
Q(V, Q)
Q) isis perfect
perfect and
IO(V,
Q): Q(V,
SZ(V,Q)i
Q)j==22ororIFIFI
= 4,
4, and
andsgn(Q)
sgn(Q)=
=+
+1.
IO(V, Q):
1 ==2,2,nn=
l.
(Hint: To
To prove
prove Q(V,
Q(V, Q) perfect
perfect use
use (1)
(1) and
and show
show rrg
rrg is contained in a
Q) for
for each
each reflection or transvection
perfect subgroup
subgroup of O(V, Q)
transvection r and each
g EE G.
G. Toward
Toward that end use Exercise
Exercise 7.5 in (2) and 22.5 in (3).)
Let G
G be
be aapermutation
permutation group
group on
of finite
finite order
order n and
and V
V the
the
7. Let
on aa set II of
permutation
module
for
G
over
F
with
G-invariant
basis
X
=
(x1:
i
E
I).
permutation module for G over F
G-invariant basis X = (xi: i E I).
on V
V by
by ff(x1,
(xi, xi)
Sij(the
(theKronecker
Kronecker delta)
delta)
Define
Define aa bilinear
bilinearform
formff on
xj) ==Sjj
. Letzz= =
CiEI
xi, U the
the core
core of
of the
the permutation
permutation module
module
for
for i,i, jj eEI.I Let
F_ZEI
V, and
and iiV =
= V/(z).
V/(z).IfIf char(F)
char(F)==2 2define
defineaaquadratic
quadratic form
form Q on U by
V,
Q(F_
a;x1)=_1
E a:
a? + Ei<j a1ap
aiaj.Prove:
Prove:
Q
( 1 aixi)
= zL.
(1)
(1) uU=z1.
(2) G 5
< OW,
O(V, ff).1.
(2)
(3) If char(F)
5422then
then
orthogonal
space.
If char(F)
(3)
char(F) #
(V,(V,
f ) f)isisanan
orthogonal
space.
If char(F)
=2 =2
f )isisaasymplectic
symplecticspace
spacepreserved
preserved by
by G
G and
and if further
further b f 2
then (U, f)
mod 4 then (U,
(U, Q)
(2) isis an
anorthogonal
orthogonal space
space preserved
preserved by
where
mod
by G, where
fi) =
=ff(u,
(u,v)
v) and
and Q(u)
~ ( i i==
) Q(u)
Q(u)for
foru,
u , vv EE U.
U.
ff(ii,
(u, v)
(4) If
If G
G=
= S6,
= 6,
6, and
andIF1
IF)== 33then
then( (U,
sign-1,O(o,
-1, O(U,f f)
_
(4)
S6, nn =
0 , f f)
) isisofofsign
)=
-Q).
((-I)
- I ) xxG,G,and
andA6
A6- Q4
QT(3).
(5) IfIf G
G ==S5,
sign -11 and
(5)
S5, n ==5,5,and
and BFI
( FI =
=2,2,then
then (U,
( 0 , Q)
(2) is of
of sign
and
0 ( 0Q)
,G0
2).
O(0,
= G4-2(0,(2).
(6)
G = S6, n = 6,
andIF
IF1I =
2, then G=O(U,
G = 0(0, f)
f )=Sp4(2).
2 Sp4(2).
(6) IfIfG=S6,n=6,
and
=2,then
If G
G ==S8,
S8, n ==8,8,and
and IFI
IF1 ==2,2,then
then G
G ==O(U,
0(0, Q)
Q)-SO6+(2).
0:(2).
(7) If
8. Let
Let(V,
(V, f)f )bebeaasymplectic
symplecticor
orunitary
unitary space
spaceover F,
F , or
or (V, Q)
Q) an
an orthogonal
orthogonal
F.Assume
Assume the
the Witt index m of the
space over F.
space
the space
space isispositive.
positive.Let
Letrr be the
of the
the space
spaceover
overII =
= {0,
polar geometry of
{0,1,
1, ...
. . .,,m
m --1)
1)and
and G the isometry
+
The classical groups
103
group of the space
space with
with m
m >2
space or (V,
(V, Q)
Q) aa hyperbolic
hyperbolic orthogonal
orthogonal space
3, 1'
of (V, Q), and G the
r the oriflamme geometry of
the subgroup
subgroup of
of O(V,
O(V, Q)
Q)
preserving the
maximal hyperbolic
hyperbolic
preserving
the equivalence
equivalencerelation
relation of
of 22.13. Let Z be a maximal
subspace ofof VV,, XX=
2m)a ahyperbolic
hyperbolic basis
basis for
i =
subspace
= (xi: 11 5
< ii <52m)
for Z, V
Vi
=
(x2j_1:
< i),
Y=
= {(x):x
LetTT =
= {K:
{Vi:115<i i5< m) if
if r
(xzj-l: 115<jj 5
i), and Y
{(x):x E
E X}.
X}. Let
is the polar geometry and T =={Vi,
< ii <
- 1)
{Vi, Vh-l: 1 5
5 m,
m, ii # m 1) if
if r is
is
the
_1 = =
(Vm_2,
Prove
(V,-2, X2m_3,
X~rn-3,x2m).
~2,). Prove
the oriflamme
oriflammegeometry,
geometry,where
whereV,VhW1
(1) If
of rI',, while
while ifif m
m>
> 11
thenGGisis2-transitive
2-transitive on
on the points
points of
If m
m ==1 1then
then G is rank 3 on these
these points.
(2) GGisisflag
on rr..
flag transitive on
(3) TT isis aa flag
of r of
I.
flag of
of type I.
(4) BB ==GT
of U with
with H
H=
GTisisthe
thesemidirect
semidirect product of
=Gy,
Gy,where
where U
U isis
the subgroup
Vi+1/ V
/ iVi,
1 <i i<<m
m- 1,
subgroup of G centralizing
centralizing V1,
Vl , Vi+1
,1 5
1, and
and
(a) Vm/Vm_i
and(V,)'
(V.)'/
V,/ Vm-l and
/ V,,
V, ifif rrisisaapolar
polarspace,
space, or
n Vm)/Vm_2
and(V,-1
(Vm_1V,)/(V,-l
+ V,,,)/(Vm_i
n vif
(b) (Vm_i
(V,-l n
V,)/V,-z
and
f l V,)
if rr is
is an
an
oriflamme
oriflamme geometry.
(5) UU isisnilpotent
nilpotent and
and H
Hisisthe
thedirect
directproduct
product of
of m
m copies
copies of
of F#
F' with
with
O(Z1-,
O(Z', ff) (or
(or O(Z1,
O(Z'? Q)).
Q)).
(6) Let
unique object
objectofoftype
typei iininrr or
or rr is
Let ii EE I.I.Then
Theneither
either U
U fixes
fixes a unique
is
a polar geometry and V is a hyperbolic orthogonal space.
(7) B
B ==NG(U).
NG(U).
(8) Assume
Assume F
F isis finite
finite of
of characteristic
characteristi9 pp and
and rr isisoriflamme
oriflamme if
if V
V isis
hyperbolic
U EE Sy1p(G).
Syl,(G).
hyperbolic orthogonal.
orthogonal. Then
Then U
(9) NG
(Y) YisisZ2wrS,
7L2wrSm
index2 2ininthat
thatgroup,
group,for
forrr a polar space
NG(y)'
oror
ofof
index
space or
or
oriflamme geometry, respectively.
oriflamme
(10) Let
I's of SS is
Let SS be
be aa flag
flag of
of corank 1 in T.
T. Then
Then either the residue rs
is
isomorphictotothe
theprojective
projectiveline
lineover
overF Fand
and(Gs)~.
(Gs)r,S- PGL2(F)
isomorphic
PGLz(F) or
L2(F),
of type
type {O,
{0,....
L2(F), or rrisisaapolar
polar geometry,
geometry, S is of
.. ,, mm -- 2},
21, I's
rsisis
isomorphic to the set of singular points
points of
of W =
= (v,-~)'/
(V,,-,)-L/ V,-l,
V,,-,, and
either
( G S ) ~S~PO(W, f)
f )(or
(orPO(W,
PO(W,Q))
Q))or
orVVisishyperbolic
hyperbolicorthogorthogeither(Gs)1'S
I's 1I =
= 2.
onal and I1 rs
spaceover
overaafield
fieldFFand
andff a nontrivial
9. Let
LetVVbe
be aa finite
finite dimensional vector space
sesquilinear
sesquilinear form on V.
V. Then
(1) IfIf char(F)
g g++
h hwhere
char(F)##22and
andffisisbilinear
bilinearthen
thenff==
whereggand
andhhare
aresymsymmetric and skew symmetric
symmetric forms
forms on
on V,
V, respectively,
respectively,and
andO(V,
O(V,f f)
<
) 5
O(V, g) n
n O(V,
OW,
OW, h).
(2) If
bilinear
If char(F)
char(F) ==2 2and
andf fis is
bilinearthere
thereexists
existsa anontrivial
nontrivial symmetric
symmetric
formgg on
on V
V with
withO(V,
O(V,ff)) <
bilinear form
iO(V,
O(V, g).
(3) Let
Let U
U=
Let char(F) ==22and
andassume
assume ffisisbilinear
bilinearand
and symmetric.
symmetric. Let
=
{x E V
(x, x)
x) =
= 0).
V::ff (x,
O}.Prove
Prove U
U isis aa subspace
subspace of V
V which
which is of codimension at most 11 if F
F isis perfect.
perfect.
+
>
104
104
Spaces with forms
forms
(4) Assume
Assumeffisissesquilinear
sesquilinearwith
withrespect
respectto
tothe
the involution
involution 08 and ff isisskew
skew
(y, x)°
x)' for all
hermitian;that
thatisisff(x,
(x,y)y)== --ff(y,
all x,
x, yy EE V. Prove
Proveff is
hermitian;
similar to a hermitian
hermitian form.
form.
(5) If
I fff isissesquilinear
sesquilinearwith
withrespect
respectto
to an
an involution
involution 0,
8, then there exists aa
nontrivial hermitian
hermitiansymmetric
symmetricform
formggon
onVVwith
withO(V,
O(V,f )f) 5
< O(V, g).
be aafield
field and
and ffaasesquilinear
sesquilinearform
formon
on VVwith
with respect
respect to
to the
the autoauto10. Let FF be
morphism
ofF,
F, such
such that
thatfor
forall
allx,x,yyEe V,
(x, yy)) =
= 00 ifif and
V, ff (x,
and only
only if
morphism 80of
ff (y,
(y, x)
x) ==0.0.Prove
Provethat
that either
either
(1) ff(x,
x)
=
0
for
all
x
(x, x) = 0 for all x EE V,
V, 08 ==1,1,and
andffisisskew
skewsymmetric,
symmetric, or
(2) there
there exists x E
EV
V with
with ff(x,
(x,x)
x) 0#00and
andone
oneofofthe
thefollowing
following holds:
holds:
(a) 08 ==11and
and ffisissymmetric.
symmetric.
= 2 and ff isissimilar
(b) 191
181 =
similartotoaahermitian
hermitiansymmetric
symmetricform.
form.
(c) X01
>
2
and
Rad(V)
is
of
codimension
1
in
V.
181
codimension 1 V.
p-groups
p-groups
Chapter
investigates p-groups
through a study of
Chapter88 investigates
p-groupsfrom
fromtwo
twopoints
pointsof
of view:
view: first
first through
p-groups
parameters (usually
(usually
p-groups which
whichare
areextremal
extremalwith
with respect
respectto
to one
one of several parameters
connected with p-rank)
connected
p-rank) and
and second
second through
through aa study
study of the automorphism group
of the p-group.
Recall that if p is
is aa prime
prime then
then the
the p-rank
p-rank of
ofaafinite
finitegroup
groupis
is the
the maximum
maximum
dimension of an elementary abelian p-subgroup, regarded as a vector space
over GF(p). Section 23 determines p-groups
p-groups of
of p-rank
p-rank 1, p-groups in which
for pp odd, p-groups in which each
each normal abelian subgroup
subgroup is cyclic, and, for
of p-rank at most 2. Perhaps most important, the
normal abelian subgroup is of
p-groups of
of symplectic
symplectic type are determined
determined (a p-group is
is of
of symplectic
symplectic type
if each of its characteristic
characteristic abelian subgroups is cyclic).
automorThe Frattini subgroup is introduced to study p-groups and their automorfocusedon
onp'-groups
p'-groups of
of automorphisms
automorphisms of
of p-groups;
p-groups;
phisms. Most attention is focused
a variety of results on the action of
of p'-groups on
on p-groups
p-groups appear
appear in
in section
section
24. One very useful result is the Thompson
Thompson A,x
A,x B
Lemma.
Also
of
importance
B Lemma. Also of importance
concept of aa critical
is the concept
critical subgroup.
subgroup.
23 Extremal
Extremal p-groups
p-groups
and G is a p-group.
In this section p is a prime and
of all
The Frattini subgroup of a group H is
is defined
defined to be the intersection of
of H. (D(H)
Frattini subgroup
subgroup of
of H.
H.
maximal subgroups of
@(H)denotes the Frattini
(23.1)
@(H)char
char H.
H.
(23.1) (1)
(1) '1(H)
If X C H
= (X).
(2)
H with
with H
H ==(X,
(X,(D(H)),
@ ( H ) )then
,
H=
(X).
(2) If
If
H/(D(H)
is
cyclic,
then
H
is
cyclic.
(3)
H/@(H)
is cyclic.
(3)
(23.2)
(G) is the smallest normal
normal subgroup
subgroup H
H of G
(23.2) If G
G isisaap-group
p-groupthen
then(D
@(G)
G
G/H is
such that GIH
is elementary
elementary abelian.
Proof. If
3.2, M
M L]
4 G and
If M
M isisaamaximal
maximal subgroup
subgroup of G then, by Exercise
Exercise 3.2,
I G: :MI
MI==pp,
so, by
by 8.8,
8.8, G(')
G(1 5
< M. Hence
< @(G),
t (G), so,
(D (G)
lG
, so,
Hence G(1)
G(') 5
so,by
by8.8,
8.8,GI
G/@(G)
G/M =ZZp,
(G). Hence
is abelian. Also, as GIM
Zp,gP
gpEEM
Mfor
foreach
eachggeEG.
G.soso9°gpEE(D
@(G).
G/(D(G)
G/@(G) is elementary
elementaryabelian.
abelian.
p-groups
p-groups
106
106
G / H ==G*
G*elementary
elementary abelian.
abelian. Then
Then G*
G* =
Conversely
Converselylet
letHH a G
G with G/H
_
..
x
G*
with
Gi
=
l,,
so
setting
H,
=
(G1:
j
G
:
H,
I
=
p
GTx
...
x
G;
with
G
f
S
Z
,
,
so
setting
Hi=(Gj:
j
#
i
)
,
I
G
:
H
i
I
=
p
and
G *x x
i),
= n H1
=n Hl.
Hi.Thus
ThusH,
Hiisismaximal
maximal in G so H =
Hi >2 4)(G).
@(G).
H=
n
n
Observe that, as a consequence of
of 23.2,
23.2, aa p-group
p-group G is elementary abelian ifif
and only
only ifif @(G)
c(G) =
1.
= 1.
Recall
positive integer
integer then Qn(G)
(G)isisthe
thesubgroup
subgroup of
of G
Ggenerated
generated
Recallthat
that if n is a positive
by all elements
elements of order at most pn.
pn.
(23.3) Let G =
= (x)
(x)be
be cyclic
cyclic of
of order q =
=pn
pn >>11and
and let A =
=Aut(G), Then
Then
(23.3)
(1) The map a H
i-+m(a)
m(a) is
is an
an isomorphism
isomorphism of
of A
A with
with the
the group
group U(q) of
(1)
units of the integers
integers modulo q,
q, where
where m(a)
m(a) isisdefined
defined by
by xa
xu ==xm(a)
xm(a)for a E A.
In particular A
A isis abelian
abelianofoforder
order@(q)
O(q)==pn-'(p
pn-1(p-- 1).
1).
of order pp - 11isiscyclic
(2) The subgroup of A of
cyclicand
and faithful
faithful on
on Stl(G).
nl(G).
(3)
(3) If p isisodd
odd then
then aaSylow
Sylow p-group
p-group of
ofAAisiscyclic
cyclicand
andgenerated
generatedby
by the
the
element bb with
with m(b)
m(b)=
= pp + 1.
the subgroup
subgroup of
of AA of
of order
order pp is
element
1. In particular
particular the
generated by the element bo
b0 with
with m(bo)
m(bo)=
= pn-'
pn-1 + 1.
1.
(4)
(4) If
I f q = 22t then
h e n A ==1lwwhile
h i l eifi fqq = 44tthen
h e n A ==((c)
c ) ~=ZLL2,
2 , w where
h e r e mm( c(c)
) =_- 1-1..
(5)
(5) If p ==22 and
and qq >>44 then
then A
A=
=(b)
( b )xx (c)
(c)where
wherebbisisof
oforder
order2n-2
2n-2 with
with
m(b) =
5, and c is of order 2 with
Theinvolution
involutionb0
bo in (b)
(b)satisfies
satisfies
= 5,
with m(c)
m(c) =
=-1.1.The
= 2"-'
2n-1 + 1 and m(cbo)
m(cb0)=
= 2"-'
2n-1- 1.
m(bo)=
1.
+
+
+
Proof. II leave
ProoJ:
leave part
part (1)
(1) as
as an
an exercise
exercise and
and observe
observe also that a:
a : a i-+
H m(a) mod
of A onto
onto U(p) with kernel CA(n1(G)).
CA(QI(G)). SO,
So,
p is
is aa surjective
surjective homomorphism of
as IU(p)I
p
=
p
1=
IU(q)Ip,,
the
subgroup
of
A
of
order
p
1
is
isomorIU(p) 1 = p - 1 = IU (q) 1 the subgroup of A of order p - 1 is isomorU(p) and
1(G),while
whileker(a)
ker(a)=_ (a
{aEEA:
A:m(a)
m (a) -11mod p}
and faithful
faithful on cZ
a1(G),
p) EE
phic to U(p)
Sylp(A). Next
Next U(p) is
of the field
field of
of order p, and
Syl,(A).
is the
the multiplicative
multiplicative group of
and
cyclic, so
so (2)
(2) holds.
holds. Thus
Thuswe
wemay
maytake
takeqq>>p.
p. Evidently
Evidentlyifif m(c)
m(c)=
= -1
-1
hence cyclic,
then c is of order 2. So,
A I =2
=4, (4)
then
So, as
as JIAl=
2 ifif qq =4,
(4) holds.
holds. Thus
Thus we
we can
can assume
assume
> 1, and
and nn >
> 22 ifif p =
2. Choose
ker(a), so
nn >
=2.
Choose b as
as in
in (3)
(3) or
or (5).
(5). Then
Then bb EE ker(a),
bP"-' =
bpn-'
=1.
1. Thus if p isis odd
odd itit remains
remains to
to show
show by"-2
bpn-' ==bo
= 2 show
show
boand
andififpp=2
b2" =b0.
b2"-3=
bo.
Observe:
Observe:
-
,I,
,I
+
l)P
(kpm + 1)P
(kpm
p2-+1(p + k2
_ 1)/2)
= (1 ++kpm+1
kpm+l +
k2p2m+1(p
1)/2)mod
modpm+2
pm+2
1 + kpm+1
m > 11 or
or pp isis odd.
odd. Hence
Hence as
with s =
odd and
and
if m>
as m(b)=l
m(b) =1++ss with
= pp if pp isis odd
+
=44 ifif pp =2,
=11 + pn-1=m(bo)
s=
=2,ititfollows
followsthat
that m(bpi-z)
m(b~"-')=
pn-' = m(bo) if
if pp isisodd,
odd, while
while
m(b2"-3)
= 2n-1 + 1 = m(bo) if p = 2. So the proof is complete.
(b2"-3
) = 2"-'
1 = m(bo) if p = 2. So the proof is complete.
+
,
Extremal p-groups
107
107
Next the definition of four extremal classes of
of p-groups.
p-groups. The modularp-group
modular p-group
Modp"
p' isisthe
group X =
Mod,. of order pn
thesplit
splitextension
extensionof
of aa cyclic=xPi-2+1.
=(x)
(x)of
oforder
orderpii-1
pn-'
by a subgroup
= (y)
subgroup Y =
(y) of
of order
order p with
with xy
xY =XP"-'+I. Modpn
Mod,. is defined only
> 3, where, by 23.3 and 10.3,
10.3, ModeMod,. is well
well defined
defined and determined
when n 2
comments hold
hold for
for the
the other
other classes.
classes.IfIfpp =
= 2 and
up to isomorphism. Similar comments
nn >222 the dihedral
by YYwithxy
withxy =
= x-'
x-1 and
dihedral group D2
DT isisthe
the split
split extension of X by
and
x2^-2-1
if nn >2 44 the
isisthe
the semidihedral
semidihedral group SD2
SD2?,
thesplit
splitextension
extension with
with xy
xy ==x~~-'-'.
The
The fourth
fourth class
classis
is aa class
class of nonsplit
nonsplit extensions.
extensions. Let
Let G
G be
be the
the split
splitextension
extension
of X =
= (x)
> 44 by
by Y
= (y) of order 44 with
=x-1. Notice
(x) of
of order
order 21-1
2"-' 2
Y=
with xxy
Y =x-'.
Notice
= Z(G). Define
(x2"-', y2)
Y2) =
Define the quaternion
quaternion group Q2n
Q2. of order 2" to
to be
be the
the
group G/(x2ii-2y2).
G / ( X ~ ~y2).
-'
group
The
The modular,
modular, dihedral,
dihedral, semidihedral,
semidihedral,and
and quaternion
quaterniongroups
groupsare
arediscussed
discussed
in Exercises
Exercises 8.2
= D8.
8.2 and
and 8.3.
8.3.Observe
Observe Mod8
Mods =
D8.
(x2"2,
(23.4) Let
LetGGbe
beaanonabelian
nonabeliangroup
groupof
oforder
orderp'°
pnwith
withaacyclic
cyclicsubgroup
subgroupof
of index
index
p. Then
Then G
G -SModp,,,
Mod,. ,Den,
D2",SD2-,
SD2., or Q2'.
122".
Proof.
= (x)
ProoJ:Notice
Noticethat,
that,asasGGisisnonabelian,
nonabelian,nn >233by
by Exercise
Exercise2.4.
2.4. Let X =
(x)be
be of
of
index ppininG.
G.By
ByExercise
Exercise3.2,
3.2,X
X 4 G.
G.As
AsXXisisabelian
abelianbut
but G
G is
is not,
not, X =
=CG(X)
CG(X)
by Exercise 2.4. So yy EE G
G-X
X acts
actsnontrivially
nontrivially on X.
X. As
As yP
yp E X, y induces
an automorphism
automorphism of X of order p. By
By 23.3,
23.3, Aut(X)
Aut(X) has
has aa unique
uniquesubgroup
subgroupof
of
4
order
order p unless
unless pp ==22and
andnn>24,4 ,where
whereAut(X)
Aut(Xjhas
hasthree
threeinvolutions.
involutions.In
In the
the first
first
case by 23.3, xxy
= xz for some z of order p in X. In
Y=
In the
the remaining case p ==22
and Xxy
=x-1z',
1 1oror00and
Y =
X-'zE,where
wheres E==
andzzisisthe
theinvolution
involutionin
in X.
X.
Now if the extension splits we may choose y of order p and
and by
by definition
definition
GModP,,,
G S Mod,. D2,,,
, D2",ororSD2,,.
SD2.. So
Soassume
assumethe
theextension
extensiondoes
does not
not split.
split.Observe
Observe
CX(y)
= (x")
Cx(y) =
( x p ) ifxy
if xY=xz,
=xz,while
whileCX(y)
Cx(y)==(z)
(z)otherwise.
otherwise.Also
AlsoyP
yp EE Cx(y).
Cx(y). As G
does
X, (y, Cx(y))
Cx(y)) does
Cx(y), so, as (y, Cx(y))
does not split over X,
does not split over Cx(y),
CX(Y))is
is
abelian,
Thus Cx(y)
Cx(y) =
yP =
= xP
xy =
= xz
abelian, it is cyclic. Thus
=(yP).
(yp).Hence
Hencewe
we may take yp
xP if xY
and
= zz otherwise.
and y2
y2 =
otherwise.
Suppose
= xz. Then z =
= [x,
=
xY =xz.
[x, y]
y] centralizes
centralizes x and y, so, by 8.6, ((yx-1)P
Y X - ' ) ~=
Suppose xy
yy px-pZp(p-1)/2
= -Zp(p-1)/2
zP(P-1)/2
unless p =
~ ~ - ~ z ~ ( ~=
Z1~)( 1~ -21 )while
while
1 2 , zp(p-1)/2
==1 1unless
=2.2.So,
So,as
asG
G does
doesnot
not
x2n_2
2"-2
, (YX
=
~1.
1.
and if n 2
> 4 then, setting ii =
= 2n-3
2i-3 --1,1 (yx`
)2' )=
=2.
2 . Here
Here z =
split, p =
=x
If n =
=33 then
then xy
XY =
=x-1,
X-', which
whichwe
wehandle
handlebelow.
below.
So
x-1zE, and y2
y2=
= z.z. IfIfsE ==0,
= Q2',
SOp ==2,
2 ,xy
XY =
=x-'zE,
0 ,then
then by
by definition
definition G S
Q2., so
so
take sE =
==1,1,soSOthe
=1.
1.Then,
Then, as
as zz cE Z(G),
Z(G),(yx)2
(yx)2==y2xyx
y 2 ~=
=
y zx-1zx
ZX-'ZX
~
theextension
extension
does
does indeed
indeed split.
split.
(23.5) Let
Let G
G be
be aanonabelian
nonabelian p-group
p-groupcontaining
containingaacyclic
cyclicnormal
normalsubgroup
subgroupUU
of
of order p"
pn with
with CG(U)
CG(U)=
=U.
U. Then
Then either
either
p-groups
108
108
(1) G Z Dz.+l,
Q2,,+1,or
Den+I, Q2""+',
or SDZn+l,
SD2n+I, or
or
(2) M
(Z51(U))
=SMod
p»+1and
andEp2
Ep22- S21(M)
char G
G..
(2)
M==Cc
CG
( u l(u))
Modpgt+1
Q1( M )char
G*
Proof. Let
LetG*
G* =
=G/U.
G / U .As
AsUU ==CG(U),
CG(U),
G*==AutG(U)
AutG(U)<(Aut(U).
Aut(U).As
As G
G is nonabelian,
abelian, G*
G* 0#11and
andnn>>2.2.IfIfG*
G*isisof
of order
order pp then
then the
the lemma
lemma holds by 23.4 and
Exercise 8.2, so assume
> p.
assume IG*I
IG*l >
p. Then
Then by
by 23.3
23.3 there
thereexists
existsy*
y* EE G*
G* of
of order
order
with uY
UY =
where U
U ==(u).
(u).Let
LetM
M==(y,
( yU).
, U )By
.By23.4,
23.4,MM- S
Modp.+l
p with
= upn-'+',
up"-'+', where
Modpn+1
and, by Exercise 8.2, E ==S21(M)
Epz.ItItremains
remainstotoshow
show EE char G.
Q 1 ( M )S=Ep2.
G . By
23.3, G* is abelian and either
either G*
G* is
is cyclic,
cyclic, or
or pp ==22and
andthere
thereexists
existsg*
g*EEG*
G*
with ug = u-1. In the first case Q, (G*) = M*, so E = Q1(M) = Q1(G) char G.
with~~=~-'.InthefirstcaseQ~(G*)=M*,soE=Q~(M)=Q~(G)char
In the second Z51(U)
(u2)==([u,
([u, g])
g]) and
and as G* is abelian, G(1)
U.. Hence
Hence
U ' ( U )==(u2)
G(')(< U
G(1)=ZS1(U)
G(1),
char G.
G(')=u'(u)
or U,
U ,and
and in
ineither
eithercase
caseZ51(U)
u l ( U ) char G('), so Z31
U 1 ((U)
U )char
G.
Therefore EE ==S21(CG
(S2'(u)))
(U)))char
char G
G..
Therefore
Q (CG (Q
'
of G
G is
subgroupHH of
A critical subgroup
subgroup of
is aa characteristic
characteristic subgroup
of G
G such
such that
that
1)(H)
< ZZ(H)
> [G,
[G, HI
H] and
= ZZ(H).
that in particular
@ ( H )(
( H )2
and CG(H)
C G ( H )=
( H ) . Observe
Observe that
particular a
subgroup is of class at most
critical subgroup
most 2.
2.
(23.6) Each p-group
p-group possesses
possesses aa critical
critical subgroup.
subgroup.
Proof. Let
characteristic subgroups
subgroups H
H of
of G
Gwith
with@
(D(H)
>
Proof.
LetSSbe
be the
the set of characteristic
( H )i<ZZ(H)
( H )2
H]. Let
Let H be
of S;
S; I claim
H is
[G, HI.
be aa maximal
maximal member
member of
claim H
is aa critical
critical subsubgroup
of G. Assume
not and
and let
let K =
= CG(H),
= ZZ(H),
Assume not
C c ( H ) , ZZ =
( H ) , and define
define XX by
group of
X/Z
ThenK K
X
/ Z ==f S21(Z(G/Z))
i l ( Z ( G / Z ) )f
lfl
K /K/Z.
Z . Then
-& H and
and Z == H fl
fl K,
K , so,
so, as
as K
K a G,
G,
X 0#ZZby
by5.15.
5.15.But
Butnotice
noticeXH
XH EES,
S,contradicting
contradictingthe maximality
maximality of H.
H.
is special
special if t(G)
@ ( G=)=
Z ( G=) =
~ ( ' 1A
.Aspecial
specialp-group
p-groupisissaid
said to
to
A p-group G is
Z(G)
G(1).
be extraspecial
extraspecial if its center
center is cyclic.
cyclic.
(23.7) The center of a special
special p-group
p-group isiselementary
elementary abelian.
abelian.
Proof. Let
e G.
gP Ee @
1(G)
Let G
G be
be special
special and g, h 6
G . Then gp
( G )=
=Z(G),
Z ( G ) ,so,
so,by
by 8.6.1,
8.6.1,
1=
[gp,h]
h]_=[g,
[g,h]p.
h]P.Hence
HenceG(')
G(')isiselementary,
elementary, so, as ZZ(G)
( G )=
=GM,
G('),the
the lemma
1 =[gP,
holds.
(23.8)
Let E be
of G
G with [G,
(23.8) Let
be an
an extraspecial
extraspecial subgroup
subgroup of
[G, E]
El <
( Z(E).
Z ( E ) .Then
Then
G=
=ECG(E).
ECG(E).
Proof. Let
<C=
= CA"~(E)(E/Z),
Let Z =
=(z)
(z) ==Z(E).
Z ( E ) .As
As E/Z
E / Z (< AutG(E)
Autc(E) 5
it suffices
to show
C . Let
Let aaEECC and
and(x;
( x Z:
i Z :11 <5 ii i<nn)) a basis
basis for E/Z.
E/Z. Then
Then
to
show E/Z= C.
Extremal p-groups
Extremal
109
[xi, aa]] =
= z'i
mi << p,
<i<
[xi,
zmifor
forsome
some 00 <(mi
p, and,
and, as
as E =
=(xi:
(xi:11 5
(n)
n ) by 23.1,
23.1, aa is determined
integers (m
E/ZI , and the lemma
(mi:i :1I<
(i <n).
(n).Thus
ThusI (CI
C (<5p"
pn==J (E/Z\,
termined by the integers
holds.
holds.
A p-group
characteristic
p-group isis said
said to
to be
be of
of symplectic
symplectic type if it has no noncyclic
noncyclic characteristic
abelian subgroups.
subgroups.
type then
then G
G=
= E ** R where
(23.9) If G is of symplectic type
(1)
(1) Either
Either EE isisextraspecial
extraspecialor E ==1,1,and
and
(2)
and of
of
(2) Either
Either RR isiscyclic,
cyclic,or
or RR isisdihedral,
dihedral,semidihedral,
semidihedral, or quaternion,
quaternion, and
16.
order at least 16.
critical subgroup
subgroupHH.. Let
Let U
U=
= ZZ(H).
Proof. By 23.6,
23.6, G possesses
possesses a critical
( H ) . By hypothesis
Let ZZ be the subgroup of
of U
U of order p and G* =
= G/Z.
pothesis U is cyclic. Let
G/Z.
H,hneU,so[h,k]p=[hp,k]=1,by8.6.Thus
H(1)
<ZsoH*is
For h, k E H , hP E U , SO [h, k]P = [hp, k] = 1, by 8.6. Thus H(') 5 Z so H* is
abelian. Let
Let K*
K* =
=Q
Qi(H*)
Z(K)*
G,,
1 ( H * )and E*
E* aa complement
complement to Z
( K ) * in K*.
K". KK char G
so Z(K)
Z ( K )isiscyclic.
cyclic.Hence
Henceifif K
K isisabelian
abelianthen
then H*
H* isis cyclic,
cyclic,so,
so, by Exercise
Exercise 2.4,
= H.
holds with
with EE =
= 11 and R
R=
= G or
U=
H . Now
Now 23.5
23.5 applies
applies and says the lemma holds
or
E=G=D8
R=1.
E
= G Z D sororQ8
Qs and
and R=1.
So K is nonabelian
E,, eg =
= eu
nonabelian and then E is extraspecial.
extraspecial. For g Ee G and e E E
for some u E U
U nfl K
K,, and ep Ee Z <
(Z(G),
Z ( G ) ,soeP
so,@' ==eP9
epg ==(eu)P
(eu)p ==enup.
eJ'uP. Hence
Hence
up
= 1,
UP =
1,so
SO that [G,
[ G ,E]
El ==Z.
Z .Therefore,
Therefore,by
by 23.8,
23.8, G =
=EE ** R,
R , where
where R
R=
=CG(E).
CG(E).
Thus H =
H*
= K*
K*=
= E*
E* xx Z(K)*
=EE **CH(E).
C H ( E )Recall
Recall
.
H*isisabelian
abelianand
and S21(H*)
Q l ( H * )=
Z(K)*
with Z(K)
Z ( K )cyclic.
cyclic.We
We conclude
conclude CH(E)*
C H ( E ) *is cyclic and hence CH(E)
C H ( E )is abelian
abelian
by Exercise 2.4.
2.4. Thus
Thus as
as HH =
= ECH
ECH(E),
= U.
= CR(H)
<
( E ) ,CH(E)
CH( E )=
U . Also CR(U)
C R ( U )=
C R ( H )(
Rn
fl H ==U,
R
U ,and
andwe
wemay
may assume
assume R
R 0#U.USo
. SoJ U
IUII >
2 p2 and 23.5 applies to R.
R.
If IU
= p2
p2 then
then 7Z51(U)
Z(G). So,
So, by
by 23.5,
23.5, RR is
is dihedral, quatemion
quaternion or
IUI I=
J1(u=
) =ZZ(<Z(G).
modular of order p3, and in particular R is extraspecial. But then
then G =
= E*R
E* R is
is
also extraspecial,
U I > p2 and
and assume
assume
extraspecial,so
sothe
thelemma
lemmaholds.
holds.Thus
Thuswe
wemay
maytake
takeI IUI
R satisfies
NN==CG
char G and
R
satisfies 23.5.2.
23.5.2. Let
LetMM==CR(Z51(U)).
C R ( u l ( u ) )Then
Then
.
C G(Z51(U))
( U ' ( U ) ) char
and
N ==EM.
EM. By
Z(Q2(N))
By Exercise
Exercise8.2,
8.2,Q2(N)
Q 2 ( N=)=EQ2(M)
E Q 2 ( M=7Lp
)2 Zpx x(E( *E 7Lp2)
* Zp2)SO
SO Z(Q2(N))
is noncyclic,
noncyclic, aa contradiction.
contradiction.
(23.10) Let
Let EE be
be an
an extraspecial
extraspecialp-group,
p-group,Z Z==ZZ(E),
and,!?E==EE/Z.
(23.10)
( E ) , and
IZ.
(1)
field of
ofintegers
integersmodulo
moduloppand
and,!?
t as
(1) Regard Z as the field
as aa vector space over
Define ff :: ,!?
E x k,!?-+
on
ZZ.. Define
+ZZ by
by ff (x,
(2,y)
y)==[x,
[ x y].
,y].Then
Then ff isisaasymplectic
symplectic form on
E, so
(E,
f)
is
a
symplectic
space
over
Z.
so (,!?, f ) is a symplectic space over Z .
(2) m(E)
m(E) ==2n
2n isis even.
even.
(2)
If p
p=2
Z ZbybyQ(x)=x
Then
(3)
(3) If
= 2define
defineQ:
Q :E,!?-+
Q ( 2 )=2.x2.
ThenQQisisa aquadratic
quadraticform
form
on t,!?associated
Q)Q is
associatedto
to f,f so
, so(E,
(,!?,
) isananorthogonal
orthogonalspace
space over
over Z.
Z.
p-groups
p-groups
110
110
~
(4) Let Z
Z(
<U<
(4)
(E.
E . Then
Then U
U isis extraspecial
extraspecial or abelian if and only ifif U is
nondegenerate or totally isotropic, respectively. IfIf pp =
= 2 then U is
is elementary
abelian if and only if U
0 isis totally
totally singular.
singular.
Proof. As
kE
is is
elementary
k as
ProoJ
As ZZ ==c(E),
(P(E),
elementaryabelian,
abelian,sosoby
by12.1
12.1we
we can
can regard E
as a
vector
convention the group
group
in a natural way. Notice that under this convention
vector space
space over
over Z in
operations
k and
8.5.4, [[xy,
= [x,
z] =
=
operationson E
and ZZ are
are written
written additively.
additively. By 8.5.4,
x y ,zz]] =
[ x ,z]Y[y,
zIY[ y ,zl
[x,
z][y,
z],
with
the
latter
equality
holding
as
E
is
of
class
2.
This
says
f
[ x ,z ] [ y ,z], with the latter equality holding as E is of class 2. This says f is
linear in its first variable and a similar
similar argument
argument gives
gives linearity
linearity in the
the second
second
variable. As
As ZZ =
= ZZ(E),
_ [y,
variable.
( E ) , f isisnondegenerate.
nondegenerate. [x,
[x,y] =
[ y ,x]-1,
X I - ' , or,
or, in
in additive
additive
notation, ff (K,
(z, y)
= --ff(y,
notation,
y") =
(y", z).
2 ) .Thus
Thus(1)
(1)holds.
holds.
Notice (1) and
and 19.16
imply (2).
(2). Let
Let pp =2.
=2. By
Notice
19.16 imply
By 8.6,
8.6, (xy)2
( x ~=x2y2[x,
= )X ~
~ ~ ~ y],
y],
[ Xor,
or,
,
additivenotation,
notation,Q(K
Q(. + y")
y) =
= Q(z)
Q(K)+ Q(y)
Q(y")+ ff (i,
( 2y).
, 7).Thus
Thus(3)
(3)holds.
holds. The
The
in additive
(4) is straightforward.
proof of (4)
+
+
+
(23.11) Assume
Assume pp isisodd
oddand
andGGisisofofclass
classatatmost
most2.2.Then
ThenS21(G)
Q 1 ( G )is of exponent p.
ProoJ
and yy be
be elements
elements of G of order p. Then [x,
[ x ,y] =
Proof. Let x and
= z E Z(G).
Z(G). By
8.6.1, zP
zp =
= [x",
= 1.
[xp,y]
y]==1,1,so,
so,by
by 8.6.2,
8.6.2, (xy)p
(xy)P=xpypZp(p-1)12
=xpypzp(p-')I2=
1.
(23.12) Let
Letpp be
be odd
odd and E an extraspecial p-group. Then 01(E)
Q1( E )isisof
of exponent
exponent
most pp in
in E.
E. If Q1
01(E)
where
( E )##EEthen
thenS21(E)
Q1 ( E =)=XXxxEo
EO
whereXX=Z7Lp
Zp
p and index at most
of order pp or extraspecial, and E =
and Eo is of
=El
El **Eo
Eowith
with El
El=ZModp3.
Modp3.
Proof.
01(E).
ProoJ Let
Let YY = Q
1 ( E ) By
. By23.11,
23.11,YYisisofofexponent
exponentp.p.Suppose
SupposeI E
I E: :Y)
Y I== pp..
Then, in
in the
the notation
notationofof23.10,
23.10,ff is aa hyperplane
of E, and
Then,
hyperplane of
and hence
hence of odd
dimension,
spacesare
are of
of even
even dimension,
dimension,Y
P is degenerate.
degenerate.
dimension, so, as all symplectic
symplectic spaces
Let R
I? be a point
a d ( Y )As
.AsfiL
P. Hence,
Hence, by
point in
in ~Rad(Y).
RI is a hyperplane
hyperplane of
of EE,, I?'
Rl ==Y.
23.10.4,
Y
=
CE(R).
As
Y
is
of
exponent
p,
R
=
X
x
Z
for
some
X
23.10.4, Y = C E ( R ) .As Y is of exponent p , R = X Z for some X of
of order
order
Eo be
be aa complement
complement to R
fi in
in Y.
E. By 19.3
19.3 and 23.10.4, Eo is extraspecial
extraspecial
p. Let
Let Eo
Y=
Eo.Let
Let El
El ==CE(EO).
CE(Eo).By 19.3
19.3 and 23.10,
is
or Eo =
=Z
Z;; of course Y
= X x E0.
23.10, El is
extraspecial.
As YY## E,
E , El
El >>01(E1)
Ql(E1)so,
so,by
by23.4,
23.4,E1=
El ZModp3.
Modp3.
extraspecial. As
It remains to show IIEE::YYI
Letuu,
andUU =
= (u,
I (<p.p.Let
, vvEEEEand
(u,v).
v). ItItsuffices
suffices to
to
show IU: Q1(U)I
01(U)I (
< p.
p. If U isis abelian
(E).IfIf U
U is
is
show
abelian this
this holds
holds because
because Z =
=(D
(P(E).
nonabelian appeal to 23.4.
By 23.10 an
an extraspecial
extraspecialp-group
p-groupisisofoforder
orderp1+2n for
for some
some positive
positive integer
integer
p-group of
of exponent
p. If
If pp isisodd,
odd,denote
denoteby
byp1+2n an extraspecial p-group
exponent pp and
.21 +2denotes
order p 1+2n 21+2n
denotes any extraspecial
extraspecial2-group
2-group of
of order
order21+2n;
21+2";by 1.13
1.13
,
Extremal p-groups
111
111
such groups
groups of exponent 2. Write D8
Di Q
there are no such
Q8r for a central product of
n copies
copies of
of D8
Ds with
with m copies
copies of Q8,
Q 8 ,and
and all
all centers
centers identified.
identified.
(23.13)
Letppbe
bean
anodd
oddprime
primeand
and nn aa positive
positive integer. Then up to
(23.13) Let
to isomorphism
isomorphism
there is a unique extraspecial p-group EE of
exponent p.
p. E is
of order
orderp1+2n and exponent
is
the central product of n copies of pl+z
Proof.
By23.10
23.10and
and19.16,
19.16,EEisisaacentral
centralproduct
productof
of nn extraspecial
extraspecialsubgroups
subgroups
Proof. By
E1,
< ii (
< n, of order p3 and center Z =
E
i, 1(
=Z(E).
Z ( E ) Now
. NowExercise
Exercise8.7
8.7 completes
completes
the proof.
(23.14)
Let nn be
be aa positive
positive integer.
integer. Then up to isomorphism
isomorphism D8
D: and
and D8-1
D,"-' Q8
Q8
(23.14) Let
are the unique extraspecial
extraspecial groups of order 22n+1
22"+'. Di
while
D8 has 2-rank n + 11while
D:-'
isomorphic.
D8-1 Q8 has
has 2-rank nn,, so the groups are not isomorphic.
+
Proof. By 23.10
23.10 and
and 21.2,
21.2, E
E isisaacentral
centralproduct
product of
of nnextraspecial
extraspecialgroups
groups
Ei,
n, of order8with
8 with Z(E;)=Z(E)=Z.
Z ( E i )= Z ( E ) =Z .isZ of
is of
order2 2sosoAut(Z)
Aut(Z) =
=1.
1.
E1, 11(<ii 5
<n,oforder
order
Hence 11.2
11.2 says E isis determined
determined up
up to
to isomorphism
isomorphism by the groups Ej.
Ei. Again
Again
by 23.10
and 21.2,
21.2, we
we can
can choose
choose((E,,
hyperbolicfor
fori i<<n
and ((En,
E ,, QQ)) hyperbolic
n and
E n , Q)
Q)
23.10 and
either hyperbolic or definite. By
By 1.13,
is not of exponent 2, so, by
1.13, EZ
Ei is
by 23.4
23.4
EI G
- D8
Ei
D8or
orQ8.
Q 8 By
.By 23.10.4,
23.10.4, (Ei,
( E i Q)
, Q )isishyperbolic
hyperbolicininthe
thefirst
firstcase
case and
and definite
definite
in the second.
2-rank and
second. Finally 23.10 and 21.2 imply the remark about the Zrank
complete
complete the
the proof.
proof.
normal subgroup
subgroupof
of G
G and
and Z
Z=
= S2(A).
(23.15) Let A
A be a maximal abelian normal
Q1(A).
Then
A =CG(A).
(1) A=CA(A).
(2)
(CG(A/Z)
(2) (cG
( A / z )n C(Z)(1)
C(Z)(')5<A.
A.
< CG(A/Z).
(3) If p is
is odd
odd then Qj(CG(Z))
Q l ( C G ( Z ) )5
CG(A/Z).
Proof. Let C
so ifif CC ## AA there
there isis DD/A
of order
order pp in
in
ProoJ
C=
=Cc
CG((A).
A ) . AA5< CCL!<GG,, so
/ A of
Z(G/A)
Z
( G / A ) nnC/A.
C/A.Then
Then D
D <i?GG and
andDDisisabelian
abelianby
byExercise
Exercise2.4,
2.4,contradictcontradicting the maximality
maximality of
of A.
A. AAstraightforward
straightforward calculation
calculation shows
shows (CG(A/Z)
(CG(A/z>n
C(Z))('
(2). Let
Let p be odd, x of
C
( Z ) ) ( '<)5C(A),
C ( A ) ,so
so (1)
(1) implies
implies (2).
of order
order p in
in CG(Z)
CG(Z)
and X =
(x,
A).
Let
Y
=
(x,
CA
((X,
Z)/Z)).
Then
Y
is
of
class
at
most
2, so,
= ( x , A ) . Let Y = ( x , C A ( ( x Z
, )/Z)).
Y
by 23.11,
= 01(Y)
= (x,
23.1 1, W =
Q 1 ( Y )is of exponent
exponent p. Thus
Thus W
W=
( x ,Z).
Z ) .But
But W
W char
char YY so
so
Nxx ((Y)
<N
Nx
=Y
Y,, so
so Y
Y=
= X and (3) holds.
N
Y)(
x ((W)
W )=
(23.16) Let p be
be an
an odd
oddprime
prime and
and ZZ aamaximal
maximalelementary
elementary abelian
abelian normal
normal
subgroup of G.
G . Then Z ==01(CG
Q 1 ( C G(Z)).
(Z)).
p-groups
112
112
Proof. Let
I'll show
show X
X is of exponent
exponent p.
p. Hence if X # Z
Let X
X ==S21(CG(Z)).
Q1(CG(Z)).
I'll
then there
thereisisDD/Z
order pp in Z(G/Z)
then
/ Z of order
Z ( G / Z )nnX/Z
X / Zand,
and,by
byExercise
Exercise 2.4,
2.4, D is
of ZZ..
elementary abelian, contradicting the maximality of
of G containing
Let A
A be aa maximal
maximal abelian
abelian normal subgroup of
containing Z.
Z . Then
Then
Z ==S21(A)
by
maximality
of
Z.
By
23.15.3,
[X,
A]
<
Z,
so
by
23.15.2,
Q 1 ( A )by maximality of Z . By 23.15.3, [ X , A] 5 Z , so by 23.15.2,
XM 5
< A.
A. Choose
Choose U < X of minimal
subject to
to U =
=Q
Q,1(U)
and U not
x(')
minimal order subject
( U )and
exponent p.
p. Then
Then there
there exist
existxx and
and yy in
in U
U of
of order
order pp with
with xy
xy not of
of order
of exponent
p. By minimality
of U
U,, U
U=
= ((x,
By 7.2,
7.2, VV =
= (xU)
0 U,
x ,y). By
( x u )#
U ,so
soVV isisof
of expoexpominimality of
nent p. Hence [[x,
y]
E
V
is
of
order
at
most
p,
so,
as
X"
<
A,
[x,
y]
E
Z.
x , E V is of order at most p, so, as x(')( A , [ x ,y] E As
XI
<.C(Z),
C(Z)U
,Uisisof
of exponent
exponent pp by
by23.11,
23.1 1,contrary
contraryto
to the
the choice
choice of
of U.
U.
(23.17) Let
Let pp be
(23.17)
be an
an odd
odd prime
prime and
and assume
assume G
G contains
contains no
no normal
normal abelian
abelian
of rank
rank 3. Then
Then G is of p-rank at
at most
most 2.
subgroup of
Proof.
By Exercise
8.4 we
we may
= ZZ 9
aG
G.. Let
Proof. By
Exercise 8.4
may assume
assume Ep2
Epz Z
Let HH ==CG(Z)
CG(Z)
and Ep3
= AA5< GG.. Then
Then ]IA:
HI5<pp and
and hence
hence m
m((A
nH
H)Z)
> 3. Thus
and
Ep3 S
A :AAnnHI
( ( An
) Z )2
m(H)
m ( H )>23.3.However
However by
by hypothesis
hypothesis ZZ isis aa maximal
maximal elementary
elementaryabelian
abelian normal
normal
of G
G,, so Z =
subgroup of
=c21(H)
Q 1 ( H )by 23.16.
23.16.
24 Coprime
p-groups
Coprime action
action on p-groups
In this section p is a prime, G is a p-group, and
and A
A is a p'-group of
of automorautomorphisms of G,
G , unless
unless the conditions
conditions are
are explicitly relaxed as in the Thompson
AxB
A
B Lemma.
Lemma.
(24.1) A
(D (G).
A isisfaithful
faithfulon
onGI
G/@(G).
Proof. Suppose
(D (G). We wish
wish to
to show
show bb=
= 1. If
If not
not there
there
Proof.
SupposebbcEAAcentralizes
centralizesGI
G/@(G).
is a prime q and a nontrivial power of b which is a q-element and centralizes
G/(D(G),
loss bb isis aaq-element.
q-element.Let
LetBB=
= (b)
(b) and
and gg Ec G. Then B acts
G / @ ( G )so
, without loss
on the coset X =g@(G).
=g(D(G). By
m is the number of
By 5.14,
5.14, m
m ==JXJmod
IXlmod q, where rn
fixed
points of B on X, and, as X j = J(D(G) l is a power of p, X i 0 Omod
0 mod q,so
q, so
fixedpointsofBonX,and,as/XI=~@(G)~isapowerofp,~X~
B centralizes
centralizes some
some x EE X. Hence
Hence BB centralizes
centralizesaaset
setYY of
of coset
coset representatives
representatives
for @
c(G)
= 1,
( Gin
)inG,
G ,so,
so,by
by 23.1.2,
23.1.2, G
G ==(Y)
( Y )<5CG(B).
CG(B).Hence B =
1, completing the
proof.
+
(24.2) (Thompson
(Thompson A xx BBLemma).
Lemma).Let
LetAB
AB be
be aa finite
finite group
group represented as a
of automorphisms
automorphismsof
ofaap-group
p-group G
G,, with
with [[A,
B]=
= 11=
group of
A ,B]
=[A,
[ A ,CG(B)],
C G ( B ) ]B, a
= OP(A).
G] =
= 1.
p-group, and A =
Op(A).Then [A,
[ A ,GI
1.
Proof. Form
Formthe
thesemidirect
semidirectproduct
productHHofofGGbybyAB
ABand
andidentify
identifyAB
AB and
and G
G with
with
Proof.
subgroups
of H.
H. We
We may
may assume
assume [[A,
G]#0 11 so,
so, as
as AA=
=OP(A),
OP(A),[[X,
G]#0 11
A ,GI
X , GI
subgroups of
Coprime action on p-groups
113
pl-subgroup X of
of A,
A, and
and replacing A
A by
for some p'-subgroup
by X we may assume A
A is a
p'-group.
pf-group.
G9
GBA with
<H
H=
= GBA
with AA5<NH(B),
NH(B),so
so GB
GBisis aa normal
normal p-subgroup
p-subgroup of
of H,
H,
by GB,
GB, we
we may
may assume
assumeBB<(G.
G.Then
ThenBB<(QQ==
CG(A),
so
and, replacing G by
CG
(A), so
CG(Q)
< Q by hypothesis, so Cc(Q)
CG(Q) (
< Q.
CG(Q) <
(CG(B).
CG(B). Also CG(B)
CG(B) (
Exercise2.2,
2.2,QQisisproperly
properlycontained
contained in
in
As [A, G
G]I 0#1,1,QQ0 #G,G,
so,so,
bybyExercise
NG(Q).
of Q, [A, NG(Q)] 0# 1,1,and
NG(Q). So,
SO,by definition of
andhence
hencewe
wemay
may assume
assume
Q
Q g<G.
G.
Let G*
= G/Q. As A
A is aa pl-group,
p'-group, CG*(A)
CG(A) =
= CG(A)* = Q* =
G* =
1, by 18.7.4.
18.7.4.
= 1,
Thus [G,
[G, A]
A] -$ Q.
Q.
As Q<
=1,1,so
=1.1.Hence
Q 9G,
G,[G,
[G, Q,
Q,A]
A] <([Q,
[Q, A]
A] =
so[G,
[G, Q,
Q, A]
A] =
=[Q,
[Q, A,
A, G]
GI =
Hence
Thus[A,
[A, G]
G I<(CG
CG(Q)
by theThree-SubgroupLemma,
the Three-Subgroup Lemma, 8.7,
8.7, [A,
[A, G,
G, Q] =
=1.1.Thus
(Q) (
< Q,
by paragraph 2,
2, contrary
contraryto
to the
the last
last paragraph.
paragraph.
(24.3)
(24.3) IfIf GGisisabelian
abelianthen
thenAAisisfaithful
faithfulon
onc21(G).
Ql(G).
Proof. Without
S21(G).Let
LetXXbe
beofof order
orderpp in G and
Without loss, A
A centralizes
centralizes Q1(G).
and
G* =
=G/X.
G/X. By
By Exercise
Exercise3.1,
3.1, A
A is
is faithful
faithful on
on G*,
G*, so,
so, by induction on the order
G,
A
is
faithful
on
Q1(G*),
and
hence
without
loss G*
G* =
=Q,
Q1(G*).
of G, A is faithful on Q, (G*),
loss
(G*). Now, by
12.1
Exercise 4.1.1,
4.1.1, we may take CG*
CG.(A)
(A)=
=1.
(A),
1.Thus,
Thus, by 18.7.4,
18.7.4,X ==CG
Cc(A),
12.1 and Exercise
so X
Hence, as
as G
G is abelian, 1.11 implies G is
X=
=S21(G).
Q1(G). Hence,
is cyclic.
cyclic. Now
Now 23.3
23.3
supplies aa contradiction.
contradiction.
supplies
!
(24.4) G =
=[G,
[G, A]CG(A).
A]CG(A).
(24.4)
Proof.
ByBy
23.2,
Proof. Let
LetG*
G* =G/4(G).
= G/Q(G).
23.2,G*G*is isananelementary
elementaryabelian
abelianp-group,
p-group,so,
so,
by Exercise
Exercise 4.1,
4.1, G*
G*==[G*,
[G*, A]
A] X
x CG.(A).
CG*(A). By 8.5.3,
8.5.3, [G*,
[G*, A]
A] =
=[G,
[G, A]*
A]* and,
and,
by 18.7.4,
(G)), so, by 23.1,
CG*(A) ==CG(A)*.
CG(A)*. Hence
Hence G
G ==([G,
([G,A],
A],CG(A),
CG(A),(D
Q(G)),
18.7.4,CG.(A)
G=
Finally, by
by 8.5.6,
8.5.6, [G, A]
Al 9
<G,
=([G,
([G, A],
A], CG(A)). Finally,
G, so
so([G,
([G,A],
A],CG(A))
CG(A)) _
=
[G,
[G, A]CG(A)
A]CG(A) by 1.7.2.
1.7.2.
(24.5)
= [G, A,
A, A].
Al
(24.5) [G,
[G, A]
A] =
Proof.
[G,
AlA]:<9H.H.
Thus
(A) acts
Proof.Let
LetHH==
[G,A].
A].By
By8.5.6,
8.5.6,HH<9GGand
and[H,
[H,
ThusCG
CG(A)
acts on
[H,
so [H, A]
(A) = G.
Next H
(A) so [G, A]
[H, A],
A],so[H,
A] <9HCG
HCG(A)=
G.Next
H==[H,
[H,A]CH
A]CH(A)so[G,
A] <([H,
[H,A]
A]
by 8.5. But of course [H, A]
A] (
< [G, A]
H<
A] as H
(G.
G.
(24.6)
(A).
(24.6) If
If GGisisabelian
abelianthen
thenGG==[G,
[G,A]
A]XxCG
Cc(A).
p-groups
114
114
Proof. Let
Let GGbe
bea aminimal
minimalcounterexample
counterexampleand
andXX==[S21(G),
[Q1(G),A].
A]. By
By 24.3,
24.3,
X#
0 1 and, by 12.1
Cx(A)
= 1.
of G,
X
12.1 and
and Exercise
Exercise 4.1.1,
4.1.1, Cx
(A) =
1. By minimality
minimality of
C([G,AI/x)(A)==1,
so CIG,Al(A)
C[G,AI(A)==1.
Now 24.4 completes
completes the proof.
C([G,Allx)(A)
1, so
1. Now
proof.
(24.7)
=[G,
[G,A]
A] and
and A
A centralizes
centralizesevery
every characteristic
characteristic abelian subgroup
(24.7) If G =
of G,
then
G
is
special
and
Z(G)
=
CG
(A).
G, then G is special and Z(G) = CG(A).
Proof. As
As A
Acentralizes
centralizes each
each characteristic
characteristic abelian subgroup of G, so
so does
does
= [G, A]
A] by
by Exercise
Exercise 3.6.
3.6. Thus
Thus Z =
= Z(G) is the
G=
the unique
unique maximal
maximal characteristic abelian subgroup. [Z2(G),
= 1,
[Z2(G),G, G]
GI =
1, so,
so, by
by the
theThree-Subgroup
Three-Subgroup
Lemma, Z2(G) centralizes GM.
Z2(G)
GM is abelian, and there~ ( ' 1Hence
.
Z2(G) f1
f l G(')
fore contained
containedin
in Z,
Z, so
so G(')
Gf' <(Z.
= (Z/Gf1)
Z.By
By24.6,
24.6,G/G1>
GIG(') =
(z/G(')) xx [GIG('),
[GIG('),A]
A]
so, as G
= GO).
Finally suppose
supposeGG has
has exponent
exponent pn
p" >
> p.
p. Let
G ==[G,
[G, A], Z =
G('). Finally
1,
so 73'
1 (G) is abelian and
g, h E G. By 8.6, [gp"-'
V_'] _=[gp°,
= 1, SO
[gp"-',, hpn-I]
tgp",hp"-Z]
hpn-'1 =
W-'(G)
?5n-1(G)
Z.But
Butthen
thenG/Zis
G/Z isofofexponent
exponentp.p.So
SoZZ==@(G).
I(G).
hence U
""(G)
5<Z.
(24.8) If
If ppisisodd
oddthen
thenAAisisfaithful
faithfulon
onS21(G).
Q1(G).
Proof.
Proof. Choose
ChooseGGtotobe
bea aminimal
minimalcounterexample
counterexampleand
andlet
letaaEEA#
A' centralize
centralize
S21(G).
By24.5
24.5 and
and minimality
minimality of
of G, G =
= [G, a].
a]. By 24.3,
Q1(G). By
24.3, aa centralizes
centralizes
each characteristic
abeliansubgroup
subgroupofofG,
G, so,
so, by
by 24.7,
24.7, G is special
special with
with
characteristic abelian
Z=
Z(G)
=
CG(a).
By
23.7,
Z
=
S21(G).
Let
g
E
G
Z,
z
=
gp
and
=
=
=
By 23.7, = Q1(G). Let g E G z = gp and vv =
g-°]. Then
z, vv EE ZZ ==S21(G),
=1.
Z=
= CG(a),
CG (a), (g-')P
(g-Q)p=
=
[g, gVa].
Thenz,
Ql(G), so vvp
p=
1.Notice
Notice that, as Z
gg_a
vp(p-1)/2
and gg-a =
hp =
= ~ zzz-1
z-', Z-1,and
=hh 04 ZZby
by 18.7.4.
18.7.4.Now,
Now, by 8.6, hP
-'vp(p-')/~ =
=1,
1,contracontradicting
dictingZZ==SZ
Q1(G).
(G).
'
(24.9)
(24.9) Let H be
be aa critical
critical subgroup
subgroup of G. Then
(1)
on
H.
A
is
faithful
on
H.
(1)
is odd
odd then
then A
A is
is faithful
faithfulon
on S21(H),
Ql(H), and there exists a critical sub(2)
(2) If p is
group H
1(H)).
group
H of
of GGsuch
suchthat
thatSZ1(H)
Q1(H)contains
containseach
eachelement
elementof
of order
orderpin
p inCG(SZ
CG(Q1
(H)).
Proof.
Proof. By
By definition
definition of H, CG(H)
CG(H)<5 H,
H,so
soby
by the
the Thompson
Thompson A x BB Lemma
Lemma
(H) ==1.1.Thus
(applied to `A'
'A' =
=CA(H)
CA(H)and
and `B'
'B' ==H),
H),CA
CA(H)
Thus(1)
(1) holds.
holds. Part
Part (1)
(1)
prove the
the second, choose H with
and 24.8 imply the first statement in (2). To prove
with
L=
Y=
= Q~(CG(L))
01(CG (L)) I
< L. Assume
=S21(H)
Ql(H) maximal. It suffices to show Y
Assume not and
of Y.
let V
V be
be aamaximal
maximal elementary
elementary abelian
abelian normal
normal subgroup
subgroup of
Y. By 23.16,
=
V=
= S21(Cy(V)),
so, as
V
Q1(CY(V)),SO,
as Y
Y -$L,
L, V
V $ L. Thus vVf1
f l Z2(Y)
Zz(Y) $ L, so
so S21(Z2(Y))
Q1(Z2(Y))=
whereX/Z(L)==Z(G/Z(L))
Z(G/Z(L))l
K $ L. By 23.1
23.11,
exponent p, so X $ L, whereX/Z(L)
1, K is of exponent
fl
S so, by
by the proof
proof
(K/Z(L)). Now
Now define
define SS as
as in
in the
the proof
proof of
of 23.6.
23.6. Then
Then XL
XL E S
XL
SZQI1(C),
criticalsubgroup
subgroupCCofofG.G.But
ButL L< <
XL<(
(C),
of 23.6, XL is contained in aa critical
contradicting the
the choice
choice of
of H.
Coprime action
action on
on p-groups
p-groups
Coprime
115
Remarks.
Remarks. The
Thediscussion
discussionofofp-groups
p-groupsininthis
thischapter
chapterisisessentially
essentiallythe
thesame
same
as
asGorenstein's
Gorenstein'streatment
treatmentof
ofp-groups
p-groups[Gor
[Gor4],
41,which
whichwas
wasinfluenced
influencedininturn
turn
by lecture
lecturenotes
notesof
ofPhillip
PhillipHall.
Hall.
by
P.
P. Hall
Hall originally
originallyclassified
classified the
the p-groups
p-groupsof
ofsymplectic
symplectictype.
type.The
Thenotion
notionof
ofaa
`critical
AxB
'critical subgroup'
subgroup' isisdue
dueto
to J.J. Thompson
Thompson as
as is of course
course the Thompson A
B
Lemma.
Lemma.
Almost
Almostall
allof
ofthe
thematerial
materialininthis
thischapter
chapterisisbasic
basicand
andbelongs
belongsin
inthe
therepertoire
repertoire
of
of any
any finite
finitegroup
grouptheorist.
theorist. For
For the
thesimple
simplegroup
grouptheorist
theoristititrepresents
representsan
an
important
important part of
of the
the foundation
foundation of
of the
the local
localgroup
grouptheory
theory involved
involved in
in the
the
classification.
classification.For
For example
example the importance of p-groups of
of symplectic
symplectic type
type is
is
reflected
reflectedin
in the
thesecond
secondcase
caseof
of Theorem
Theorem48.3.
48.3. More
Moregenerally
generally the
the results
resultsof
of this
this
chapter
chapterwill
willbe
be used
used repeatedly
repeatedly in
in chapters
chapters10
10through
through 16.
16.
Exercises
Exercises for chapter
chapter88
1.
1.Let
Letqqbe
beaaprime
primeand
andAA an
anelementary
elementaryabelian
abelianq-group
q-groupacting
actingon
onaa q'-group
q'-group
G.
=q).
G. Prove
ProveGG==(CG(B):
(CG(B):IA:
!A:BI =
q ) .(Hint:
(Hint:Use
Use18.7
18.7totoreduce
reducetotothe
thecase
case
G
G aap-group.
p-group.Then
Thenuse
useExercise
Exercise4.1
4.1 and
and 23.1.)
23.1 .)
2.
(x)<I4G,
G,yyof
oforder
orderpp in
in G
G- X,
2. Let
LetGG=SModel.
Mod,. ,,n >23,
3, with
with 7Lp"-1
Z,.-I S=XX==(x)
X,
and xy
X Y =xp-1.
= X P - ~ .Prove
Prove
and
(1)
withZ(G)
Z(G)=_@(G)
t(G) =
-l Zpn-2.
pf-2.
=(xP)
(xp) Z
(1) GGisisof
of class
class 22 with
(2)G(1)
G("==(xpn
(xpn-')
Z Z,.
(2)
2) =7L
p.
(3)
= (xp°
m,y)
y)Z-7Lpm
7Lpfor
for.p
- 1,
(3) SZm(G)
Q,(G) =
(XP"-~,
Zpm xx Zp
&
<
I<mmc<n
n1,unless
unless p°
pn==8.8.
3.
, n > 3, orOSDzn
n >~ 4.~ Let
3. Let
LetGG=2Dzn
D 2,"n, n> >2,2Qzn
, Q2.,n23,
~ S,D
, ~ 7L2--1
> ~ . L ~X~=Z(x)
~ 4~ G
-LZX
and
y
(=G
X
with
y
an
involution
if
G
is
dihedral
or
semidihedral
and
and y E G an
is dihedral or semidihedral and
of order
order44ifif GGisisquaternion.
quaternion.Prove
Prove
yy of
(1) GO)
G(')==(D(G)
@(G)==(x2)
(x2)
S Z2n-2.
(1)
=7L2
-2.
(2)
-2) isisofoforder
(2) Either
EitherGGisisdihedral
dihedralofoforder
order44ororZ(G)
Z(G)==(x2
(x2'-')
order2.2.
(3)
G
classnn (3) G isisof
of class
- 1.1.
(4)
(4) XXisisthe
theunique
uniquecyclic
cyclicsubgroup
subgroupof
of G
Gof
of index
index p,
p,unless
unlessGGisisdihedral
dihedral
of
of order
order44 or
orquaternion
quaternionof
of order
order8.8.
(5)
(5) GG--XXisisthe
theunion
unionofoftwo
twoconjugacy
conjugacyclasses
classesof
of G
Gwith
with representatives
representatives
yy and
yx.
Each
member
of
G
X
is
an
involution
if
dihedral,
and yx. Each member G - X is an involution ifGGisisdihedral,
each
each is
is of
of order
order 44 if G
G isis quaternion,
quaternion, while
while if G
G isis semidihedral
semidihedral then
isof
of order
order22 and
and xy
xy of
of order
order4.4.
yy is
(6) GG has
from X.
X. If G is
hastwo
twomaximal
maximal subgroups
subgroups distinct from
is dihedral
dihedral of
order
at
least
8,
both
are
dihedral.
If
G
is
quaternion
of
order
order at least 8, both are dihedral. G is quaternion of order at
at least
least
16,
16, both are quaternion.
quaternion. If G is
is semidihedral
semidihedral then one is dihedral
dihedral and
the other
otherquaternion.
quaternion.
the
(7) Quaternion
Quaterniongroups
groupshave
haveaaunique
uniqueinvolution.
involution.
4.
4. Let
LetGGbe
beaap-group
p-groupwith
withno
nononcyclic
noncyclicnormal
normalabelian
abeliansubgroups.
subgroups. Prove
Prove G
G
is cyclic,
I GII >>8.8.
cyclic,quaternion,
quaternion,semidihedral,
semidihedral,orordihedral,
dihedral,and
andininthe
thelast
lastcase
caseI G
116
116
p-groups
of order
order p, prove
prove H is cyclic or
If H is
is aa p-group
p-group with
with just one subgroup of
quaternion.
quaternion.
5. Let
Let G
G be
be aanonabelian
nonabelian p-group
p-group of
of symplectic
symplectic type
type and
and exponent
exponent p or
or 4.
4.
Set Z =
= Z(G), G,
G, ==G/Z,
G/Z,AA==Aut(G),
Aut(G),and
andA*
A* ==Out(G).
Out(G). Prove
(1) Inn(G)
Inn(G)==CA((3).
cA(G).
(2) CA(Z)*
= Sp(O)
CA(Z)*=
sp(G) and
and A*
A* is the group of all similarities of some symplectic form on G if p isis odd.
odd.
(3) If
Di-1
and
OZ
(2), OZn(2),
If pp ==22 then
then G
G -ED',
Dn,
D"-'Q,Q,ororZ4Z4**D',
Dn,
andA*
A*- E
0L(2),
0,(2),
or SP2n(2),
respectively.
Sp2,(2), respectively.
containingan
aninvolution
involutionx xwith
withCG(x)
CG(x)E-E4.
E. Then G
6. Let G
G be aa 2-group containing
is dihedral or semidihedral.
semidihedral.
Letpp be
bean
anodd
oddprime.
prime. Prove
Prove
7. Let
p3
(1) Up
Up to
to isomorphism
isomorphism there
there is a unique extraspecial group E of order p3
and exponent p
p..
(2) A u ~ A ~ ~ ( E ) ( ZE( Aut(Z(E)).
E))
of nn copies of
of E
E
(3) Up
Up to
to isomorphism
isomorphism there is a unique central product of
with identified centers.
centers.
8. Let
LetAAbe
beaa7r'-group
nl-group acting
acting on
on aa 7r-group
n-group G. Prove
(1) G
G ==[G,
[G,A]CG(A),
A]CG(A),and
(2) [G,
[G,A]
A1 _=[G,
[G,A,
A, A].
A].
9. Let
Let rr be
beaaprime,
prime,AAan
anelementary
elementaryabelian
abelianr-group
r-group acting
actingon
onaasolvable
solvable
r'-group
rl-group G,
G, D
D <5 A,
A, and
and BB aanoncyclic
noncyclic subgroup
subgroup of A.
A. Prove
Prove [G,
[G, D]
Dl =
([CG(b),
D]:
b
E
B#).
([CG(b), Dl:
B') .
10. Let A
A be aa p'-group
p'-group with
with aa unique
unique minimal
minimal normal subgroup B, assume
faithfully on
on aa nontrivial
nontrivial p-group
p-group PP,, and assume A is faithful on
A acts faithfully
no proper subgroup of
of P
P.. Prove
Prove that either
(1) PPisiselementary
on PP,, or
elementaryabelian
abelian and
and A
A is irreducible on
(2) PP ==[P,
[P,B]
B]isisspecial,
special,[B,
[B,Z(P)]
Z(P)]==1,1,and
andAAisisirreducible
irreducible on P/Z(P).
If [A, Z(P)]
Z(P)]==11and
andAP
APpossesses
possessesaa faithful
faithful irreducible
irreducible representation
representation
field, then
then P is extraspecial.
over some field,
extraspecial.
11. Let p be
3. Prove
Prove G
G has
has aa
11.
be an
an odd
odd prime and G a p-group
p-group with m(G) > 3.
normal
abelian subgroup
subgroupof
of p-rank
p-rank at
at least
least 4.
4. (Hint:
(Hint: Let
Let G be a councounnormal abelian
terexample,
V an elementary
normal subgroup
subgroup of
of G of
terexample, V
elementary abelian
abelian normal
of maxmaximal rank, HH==CG(V),
V =V
S21(H)
EPEps,
3,
CG(V),and
andEp4
EP4- A
E <AG.
( Show
G. Show
= a l (_2-H) E
m(Af H)
(A) and
m(An
H )==2,2,and
andAAisisthe
theunique
uniqueEp4-subgroup
Ep4-subgroupofAV.
ofAV. Let K ==NG
NG(A)
g E NG(K)
NG(K) --K.
K.Show
ShowAA9
A A g is
isof
ofclass
classatatmost
most22and
andAA9
A A g <AV.)
(AV.)
< i <5 n)
12. Let
Let G
G be
be aa p-group
p-group and
and(Gi
(Gi:: 11 5
n) aa family
family of
of subgroups
subgroups of G
G which
which
generates G. Then, for any
n) of
of elements of
of G, G is
any family
family (xi
(xi:: 115<ii 5< n)
is
generated
by
((G):
1
<
i
<
n).
((Gi)X1
:
1
5
i
5
generated by
9
field of
of aa linear
linear representation
Change of field
E a subfield
subfieldof
of FF,, and
and K
K an
Let njr:: G -->
+ GL(V,
GL(V, F)
F)be
be an
an FG-representation, E
extension field
fieldof
ofFF.. Then
Then V is
is also
also aa vector
vectorspace
spaceover
overEEwith
withGE(V,
GL(V,F)
F) 5<
extension
GL(V, E), so
so 7r
n also
also defines
defines an
an EG-representation.
EG-representation. Further, by aa tensoring
tensoring
process discussed in section
inducesaaKG-representation
KG-representation 7rK
n Kon a KKsection 25, nninduces
space VK.
VK. This chapter
chapter investigates
investigates the relationship among these representations. ItIt will often
often be
be very
very useful
useful to
to extend
extend F
F to K
tations.
K by
by passing
passing from
from 7r
n to
to
example several
K . For example
several results
results at the end of chapter 99 are
are established
established in this
this
n7r K.
way.
said to
to be
be absolutely
absolutely irreducible
irreducible ifif 7r
irreducible for
for each
each extension
extension
n Kisisirreducible
nTrisissaid
K of
saidtotobebea asplitting
splittingfield
jieldfor
forGGififevery
everyirreducible
irreducible FGFGof F,
F ,and
andFFisissaid
representation isis absolutely
absolutely irreducible.
irreducible.ItIt develops
developsin
in section
section 25
25 that
that n
7r is
is
representation
precisely when
when F
F=
27 that
that ifif
absolutely irreducible precisely
=EndFG(V)
EndFc(V) and in section 27
G is finite then a splitting
splitting field is obtained
obtained by adjoining
adjoining a suitable
suitable root
root of unity to
F.
section 27
F.It's
It's particularly
particularlynice
nicetotowork
workover
overaa splitting
splittingfield.
field. For example
example in section
shown that,
that, over
over aasplitting
splittingfield,
field,the
theirredycible
irreduciblerepresentations
representations of
of the
thedirect
direct
it is shown
product
product of groups
groups are
are just the tensor products
products of irreducible
irreduciblerepresentations
representationsof
of
the factors.
over finite
finite fields,
fields, where
where change
change of
of
Section 26 investigates representations over
field goes very smoothly. Lemma 26.6 summarizes many of the relationships
relationships
involved.
Section 27 introduces the minimal
minimal polynomial
polynomial of
of aa linear
linear transformation.
transformation.
and unipotent
unipotent elements
elements are
are discussed
discussedand
andititisisshown
shownthat
thatififFF is
Semisimple and
GL(V) admits
admits aa Jordan
Jordan decomposition;
decomposition;that
that is
is
perfect then each member g of GL(V)
g can be written
written uniquely as
as the commuting
commuting product of aa semisimple
semisimpleelement
element
unipotent element.
and a unipotent
element.
25 Tensor products
25
In this section G is aa group,
group, FFisisaafield,
field,and
andVVaafinite
finitedimensional
dimensionalvector
vector
F.
space over F.
(V1:005< ii (
< m)
Let (Vi:
m) be
be vector
vector spaces
spaces over F,
F,and
and denote
denoteby
by L(V1,
L(Vl, ...
. . ., ,Vm;
Vm;
x
V.
-,
Vo
such
that
for
each
i,
1
Vo)
the
set
of
all
maps
a:
Vl
x
Vo) the set of
a : Vl . . - Vm -+ Vo such that for each i, 1 <
5
+ Vo
Vo defined by
i <I
m,m,and
.
andeach
eachchoice
choiceofofVv jj E Vj, j # i,
i , the
the map
mapB8:: Vi --+
v18
= (v1.....
vie =
(vl, . . .vm)a
, vm)aisisananF-linear
F-lineartransformation.
transformation.Such
Suchmaps
maps are
are called
called
118
118
Change of
field ofof aa linear
linear representation
ofJield
m-linear. L(Vl,
L(VI, ....
Vo)isisaavector
vectorspace
spaceunder
under F
F via
. .,, Vm;
V,; Vo)
v(a+0)=va+v0
v(act) = a(vu)
a, P E L(Vi, ..., V.; Vo).
A tensorproduct
tensorproduct of
F-space TT together
7r EE L(V1,
L( Vl ,....
of V1,
Vl , ....,
. .Vm
, V, is
is an F-space
together with n
. . ,,
Vm;
T)with
with the
the following
following universal
universal property:
property: whenever
whenever U
U is an F-space
V,; T)
F-space and
and
aa EE L(Vi,
exists aaunique
unique/33EEHomF(T,
HomF(T,U)
U)with
withn/
n$
= a.
a.
L(V1,...
. , Vm;
, V,; U), there exists
3=
..
(25.1) Tensor products exist
exist and are
are unique up to
to isomorphism.
isomorphism.
-(25.1)
Prooj See
Seefor
forexample
examplepage
page408
408 in
in Lang
Lang [La].
[La].
Proof.
Because of 25.1 there
there is
is aa unique
unique tensor
tensor product
product of
of V1,
Vl , ....,
. .Vm
, V, which
which isisdedeBecause
®V
Vm
V.Write
Writevlv1@®. ®
noted by
- @vm
v, for
for the
the image
image of
by Vl
Vl ®
@ ... . @
, oror ®m 1&.
under the
the map
map (denoted
(denotedby
by nr above)
(vl, ...,
.. . ,vm)
v,) under
above) associated
associated to the tensor
tensor
(vi,
product.
Vm,
fundamental tensors.
product. The
Theelements
elementsvlvl®@.... .®. @
v, V1
, vi EE V1.
Vi, are calledfundamental
tensors.
It is easy to verify
verify from
from the
the universal
universal property
property that:
that:
-
n
generated as
as an
an F-space
F-space by the fundamental
Vl ®
@ .. -. .. ®
@ Vm
V, isisgenerated
fundamentaltensors.
tensors.
(25.2) Vl
(25.2)
some elementary
elementary properties
properties of the tensor
tensor product;
product; they can be found
Here are some
example in Lang, Chapter
Chapter 16.
16.
for example
(&: 11 5< i <5 m)
m) be
be F-spaces.
F-spaces. Then
Then
(25.3) Let (Vi:
(1)
V1.
(1)V1
v l®@V2
v 2=~V2
v 2®@
Vl.
(2) (Vl
@ V2) @ v3 Vl @ (V2 @ V3).
(2)
(Vi®V2)®V3V1®(V20V3)
V)for
forany
anydirect
directsum
sum ®UEIUU of
of FF(e,,, U) ®
@V
V % (DUEI(U
@,,,(U @®V)
(3) ((DUEI
spaces.
spaces.
(4) Let
2. Then
Xi be
be aa basis
basis of Vi,
Vi, i =
=1,1,2.
Then
Let Xi
e,,,
X
®X2
1 ®x2:
XI1 @
X2 =={X{XI
8 x 2Xi:EE~X1
Xi}
~}
basis for
forVl
Vl®@V2.
V2.
is aa basis
(5) Let al
Thenthere
there
exists
a unique
map
al ®
ai EEndF(V1).
E EndF(Vi). Then
exists
a unique
map
a1 @
. - .®@an
a, EE
EndF(Vl
®...(9
Vm)With (VI
®...®vm)(a1(9 ...®am)==vial
vial @...@v,a,
®...®vmam..
End~(V1@...
@V,)with(vl
@..-@v,)(al@...@a,)
(6) Forvl,u1
Forvi,ui E Q , i = 1,
1,2,anda
E F:
F:
2, and aE
(VI + ul) ® V2 = (Vi (9 V2) + (ul ® V2),
and
and
+
=
+
V l ® (V2
(212 + u2)
u2) = (VI
(211(9
€3 V2)
212) + (v]
(211 €3
u2),
VI
0 u2),
8
a(vl (9 V2) = avl ® V2 = V1 ® avg.
Tensor products
119
If 7r1:
--).GL(Vj),
GL(V; ),1 15<i i < m,
m, are FG-representations,
FG-representations, then by
ni: GG-+
by 25.3.5
25.3.5 there
there
® nm
zr,,, of
of G on
FG-representation7r1
nl 8
on V1
Vl ®
@ .. .. .®
@V,,
Vmdefined
defined by
by
an FG-representation
is an
®..
.. . @
g(nl
(9 n7r,,,)
®g7r,,,,
® ®7rm is the tensor
g(n1 ®.
8.--@
m ) = ggJr1
n@
l ®...
...@
g n m , ffor
o r ggtEGG.
. n lnl@...@n,isthetensor
product
productof
ofthe
therepresentations
representations7r1,
nl,.... .,.7r,,,
, n,. .
A special case of these constructions is of particular interest.
interest. Let
Let K
K be an
extension
field of
of F.
F. Then K is
extension field
is aa vector
vector space over F,
F , so
so the
the tensor
tensor product
K®
@ U can
can be formed for any F-space U. Let X and B be
be bases for U and K
F , respectively.
respectively. By 25.3 each member of K ®
@U
U can
can be
be written
written uniquely
uniquely
over F,
as
ab,x(b@(9 x),withab,x
with ab,xEEF.
F.AS
Asab,x(b
ab,x(b@(9X)
x)=
= ((ab,x)b)
® Xx with
((ab,x)b) @
as F-(b,x)EBxx
C(b,x)EBxX
ab.x(b
(ab,x
K, ititfollows
follows that
that eachmember
each member of K
K8
® U is of the
the form
form ~,,,(c,
Ex Ex (c® @
(ab,,)b)bEEK,
(9 x),
x),
cx
c, E
t K.
K.Indeed
Indeeditit turns
turns out
out K
K®
@U
Ucan
canbe
bemade
made into
into aa vector space K ®
@F U =
=
UK
over K by defining scalar
lJKover
scalar multiplication
multiplicationvia:
a
1:(cx ®x)) _ (acx
(XEX
®x)
a, cx E K, X E X.
XEX
These
These remarks
remarks are
are summarized
summarized in the
the following
following lemma;
lemma; see
see chapter
chapter 16,
16,
section 3 in
in Lang
Lang [La]
[La]for
for example.
example.
and X
X aa basis
basis for
for a vector space U over
LetKKbe
bean
anextension
extensionfield
field of FFand
(25.4) Let
F.
Then UK =
is a vector space over K with 1 @
®X=
® xx:: x E X)
X}
F.ThenuK
=KKOF
m FUUi~avectorspaceoverKwith1
={(11 @
a basis for
for UK.
uK.
I"i
It will be useful to have the following
following well known property
property of this construction,
construction,
419 in
in Lang
Lang [La].
[La].
found on
on page
page 419
which can be found
(25.5)
If L >>K
(25.5) If
K >>FFisisa atower
towerofoffields
fieldsand
and UUan
anF-space
F-spacethen
then LL ®F U
UZ
L ®K
@K(K
@F
U).
(K OF U)
Notice that, for gg Et EndF(U),
® g is the
EndF(U), 11®
@gg EtEndK(UK),
E ~ ~ K ( u where
~ ) , 11 8
the map
map
defined
25.3.5 with
with respect
respect to
to the
the identity map
map 1 on
on K. That is 11 ®
defined in 25.3.5
@ g: a ®
@
xx H
a
®
xg.
In
this
way
EndF(U)
is
identified
with
a
subalgebra
of
EndK(UK).
++a @ xg. In this way EndF(U) is identified with a subalgebra of ~ n d ~ ( ~ ~
Further
then we
we obtain
Further ifif jr:
n: G ---+ GL(U)
GL(U) is an
an FG-representation
FG-representation then
obtain a KGKGrepresentation n.rrK:
--±GGL(UK)
definedbybyn7rK==11@®7r,
where 11 is the
K :GG-+
L ( U ~ )defined
n , where
Equivalently (1
) ( ~ ==n110
~
@xg7r
xgn
) for
for
trivial representation
representation of
of G
G on
on K. Equivalently
(1@
(9~x)(girK)
each x E X, gg EE G.
then Mx(gn)
Mx(gir) =
G. Observe
Observe that
that if U
U isisfinite
finite dimensional
dimensional then
M(l@x,(gn
K
>
M(i®x)(gJrK)
Recall the
the definition
definition of
of enveloping
enveloping algebra
algebrain
in section
section12.
12.
Recall
(25.6)
G -+
--- EndF(V) =
(25.6) Let 7r:
n:G
=EEbe
bean
anFG-representation,
FG-representation,AA the
the enveloping
enveloping
algebra of
of n7r in
in E, and K an
extension
of
F.
Then
the
enveloping
an extension F. Then the enveloping algebra of
7rK in
in ~EndK(VK)
isomorphic to
to AK as
as aa K-space.
n
ndK
(vK)isisisomorphic
of field of
ofaa linear
linear representation
representation
Change ofjield
120
120
Prooj
We may
may regard
E~ asasthe
therings
ringsFnxn
FnXn
andKnxn
KnXn,
Proof. We
regard E
E and
and EndK(vK)
Endx(V') ==EK
and
respectively, with
with EE the
the set of
of matrices
matricesininEEx
are in
in F.
F. Now
K whose entries are
ag(g7r):ag E F}
A
so
bg(g7r):bg c K .
AK =
gEG
gEG
}
n and
n KKare
But, as matrices, gg7r
and gg7r
arethe
thesame,
same,so,
so,asasGGisis aa group,
group, AK
Ax is the
of E
Ex generated
subalgebra of
generated by G7rx.
G n K .That
That is
is Ax isisthe
the enveloping
enveloping algebra
algebra of
x.
nK
7r
Let G1
G1 and
and G2
G2 be
be groups
groupsand
and7ri
ni an
an FGi-representation.
FGi-representation.Denote
Denotebyby7r1
nl ®7r2
8 n2
the tensor product
® it2
n2 where iti
ni is
ri
product ititl1 @
is the
the representation
representationof
ofGG11 xx G2
G2with
with: fti
restricted
ni restricted
restricted to Gi equal to
to 7ri
ni and iti
restricted to G3_i
G 3 4 trivial. This is aa small
small
of notation which will hopefully cause
abuse of
cause no
no problem.
problem. The
The convention
convention is
is
used in the proof of the
the next
next lemma.
lemma.
Notice that if G1
G 1E
= Gz
G2 E
= GGthen
thenGGisisdiagonally
diagonallyembedded
embeddedin
in G
G11 x G2
G2via
via
the map g H
H(g,
(g,g),
g),and
andififwe
weidentify
identifyGGwith
withthis
thisdiagonal
diagonalsubgroup
subgroupvia
via the
the
the tensor
tensorproduct
product representation
representation7r,
nl ®87r2:
n2: G
GL(Vl 8
isomorphism, then the
G -+
-> GL(Vi
V2)
of G
G isis the
the restriction
restriction of
of the
the tensor
tensor product
product representation
representation of
of G
G1
V2) of
I xx G2
G2 to
this diagonal
diagonal subgroup.
subgroup.
(25.7) Let K be a Galois extension of
r the Galois group ofof K over FF,, and
and
of F,
F, F
V an FG-module. Then
( r xxG)-representation
G)-representationon
on VK
VK with
There is aunique
(1) There
a uniqueFF(r
with(y,
(y, g):
g):aa @
® vv H
H
aayy 0@vg
v gfor
f oeach
r e aac h K,
a ~vK , vV,y
~ VE, yF,~ g, ~G.E G .
W is
is a KG-submodule of
=W,
W, then
then W
W ==UK
uKfor
for some
(2) IfIf W
of VK
VKwith
withW
Wrr =
FG-submodule U of V.
Proof.
Prooj The
Therepresentation
representationinin(1)
(1)isisjust
just the
thetensor
tensorproduct
product representation
representationaa ®
8fiB
of rr xxGGwhere
where aaisisthe
theaction
action of
of rron
onK,
K,and
and$,f3isisthe
therepresentation
representation of
of G
G
on V.
V.
Assume
W isis as
as in
in (2), and
and extend
extend aa basis
basis ZZ =
= (zi: 11i
< ii _(
<m
m)) for W
W
Assume W
insideofofX'X'UUZZtotoaabasis
basisYY=
(zl,...
inside
= (zl,
. . . ,, zm,
. . ,,xn)
x:) of Vx,
v K ,where
where
zm, x;,,,
xm+1 1. ...
xi ==11(9
= {xi:
< i <5n)n)isisan
For ii <
xf
@ xi, and
and XX =
{xi: 11 5
anF-basis
F-basis for
for V. For
5 m,
m,
xi =
j>m(aijxl + wi),
y eE F.
xf
=ECj,m(aijxl
wi), for
for wi EE W
W and aij E
E K. Let y
r . Then
Then
+
MY = xi - 1:(aij)Yx' =
i>m
(aij - aijy)x + wi.
j>m
y Ee W,
y for
But by
by hypothesis
hypothesiswi
wiy
W,so,
so,asasYYisisaabasis
basisfor
forvK,
Vx,aijaij==aij
aijy
forall
alli,i,j.j.
Tensor products
121
Hence
aij E Fix(r)
Fix([') =
< ii <_(m}
Hence ai,
=F.
F .As
AsX'
X'isisaabasis
basisfor
forVK,
vK,(wi:
(w,: 1 5
m )is
is aa basis
basis
for W and we have shown w,
wi =
= 11®
@ vi,
vi, where
where
vi=xi
V.
j>m
generatedby
by(vi:
(vi:11_(<ii (
< m).
Thus W =
=UK
uKwhere U is the subspace ofof VV generated
An irreducible FG-module V
vKremains
remains irreirreV isis absolutely
absolutely irreducible
irreducible ifif VK
extension K
K of
of FF..
ducible for each extension
(25.8) Let V
V be
be an
an irreducible
irreducibleFG-module.
FG-module. Then
Then VVisisabsolutely
absolutelyirreducible
irreducible
if and only ifif F
F ==EndFG(V).
EndFG(V).
Proof. Assume
Proof.
AssumeFF==EndFG(V).
EndFG(V).Then,
Then,by
by 12.16,
12.16, E ==EndF(V)
EndF(V)isisthe
theenvelopenvelopfor G on V. So, if K is
is an
an extension
extension of F
F and
and A'
A' the
the enveloping
enveloping
ing algebra for
algebra
of G
G in E'
_
algebra of
E' ==EndK(VK),
~ n d ~ ( Vthen,
then,
~ ) , by
by 25.4
25.4 and
and 25.6,
25.6, dimK(A')
dimK(Af) =
= n2
dimF(A) =
n2 =
=dimK(E'),
dimK(E1),so
so A'
A' ==E.E'.InInparticular
particularA',
A',and
andhence
hencealso
also G,
G,
is irreducible
irreducible on
on VK.
vK.
Conversely assume
assume V is absolutely
Conversely
absolutely irreducible.
irreducible. Then VK
vK is irreducible
irreducible
where K
K is the
closureofof FF.. By
By 12.17, E'
E' ==EndK(VK)
where
the algebraic
algebraic closure
~ n d K ( v is
~ )the
the
enveloping algebra
algebrafor
forGGininE',
E', so,
so, by
by 25.6,
25.6,n2
n2=
= dimK(Ef)
dimK(E') =
enveloping
=dimF(A).
dimF(A). Let
Let
'r
D ==EndFG(V).
A2
= Dmxm
wheremm =
= dimD(V).
Thennn =
=
EndFG(V).By 12.16,
12.16, A
DmXm
where
dimD(V). Then
dimF(V) =
= mk,
= m2k. So
So m2k2
m2k2==n2
= m2k
dimF(V)
mk, k =
=dimF(D)
dimF(D) and
and dimF(A)
dimF(A) =
n2 =
m2k
and hence
hencekk=
= 1. That
ThatisisFF =
= D.
D.
-> GL(V)
GL(V) is
is an
anFG-representation,
FG-representation, X
X is
is abasis
a basisfor
forV,
V,and
andaa EE Aut(F),
Aut(F),
If n7r: :GG+
then n7r°:
- GL(V)
then
u : GG +
GL(V)isisthe
theFG-representation
FG-representation with
with Mx(g7r°)
Mx(gnU)=
=Mx(g7r)°.
Mx(gn)".
Here ifif AA =
= (aij) E
F""" then A°
). Notice
Here
E Fnxn
A" =
=(a
(a;).
Notice that
that if
if n°
it" isisthe
therepresenrepresentation defined
defined with
with respect
respect to
to aa different
different basis
basis of
of V,
V, then
then it"
ir° isis equivalent
tation
equivalent
to 7r°
n u by aa remark
remark after
after 13.1.
13.1. So
So 7r°
n u isis independent
independent of X,
X, up
up to
toequivaequivasometimes write V°
V" for
for V
V regarded
regarded as
as an
an FG-module
FG-module with
with respect
respect
lence. I'll sometimes
nu.
to 7r°.
is the
the function
function X:
X:GG-+-± F
F
Recall the character of an FG-representation 7r
n is
).
defined by X
~ (g)
( g==
) Tr(g7r
Tr(gn).
with F
F <5kk<5EndFG(V).
Let V be an FG-module
FG-module and k a field with
EndF&). Then the
k-space. Further
Further that
that k-space
k-space structure
structure extends
extends
action of k on V makes V into a k-space.
by G, so we can regard
the F-space structure
structure and is preserved
preserved by
regard V
V as
as aa kGkGmodule. Similarly ifif K is a subfield of
of F then V
is
certainly
a
K-space
and
G
V
G
preserves that K-structure. So V is also a KG-module.
KG-module.
Change
field of
of aa linear
linear representation
Change of
ofjeld
representation
122
122
(25.9)
that K
K=
(25.9) Let
Let a:
a:G
G -+
-+ GL(V)
GL(V)be
bean
anirreducible
irreducible FG-representation
FG-representation such that
=
field, and
and let
let p,8 be
be the
the representation
representation of
of G
G on V regarded as
EndFG(V) isis aa field,
as aa
KG-module,
KG-module, and
and X
x the character
character of P.
p. Then
Then
(1) K
F[x],where
where F[X]
F[x]isisthe
theF-subalgebra
F-subalgebra of
of KKgenerated
generatedby
by the
the
K ==F[X],
elements X
(g), gg E
E G.
x (g),
elements
(2)
of K
K and aa EE Gal(K/F)#.
Gal(K/F)#. Then
Then
(2) Assume
Assume LL is aa normal
normal extension
extension of
L
G-isomorphic toto
LL
®K
V V.
BKV
V"° is not
not LLG-isomorphic
BK
L ®K
1x1,
Proof.
I X1, and
Proof.Let
LetXXbe
beaabasis
basisfor
forVVover
overK,
K,mm==
andM
Mthe
theenveloping
envelopingalgebra
algebra
in EndF(V).
EndF(V). By
By definition
definition gga
a=
gp for
of a in
= g18
foreach
eachggEEGG and
and M
M isis the
the FFsubalgebra
generatedbybyGa.
Ga. Thus
Thus M
M consists
consists of
of the
the elements
elementsEEag(ga),
subalgebra generated
U,(~~),
ag
the map
map E
>ag(ga)
a, E
E F.
F.By
By 12.16,
12.16, the
a, (ga)i-+
H>agMx(g,8)
E a, Mx(g/?)isisananisomorphism
isomorphism of
KmXm
of
M with the ring Kmxm
of all
allmmby
bymmmatrices
matricesover
overK.
K.Hence
Hencefor
foreach
eachxx Ec K
there is A EE M with
with Tr(A)
Tr(A)=
= xx,, and
and AA =
= E,,,
EgEGa,Mx(gj?)
agMx(gf) for
for some
somea,agEc F.
F.
Now xx ==EgEG
agX(g)
CgEG
a,x(g) EEF[X].
F [ X So
]SO(1)
(I)isisestablished.
established.
Assume the hypothesis of (2)
# 1,
bethe
thefixed
fixedfield
fieldof
of or.
a. As
As Cr
a#
1,
(2) and let
let EEbe
E 0#K.
EndFG(V)
we
K.As
AsKK==
EndEG(V)
wemay
may assume
assume E ==F.F.Let
Letyybe
bean
anextension
extension of
a to
=LL®K
gK
V. Then
Then LL ®K
BK V°
V" ==UY
UY and
andthe
thecharacter
characterof
of GGon
onUUisis
or
to L
L and
and U
U=
V.
still the character
character X
x of p.
LG-isomorphicto
to UY
UY then X
x ==XY,
x Y , so
SO
still
P. Thus if U is LG-isomorphic
X
(g)isiscontained
containedininthe
thefixed
fixedfield
field
k of
But
kK
n K==FF,
so,by
by(I),
(1),KK== FF,,
x (g)
k of
y. y.
But
kn
, so,
aa contradiction.
contradiction.
(25.10) Let V
V be
be an
an irreducible
irreducible FG-module, k a Galois extension
(25.10)
extension of
of FF,, and
K ==EndFG(V).
EndFG(V).Then
Then
Wafor
forsome
someirreducible
irreducible kG-module
kG-module W and some A C
(1) VK
VK ==®aEA Wa
(1)
with rr==ANr(W),
ANr(W),where
whereNr(W)
Nr(W) =={y
{yEEFr : :WY
WY G
W).
Gal(k/F) ==Frwith
= W).
LetUUbe
(2) Let
bean
anirreducible
irreduciblekG-module.
kG-module. Then
Then V
V is
is an
an FG-submodule
FG-submodule of
of UU
kG-isomorphic to
(regarded as an FG-module) precisely
precisely when
when U
U is kG-isomorphic
to W"
W° for
some a
or EE r.
r.
thenAA==I' rand
andW°
W"is iskG-isomorphic
kG-isomorphictotoVVfor
forsome
someor
aE
E F.
r.
(3) IfIf kk <5KKthen
Galoisextension
extension of
of FFthen
thenAAisisaaset
setof
ofcoset
cosetrepresentatives
representatives
(4) IfIf KKisisaaGalois
for
for Nr(W)
Nr (W) in 1.
.'I
eaEA
Let Fr ==Gal(k/F).
Gal(k/F).By
By25.7,
25.7,FrxxGGisisrepresented
represented on
on VK.
VK.Let
Let W
W be
Proof. Let
an irreducible kG-submodule
an
kG-submoduleofofvk
Vkand
andMM==(Wy:
(Wy:yy E
E r).
F). If
If M 0#Vk
Vkthen,
then,
25.7.2, M
uk,for
for some
some FG-submodule
Vk,
by 25.7.2,
M=
= Uk,
FG-submoduleU UofofV.V.AsAs0 0#0M
M#
# Vk,
00 ##UU# #
V contradicting
V contradictingthe
theirreducibility
irreducibilityof
of V.
V. Hence
Hence Vk
vk ==MMand
andthen
then
(1) follows
from 12.5. Also
Also Vk
generated as
as an
an F-module by the
follows from
vk isis generated
the copies
copies
(a
V:
a
E
V)
of
V,
so
Vk
is
a
homogeneous
FG-module
and
hence
(aV:a E k#) of V, so vk is a homogeneous FG-module and hence each
each
summand W°
Wa is the sum of
of copies
copies of V,
V, as
as an
an FG-module.
FG-module. This
This gives
gives half
of (2).
(2).
Representations over finite fields
123
Assume U is
is an
an irreducible
irreducible kG-module and V an FG-submodule of U.
U . Let
X be an F-basis
F-basis of
of V.
V. As
As UUisisirreducible
irreducibleand
and V
V an
an FG-submodule
FG-submodule of U,
U ,X
generates U as
as aa kG-module.
kG-module.Also
Also 11(9
BX
X isisaabasis
basisfor
forVk
vkover
overk,
k, so
SO we
we can
can
define:
define:
a:Vk --> U
ax(1
XEX
x) H
axx
ax E k.
XEX
Then a is a surjective kG-homomorphism,
kG-homomorphism, so
so Vk/ker(a)
Vk/ker(a) =="UUasasaakG-module.
kG-module.
Hence (1) and 12.5
12.5 imply
imply UU ="
= W°
W" for
for some
some oa EE A.
A. So
So (2)
(2) holds.
In (3) we may
may regard
regard V
V as
as an
anirreducible
irreduciblekG-module,
kG-module,so,
so,byby(2),
(2),VV%- W°
W"
IFI1 so
for some
some aQ EE A.
A. Then,
Then,by
by(I),
(1),IAl
1Al==dim~(V)/
dimF(V)/dimk(V)
FlI = Il?
dimk(V) =_ Ik:
Ik : F
= Fl?and
A=
and(3)
(3)isis established.
established.
To prove (4) let L be the extension generated by
by k and K and
and assume
assume K
K isis
Galois over F.
F.Then
Then LL isis Galois
Galoisover
over K,
K,and,
and,by
by (3),
(3),25.5,
25.5, and
and 25.3.3:
25.3.3:
V L- L ®K V K= ® L ®K U°
°eGal(K/F)
with U KG-isomorphic
to V. Now
KG-isomorphic to
Now L ®K
B K U°
U" isisan
anirreducible
irreducibleLG-module
LG-module
with
a EE Gal(K/F),
for each a
Gal(K/F),since,
since,byby25.8,
25.8,VVisisabsolutely
absolutelyirreducible
irreducibleas
asaaKGKGmodule. Further
Furtherififao #
# rsthen
U`Tby
module.
then LL ®K
B K U°
U" 9L ®K
BK U
by 25.9.2.
25.9.2. On the other
members of
of A with
with W
W°a S
= Wb
hand if a and
and b are
are distinct members
wbthen
then LL ®k
Bk Wa
W a Ll
2
L ®kWb
so some irreducible occurs in
B kwb SO
in VL''with
vL%ith multiplicity
multiplicity greater
greater than
than 1,1,
a contradiction.
contradiction. This
This completes
completesthe
the proof
proof of
of (4).
(4).
field for a finite group G is a field F
Fwith
with the
theproperty
property that
that every
every
A splitting jield
FG-representation isis absolutely
absolutely irreducible.
irreducible. Notice
Notice that
that by
by 25.8
irreducible FG-representation
FG-module V
V is
is absolutely
absolutely irreducible
irreducibleprecisely
preciselywhen
whenFF =
=
an irreducible
irreducible FG-module
EndFG(V).
Hence,
by
12.17:
EndFc(V).
12.17:
(25.11)
algebraically closed
closed then
then FFisisaasplitting
splittingfield
fieldfor
foreach
eachfinite
finite
(25.11) If FFisisalgebraically
group.
group.
It will turn out in section 27 that if G is
is aa finite group
group then a splitting
splitting field for
G is obtained
of unity
unity to
to GF(p)
GF(p) for any prime
prime pp..
obtained by adjoining a suitable root of
26 Representations
Representationsover
overfinite
finite fields
fields
hypotheses of section
The hypotheses
section 25 are continued
continued in this section.
section. In addition
addition assume
assume
F
F isis of
of finite
finiteorder.
order.
observations make
make things
things go
go particularly
particularly smoothly
smoothlywhen
when F
F
The following observations
is finite.
finite.
Change of
field of
ofaa linear
linear representation
ofjield
124
(26.1) (1)
(1) Each
Eachfinite
finitedimensional
dimensionaldivision
divisionalgebra
algebraover
overFFisisaafinite
finitefield,
field,and
and
hence a Galois extension of
of F.
(2) If
If VVisisan
anirreducible
irreducibleFG-module
FG-modulethen
thenEndFG(V)
EndFG(V)is
is aa finite
finite Galois
Galois exextension of F.
F.
The first remark follows from the well
well known
known facts
facts that
that finite
finite division
division rings
are fields, and that every finite field is Galois over each of its subfields.
subfields. The
second remark
remark is a consequence
of the first and the hypothesis
consequence of
hypothesis that V is
is of
of
finite dimension over F.
F.
(26.2)
of F,
(26.2) Let V be
be an
an irreducible
irreducible FG-module,
FG-module, k a finite
finite extension
extension of
F , and
and
Pr ==Gal(k/F).
Then
Gal(k/F). Then
(1) Vk
vk==®QEA W°
WOfor
for some
some irreducible
irreducible kG-module
kG-module W and any
any set A
A of
of
coset representatives
representativesfor
for Nr(W)
Nr(W) in r.
P.
(2) Let
LetUUbe
bean
anirreducible
irreduciblekG-module.
kG-module. Then
Then V
V is
is an
an FG-submodule
FG-submodule of
of U
U
precisely when U is kG-isomorphic to
to Wu
W° for some a
a EE F.
r.
eaGA
Proof.
Proof.This
Thisisisaadirect
directconsequence
consequenceof
of 26.1
26.1 and
and 25.10.
25.10.
n : G -*
-+ GL(V)
GL(V)be
bean
anFG-representation
FG-representationand
and K
K aa subfield
subfield of
Let jr:
of F
F.. We say nn
can
X of
of V
Vsuch
such that
that each
each entry
entry of
can be written over K if there exists an F-basis X
MX(gr)isinKforeach
Mx(gn) is in K for eachggEE G.
G.
(26.3)
GL(V)
(26.3) Let
Let jr:
n: G
G ---,
GL(V)bebeananirreducible
irreducibleFG-representation,
FG-representation,KK aasubfield
subfield
of
==Gal(F/K).
Then
thethe
following
of F,
F ,and
and (a)
(a)
Gal(F/K).
Then
followingare
areequivalent:
equivalent:
(1) rrn can
can be
be written
written over
over K.
K.
@ K Ufor
forsome
someirreducible
irreducibleKG-submodule
KG-submoduleU
U of
of V.
V.
(2) VV==FF®KU
(2)
(3)
(3) VVisisFG-isomorphic
FG-isomorphicto
to V°.
V".
The equivalence
equivalence of (1) and
and (2)
(2) is
is trivial
trivial as
as is
is the
the implication
implication (1) imProof. The
U be
be an
an irreducible
irreducibleKG-submodule
KG-submodule of V. Then
plies (3). Assume (3) and let U
Gal(F/K)==
Nr(V)asasf'r==(a)( mand
) and
V".Hence,
Hence,by
by26.2,
26.2, OF
uF==V.
V.
Fr ==Gal(F/K)
NF(V)
VV= ZV°.
(3) implies
implies (2).
(2).
That is
is (3)
That
LetVVbe
bean
an irreducible
irreducibleFG-module,
FG-module,K
K aasubfield
subfield of F,
F ,UU an
anirreducible
irreducible
(26.4) Let
KG-submodule ofofV,
F ®E
B E U.
KG-submodule
V,and
andEE=
= NF(U).
NF(U). Then
Then V
V=
=F
eaEA
uF
E UU and
By 26.2,
26.2, UF
uF= =®QEA VQ
Va,
Proof. Let U
F=
=FB
®E
and rh =
= Gal(F/E). By
where A
coset representatives
be
where
A is
is a set of coset
representativesfor
forAA== Nr(V)
NF(V) in
in r.
F. Let L be
fixed field
and W
Wan
anirreducible
irreducible LG-submodule
LG-submodule of
V isis
the fixed
field of
of A and
of V;
V; as V
Representations
finite jelds
fields
epresentations over
overjnite
125
a homogeneous
EG-modulewe
wemay
mayassume
assumeUUI
< W.
homogeneous EG-module
W. If L # F
F then,
then, by
by
induction
onIF
IF::EE1,
®E
W, while,
while,byby26.3,
26.3,VVZ= FF ®L
induction on
1, L LgE
U U2= W,
gLW,
W, So
V=
Thus we may take L = F.ButthenA
F. But then A=
= 1I so
Z FF®E
g EU.
U.ThuswemaytakeL=
so
dimE(U) = dimF(UF) = I F : EI dimF(V) = dimE(V),
so U =
=V.
V. Hence
Hence FF ==NF(U)
NF(U)==E,E,and
andthe
thelemma
lemmaholds.
holds.
(26.5) Let
Let VVbe
bean
anirreducible
irreducibleFG-module.
FG-module.Then
Thenthe
thefollowing
followingare
areequivalent:
equivalent:
(1) VVcan
of F.
F.
canbe
bewritten
written over
over no proper subfield of
(2) VVisisan
subfield K
K of
of FF..
anirreducible
irreducibleKG-module
KG-module for each subfield
(3) NAut(F)(V)
= 1.
NAU~(F)(V)
=
1.
Proof.
ProoJ This
Thisfollows
followsfrom
from26.3
26.3 and
and 26.4.
26.4.
FG-module V
irreducible and can be
be written
written
An FG-module
V is
is condensed if V is absolutely irreducible
over no proper subfield of
of F
F..
Theorem 26.6.
p its
Theorem
26.6. Let pp be
be aaprime,
prime, Fp
Fpthe
thefield
field of
of order
order p,
p,FFp
itsalgebraic
algebraic
subfieldsof
ofE,,
Pp, and
and a
Il the set of pairs
pairs (F,
(F, V) where
closure, A the set of finite subfields
F EE A
A and
andVVisisanan(isomorphism
(isomorphismtype
typeofofan)
an)irreducible
irreduciblefinite
finitedimensional
dimensional
Defineaarelation
relation,,
on01 by
by (F,
(F, V)r(K,
V)T(K, U) ifif FF <5 KKand
r on
andVVis
is an
an
FG-module. Define
of U.
U. Let -- be
bethe
theequivalence
equivaleqcerelation on 1 generated
generatedby
by T.
r.
FG-submodule of
Then
(1) (F,
(F,V)T(K,
V)r(K,U)
U)ifif and
and only
only if FF<(.KKand
andUUisisaasummand
summandof
of K
K OF
@,P VV.
(2) Let A
classofof--.
^-.Then,
Then,for
foreach
eachFFEEA,
A,AF
AF =
=
A be
be an
an equivalence
equivalence class
{(F, V)
V)EEA)
A}isisnonempty
nonemptyand
andAut(F)
Aut(F) isis transitive
transitive on
on AF
AF. In
((F,
In particular
particular
AFp 1 I== 11 and
and the
the map
map A H
lAFp
I+ AFp
AFpisis aabijection
bijection between
between the set
set of
of equivaequivalence classes of -- and the isomorphism
isomorphism classes of irreducible
irreducible finite
finite dimendimensional FpG
modules.
FpG modules.
(3)
(K, U)
U) EE 1 with F
F <5 KK then
then (F,
(F, V)
V) ---(K,
(K,U)
U)ififand
and only ifif
(3) If
If (F,
(F, V),
V), (K,
(F, V°)/(K,
Vu)r(K,U),
U),for
for some
some aa eE Aut(F).
Aut(F).
unique F
F EE A
(4) In
In each
each equivalence
equivalence class A of ^-- there exists a unique
A such
such that
that
the members of
of AF are condensed. Indeed
Indeed for
for (Fp
(Fp,, V) E A, FF==EndFpG(V)
EndFpG(V)
(F, V)
and (F,
V) EE AAwith
with VVaacondensed
condensedFG-module.
FG-module.
thenthere
there is
is an
an
Proof.
from 26.2.
26.2.Let
Let(F,
(F, V)
V) EE a.
Q. IfIf E
E < FFthen
Proof. Part
Part (1)
(I) follows from
irreducible EG-module
VE of V, and we saw during
during the proof of
of 25.10
25.10 that
that VV
irreducible
EG-module VE
homogeneous EG-module,
isomorphism. Write
is a homogeneous
EG-module, so
so VE
VE is determined up to isomorphism.
Vp
for VFp.
V, for
VF,.
Let (K,
(K, U) E
E a.
0. Claim
Let
Claim (F, V)
V) -- (K,
(K, U)
U) ifif and
and only
only if Vp
Vp =
=Up.
Up.The
Thesuffisuffi= Up
ciency of Vp =
Up is
is immediate
immediate from the definition of --; to prove necessity
of fieldof
ofaalinear
linear representation
representation
Change offield
126
126
it suffices to
to take
take(F,
(F, V)r(K,
V)/(K, U)
U)and
andto
toshow
show Vp
Vp =
2 Up.
Up.But
But this follows from
the last
last paragraph.
paragraph.
Let A be an equivalence
equivalenceclass
classof
of--- and (Fp,
(F,, V)
V) EE A.
A. By
By 26.2,
26.2, for each F
FEE
Wafor
forsome
someset
setAAof
ofcoset
cosetrepresentatives
representatives for NA
t(F)(W) in
A, VF
VF ==(DaEA Wa,
NA"~(F)(W)
Aut(F).
Aut(F). By
By (1)
(1)and
and the
the claim,
claim,
eaEA
AF =={(F,
a EE Aut(F)}.
{(F,W°):
Wa):a
Aut(F)}.
That
That is (2)
(2) holds.
holds.
To prove
prove(3)
(3)observe
observethat
thatif if(K,
(K,U)U)E EC20and
andFF5< K
K then
then(F,
(F, UF)r(K,
UF)/(K, U).
U).
Then, by
by (2),
(2),(F,
(F, W)
W)-- (K,
if W"
W° =
(K,U)
U)ifif and
and only if
=UF
UF for some aa EE Aut(F).
Aut(F).
So (3) holds.
Then (F, V) EE A
Let F
F ==EndFPG(V).
EndFp&').
Then
A and,
and, by
by 25.8,
25.8, V
V is
is an
an absolutely
absolutely
irreducible FG-module. By 25.10.3, 25.10.4, and 26.5, V
V can
can be
be written
written over
over
no proper subfield of
of F
F,, so
so V
V isis condensed.
condensed.
Finally suppose (K, U) is
is another
another condensed
condensed member
member of A.
A. To
To complete
complete
proof of
of (4)
(4) we
wemust
mustshow
showKK=
= F.
F. Let k be the subfield
subfieldof
of FF., generated
the proof
generated
by K and
and F.
F.Use
Use26.5
26.5 and
and the
the fact
fact that
that V
V isis condensed
condensed as an FG-module to
5t V"
V° for
for aa EE Aut(F)#;
conclude V 2
Aut(F)#;then
then by
by 26.2,
F ®GF(p) V = ® VQ.
a EAut(F)
Then by 25.3 and 25.5,
k ®GF(p) V = k ®GF(p) (F ®GF(p) V) = k OF ( ® V y 1
yEAu[(F)
///
® (k OFV').
yEAut(F)
Y is irreducible
@FVVY
irreducible for
for
As V
V is
is absolutely
absolutely irreducible
irreducibleas
asan
anFG-module,
FG-module,kkOF
each y.
y. Hence
Hence kk ®FP
gFpV has exactly IF
IF ::Fp
FpI irreducible summands. But, by
irreducible summands,
summands,
symmetry between K
K and
and F,
F ,k®FP
kBFpVValso
alsohas
has(K
IK::Fp
F,JI irreducible
soK=F.
so
K = F.
dimensional repreTheorem 26.6 defines an equivalence relation on the finite dimensional
of characteristic
characteristic p.
p. This equivalence relation
relation
sentations of G over finite fields of
has the property that each class
class contains
contains aa representative over each
each finite field
of characteristic p. Hence we can
can think
think of
of such
such aarepresentation
representationas
as written
written
finite field
field of
of characteristic
characteristic p.
p. However
However the
the lemma
lemma suggests
suggests that
that to
to
over any finite
associated a field
field F over
each class there is associated
over which the representation is best
written: namely the unique field over which the representation is condensed.
I will
F can also
will refer to FFasasthe
thefield
field of
of definition
definition of the representation.
representation. F
be described as the
th smallest field over which the representation is absolutely
Minimal polynomials
127
127
irreducible. Equivalently F
F ==EndF,G(V
), where V is the unique
unique F,G-module
FpG-module
irreducible.
EndF,,c(V),
in the class.
27 Minimal
Minimal polynomials
I
F is
dimensional vector
vector space
spaceover
overFF,, and
and G
In section 27 F
is aa field, V is a finite dimensional
is a group.
group.
Suppose for the moment that A is a finite dimensional algebra
algebra over
over F
F and
F[x]
o b;x`
F[x] isis aapolynomial
polynomial ring. For a EE A and ff (x)
(x) _=Ym
Cy=o
bixi EE F[x],
F[x],define
define
suchthat
thata,:aa:f fH
H ff (a),
ff (a) = CrZ0
o biai
bra` E
E A. Then the
the map
map a,:
aa: F
F[x]
[x] + A such
is an F-algebra
F-algebra homomorphism.
homomorphism. The
The assumption
assumption that
that A
A isis finite
finite dimensional
dimensional
forces
ker(aa)
54
0
since
F[x]
is
of
infinite
dimension.
As
F[x]
forces ker(a,) # 0 since F[x]
dimension. As F[x] is aa prinprincipal ideal domain
domain (PID), ker(aa)
ker(a,) is
is aa principal
principal ideal,
ideal, and
and indeed
indeed there is
a unique monic
polynomialf,(x)
fa(x) with
with ker(a,)
ker(aa) =
= (f,).
(fa). By
fa is
monic polynomial
By definition
definition fa
is
polynomial of
of a. By
fa divides
divideseach
eachpolynomial
polynomial
the minimal polynomial
By construction
construction fa
a. Further fa
fa is
is monic
monic of
of degree
degree atat most
most dimF(A),
dimF(A), since
since
which annihilates a.
> dim~(F[xIl(f,))
dimF(F[x]/(fa)) =
ddimF(A)
i m ~ ( A )>
=deg(fa).
deg(f,).
Applying these observations to
to the n2-dimensional F-algebra
F-algebra EndF(V), it
that each
eachgg EE EndF(V)
EndF(V) has
has aaminimal
minimalpolynomial
polynomialmin(g)
min(g)==min(g,
min(g, FF,,
follows that
B
is
an
F-algebra
isomorphism
then
andaan
that if n:
V). Observe that
n :A
A+
isomorphism then aa and
n
minimal polynomial.
polynomial.In
Inparticular
particularififXX isis aa basis
basis for
for V then
have the same minimal
MX:
EndF(V) +
-+ Fn"n
is such an isomorphism, so
somin(g)
min(g)=
= min(Mx(g)). If
Mx: EndF(V)
FnXn
MIBx(l(9
@ g) =
=MX(g),
Mx(g), so
so the
the minimal
minimal polypolyK is
is an
an extension
extension of
of F,
F ,then
thenMl®X(1
polynomial of
of g over F.
F. Indeed
nomial of 1 ®
@ g over K divides
divides the minimal polynomial
Indeed
an easy application
application of rational
rational canonical
canonical form
form shows
shows the two
two minimal
minimal polypolynomials are equal. (Reduce to the case where V is a cyclic FG-module. Then
there is
is aa basis
basisXX=
= (xi:
(x;: 11 (
< ii <
which xlg
xjg =
= xi+l
xj+1for
fori i <
< n and
(n) for V in which
rz=o
En
o
a,x`
is
the
minimal
polyno
xng== -aj+lxj, where
whereff(x)
(x)=
= xn
x" +
xng
ai+lxi,
alxi is the minimal polynomial for g over FF.. Then ff (1 (9
g) =
= 00 and if h(x) E K[x]
@ g)
K[x] properly
properly divides f
Oas
thenh(1
then h(l (9
@ g) # 0
as
~yzi
+~ : i i
[1
® x1:1
< ii (deg(h)+
< deg(h) + 1}
(1 @xi:
15
1)
is linearly independent and 1 @
®xj+1
xi+l =
=(1
(1(9
@ xl)g`.)
x ~ ) ~ Thus
'Thus
. ) we
we have
have shown:
shown:
(27.1) The minimal polynomial
of a linear transformation is unchanged
unchanged by
by
(27.1)
polynomial of
extension
( V ),
andg gE EEndF
EndF(V),
extensionof
of the
the base
basefield.
field. That
Thatis,
is,ifif KKisisan
anextension
extensionofofFFand
then
min(g, F,
min(1 (9
thenmin(g,
F, V)
V) ==min(l
@g,g ,K,
K,VK).
vK).
Let g EE EndF(V) and a EE F.
F.We
We say
say aa isis aa characteristic
characteristic value
value for g if there
with vg
vg =
= av.
exists v E V#
V# with
av.We
We call
call vv aa characteristic
characteristic vector
vector for a.
a.
(27.2) Let g E
E EndF(V) and a E
E F.
F.Then
Then aa isisaacharacteristic
characteristicvalue
value for g if
if
(27.2)
and only if a is
is aa root
root of the minimal polynomial of g.
g.
Change of
field of
of aa linear representation
ofjield
representation
128
128
Proof. Let
Let ff (x)
(x)==min(g).
min(g).IfIfa aisisnot
notaaroot
rootofofff then
then (f,
(f, xx - a) ==11 so
Proof.
so
there exist
existr,r,ssEEF[x]
F[x] with
withffrr + (x
(x - a)s
vg =
= av
a)s==1.1.Now
Nowifif vv EE V#
V' with
with vg
then
= vv1
= v(
v(ff (g)r(g)
(g)r(g) + (g
(g - aI)s(g))
(g)=
=0
= v(g
v(g-- aI).
1=
aI)s(g)) ==0,
0,since
since f (g)
0=
a I).
then vv =
This contradiction shows characteristic
characteristicvalues
valuesof
ofggare
areroots
rootsofofff..
Conversely
rooto foff ,f,then
thenf f=
= (x
(x - a)h,
Conversely ififaaisisaaroot
a)h,for
for some
some h E F[x].
F[x]. If
If a is
not acharacteristic
a characteristicvalue
valueofofggthen
thenker(g
ker(g--aaII)) =
= 00 so (g-a
(g -a I)-1
I)-' exists.
exists.Hence,
Hence,
as 0 =
= ff (g)
(g) ==(g
(g --aI)h(g),
aI)h(g),we
wealso
alsohave
have h(g)
h(g) ==0.0.But
But then
then ff divides
divides h, a
contradiction.
contradiction.
+
+
g EE EndF(V)
EndF(V) is
is semisimple
semisimple if the minimal polynomial of g has
has no
no repeated
repeated
roots.
roots.
(27.3) Let
F. Then
semisimple
(27.3)
Let g EE EndF(V)
EndF(V)and
andassume
assumemin(g)
min(g)splits
splitsover F.
Then g is semisimple
if and only if g is
is diagonalizable.
diagonalizable.
Proof. This
Thisisis Exercise
Exercise9.2.
9.2.
subset of commuting
commuting elements
EndF (V), and assume
(27.4) Let
Let SS be
be aa finite
finite subset
elementsof EndF(V),
the minimal polynomial of each number of S splits over F
F.. Then
Then there
there exists
exists
a basis X of V
such
that
for
each
s
E
S
the
following
hold:
V such that for each s E S the following hold:
(1) MX(s)
Mx(s) is
is lower
lower triangular.
Mx(s) are
of min(s).
(2) The
The entries
entries on
on the main diagonal of Mx(s)
are the eigenvalues of
(3) Ifs
If sisissemisimple
semisimplethen
then Mx(s)
Mx(s) isis diagonal.
diagonal.
+
= 1 the result is
is trivial,
trivial,so
sotake
takenn >
> 1. If some
Proof. Induct
Induct on
on n + IS
ISI.1.IfIfnn=
g EE SSisisaascalar
scalartransformation
transformationthen
thenby
byinduction
inductionon
onBSI,
IS/, the result holds for
(g},and
and then
then also
also for
for S.
S.So
Sono
no member
member of
of SSisisaascalar
scalartransformation.
transformation.
S --{g},
Let g EE S.
S. By
By 27.2,
27.2, ggpossesses
possessesaacharacteristic
characteristicvalue
value a1.
al. Let
Let V1
Vl be
be the
the
eigenspace
C(g) 5
< N(V1).
V1#0 V,
eigenspace of aa1l for g. Then S C_C(g)
N(Vl). As g is not a scalar, Vl
1 for
(SI v) as claimed in
forV1
Vl with
with MX,
Mx, (SIv,)
so, by induction
induction on
on nn there
thereisisaabasis
basisXX1
the lemma.
lemma. In particular
particular there
there is
is aa 1-dimensional
1-dimensionalsubspace
subspaceUUofofV1
Vl fixed
fixed by S.
S.
If g is
is the
the eigenspace
eigenspace
Vi is
is semisimple,
semisimple,then,
then,by
by 27.3,
27.3,VV==®im-t1 Vi, where Vi
of the characteristic value
value ai
ai of g. Now,
Now, choosing a basis Xi for
for V,
Vi as
as in
in the
the
of
last paragraph, we see that
that X
X=
= U`
1 Xi
UyZl
Xiisisaabasis
basiswith
withthe
thedesired
desiredproperties.
properties.
So we can assume no member of S is semisimple.
semisimple. Thus (3) is established.
established.
Finally S acts on V/
U=
V/U
=V
V and,
and,by
by induction
induction on
on n, there
there is a basis X for V
V
with MX(SIv)
triangular.Now
Nowpick
pickXX==(xi:
(xi:115<i i5<n)
n) with
with U
U=
= (xi)
Mz(SIv) triangular.
( X I ) and
and
(ii: 11 <5 ii <5n)n)==X;X then
(fi:
;thenMx(s)
Mx(s)isistriangular
triangularfor
for each s E
E S.
g EE EndF(V)
EndF(V) isis nilpotent
nilpotentififg"
g' ==00for
m, and
and gg is
forsome
somepositive
positive integer
integer m,
+
unipotent ififgg =
= II + hhfor
forsome
some nilpotent
nilpotent h E
E EndF(V).
unipotent
Minimalpolynomials
polynomials
Minimal
129
(27.5)
Then
(27.5) Let
Let gg EE EndF(V)
EndF(V) and
and n ==dimF(V).
dim~(V).
Then
(1) The
Thefollowing
followingare
areequivalent:
equivalent:
(1)
(i)
(i) ggisisnilpotent,
nilpotent,unipotent,
unipotent,respectively.
respectively.
min(g)==x',
xm,
- l)m,
respectively,
some
positiveinteger
integerm.
m.
(ii) min(g)
(ii)
(x (x
- 1)',
respectively,
forfor
some
positive
(iii) There
Thereexists
existsaabasis
basisXXfor
forVVsuch
suchthat
thatMX(g)
Mx(g)isislower
lowertriangular
triangular
(iii)
and
ent'ries on
on the
the main
main diagonal
diagonalare
are0,
0,1,1,respectively.
respectively.
and all
all entries
(2)
(2) Unipotent
Unipotentelements
elementsare
areofofdeterminant
determinant1,1,and
andhence
hencenonsingular.
nonsingular.
(3) IfIfggisisnilpotent
nilpotentand
and semisimple
semisimple then g =
=0.
0.
(3)
(4) IfIfggisisunipotent
unipotentand
and semisimple
semisimple then g =
=I.I .
(4)
(5)
1 1if ifg gis is
unipotent.
(5) Let
Let char(F)
char(F) ==pp>>0.0.Then
ThengP"
gP"==
unipotent.Conversely
Conversely ifif
gP"' 1=for
1 for
some
positive
integerm,m,then
thenggisisunipotent.
unipotent.
gP=
some
positive
integer
(6)
0 0and
char(F)==
andg gE EGL(V)
GL(V)isisofoffinite
finiteorder
orderthen
thenggisissemisimple.
semisimple.
(6) IfIfchar(F)
(7)
p >p 0>and
g E GL(V)
is of
order
mmthen
(7) IfIfchar(F)
char(F)= =
0 andg
E GL(V)
is finite
of finite
order
thenggisissemisimsemisimple ifif and
and only if (m, p) =
=1.1.
ple
Proof.
Proof. All
Allparts
partsofofthe
thelemma
lemmaare
arereasonably
reasonablystraightforward,
straightforward, but I'll make
make
aa couple
couple of
of remarks
remarks anyway.
anyway. If
If gg EE GL(V)
GL(V)isisofoffinite
finiteorder
ordermmthe
theminimal
minimal
polynomial
x'" -1,1,which
roots ifif(m,
(m,char(F))
char(F)) =
polynomialof
of g divides
divides xm
whichhas
has no multiple roots
=1.1.
Hence
Hence (6)
(6) and
and half
half of
of (7)
(7)hold.
hold.Parts
Parts (4)
(4) and
and (5)
(5)imply
imply the
theremaining
remaining half
half of
of
(7),
(7),since
sinceany
anypower
power of
of aasemisimple
semisimpleelement
elementisissemisimple.
semisimple.
Recall
field FF is perfect
perfectififchar(F)
char(F)==00or
orchar(F)
char(F)== pp >
> 0 and
and F =
=FP,
FP,
Recall a field
where
where FP
F P is
isthe
theimage
image of
of FFunder
underthe
thep-power
p-powermap
mapaai-+
HaP.
up. For
For example
example
finite
following
finitefields
fieldsare
are perfect
perfect as
as are
are algebraically
algebraically closed
closedfields.
fields. We need the following
elementary
elementary fact,
fact, which
which appears
appears for
for example
example as the Corollary
Corollary on page 190
190 of
of
Lang [La].
[La].
Lang
(27.6)
(27.6) IfIf FFisisperfect
perfectthen
thenevery
every polynomial
polynomial in F[x]
F [x]isis separable.
separable.
(27.7) If
perfect and a EE EndF(V),
EndF(V),then
then there exists ff EE F[x]
F [x]and
andaaposiposiIf FFisisperfect
tive integer
integereewith
with(f,(f,ff')
= 11 (where ff''is
derivativeoof
andffe(a)
= 0.
') =
is the derivative
f f )f)and
e(a) =
If ,B
EndF(V) with
B EE EndF(V)
with ff(,B)
(B) ==00then
then,B
B is semisimple.
semisimple.
ny!Jei
Proof.
Proof.Let
Letmin(a)
min(a)==flm 1 fee' with
withfif;irreducible.
irreducible.Let
Let
=nfi.
m
m
f=f
ee=max(e;:l
= max{el: l ii
<i <i m
ml} and
and f
i=l
(a) =
Then ff ee(a)
=00and,
and,by
by 27.6,
27.6, fifihas
hasno
no repeated
repeated roots.
roots. Hence ff has
hasno
norepeated
repeated
roots, so (f, f')
f ')==1.1.Also,
Also,ififf f(,B)
(B) ==00then
then min(p)
min(B)divides
divides f,
f ,and
andhence
hence has
has
no repeated
repeatedroots,
roots,so
so,B
B isis semisimple.
semisimple.
no
of field of
ofaalinear
linear representation
representation
Change ofjeld
130
130
The proof of the following
following lemma
lemma comes
comes from
from page
page 71
71 of
of [Ch
[Ch 2].
21.
(27.8) Let F
F be
(27.8)
be perfect
perfect and aa EE EndF(V).
EndF(V). Then
Then there
there exist j3, yY EE EndF(V)
j3 + y,
y,,B
j3 semisimple, and
nilpotent. Further ,B
j3 =
for some
some
with aa ==,B
and y nilpotent.
= t(a) for
t(x) E F[x].
+
Proof. Choose
E F[x]
Choose ff and
andeeasasininthe
thelast
lastlemma.
lemma. Then
Then there
there exist h, h1
hl E
with 1 =
= f'h + f fh hl.
1. Define
F-algebrahomomorphism4:
homomorphism o:F[x]
F[x]+
-+ F[x]
F[x]by
Define an F-algebra
by
(go)(x)
f h) h)
mod
(g@)(x)==g(x
g(x --f (x)h(x)).
f (x)h(x)).Observe
Observego
g@_ (g-g'
(g-g'f
modf 2,
f 2,sosoinparticular
in particular
f4
O=
= ff --f'ffh'fh= f=- ff (1
f h-1)f h=l 0) =
mod
f 2.fThen
proceeding
- f- (1
Omod
2. Then
proceedingby
byinducinduction on
on m
m and
andusing
usingthe
thefact
factthat
that@0isisaahomomorphism,
homomorphism,
f 0'=
= 0 mod
modff2-.
f @m
2m.
tion
Choose
>e.e .Then:
Then:
Choose mmwith
with2m
2' >
+
(27.8.1)
(27.8.1)
fo'
f 4"=_=00mod
mod ffe,',
so
so
(f¢')(a)
(f @")(a)==0.
0.
xi
Next,for
forg(x)
g(x)== Tj aixi
a,x' E F[x],
Next,
F[xl,
(aixi)
gok =
ok
= >ai(xgk)` = g(xok),
@k isisaahomomorphism,
so
as q5k
homomorphism, so
(27.8.2)
gok
g@k
F[x].
= g(xok)
=
g ( ~ ~ kfor) each gg E F[x].
proceeding by induction
Also xxo
4=
mod ff,, so proceeding
induction on k, and using the
Also
= xx - ffh ==xx mod
with ff @0 =
=00mod
mod f , we conclude:
fact that @ is a homomorphism with
(27.8.3)
xxqk
@=
=~xx mod f.f.
We can
can now
nowcomplete
completethe
theproof
proofofofthe
thelemma.
lemma.Let
Lett t==x@",
x0"', j3,B==t(a),
t(a), and
= (fom)(a)
y=
=aa--P.j3.Then
Then.ff(fl)
(j3)==.ff(t)(a)
(t)(a) ==.ff(xom)(a)
(x@")(a) =
(f @")(a)==0,0,by
by27.8.2
27.8.2
Hence, by 27.7, ,B
and 27.8.1, respectively. Hence,
j3 is semisimple. Also, by
by 27.8.3,
27.8.3,
(x - t)t)==00mod
= (a
(X
mod f , so ye =
(a --,B)e
j3le =
=(x
(x --t)e(a)
t)e(a)==0.0.Thus
Thusyyisisnilpotent.
nilpotent.
(27.9) Let F
F be perfect and aa! EE GL(V).
GL(V). Then
(27.9)
(1) There exist a,,
as, a,
a EEGL(V)
GL(V) with
with as,
a,, semisimple,
semisimple, a,
a, unipotent,
unipotent, and
and
aa =
= asau
asa =
= auas.
a=
and p
µ unipotent,
= cp
µ ==µpcwith
p pc GL(V),
(2) If a
with, c,
E GL(V), ( semisimple, and
unipotent,
then=
asa,and
then ( =
and/-t
p=
=au.
a,.
(3) There
exist polynomialst(x),v(x)
v(x)E E
F[x]
withal==t(a)
t(a)andau
= v(a).
Thereexistpolynomialst(x),
F[x]
witha,
anda, =
Proof. By
27.8,aa =
= +j3 y,
P, j3,
y cy EndF(V),
B jsemisimple,
By 27.8,
+ with
y, with
E EndF(V),
3 semisimple,yynilpotent,
nilpotent,
and j3 =
= t(a),
=t(a),
t(a),for
forsome
some t(x)
t(x) EE F[x].
F[x].As
As,B
j3 =
t(a), $j3EEZ(C(a)
Z(C(cf)flf EndF(V)).
l EndF(V)).
Let F be
of F.
F. By
there is
is aa basis
basis X
X of
of
be the
the algebraic
algebraic closure of
By 27.4
27.4 and 27.5
27.5 there
diagonaland
andMx(y)
Mx(y) strictly lower triangular.
triangular. From
From this
this itit is
F with
with MX(,B)
Mx(j3) diagonal
evident that det(,B)
= det(er).
evident
det(j3) =
det(a). Thus, as aaisisnonsingular,
nonsingular,so
sois,
is P.
j3.
131
131
Minimal polynomials
j3 and
+ j3-'P-1 y. Asp-'
As j3-'and
andy ycommute
commuteand
andyyisisnilpotent,
nilpotent,
Let a,
as =
= ,B
anda,a.==II-Iso a,
a isisunipotent.
By construction
constructiona a==a,a,
asa =
= auas.So
,B-1y
is nilpotent,
nilpotent, so
j3-'
y is
unipotent. By
(1) and (3) hold.
µ are
are as
as in
in (2).
(2). As
As ,B
j3 EE Z(C(a)), 6 and µpcommute
commutewith
with,Bj3
Suppose 6 and p
a,,. By
By 27.4,
27.4, j3,Band
and6 can
simultaneously diagonalized over F
F,, and hence
and a,.
can be simultaneously
semisimple by 27.3. Similarly
Similarly a,p-'
,B-1 is diagonalizable
diagonalizable over F
F,, so,8-1
j3-'6
so j3-'6 is semisimple
is unipotent. Finally as Pa, = a = c p , j3-'6 = a&-' is both semisimple and
unipotent,
= 6 and
j3 =
and a,a =
= µ.
p.
unipotent, so,
so, by
by 27.5.4,
27.5.4, ,B
as
part of a and
part of
of aa,,
as and
and aa,are
arecalled
calledthe
thesemisimple
semisimple part
and the
the unipotent
unipotent part
Jordan derespectively,
andthe
thedecomposition
decomposition
a =a,a,
asa =
= auasis called the Jordan
respectively, and
a=
composition of a.
a.As
As aa consequence
consequenceof
of 27.5
27.5 and
and 27.9
27.9 we
we have:
have:
(27.10) Let
Let aa E
GL(V) be
be of
of finite
finiteorder.
order.Then
Thenas
asand
anda,a are powers
of aa..
(27.10)
E GL(V)
powers of
If char(F)
= 0 then
= as,
char(F)== pp > 0 then
then lasl
Ias _=aI
p, and
char(F) =
then aa =
a,, while
while ifif char(F)
lalp~
and
laul
= Ialp.
Iaul =
Ialp.
(27.11) Let F
of aa in FF,,
(27.11)
F be
be perfect and aa cE EndF(V),
EndF(V),bbaacharacteristic
characteristic value of
U the eigenspace of
of bb for
for aa in V,
V, and
and K
K an
an extension
extension of F.
F . Then
Then UK
uKis the
eigenspace
eigenspace of bb for
for 11 ®
€3 aa in
inVK.
vK.
/.
Proof. This
Thisisisessentially
essentiallyan
anapplication
applicationof
of Jordan
Jordan Form;
Form; I sketch a proof.
Recall the
the map
mapff H
is isananF-algebra
Hf(a)
f (a)
F-algebrarepresentation
representation of
of F[x]
F[x]on
on VVwith
with
kernel (M(x)), where M(x) =
=min(a).
rnin(a).Let
LetM(x)
M(x)==Fl'= 1 pi(x)el
pi(x)" be
be the
the prime
factorization of
of M.
M. From the theory of modules
factorization
modules over a principal ideal domain (cf. Theorems 3 and 6 on pages 390
390 and
and 397
397 of
of Lang
Lang [La])
[La])we
we know
know
V ==®i-1
where
V V(i)
(i) = =
ker(pi
(a)ej ). Indeed
(x )e'
@;=, V(i),
V(i),
where
ker(~i(a)~l).
Indeedasasthe
thepolynomials
polynomialsp;pi(x)''
prime the
the same
same holds
holds in VK, SO
so as V(i)K
in the
are relatively
relatively prime
~ ( i is) contained
contained
~
the
kernel
weconclude
conclude ~V(i)K
=) (VK(i)).
(vK)(i)of p{(x)ei
pi(x)" on vVK,
K ,we
( i=
(vK
~ (i)). Thus without
without
kernel (VK)(i)
loss M
M=
= (x - b)e.
loss
b)'.
Again
from
the
theory of modules over
over a PID, V ==®l=1
Again
@:=, Vi,
Vi,where
where V,
Vi =
v,F[x]
viF[x] is
is aa cyclic
cyclic module
module for F[x]
F[x]with
with annihilator
annihilator (x
(x --b)ei,
b)" ,ee==e1
el >2e2
ez >
L
=
- . L> e5
e, >
31,1,and
and the
the invariants
invariants el
ei are
are uniquely
uniquely determined.
determined. As VK
vK=
VK isis aamodule
module with
with invariants ei,
e1,we
we conclude
conclude the
the ei
el are also the invariants of
viK
F[x]
= UUflnV,
F[x]on
onVK,
v K ,and
and itit remains
remains to
to observe
observe that
that U
U ==®;=1 U;,
Ui,with
withUU
Ui =
Vi
of dimension
dimension 1.
1.
niZl
(27.12)
G+
-+ GL(V) be an FG-representation and
and K an extension of
of
(27.12) Let 7r:
n: G
K))
=
dimF
(Cv
(g7r
))
for
each
g
c
G.
F. Then
(CVK (g7r
F.
ThendimK
dimK(Cv~(gnK))
= dimF(Cv(gn)) for each g E
of field of
ofaa linear
linear representation
Change ofjield
132
Proof. This
Thisisis aadirect
direct consequence
consequenceof 27.11.
27.1 1.
Recall the definition of a splitting
splitting field in section 25.
(27.13)
Let G be a finite
groupofofexponent
exponentmmand
andn:
r:GG+
-- GL(V)
(27.13) Let
finite group
GL(V) an FGFGrepresentation.
Letkk== m
mififchar(F)
char(F) =
= 0 and
andkk =
= m,~
m p,ififchar(F)
char(F)== pp >
representation. Let
> 0.
0. Let
n ==dimF(V).
dimF(V). Then
Then
(1) Let
of nr and
x be the character of
and gg EE G.
G.Then
Then X(g)
~ ( gis
is) aa sum
sum of
of nn kth
kth
Let X
roots of unity.
unity.
(2) FFisisaasplitting
splittingfield
fieldfor
forGGififFFisisfinite
finiteand
andcontains
containsaaprimitive
primitivekth
kth root
root
of unity.
unity.
Proof.
Proof. Let
LetFFbe
bethe
thealgebraic
algebraicclosure
closureof
ofF.
F .By
By27.4
27.4there
thereisisaabasis
basisXXfor
forVF
V'
such that Mx(g) is
with
is triangular. By 27.10, g ==gsg
gsgu
withgsg,semisimple
semisimpleand
and1gsI
Igsl
dividing k, with
with g,
g, unipotent,
gs and
and g,
g powers
unipotent, and with g,
powers of
of g.
g. Thus
Thus the
the entries
entries
Mx(g) are
on the main diagonal of Mx(g,) are 11 and so the entries of Mx(g)
are the
the same
same
as those of Mx(gs).
MX(gs).In
Inparticular
particularififaa isis such
such an
anentry
entrythen
thenaak
= 1 as (gs)k =
= 1.
k=
1.
So (1)
(1) holds.
holds.
Assume
irreducible
finite,FFcontains
containsco,
o,aa primitive
primitive kth
kth root
root of 1,
1, V
V is an irreducible
AssumeFFisisfinite,
FG-module,
EndFG(V).By
By 26.1.2,
26.1.2,K
K isis aa finite
finite field extending F
FG-module, and
and let K ==EndFG(V).
and hence containing
containing w
co.. Let
Let r be the character
character of
of V regarded
regarded as
as aa KG-module.
KG-module.
The argument of
of (1)
(1) shows
shows +(g)
Vr(g)EEFFfor
foreach
eachgg Ee G. As i/r(g)
EF
+(g) E
F for
for each
each
g eE G,
25.9.1 that
that K
K=
= F.
G,we
we conclude
conclude from 25.9.1
F.Thus
ThusFFisisaasplitting
splittingfield
field for G
by 25.8,
25.8, completing
completingthe
the proof.
proof.
+
Recall that given
representationscra and
and BP of groups G and
and H,
H,respectively,
respectively,
given representations
there
®P
there is a representation
representation aa @I
B of GG xx H.H.The
Thedefinition
definition of
of acr ®
@IfBappears
appears
just before
before the
the statement
statement of
of Lemma
Lemma 25.7.
25.7. This
This representation
representation appears in the
statement
statement of
of the
the next
next two
two lemmas.
lemmas.
(27.14)
), M
GL(V),
M==CGL(V)(G),
CcL(v)(G), and assume V is
is aa homogeneous
homogeneous
(27.14) Let
Let G
G <5GL(V
FG-module. Write I ==Irr(G,
Irr(G,V,
V,F)
F )for
forthe
theset
setof
of irreducible
irreducible FG-submodules
of V
Vi Ee I,
1, 1 5
< ii <5m,
Let
EndFG
(VI )
v and choose %
m,with
withVV==®ml Vi.6.
LetKK= =
EndFG(V1)
be aa field and A =
=HomFG(VI,
HomFG(V1,V). Then
and a
1=
1.
(1) There
exists YY== (ai:
(ai: 11 5< i <
a1
= 1.
There exists
5 m)
m) cEAA with
with Viai
VIai ==Vi and
(2) A
A is
is aaKG-module
KG-module and
and Y
Y is aa K-basis
K-basis for
for A.
A. Y
Y induces
induces a unique Kspace structure
structure on V
V extending
extendingthe
the F-structure
F-structuresuch
suchthat
thataicriEEHomKG(VI,
Hom~G(v1,
Vi)
space
Vi)
This structure
structure is
is preserved by G.
G.
for each i.i . This
(3) The
n: M -+
GL(A;
The map n:
GL(A;K)
K)defined
definedby
byxir:,8
xn: B HH18x,
Bx,xx eEM,
M,PB EEA,
A,
is an isomorphism.
isomorphism. M preserves the
the K-space structure
structure on V.
Minimal polynomials
133
133
+:
+:
(4)
A#+
± II defined
Vl,8
(4)The
The map
map 7/r:A#
defined by 1:,8,B H
H
V1,Bisisaasurjection
surjectionand
and defines
defines
a bijection
0: S(A)
SG(V)
BH(b:beB)
between the
the set S(A) of all K-subspaces of A and the set SG(V)
of all FGbetween
SG(V) of
submodules of V. 04 is
submodules
is aa permutation
permutation equivalence
equivalence of the actions of M on S(A)
S(A)
and SG(V).
(5) The
M x G -+
The map
map B:
8:M
+GL(V,
GL(V,K)
K)defined
defined by
(x, g)8: uv H
H vxg
uxg
is a K(M xx G)-representation
G)-representationwhose
whose image
image is
is GM
GM and
and which
which is
is equivalent to
of the
the representations
representations of
of M
M on
on A and
and G
G on
on Vl
Vl over
over K.
K.
the tensor product of
Proof. As
al: Vl
-+ Vi
As VVisishomogeneous
homogeneous there
there exists
exists an
an isomorphism
isomorphism al:
Vl +
Vi of
FG-modules. Composing ai
ai with the inclusion
inclusion V
Vi
cV
ai as a
iE
V we may regard ai
Choose al
ar =
member of A. Choose
=1.1.Then
Then(1)
(1) holds.
holds. As K ==EndFG(VI)
EndFG(Vl)and ai
aiis
is an
an
FG-isomorphism,
(Vi) for
ai is also aaKG-isomorphism.
KG-isomorphism.
EndFG(X)
for each i and ai
FG-isomorphism,KK==EndFG
Thus we have a unique scalar multiplication of
of K
K on Vi extending that
that of
of F
ai eEHomKG(V1,
Vi), so
sothere
thereisisaa K-space
K-space structure on V extending
such that ai
HomKG(Vl,Vi),
F. ItIt is
that on F.
is defined
defined by:
m
a
(uai) =
i=1
m
F
a(uai), u e Vi, a E K.
i=1
HOmFG(Vi,
to
Next AA ==HomFG
(V1, ®m 1 Vi)
= ®m j HomFG
(Vi ,Vi)
Vi)isis isomorphic to
H0mFG(V1,
vi) =
Km as an F-space.
V)isisisomorphic
isomorphictoto Km
Km as
as aa KF-space. Similarly
Similarly HomKG(V1,
HomKG(Vl,V)
V) isis also
also aa K-space.
space and is an F-subspace of
of A,
A, so
soAA==HomKG(V1,
HomKG(Vl,V)
Also, as Y is a K-linearly
m, Y
is a K-basis
K-linearly independent
independent subset of A of order m,
Y is
for A, so (2) holds.
Evidently,
for ,BP E A and x cE M,
M,the
thecomposition
composition ,8x
,Bx is also in A,
A, so
so the
the
Evidently, for
map n
r in
KM-representation.IfIfsix
aix=
=a
aii then, as G is
in(3)
(3)isis aa well-defined
well-defined KM-representation.
map
irreducible
on Vi,
Vi,VVi
Cv(x). Hence
Hencenr is faithful.
faithful.Let
Let(uj:
(vj:11(< jj < d)
irreducible on
i 5< Cv(x).
d) be
be aa
K-basis
K
-basis for
for V1.
Vl . Then
X = (vjai: 1 < j < d, l < i < m)
is a K-basis
K-basis for
for V.
V. An
An element
element of
of the
the general
general linear
linear group
group GL(A,
GL(A, K)
K)on
on A
A(re(reK-space) may
(aid) with
with respect
respect to
to
garded as a K-space)
may be regarded
regarded as
as an m by m matrix (aij)
the basis Y
Y of
of A.
A.Given
Givensuch
suchananelement
elementdefine
define
byU vjai
= Ek vjaikak.
x xE eMMby
j f f i X x=
ujaikffk.
Then x cEM
= (aid),
preserves the
the K-structure
K-structure on
on V.
V. Hence
Hencenr
Mwith
withxx7r
n=
(aij), and xx preserves
is an isomorphism
isomorphismand
and (3)
(3) holds.
holds.
xk
Change of field
of aa linear
linear representation
Id of
representation
134
Evidently *maps A# into I and the induced map 0 takes S(A) into SG(V) and
preserves inclusion. It is also clear that 0q5 is
is an
an injection
injection from
from the
the set
set S1(A)
Sl (A) of
1I-dimensional
-dimensionalsubspaces
subspacesof
ofA
Ainto
intoI.
I. Let
Let W
W E II and
W+
-k V,
and let
let 7ri:
ni: W
Vi be
bethe
theii th
projection. 7ri
is
trivial
or
an
isomorphism
by
Schur's
Lemma,
and
there
ni is trivial or an isomorphism
Lemma, and there exists
exists
an isomorphism B:
f: Vl
W. Then
Thenaiai==Bnia;'
ftiai 1 EE K and
Vl + W.
and a =
= >2aiai
C a p i EE A
A
with a$
a* ==Via
Vla==W,
W,soso0:q5:S1(A)
Sl(A)-k
+I Iisisa abijection.
bijection.For
For each
each B,
B, D
D EE S(A),
S(A),
(B + D)O
D)@==BO
Bq5+ DO;
Dq5;from
fromthis
thisremark
remarkand
andits
itspredecessor
predecessorititisis not
not difficult
difficult
to complete
the
proof
of
(4).
complete
(4).
Finally, by (3), M preserves the K-space structure on V. Hence the map 08
in (5) is indeed
indeed aa well-defined
well-definedK(M
K(M xx G)-representation
G)-representationwhose
whose image
image is GM.
GM.
The map ujai
vjai H
++vjuj®€9aiaiinduces
inducesananequivalence
equivalenceofof08with
withthe
thetensor
tensor product
product
representation
representation
+
+
(x, g): ai ® vi ra aix ® vj g
of M
M xxGGon
onAA®€9V1,
Vl,so
so(5)
(5)holds.
holds.
Gi, ii ==1,1,2,
groups,FFa asplitting
splittingfield
fieldfor
forG1
G 1and
and G2,
G2, and
and Ai
Ai
(27.15) Let Gi,
(27.15)
2, bebegroups,
collection of representatives
representativesfor
for the
the equivalence
equivalence classes
classes of finite
finite dimensional
dimensional
aa collection
irreducibleFFGi
Gi -representations.
(771,772)
€9 772 is a bijecirreducible
-representations. Then the map (7r1,
7r2)t+
H n1
ni ®7r2
tion between
between A
A1 x L2
A2 and
andthe
theset
setof
of equivalence
equivalenceclasses
classesof
of finite
finite dimensional
dimensional
irreducible F(Gl
F(G1 xx G2)-modules.
G2)-modules.
Proof. Let7ri
Let nicELiAi
with
moduleVi.Vi.By
By27.14.4
27.14.4and
and27.14.5
27.14.5there
thereis
is aa bijection
bijection
Proof.
with
module
between the F(G1
F(Gl xxG2)-submodules
G2)-submodulesofofV1
Vl ®€9V2
V2 and
and the
the FG2-submodules
FG2-submodules
V2, so,
SO,as 7r2
772 is
so is
is 7r1
771 €
3772. Conversely
: G1
of V2,
is irreducible, so
®7r2.
Conversely let
let n7r:
G1xx G2 +
-*
GL(V) be an irreducible
irreducible FG-representation.
FG-representation.By
By Clifford's
Clifford's Theorem,
Theorem,12.13,
12.13,V
V
GL(V)
is a homogeneous FGi-module,
FGi-module,so,
so,by
by27.14.5,
27.14.5,7rIT isisequivalent
equivalentto
to7ri
ni ®7r2
€9
for
ni EE Ai. Indeed 7ri
ni is determined up to equivalence by the equivalence
equivalence
some 7ri
irreducible FGi-submodules
FGi-submodules of
of V.
V.So
Sothe
thelemma
lemmaholds.
holds.
class of irreducible
LetGGbe
be aa finite group and n7r: :GG+
(27.16) Let
-- GL(V)
GL(V) an
an irreducible
irreducible FG-represenFG-representation. Then Z(G7r)
Z(Gn) =
(z)isisaacyclic
cyclicgroup
groupof
of order
orderrelatively
relatively prime to the
the
tation.
= (z)
characteristic of F
F and,
and, ifif FFcontains
containsaaprimitive
primitiveIzIth
1~1th
root of
of unity
unity w,
w , then
then z
characteristic
root
wk of
of w
w with (Iz1,
(lzl, k)
1.
acts on
on V
V by
by scalar
scalarmultiplication
multiplicationvia
viaaapower
powerwk
acts
k) =
= 1.
Proof.By
By12.15,
12.15,Z(Gn)
cyclic,say
say Z(G7r)
Z(Gn) ==(z).
(z).By
ByExercise
Exercise4.3,
4.3,nn ==JzI
lzl
Proof.
Z(G7r) isiscyclic,
char(F).So
Sowe
wecan
canassume
assumewoisisaaprimitive
primitiventh
nthroot
rootof
of11
is relatively prime to char(F).
F.Now
Now z satisfies the
x) =
x n --1 1sosoits
itsminimal
minimalpolynomial
polynomial
in F.
thepolynomial
polynomialff ((x)
= x"
divides ff and
andhence
hencehas
hasroots
rootspowers
powersof
ofw.
w .So
Soby
by27.2
27.2wk
wkis
is aacharacteristic
characteristic
value for z for some 00 <
<k <
< n, and then by Clifford's Theorem
Theorem z acts by scalar
wkon
on V.
V.Thus
Thusnn==
lok1,
1, so
multiplicationvia
via wk
multiplication
1zlzI 1 _=I wk
so (k,
(k, n)
n) =
= 1.1.
Minimal polynomials
135
(27.17) Let V be an
(27.17)
an irreducible FG-module. Assume G is finite and a semidirect product
product ofof H 2
a_GGby
by X
X of
direct
of prime
prime order
order p.
p. Assume
Assume dim(V)
dim(V) #
FH-module and
and ifif F is finite
p(dim(Cv(X))). Then V is
is aa homogeneous
homogeneous FH-module
finite of
order prime to pp then
then VVisis an
an irreducible
irreducible FH-module.
FH-module.
Proof.
Proof.By
ByClifford's
Clifford'sTheorem,
Theorem,12.13,
12.13,VVisisthe
thedirect
directsum
sum of
of the
the homogeneous
components
(V,:115< ii <I
r)r )ofofHHononV,V,and
andXXpermutes
permutesthese
thesecomponents
components
components (Vi:
transitively.
Soas
as X
X=
= (x) is of
transitively. So
of prime
prime order
order p either
either VV isisaahomogeneous
homogeneous
FH-module or
or p ==rr and
FH-module
andXXisisregular
regularon
onthe
thecomponents.
components. But in the latlatter case CV(X)
Cv(X) =
_
1 vxi:
vx`:vvEc V,),
V,},SO
so dim(V)
dim(V)=
= ppdim(Cv(X))
dim(Cv(X))contrary
contrary
to hypothesis.
hypothesis.
So assume F
F is
is finite
finite of order q prime
prime to p.
p. By
By 27.13
27.13 there
there exists
exists a finite
extension
K of FFwhich
extension K
which isisaasplitting
splittingfield
field for
for H
H and
andcontains
containsaaprimitive
primitive
pth root
W°for
forsome
someAAGCGal(K/F).
Gal(K/F). By
vK== ®aEA Wa
By 27.12,
27.12,
root of
of 1.
1. By 26.2, VK
dim(CvK(X))=dim(Cv(X)).
So dim(
dim(W);
dim(Cv~(X)) = dim(Cv(X)). So
W) #pdim(CW(X)).
p dim(Cw(X)). Further
Further ifif H isis
irreducible on W
Wa:aa EE A)
A) is
is the set of
W then,
then, by
by 26.2,
26.2, {(Wa:
of H-homogeneous
H-homogeneous
components of VK.
components
vK.But if 0 # U is an FH-submodule of VV then 0 # uK
UK is a
Gal(K/F)-invariant
Gal(K/F)-invariant KH-submodule
KH-submoduleof
of VK,
v K ,so
SO Wa <
5 UK
uKfor some aa,, and then
VK =
= UK.
That is,
is, H
H is irreducible on
on V.
V. So,
So, replacing
replacing(V,
(V, FF)) by
by (W,
(W, K), we
vK
uK.That
may assume F
F is
is aa splitting
splitting field
field for H and
and contains
contains aa primitive
primitive pth root
root of
of
existence of
of the pth root
1. The existence
root forces
forces qq =_
E 1 modp.
the
mod p. Then by 27.14.4 the
set II of
H-submodulesofofVV!isIsofoforder
order(qm
(qm--l)/(q
1)/(q-- 1)
of irreducible
irreducible H-submodules
1) =
set
m mod
= m dim(U)
dim(U) for
for U
U EE I.
I. As
m
mod p,
p, where
where dim(V)
dim(V) =
As V
V isisan
anirreducible
irreducible
FG-module, V
V=
= (Ux)
sum of
of at
at most
mostpp conjugates
conjugatesofofU,
U,so
somm(
< p,
FG-module,
(uX)is the sum
with
uX'.
out by
by an
an argument
argument in
in
with equality
equality only
only ifif V
V=
= ®p 1 Ux'
. This last case is out
m<
< p. Further as V is an irreducible FG-module either
the last paragraph, so m
{V
I, and
(V)}== II or X is fixed point free on I,
and we
we may assume the latter. But then,
by 5.14,
5.14, mm =
=III
by
1 11 ==00mod
modp,p,contradicting
contradicting m < p.
(EL,
eaEA
,$:
primes with
withqq >
> p,
oforder
orderpq,
pq, X
X EE Syl,(G),
Syl p (G),
p, G
G a group of
(27.18) Let
Let pp and
and q be primes
faithfulFG-module
FG-modulewith
with(pq,
(pq,char(F))
char(F)) =
= 1 and
and CV(X)
Cv(X) =
= 0.
and V a faithful
0. Then G
cyclic.
is cyclic.
Proof. Extending
that F contains
Extending FFififnecessary,
necessary, we
we may
may assume with 27.12 that
qth root of 1.
a primitive
primitive qth
1. By Exercise 2.5, G has
has aa normal
normal Sylow
Sylow q-group
H. As
As (pq,
(pq,char(F))
char(F))==1,1,VVisisthe
thedirect
directsum
sumofofirreducible
irreducibleFG-modules
FG-modules
by Maschke's
so H isis faithful
Maschke's Theorem,
Theorem, so
faithful on
on one
one of
of these
these irreducibles,
irreducibles, and
hence we may assume V is an irreducible FG-module. So, by
by 27.17, V is a
homogeneous FH-module. Hence 27.16 says H acts
acts by scalar multiplication
on V, so H <(Z(G).
Z(G).Thus
Thus G
G ==HX
HXisiscyclic.
cyclic.
136
136
Change ofjield
of field of
ofaa linear
linear representation
Remarks.
Remarks. The
Theclassical
classicaltheory
theoryof
oflinear
linearrepresentations
representationsof
offinite
finitegroups
groupsconcongo relatively
relatively
siders representations over the complex numbers where things go
smoothly. Unfortunately many questions about finite groups require consideration of representations
representations over fields of prime characteristic,
characteristic, particularly finite
For example
example we've
we've seen that the study of
of representations
representations in the
the category
category
fields. For
group G
G on
on an
anelementary
elementary abelian
abelian p-group
p-group E is equivalent to the
of groups of aa group
of GF(p)G-representations on E
E regarded
regarded as a GF(p)-space.
GF(p)-space. Represenstudy of
requires the
the kind
kind of
of techniques
techniques
tation theory over such less well behaved fields requires
introduced in
in this chapter.
Lemma 27.18
27.18 provides
provides one application
of these
introduced
chapter. Lemma
application of
36.
techniques, and we will encounter
encounter others
others in section
section 36.
Little
decomposition studied
Little use
use is made
made in this book of the Jordan decomposition
studiedin
in section
section
is however
however fundamental
fundamental to the study of groups
groups of Lie
Lie type
type as
as linear
linear groups
groups
27. It is
or algebraic groups.
groups.
Exercises for chapter
chapter 9
1. Let
Let U,
U, V,
V, and
and W
W be FG-modules.
FG-modules.
(1) Define
by v(u(a4)))
_
Define 0:
4: L(U, V;
V; W) ->
+ HomF((U,
HOmF((U, HomF(V,
HOmF(V, W))) by
V(u(a!@))=
(u, v)a
v)a!for
for uu EE U,
U,Vv EE V,
V, and
and aa! EEL(U,
L(U,V;
V;W).
W).Prove
Prove04isisan
an isomorisomorphism of F-spaces.
F-spaces.
(2) Prove
Prove 04 is
is an
an isomorphism
isomorphism of L(U, V; F)
F ) with
with HomF(U,
HOmF(U,V*),
V*), where
V* is the dual of V.
V.
(3) G
G preserves
preserves ff EE L(U,
L(U, VV;; FF)) if ff (ug, vg) =
=ff(u,
(u,v)
v) for
for each g E G.
Prove G preserves
preserves ff ifif and
and only if f0
f 4EEHomFG(U,
HomFG(U,V*).
V*).
(V, Vv')
B) the
bean
anautomorphism
automorphismofofFFofoforder
orderatatmost
most2 2and
andLG
LG(V,
(4) Let
Let06be
set of sesquilinear
sesquilinear forms on V with respect to 06 which are preserved
LG(V,V9)
v') # 0
by G. Assume V
V is an irreducible
irreducible FG-module.
FG-module.Prove
ProveLG(V,
if and only if V is isomorphic to (Vs)*
(Ve)* as an FG-module, in which
case each member of LG(V,
9)# is
is nondegenerate.
nondegenerate. If V is absolutely
LG(V,V
ve)#
absolutely
LG(V,Vv')B) are
each
irreducibleprove
prove the
the members
membersof
of LG(V,
irreducible
are similar,
similar, ifif60 =
= 11each
= 2 some
member is symmetric
symmetric or
or each
each is skew symmetric,
symmetric,and
and if 101
161 =
member is hermitian symmetric.
ProveLemma
Lemma27.3.
27.3. (Hint:
(Hint:Use
Usethe
thetheory
theoryof
of modules
modulesover
overaaPID
PIDas
asin
in the
the
2. Prove
proof of 27.11.)
27.11.)
3. Let
Let n1
nl and
and n2
n2 be
be FG-representations,
FG-representations, Xi
xi the
the character
character of nl,
ni, and
and Xx the
the
nl 8
n2. Prove
Prove Xx ==X1
~ 1X2,
x 2that
; is,
E G, X(g)
~ ( g=
character of 71
character
0 72.
that
is, for
for each
each gg E
_)
X1(g)X2(g)
x1(g)x2(g>.
character of a complex G-representation.
G-representation.
4. Let
LetGGbe
beaafinite
finitegroup
groupand
and X
x the character
Prove X(g)
= X*(g)
= X(g-1)
X(g) =
x*(g) =
x ( ~ - ' )for
for each
each gg EEG,
G,where
whereX*
X * is
is the
the character
character
thecomplex
complexconjugate
conjugateof
ofX(g).
~(g).
of the dual representation
representation and
andX(g)
ft) isisthe
5. (Spectral
(Spectral Theorem) Let V
V be
be aafinite
finitedimensional
dimensional vector
vector space
space over the
complex numbers
numbers and
and ff aapositive
positive definite
definite unitary form on V. Then for
Minimal polynomials
137
each g EE O(V,
O(V, f)
f )there
thereexists
existsan
an orthonormal
orthonormal basis for (V, ff)) consisting
consisting
of characteristic vectors
vectors for
for g. In particular every
every element
element of
of O(V,
O(V, ff)) is
is
semisimple.
semisimple.
6.
), ii==1,1,2,
2, bebe2-dimensional
(K, fifi),
2-dimensionalsymplectic
symplecticspaces
spaces over
over a field F
F
6. Let
Let(Vi,
and let
be the
the group
group of
of similarities gg of K; that
let V
V ==V1
Vl ®
@V2.
V2. Let O(V1)
A(Vi) be
is g EE GL(VV)
withfi(ng,
fi(xg, yg)
yg) =
= X(g)
fi(x, y) for
GL(K) with
h(g)fi(n,
for all x,
n, yy EE Vi,
K, and
and some
some
h(g) EE F'P.
. Prove
Prove
X(g)
(1) There
symmetricbilinear
bilinearform
formff =
There exists
exists aa unique
unique nondegenerate
nondegenerate symmetric
=
f,
fl ®
@f2
f2 on
onVVsuch
such that
that
f(vl ® v2, U1 ® u2) = fi(vi, ui).f2(v2, u2), ui, vi E V.
(2) There
form QQ on
on V associated
to ff with
with
There is aa unique
unique quadratic
quadratic form
associated to
Q(vl ®v2)
@ v2)==00for
forall
all vi
vi EE Vi.
V,.
(V,Q)
Q)isisaa4-dimensional
4-dimensionalhyperbolic
hyperbolicorthogonal
orthogonalspace.
space.
(3) (V,
Let Di
A, ==O(Vi,
A(Vi,fi),
fi), Gi
Gi==O(Vi,
O(V,,fi),
f,),and
anditnthe
thetensor
tensorproduct
product reprep(4) Let
Al Xx A2
A2on
on VV(cf.
(cf.the
theconvention
conventionbefore
before 25.7).
25.7).
resentation of A ==O1
resentation
ProveA7r
< O(V,
A(V, Q)with(gi,
Q) with (gl,g2)ir
g2)nEEO(V,
O(V,Q)
Q)ififand
and only ififk(gl)
h(g1) =
Prove An 5
{(,LI,-1I):.l
~ ( ~ 2 ) -ker(n)
'.
=
I): h E F#}.
F'}.
)1(g2)-1.
ker(ir)
= (@I, h-'
(5) Let
Let a:a :(V1,
(Vl, fl) ->
+(V2,
(V2,f2) be
be an
an isometry.
isometry. Prove
Prove there
there is
is aa unique
unique
t EE GL(V)
va)t =
= vv ®
GL(V) with
with (u (&
@ va)t
@ ua.
ua.Prove
Prove tt isisaatransvection
transvection or re(Aln)' =
A2n, and (GI7r)`
(Gin)' =
=G27r.
G2n.
flection in O(V, Q), (O17r)`
= O2ir,
(6)
Q)== (An)(t).
(on)(t)
(6) O(V,
A(V, Q)
(7) Q(V,
O(V,Q)
Q)==(G1G2)ir
(G1G2)n=ZSL2(F)
SL2(F)**SL2(F),
SL2(F),unless
unlessIFS
IFl ==2.2.
7. IfIf itn isisananirreducible
irreducibleFG-representation
FG-representationand
andOr
a E Aut(F),
Aut(F), then
then 7r°
n u is
is an
an
irreducible FG-representation. IfIf xX is
is the character of
of nit then X°
xu isis the
the
character of
of n7r°,
where xU(g)
X°(g) =
= (X
( ~ ((g))'
g ) ) "for g E G.
G.
u , where
8. Let
V
be
an
n-dimensional
vector
space
over
a
field
Let V be an n-dimensional vector space over a field FFof
ofprime
primecharaccharacteristic p and
and xn an
anelement
element of
of order
order pp ininGL(V).
GL(V).Assume
Assume nn >>p.
p.Prove
Prove
dim(Cv(n)) >>1.1.
dim(Cv(x))
9. Let
vector space
space over
overaafield
fieldFF,, ff aanontrivial
Let VVbe
beaafinite
finite dimensional
dimensional vector
nontrivial
sesquilinear
form
on
V
with
respect
to
an
automorphism
9
of
finite
sesquilinear form
V with respect to an automorphism 6 of finite order
order
m,
m, and G ==O(V,
O(V,f).
f ).Assume
AssumeGGisisirreducible
irreducibleon
onV.
V.Prove
Prove that
that either
(1) VVisisFG-isomorphic
FG-isomorphictotoVe
V' and
andGGpreserves
preservesaanondegenerate
nondegenerate bilinear
V, or
form on V,
(2) m
rn isis even
even and
and VVisisFG-isomorphic
FG-isomorphictotoV02
V" but
but not
nottotoVB.
v'. Further
Further
V=
=FF®K
mKUUand
andGGpreserves
preservesaanondegenerate
nondegeneratehermitian
hermitian symmetric
symmetric
form on U,
U ,where
where K
K isisthe
thefixed
fixed field
field of
of 92
o2 and
and U
U isisan
anirreducible
irreducible
KG-submodule
V.
KG-submodule of V.
(Hint: Use
Use Exercise
Exercise9.1
9.1 and
and the
the fact that (V*)*
(Hint:
(V*)* is
is FG-isomorphic
FG-isomorphic to V.)
bean
anirreducible
irreducibleCG-representation
CG-representation and
and or
a a 1-dimensional
I-dimensional CG10. Let itn be
representation.
CG-representation.
representation.Prove it
n®
@ or
a is an irreducible CG-representation.
10
Presentations of groups
A group
group FFisisfree
freewith
withfree
freegenerating
generating set
set XXififititpossesses
possessesthe
thefollowing
following
universal property:
property: each
each function
function aa:: X + H of
universal
of X
X into
into aa group
group H
H extends
extends
uniquely to a homomorphism of F
F into
intoH.
H.We
Wefind
findin
in section
section28
28 that
that for
for each
each
cardinal C there exists (up to isomorphism) a unique
unique free
free group
group F with
with free
generating
cardinality C.
the largest
largest group
group generated
generated
generating set
set of cardinality
C. Less
Less precisely: FFisisthe
by X.
X.
If W
alphabet X
U X-',
X-1, it develops
develops that there
there is
is also
also aa
W is a set of words in the alphabet
XU
largest group G generated by
by X
X with
with ww =
= 11in G for each w EE W.
W. This
This is the
group Grp(X :: W)
W) generated
generated by
by X
X subject
subject to
to the
the relations
relationsww =
= 11 for w EE W.
W.
In section 29 we investigate Grp(X:
Grp(X: W) when
when X
X=
= {x1,
{XI,...
. . ., ,x,x,)} isis finite
finite and
W consists
consists of the words
words (xi xj)'"ii =
= 1,1,for
for suitable
suitableintegral
integralmatrices
matrices(mid).
(mij). Such
Such
a group is called
called aa Coxeter
Coxeter group.
group. For
For example
example finite
finite symmetric groups are
Coxeter groups. We find
find that
that Coxeter
Coxeter groups
groups admit
admit aa representation
representation nir:: G +
O(V,
Xirn consists
O(V, Q)
Q)where
where(V,
(V,Q)
Q)isisan
anorthogonal
orthogonalspace
spaceover
over the reals and X
consistsof
of
Q) turns
turns out
out to
to be Euclidean space. Finite
Finite Coxeter
Coxeter
reflections. If G is finite (V, Q)
groups are investigated via this representation
30, which
which develops
develops
representation in section
section 30,
the elementary
elementary theory of root
root systems.
systems.
The theory of Coxeter groups will be used extensively in
in chapter
chapter 14 to
to study
study
the classical
classical groups from
from aa geometric
geometricpoint
point of
of view.
view.
28 Free
Free groups
groups
free with free
free generating
An object G in an algebraic
algebraic category A
A is
is said
said to be
befree
generating
in AA and
and a:
a: X + H
set X if X is a subset of G and, whenever H is an object in
exists a unique morphism B:
P: G + H of G
is a function from X into H,
H,there,
there,exists
G
into H extending
extending a.
a.This
Thissection
sectiondiscusses
discusses free
free groups.
groups.
recall that a monoid is
is a set G together with an associative binary
But first recall
operation on G possessing an identity 1.
1. Here's an
an example of a monoid. Let
xi in X; n is the
~ 1 x ...
.2. .X,
X, with xi
X be a set.
set. A word in X is
is aa finite
finite sequence
sequence xlx2
length of the word. The empty sequence is allowed and denoted by 1. Let M
x1 ..... .x"
be the set of words in X and
and define
define the product of two words xl
x, and
and
ymto
tobe
bethe
theword
wordxlxl....
x"yl .... .Ym
of length
lengthnn + m. Observe that
that M
M is
is
yl
. .x,yl.
y, of
yi . . . y,
a monoid with identity the empty sequence
sequence 1.
1. Indeed
Indeed
+
(28.1) M
M is
is aa free
free monoid with free generating set X.
Free groups
139
H isisaamonoid
monoid and
and a:
a:X
X + H is
is aa function
function then
then aa can
canbe
be extended
extended
For if H
to
to aa morphism
morphism/3:
fi:M
M + H defined
defined by (xl ...
. . .x7f)p
x,)fi =
=xla
xla.... .x7za.
.x,a. /3fi isiswell
well
defined as each word has a unique representation as a product of members of
X. Evidently
Evidently/3
fi is the unique extension
extensionof a.
a.Indeed
Indeedin
ingeneral
generalin
inany
anyalgebraic
algebraic
X.
is aa generating
generating set for an object G and aa:: X + H isis aa function
function
category if X is
to aa morphism of G
G into
into H.
H .This
Thisisis
then there is at most one extension fi of a to
anotherextension
extension then
then K
K =={g{gEEG:
G:gj3
gfi ==g/3'}
gfi'} isis aa subobject
subobject
because if f'fi'isisanother
X.
G containing
containing X.
of G
=YY U
U Y-'
l Y-'
and yy i-+
H y-1
y-l isis aa bijection of
Next assume X =
Y-1 with
with YYffl
Y-1=
=0
0 and
Y with
for each y E Y; thus
X-l is
isaapermutation
permutation
Y
with Y-'.
Y-1. Set
Set(y-l)-l
(y-1)-1 =
= yy for
thus xx H
H x-1
of X
X of
of order
order2.
2. Define
Define two
two words
words uu and
and w
w to
to be
be adjacent
adjacentififthere
thereexist
existwords
words
Mand
and xx EE XXsuch
such that
that u ==axx-lb
axx-'b and
andww==ab,
ab,ororvice
viceversa.
versa.
a , bb EE M
a,
Thus
Thus adjacency
adjacency is
is aareflexive
reflexive and
and symmetric
symmetricrelation.
relation.Define
Define an
anequivalence
equivalence
existsaasequence
sequence
u1,. ....
relation -- on M by uu -- w if there
there exists
u u==ul,
. , u,, u ==
relation
w of
for each
each ii,, 11 5< ii <
of words
words such that
that u,
ui and
and u;+l
ui+l are
are adjacent
adjacent for
< n. That
That
is ---isisthethetransitive
transitiveextension
extensionof
ofthe
theadjacency
adjacency relation.
relation. Write
Write w for
for the
the
equivalence
classofofaa word
wordwwunder
under--- and
equivalence class
and let
let FFbe
bethe
theset
setof
ofequivalence
equivalence
classes.
classes.
thenuw
uw--- vw and
wu--- wv.
(28.2)
(28.2) If u,
u , v,
v, w EE M
M with
with uu -- v then
and wu
wv.
There is a sequence
sequence uu =
u 1, ...
. . .,,u,u, =+=
wordswith
withu,uiadjacent
adjacentto
to
Proof. There
= u1,
v vofofwords
ui+1.
= wi
and uw
uw =
= w1,
VW,
ui+l. Observe Uuiw
~ W=
wi isis adjacent
adjacent to w;+1
wi+l and
wl, ...
. . ., ,ww,= =
vw,SO
so
UW "-VVW.
W.
UW
by uv
ii3==ii-D.
iZ.
28.2 this product is
is well
well defined.
defined.
Now define a product on FFby
By 28.2
Further
Further the
the product
product of
of the
the equivalence
equivalence classes
classes of
of the
the elements
elementsxn
x;', 1, ..... .,,xi
x c1l isis
an inverse
F is
X1 ..... .Xn,
X,, so F
is aa group.
group.Hence
Hence
inverse for
for XI
groupand
andwwi-+
Hw
Wisisaasurjective
surjectivemonoid
monoid homomorphism
homomorphism of M
(28.3) FFisisaagroup
F.
onto F.
(28.4) FFisisaafree
freegroup
groupwith
withfree
freegenerating
generatingset
setY.
P.
Proof. Observe
Observefirst
firstthat
that Y
Y generates
generates F.
F.This
Thisfollows
follows from
from 28.3
28.3 together with
P-'.
the fact that X generates M
M and
and XX==Yy UUY-1.
=
Now let H be a group and
andaa:: EY +
- HHa function.
Define
a function.
Define/B:
fi: XX + H by y/3
yfi =
y-l Q =
= (ya)-1
ya and
and y-lfi
(ya)-' for
foryyEEY.Y.As
AsMMisisaafree
freemonoid
monoidon
onX,
X,/3
fi extends to a
ya
by 08 ==wy.Imustshow6iswell
wy. I must show S is well
Define S: F + H
morphism
y: M + H.
morphismy:M
H.Define6:F
HbyW6
defined; that
that isis ififuu -- vv then
then uy
u y ==vy.
v yIt. Itsuffices
sufficestotoassume
assumeuuisis adjacent
adjacent to v,
say
u = axx-1b and v = ab. Then uy=(axx-'b)y
= (axx-1b)y=ayxfi(xfi)-'by
= ayx/B(xp)-l by==a ayby,
sayu=axx-'bandv=ab.Thenuy
yby,
Presentations of
of groups
140
140
as desired. Evidently
Evidently 6bisisaahomomorphism
homomorphismextending
extendinga a.
AsYf generates
generates FF,, an
. As
earlier remark shows 6b is the unique extension of
of a.
a.
Lemma
24.8 shows
shows that
that for
for each
each set S there exists
Lemma 24.8
exists a free group
group with
with free
generating
S. The
The universal
universal property
property implies:
implies:
generating set S.
(28.5) Up
Upto
to isomorphism
isomorphismthere
thereexists
exists aa unique
unique free group
group with free generating
set of cardinality C for each cardinal C.
C.
If W C
Grp(Y ::W)
W)for
forthe
thegroup
groupFF/N,
2M
M isis aa set
set of words in X, write Grp(Y
I N , where
N ==(WF)
isisthe
of F.
(wF)
thenormal
normalsubgroup
subgroup of
of FFgenerated
generated by
by the
the subset
subset W
w of
F.
Grp(Y::W)
W)isisthe
thegroup
groupgenerated
generatedbybyYYsubject
subjecttotothe
therelations
relationsww== 1 for
for
Grp(Y
w cEW.
W.That
ThatisisGrp(Y
Grp(Y::W)
W) is
is the
the largest
largest group
group generated
generated by the set Y in which
w=
1
for
each
w
c
W.
To
be
more
precise
Grp(Y
: W)==FF/N
=G
= 1 for
E W. To
more precise Grp(Y : W)
IN =
G with
with Y
and W identified with
(yN:
(w- N: wEc W),
(jiN: y cE Y)
Y) and (wN:w
respectively.
AsYYgenerates
generatesFF,
generatesG.G.AsAsw wEEN,
N,ww =
= 11 in G
, YYgenerates
G for
for
respectively. As
W. So
So G
G is
is generated
generated by
by Y
Y and each of the words in
each w EE W.
in W is
is trivial.
trivial.I'll
I'll
say the
the relation
relation w
w=
= 11is
indicate that
that w
w=
= 11in G.
also say
is satisfied
satisJied in G to indicate
G. G is
is
the largest
largest group
group with these properties
properties in
in the
the following
followingsense:
sense:
(28.6) Let
Let aa:: Y
Y4
- YaY be
a bea function
a functionofofYYonto
ontoaaset
set Ya,
Ya,HHaagroup
groupgenerated
generated
(28.6)
by
Ya,
and
W
a
set
of
words
w
=
y8,
...
ynn
in
Y
UY-l
with
wa
=
(y,
a)" ...
byYa,andWasetofwordsw = y:' . . . y F i n Y ~ Y - lwithwa = (yla)"
...
(yna)8n =
= 11 in H for each w E W.
Yaa and satisfies
(y,a)"
W. (That
(That is H is
is generated by Y
the relations
relationsww =
= 11for
for wwEEW.)
W.)Then
Thena aextends
extendsuniquely
uniquelyto
toaasurjective
surjective
homomorphism
H.
homomorphism of Grp(Y :: W) onto H.
Proof. Let
exists aaunique
uniquehomomorphism
homomorphism
Proof.
LetFFbe
bethe
thefree
freegroup
groupon
on Y.
Y. Then there exists
8: FF -4H HofofF F
onto
H vN
j?:
ontoHHextending
extending a.
a .Let
Let N ==(WF)
(WF)and
and b:
6: vv ++
vN the
the
natural
mapofofFFonto
ontoGG== FF/N.
N=
For ww =
= xl ..... .xn
natural map
I N . Then
Then N
=ker(b).
ker(6). For
x, E
EW
W
withxixtEEX,
X,11==wa
wa =
= xxla
....x,a
xna == xlj?
xl/'....
xn _= (XI,
xnV==wj?,
4,
with
l a ..
. .x,j?
(XI,....
. .xn)j?
ker(j?), as
as N
N isisthe
thesmallest
smallest normal
normal subgroup
subgroup of F
F
so w EEker(j?).
ker(b). Thus
Thus NN 5
< ker(8),
W. But
Butas
asNN (
< ker(j?),
ker(8), ?j induces
inducesaahomomorphism
homomorphismy :y:GG4- H
containing W.
H with
with
by. As is surjective so isy.Also,fory
y. Also, for Ey Y,
c Y,
= yay
yby =
= yy.
j? =
=6y.Asj?issurjectivesois
y a ya
==
yj?y=
presentation for
for aa group
group G
G is a set Y
of G together
with a
A presentation
Y of generators of
together with
set W of
of words
in Y
YU
U Y-'
Y-l such that
that the
the relation
relationww =
= 11isis satisfied
words in
satisfied in G
for each w EE W
W) onto
onto G
G described
described in
in
Wand
and the
the homomorphism
homomorphism of
of Grp(Y
Grp(Y : W)
28.6 is an isomorphism.
isomorphism. I'll
I'll summarize this
this setup with
with the
the statement
statement
lemma 28.6
G ==Grp(Y
Grp(Y::W).
W). Every
Every group
group has
has at
at least
least one
one presentation;
presentation; namely:
Coxeter groups
141
(28.7)
(28.7) For
For each
each group
group G,
G,
G ~ ~ ( xy(xy)-1
G:X~(X
X,
- ' yy EE G)
G)
G ==Grp(G:
==~1,1,)x,
presentation for
for G.
G.
is a presentation
Proof.
Proof. Let
Let gg H++k gbebea bijection
a bijectionofofGGwith
withaaset
setG,
G,let
letFFbebethe
thefree
freegroup
group
on G,
set of
of words
words, ?xy(xy)-1,
G, and
andlet
letNN =
= (WF).
G , let W
W be the
the set
Y E ) - ' , x ,x,y yEEG,
(WF).
Evidently
by W
W so
sothe
themap
mapgg++
H gg extends
extends to
to a
Evidently G
G satisfies
satisfies the relations defined by
homomorphism
F onto G with N
N<
toshow
showNN =
= ker(a).
homomorphismaa of F
_( ker(a).
ker(a). It remains to
ker(a) - NNofofminimal
Assume otherwise
be a word
word in ker(a)
minimal
Assume
otherwiseand
andlet
letvv== Zl
1 . . .Zn be
length n.
n. As
As 11EEN,
N,nn>> 0.
0.IfIfnn== 1 then
thenxlx1==11and
andv v==jlla =
= 1.1x1
flx12;'1 EE W,
W,
contrary
v. Hence
Hencenn >
> 22 so
length
contrary to
to the choice of v.
so v ==x1x2u
ZIZzufor
for some
some word u of length
N,, so
sow-1v
= .1x2
ker(a) -- N.
n --2.2.Now
Now w
w =x l x 2 ( ~ ) -E
E
' W
W CN
w-'v =
m uu EE ker(a)
N.
As (xix2)u
is
of
length
at
most
n
1,
the
choice
of
v
of
minimal
length
(m)u
- 1, the choice of v of minimal length is
is
contradicted.
contradicted.
Here's aa slightly
slightly more
more nontrivial
nontrivial example. The dihedral group of order 2n is
defined to be the semidirect
semidirectproduct
productof
ofaacyclic
cyclicgroup
groupXX=
= (x)
(x) of order n by
Y=
(y)of
oforder
order2,
2, with
with respect to the automorphism
The
a group Y
= (y)
automorphismxY
xy =
= x-'.
x-1. The
case where n ==oc
coand
andxxisisthe
theinfinite
infinitecyclic
cyclicgroup
groupisisalso
alsoallowed.
allowed.Denote
Denote
group of
of order
order 2n
2n by
by D2n.
Dzn.Dihedral
Dihedral 2-groups
2-groups have
have already
already been
been
the dihedral group
p-groups.
discussed in the chapter on p-groups.
4
(28.8) D2n = Grp(x, y : xn = y2 = 1 = xyx).
If n ==oo
cothe
therelation
relationxn
xn==11isistotobe
beignored.
ignored.The
The proof
proof of
of 28.8
28.8 is
is easy.
easy. Let
D = X Y = Dzn and
D=XY=D2nand
By 28.6
28.6 there
thereisisaahomomorphism
homomorphisma aofofGGonto
ontoDDwith
withxaxa==xxand
andY
ya
= y.
a =
y.
Then
I divides
I I,so,
so,asas3"xn==1,1,1x11 I==nn and
anda a:
- XX isisan
divides 1x1,
: jiX4
an
Then nn ==Ix1x1
isomorphism,
whereX X== (2).
(x). Similarly,
Similarly,setting
setting
(y),a:
a: Y4- YY isis an
an
isomorphism, where
Y Y==(y),
Thus
X
<G
=
(x,
y),
so
G
=
XY.
isomorphism.
= XXY
soZYV =
= (3)-'. Thus X I
3 3 Y so
!G = (x, ), so = XY.
isomorphism. 1 =
Hence,
isomorphism, a
a itself
isomorphism.
Hence, as aa restricted
restrictedto
to X
X and
and Y is an isomorphism,
itself is an isomorphism.
29 Coxeter
Coxeter groups
groups
Define a Coxeter
Coxeter matrix
matrix of size
size n to
to be
be an
an nn by
by nn symmetric
symmetricmatrix
matrixwith
with11ss on
on
the main diagonal
diagonal and
and integers
integers of
of size
size at
at least
least 22 off
off the
the main
main diagonal.
diagonal.To
To each
each
matrix M
M ==(mid)
Coxeter matrix
(mij) of
of rank
rank nn there
there isis associated
associated aa Coxeter
Coxeter diagram:
of nn nodes,
nodes, indexed
indexedby
byintegers
integers115< ii _(
< n,
n, together
together with
this diagram consists of
an edge
edge of
ofweight
weightmij
mid--22joining
joiningdistinct
distinct
nodes
i and
1 <i i<<jj 5
< n. We
nodes
i and
j, j,
1 _(
142
142
Presentations of
of groups
groups
Presentations
will
will be
be most
most concerned
concernedwith
withCoxeter
Coxetermatrices
matriceswith
withthe
thefollowing
followingdiagrams:
diagrams:
o-o2
1
An
Cn
n-1
n
o-®
1
2
n-2 n-1 n
1
2
n-2
o-o
0
n-1
Dn
o-o
n
Thus
of size nn with mij
mid =
= 3 if
Thus aa diagram
diagram of
of type
type An
A, defines a Coxeter matrix of
Jili --j jI I ==1 1and
mid
=
2
if
Ii
j
I
>
1.
Similarly
a
diagram
of
type
Cn
defines
and mij = 2 if li - j 1 > 1. Similarly a diagram of type C, defines
aamatrixwithmij
matrix with mid==3 3i fifl iJi- - jjl I =
= 1l aand
n, mn-l,n =
4, and
n d i i,, jj <<n,m,-I,,
= mn,n-,
m ,,,- 1 ==4,and
mid=2ifIi-j!>1.
m i j = 2 i f l i - j ] > 1.
A
A Coxeter
Coxeter system
system with Coxeter matrix M =
=(mid)
(mij)of size n is
is aa pair (G,
(G,S)
S)
where
G
is
a
group,
S
=
(si:
1
<
i
<
n)
a
family
of
elements
of
G,
and
where
= (si:1 _( i ( n ) a family of elements of G , and
G
1,
G ==Grp(S:
G r p ( S(sisj)m,i
: ( ~ ~ s=
~=)1,
~ 1'1j_<( ii I<nn,, 11 <5 j <( nn).
).
G
G isisaaCoxeter
Coxetergroup
group ifif there
there exists
existsaa family
family SS such
suchthat
that (G,
( G ,S)
S )isisaaCoxeter
Coxeter
system.
system.
In the
theremainder
remainderof
of this
this section
sectionlet
let (G,
(G,S)
S)be
beaaCoxeter
Coxetersystem
systemwith
with matrix
matrix
In
M
.) of
of size
sizen.n.Let
LetSS=
= (si:
(si: 1 5
< i <5n).
M==(m1,
(mij)
n).Notice
Notice
s2 = (SiSi)mii = 1.
(29.1)
(29.1) Let
Let TTbe
bethe
theset
setofofconjugates
conjugatesof
of members
members of
of SS under
under G
G and
and for
for each
each
word
rn
in
the
alphabet
S
and
each
t
E
T
define
word rr ==r1rl ...
. . . rm in the alphabet S and each t E T define
N(r, t) = (i: t = ri
< i < m}.
Then,
rn ==rlri. . . .rk
Then,ififr1r l...
. ..rm
r i in
in G,
G , we
we have
have IN(r,t)l
IN(rl,t)j
t)lmod
mod22for
for
(N(r, t)j =
- IN(r',
each
eachtt EE T.
T.
Proof.
(±1}
Proof. Let
LetAAbe
bethe
the set
set product (
f1) x TT and
and for s E
E S define ssn
n cESym(A)
Sym(A)
by
where6(s,
8(s,
by (E,
( E , tt)sn
)sn =
=(EE(s,
( ~ 6 ( st t),
), , t t''),
S ) ,where
t ) t)
=_-1-1ififs
s=
=ttand
and 8(s,
6(s,t)
t)=
=+1
+1ifif
ss##t. t.Observe
Observethat
thatsn
s nisisan
aninvolution.
involution. I'll show
show (sinsjn)mii
(sinsjn)"lj =
=11for
for all
all i, j;
j;
hence,
extendsto
to aapermutation
permutation representation
representation nn of G
G on
on A.
A. In
In
hence, by
by 28.6,
28.6, nnextends
particular
homomorphismso,
so, ifif gg ==r1rl ...
. . rm
. rmcEGGand
andr1ri EES,S,then
then
particular nnisisaahomomorphism
(E, t)gn = (e, t)rln ... rnn = E
ri,
i=1
tri...r;_1)
t8)
Coxeter groups
143
t''...ri-') =
N(r, t).
=
=-11 exactly when i E N(r,
Further 3(ri,
6(ri,tr'...'i-1)
t). So
So (e,
( E , tt)g7r
)gn =
Further
(E(-l)IN(r,t)I,
t8)
( ~ ( - l ) I ~ ( ' , ' )tg)
l , and hence IN(r,
IN(r,t)t)lI mod 2 depends
depends only
only on g and not on r.
(si7rsj7r)m"j
1;equivalently
equivalentlyaa =
= (si
rsj7r)m'j fixes
nsjn)"'l == 1;
(s,ns,
fixes each
each
It remains to show (s,
t(s'si)»")
=" (EE,
=" ~1,
( E , tt)) E A.
A. Now
Now ((e,
E ,t)a
( ~ 6t (, S ~ S ~ )=
"( l~) 6tt),
),, as ((sisj)m'j
s , ~ ~ )=
1,]where
where of
(e,
t)a ==(Es,
course
course
n)"ll
8 = 8(si, t)E(sj, ts')....
)k) _
So
t(S,sj
But8(si,
6(si,
t('1'1)~)
=-1
- 1for
forsome
some00 <5kk <<mij,
m,j,precisepreciseSo we must show
show 86 ==1.1.But
-1) 1precisely
(si(sisj)2k+1~i.
sj )2k+1 si .
ly
s j )2ksi . Also
t(S'Sj
ly when
whentt==(si(sisj)2ksi.
Also8(sj,
6(sj,
~ ( S ~)ks')
S J ) =
~ ' ~preciselywhen
whent =
t=
Further,
Exercise 10.1, (si,
divides
Further, by Exercise
(s,,sj)
s j )isiseither
eitherdihedral
dihedralof
of order
order 2m,
2m, where
where m divides
+
at most
most 2.
2.Hence
Henceifift t $2
0 ((sisj))si
terms in
in 68 are
are +1,
mij, or of order
order at
((sisj))sithen all terms
1,
while ifif t EE ((sisj))si
exactly
(sisj)dIs"'jIsi
((sisj))si
exactly (2mij)/IsisjI
(2mij)/lsisjI terms
terms are -1,
-1 ,as
ast t==
( S ~ S ~ ) ~ lsi
~'~'J
for 0 <5dd<<
(2mij
)/ Isis j 1.
(2mIJ)/lsisjl.
(29.2) The
The members
members of
of SS are
areinvolutions.
involutions.
of G into Sym(A)
Proof. In
In the
the proof
proof of
of 29.1
29.1 aa homomorphism
homomorphism 7r
n of
Sym(A) was constructed for which ss7r
was
an
involution
for
each
s
E
S.
So
2 ==Isn
n was an involution for each s E
IsnII divides
divides
is (.But
Butof
ofcourse,
course,as
as((G,
Coxetersystem,
system,s?
s? =
= 1.
Is!.
G , SS)) isis aaCoxeter
1.
If H
of hh E H
H is
is aa group
group with
with generating set R then the length of
H with
with respect
respect to
length of
of aa word
wordwwin
inthe
theabhabet
alphabetRRUUR-'
R-1such
suchthat
thatww=
=h
R is the minimal length
in H.
= lR(h).
H . Denote
Denote this
this length
length by
by 1(h)
l(h) =
l~(h).
(29.3) Let g cE GGand
with gg =
= rr in
and rr ==r1rl.... .rm
. rmaaword
wordin
in the
the alphabet
alphabet S with
in G.
G.
Define
Define
r7(r)
=)II{t
It E T:
~ ( r=
T :IN(r,
IN(r,t)t)lI ==1 1mod
mod2}1
211
29.1. Then ~1](r)
=) l(g).
l(g).
in the notation of 29.1.
( r=
while
of l(g)
l(g) there
Proof. By
B y 29.1,
29.1, if r' ==ggthen
then1](r)
~ ( r=.1](r'),
=) .q(rl),
while by
by definition
definition of
rk with
with kk =
= l(g)
loss, m
m=
= l(g).
is
= ri
ri ...
. . . ri
l(g) and
and r' ==ggininG.
G.So,
So,without
without loss,
l(g).
is r'r' =
So
evidently~ 1](r)
l(g).IfIf~1](r)
there
1 <i i5<jj <5
So evidently
( r 5) <mm==l(g).
( r <) <mmthere
areare
i , j,i, 1j, 5
rj-1 ...r~
r,_I ...ri
= rj
,, so riri+t
....rj-1
rj-1 ==Yi+1
rj and hence g =
m, with ri
r"-~...r~
=
rJ.
riri+l..
ri+l......rj
I
r1
rl ..... .r1-1ri
ri-lri ..... .rj_1rj
rj-lrj ...
..rm
. rm==r1rl.... .ri_1ri+1
. ri-lri+l..... rj_1rj+1
. rj-lrj+l.....rm
. rmisisof
of length
length at
most m
m- 2,2,contrary
contraryto
to the
the choice of r.
(29.4) Let
H be
beaagroup
groupgenerated
generated by
by aa set
set R
R of
of involutions. Then ((H,
H ,R)
R ) is
is a
Let H
Coxeter
Coxeter system
system precisely when the following Exchange
Exchange Condition
Condition is satisfied:
satisfied:
ExchangeCondition:
Condition:
If Eri E
0 <i i5<nn,
andh h== rlr1. .....r,r EE HH with
Exchange
If ri
R,R,
0 5
, and
with
1(h)
= n and
l(h) =
and l(roh)
l(roh)<5 n,
n,then
then there
there exists 1 <
5 kk <5nnsuch
suchthat
thatror1
rorl . ..
. . rk-1
rk-1 =
=
r1
rk in H.
r l ....
. .rk
Presentations
Presentations of
of groups
144
144
Proof.
and let
let ri,
rt, hh,, satisfy the
Proof. Suppose
Suppose first
first that (H,
( H ,S)
S ) is aa Coxeter
Coxeter system and
hypothesis
of
the
Exchange
Condition.
Then,
setting
r
=
rori
. . .rn,
r,, ri(r)
~ ( r<5)
hypothesis
Condition. Then, setting r = rorl ...
l(roh)
29.3, so
sothere
thereare
arei, i,j,j,005<i i<< jj <5nnwith
l(roh)<<nn + 11 by 29.3,
with r;
r,?-"..ro=
= rrj-'.'-ro
J
and
hence rir, ..... .rj_1
and hence
rj-1 =
=rt+i
ri+l ...
. . rj.
.rj.ByBy29.3,
29.3,ri(ri
v(rl ...
. . .r,)
r,) ==1(h)
l ( h )==n,
n,so
so i ==0.
0.
Thus
Thus the
the Exchange
Exchange Condition
Conditionisis satisfied.
satisfied.
Conversely
Exchange Condition.
Condition.Let
Leta!:
a: RR +
-+ X
H ,R)
R ) satisfies
satisfies the Exchange
X
Conversely suppose ((H,
be a function into
into aagroup
groupXXsuch
suchthat,
that,for
foreach
eachr,r,s sEER,
R,(rasa)lrsl
(rasa)I's' ==1.1.ItItwill
will
suffice
a extends
extends to
to aa homomorphism
homomorphism of H into
intoX.
X.
suffice to show a!
Let h EE H,
H , nn ==l(h),
l(h),and
andr1
rl ...
. . .rn
r, ==hh==s1sl ...
. .sn
. s, with
with ri,
ri,si
si cER.
R.Claim
Claim
+
'-'.__Y°
r1a
rna =
sna
r l a ....
. .r,a
=sla
sla!..... .s,a
and
and
{rt:
< nn)] =
= {s,:1
< i <5 n).
(ri:115< ii 5
(si:1 5
n].
Assume not and pick aacounterexample
Assume
counterexample with
withnnminimal.
minimal.Then
Thenl(sl
l (sih )h)==nn--11 <<
l(h),
l(h),so,
so,by
by the
theExchange
ExchangeCondition,
Condition,s1r1
slrl ...
. . .rk-1
rk-1 ==r1rl.... .rk
.rkforforsome
somek.k.Hence
Hence
s1r1
rk-lrk+l ....r,rn =
= hh= =
s1 sl....sn
rn ==S2s2..
... Sn.
slrl ....
. .rk-lrk+l..
. .s,sosor1r...l . .rk_1rk+1
.rk-lrk+l ...
.. .rn
.s,.
Thus,
rna =
= sea
Thus, by
by minimality
minimality of n,
n ,r1a
rla ...
. . .rk_1ark+la
rk-lark+la.. ...
.rna
s p ...
. . sna
. s,a and
and
{S2,
($2,...
. . .,,Sn)
s,] =
{ri:ii #
k ] .Also
Also if k <<nnthen,
then, by
by minimality
minimality of
= {ri:
0 k).
ofnn,, slurla!.
siaria .... .
rk_la
which
rk-la ==r1a
r1a.... .rka
.rkaand
and{Si,
Isl, r1,
rl , ...,
. . .rk_11
, rk-1) ==jr,
(rl...,
. .rk),
., rk],
whichcombined
combinedwith
with
the
the claim.
claim.So
Sokk=
= nn,, slrl
sir, ...
the last
last set of equalities establishes the
. . .rn-i
r,-1 ==h,h ,and
and
{ri
,
...
,
rn_i)
=
{s2,
...,
sn
)
.
Similarly
rise
...
sn_1
=
h
and
{s1,
...
,
Sn_1)
{ r l ,. . . , r n v l )= (s2,. . . , s,). Similarly rlsl . . .s,-1 = hand { s l , . . . , s,-1) =
=
{r2
{s1(sl..... .sn
), establishing
{r2...
. . ., ,rnr,).). In particular jr,
{rl....
. . .rn)
,r,)_ =
. , s,),
establishinghalf
halfthe
theclaim.
claim.
Replacing
si, ....,
Replacing r1
rl ....
. . . ,rn,
r,, and si,
sl,r1,
r l , ...
. . .,,rn-i
r,-1 and
by si,
and r1,
r l , sl,
..,
sl, ...,
. . . ,sn,
s,, by
sn_i,
s,-1, and
and continuing
continuing in this
this manner,
manner, we
we obtain
obtain (sire)n/2=(rise)n/2
(slr1)"I2=(rls1)"I2 or
or
ri(Siri)(n-1)/2 =
under a failing
r l ( ~ ~ r ~ ) ( "=
- ' )~/l~( r l s ~ ) ( ~with
with
- ~ )equality
/~,
of images under
failing in
in
so the
the order
order m
m of
of s1r1
slrl in H
the respective
respective case.
the
case. ItIt follows
follows that
that (slrl)"
(sire)n =
= 1,1, so
divides
the order
order of
of slarla
slarla divides
divides n. But by hypothesis
hypothesis the
dividesm,
m ,so
soequality
equality of
of
images
images under aa does
doeshold,
hold,aacontradiction.
contradiction.
So
So the
the claim
claim isisestablished.
established.Since
SinceCoxeter
Coxeter systems
systems satisfy
satisfy the
the Exchange
Exchange
Condition
Condition we
we can
can record:
record:
Si(risi)(n-1)/2,
(29.5)
(29.5) Let g cEGGwith
with1(g)
L(g)==mmand
andr1,
ri, ttti cESSwith
withr1
rl ...
. . .rn
rm==t1tl ...
. . .to
tm==g.
g.
Then
{rt:1
Then (ri:
1 5<i
i 5<m)={ti:1
m ) = {ti:1<5ii <m).
5 m).
Now back
back to the
the proof
proof of
of 29.4.
29.4. Define
Define a:
a : H ->.
-+ XX by
by ha
ha ==r1a
rla! ...
. . .rna,
ma!,
Now
for
for hh==r1rl ...
. . .rn
r, with
with nn ==1(h)
l ( h )and
and r,ri EE R.
R. The
The claim
claim shows
shows a to
to be
be well
well
defined.
Let'ssee
seenext
nextthat
that(rh)a
(rh)a== raha
1(h)
defined. Let's
ruhafor
forr rc ER.R.IfIfl(rh)
l ( r h=) =
l ( h )++11
this
this isis clear,
clear, so
so assume
assume not.
not. Then,
Then, by
by the
the Exchange
Exchange Condition,
Condition, rr1
rrl ...
.. .rk-1
rkPl =
=
r1
somekk5< n.
n. By
Bythe
theclaim,
claim,rurla
raria ....
rk-1a =
rl ..... .rk
rk for
for some
..rk-la
=r1a
rla ...
. . .rka.
rka.Also
Also
rh
- 1.
r h ==r1rl . .....rk-irk+1
rk-lrk+l .. ...rn
r,, is
is of
of length at most n 1.As
As 1(h)
l(h)=
=n,
n ,we
weconclude
conclude
l(rh)
l ( r h )=
=nn --1.1.So
So (rh)a
(rh)a!==r1a
rla!..... .(rk-i)a(rk+1)a
( ~ ~ - ~ ) a ( r ~....+
,arna
~ )=
a rurla!.
.raria
.
=
.... .rna
rna ==
raha,
establishing
raha!,
establishingthe
thesecond
secondclaim.
claim.
ItIt remains
showgaha
gaha ==(gh)a
remains to show
(gh)afor
forg,g,hhc EH.HAssume
. Assumenot
notand
andchoose
chooseaa
counter
minimal.By
Bythe
thelast
lastparagraph,
paragraph,1(1(g)
1,so
sogg=
= rrk,
counter example
example with 11(g)
( g )minimal.
g ) >> 1,
k,
Coxeter groups
145
145
R ,kk EE HHwith
with l(k)
l(k) =
Then (gh)a
= rakaha
rakaha==
rr EER,
= l(g) - 1.1.Then
(gh)a =
= ra(kh)a
ra(kh)a =
guhabybyminimality
minimalityofofl(g),
l(g),completing
completingthe
theproof.
proof.
gaha
Let VVbe
beanann-dimensional
n-dimensionalvector
vector space
space over
over the
the reals
reals RR with
with basis
basisXX==
Let
(xi:1
<
i
<
n)
and
let
Q
be
the
quadratic
form
on
V
with
(xi,xj)
(x,: 1 5 i 5 n) and let Q be the quadratic form on V with (x,, x,) ==
- cos(n/m,,), where
where (( , , ))isisthe
thebilinear
bilinearform
form determined
determined by Q
Q as
asininsecsec-cos(7r/mid),
(v, v)/2
v)/2 and
and
tion 19.
19. Thus, for uu,, vv E
tion
E V,
V, Q(v)
Q(v) =
_ (v,
(u, v) = Q(u) + Q(v) - Q(u + v).
(29.6)
(29.6) (1) Q(xi)
Q(x~)==1/2.
112.
(2)
(2) (xi,
(xi, xj) <500for
for ii 0#j,j,with
with(xi,
(xi,xj)
xj)==0 0ififand
andonly
onlyifif mii
mij ==2.2.
Proof.
cos(h)/2
1/2.
Similarly
Proof. mii
mii =
= 1,
1,so
so Q(xi)
Q(xi) =
=(xi,
(xi, xi)/2 =
=- cos(n)/2==
112.
Similarly ifif
ii ##j jthen
thenmil
mij>>2,2,soso(xi,
(xi,xj)
xj) = -cos(7r/mid)
cos(n/mij) <500with
withequality
equalityififand
andonly
only
mij ==2.2.
ifif mil
By 29.6
29.6and
and22.6.2
22.6.2there
thereis
on V
V with
withcenter
(xi);
By
is aauniquereflection
unique reflection riri on
center (xi)
; moreover
moreover
vri ==vv --2(v,
2(v,xi)xi
xi)xifor
for vv cEV.
V.
vri
(29.7) For i 0#j,j,rirj
rirjis isofoforder
ordermil,
mij (ri,
, (ri,rf)
r j )=E
D2m,j,
and ifif mil
mij >>22then
then
(29.7)
D2,,,,i,
and
I
(ri,rj)
r j )isisirreducible
irreducibleon
on (xi,
(xi,xj).
xi).
(ri,
'
Proof.
(ri,(ri,
rj),rj),
mm
==
mid,
7r/m.
Proof. Let
LetUU ==(xi,
(xi,xj),
xj),DD==
mij,and
and9 8= =
n l mObserve
. Observethat,
that,
for a,
a ,bb cEIII,
R,
for
+
+
9)2 >
2Q(axl
2Q(axl + bx2)
bx2) =
=a2
a2--2ab
2abcos
cos 98 + b2
b2 =
=(a
(a --bbcos
cos0)2
812++b2(sin
b2(sin8)2
2 0,
0,
with equality
equality precisely when a ==bb==0.0.Thus
ThusQQisisa apositive
positivedefinite
definitequadratic
quadratic
with
U ,so
soin
inparticular
particular UU isisaanondegenerate
nondegeneratesubspace
subspace of V
V and
and hence
hence
form on U,
form
V=
=U
U®
@ U1.
u'. But
But U1
U' <
5 xk
xk <5Cv(rk)
CV(rk)for
for k =
=iiand
and j,j,so
so U1
U' <5Cv(D).
Cv(D).
V
Hence D
D isisfaithful
faithfulon
on U.
U.
Hence
As Q is positivedefiniteon
U ,U,
(xk,
xk)xk)
= 1,
xj)xj)=_--cos9,
cos 8,(U,
( U ,Q)
Q)is
is
As
positive definite on
(xk,
= and(xi,
1,and(xi,
isometricto
to 2-dimensional
2-dimensionalEuclidean
Euclideanspace
spaceR2
R2with
with the
the standard
standardinner
inner product
isometric
(1,O)
andx,
(cos(n -0),
-@),sin(7r
sin(n-0))
-8)) ininthe
thestandard
standardcoordinate
and withxi
and
with xi =
= (1,
0) and
xj ==(cos(7r
coordinate
system. Thus ri
r, and
and rjr, are
arethe
thereflections
reflectionson
onR2
R2through
through the
the vertical
vertical axis and
system.
/ 2 - 8,respectively.
respectively.Hence
Henceririr,
the rotation
rotation through
the axis determined by
the
by n7r/2-0,
rj isisthe
-2n/m, and
andtherefore
thereforeisisof
oforder
orderm
mas
asdesired.
desired.
the angle -27r/m,
Thus the
the first
first claim
claim of 29.7
29.7 is established
established and
and the
the second
second is
is aa consequence
consequence
Thus
the first
first and
and Exercise
Exercise 10.1.
10.1. (Xk)
(xk)and
are the
the only
only nontrivial
nontrivial proper
of the
and xk
xk ffll U are
U ,so
so D
D
subspaces of
of U
U fixed
fixedby
byrkrkand
andififmm>>22then,
then,byby29.6.2,
29.6.2,(x,)
(xi)#0xfx flf?U,
irreducible on
on U.
U.
is irreducible
Presentations
Presentations of
of groups
146
(29.8) Let
Let W
W=
= (ri:
n)nbe
thethesubgroup
by
(29.8)
(ri:1 <(i i<(
) be
subgroupofofO(V,
O ( V ,Q)
Q )generated
generated by
the reflections
(ri: 11 5
< ii <(n).
reflections (ri:
n).Then
Thenthere
thereexists
existsaasurjective
surjectivehomomorphism
homomorphism
a:
a : G -+
+W
W with
with sia
sia ==ririfor
foreach
eachi.i.InInparticular
particularaaisisan
anRG-representation
RG-representation
which identifies S with a set of reflections in
in O(V,
O(V, Q).
Q).
Proof. This
This isis immediate
immediatefrom
from 29.7
29.7 and 28.6.
(29.9) (1)
(1) S is of order nn..
j I ==mimij
j and
(2)
For
each
each i # j,
j, Isis
Jsisj
and(si,
(si,ssj)j )= D2m;j
DZmij.
(2)
Proof. The
The map a:
a :(si,
(si,s jj)) -+
+(ri,
(ri,r rj)j )induced
induced by
by the
the map
map of
of 29.8
29.8 is
is aa sursurjective homomorphism.
homomorphism.By
ByExercise
Exercise10.1
10.1and
and29.7,
29.7,(ri(ri,
=Grp(ri,
r: r?
r? =
=
, r jr)j)=Grp(ri
, rj:
rr:= =
(ri(rr ji r)mii)
28.6 says there is a hoj ) m ~so,
SO,
l ) as
as (si,
(si,ss j)
j ) satisfies
satisfies these relations, 28.6
of (ri,
= Sk.
Thenp/3==a-'
a-1 so
so aa is
(ri,rrj)
j )onto
onto (si,
(si,ssj)
j ) with
with rrkp
kp =
sk. Then
momorphism pY of
an isomorphism
isomorphism and 29.7
29.7 implies
implies (2).
(2).As
As r,ri # rj for
for i # j,j,(1)
(1)holds.
holds.
0 =={1,
of (G, S). The
The
{ I ,.... .,.n}
, n )bebethe
theset
setofofnodes
nodesof
of the
the Coxeter
Coxeter diagram of
Let A
graph of
of the
the diagram
diagram isis the
the graph
graphon
on A0 obtained by
by joining
joining ii to j ifif the
the edge
edge
between i and jj in
inthe
theCoxeter
Coxeterdiagram
diagramisis of
of weight
weight at
at least
least 1,
1,or
or equivalently
equivalently
ifmij>3.
mij 2 3.
if
(29.10) Let (Ok:
< k <5r)r )be
( A k :1 5
bethe
theconnected
connectedcomponents
componentsof
of the graph of the
Coxeter diagram
diagram A0 of (G,
(G, S)
S ) and
and let Gk
Gk=
=(Si:
(si:i EE Ok).
A k ) .Then
Then
ofthe
thesubgroups
subgroupsGGk,
< rr..
(1) G
G=
= G1
(1)
G1 x . . x G, isis the
the direct product of
k ,115<kk 5
(2) V is the
= [Gka,
(2)
the orthogonal
orthogonal direct sum of the subspaces Vk
Vk =
[ G k aV],
,V ] 1,1 (
<< r.
kk 5
Proof. If i and
of A0 then
mij =
= 2,
and jjare
areinindistinct
distinct components
components of
then Isis
/ s i sjj /I== mij
2, so
so
[si,
of the
thesubgroups
subgroupsGGk,
[si,ssj]
j]=
=1.1.Thus
Thus G
G isis the
the central
central product
product of
k ,1 1(<kk5< rr,,
homomorphism pP of GI
G1 x . . - x Gr
and hence there is a surjective homomorphism
G, =
=D
D onto
onto
G with sip ==sisifor
foreach
eachi. i.Conversely
ConverselySSsatisfies
satisfiesthe
theCoxeter
Coxeter relations
relations in
D, so by 28.6
there is
is a homomorphism
homomorphismyy of
of G
G onto
onto D
D with
with sisiyy =
= si.
28.6 there
si. Then
Then
y ==P-1,
Similarly,by
by29.6.2,
29.6.2,xjxj EE xx;l ,
p-', soso/3,9isisan
an isomorphism
isomorphism and (1)
(1)holds. Similarly,
so, ifif jj EE Da
A, and
and i EE Ab,
Ab, then
v, = ( [ V ,r k ] :k E A a ) = ( ~ kk: E A,) ( v:,
Va=([V,rk]:kEA.)=(xk:kEDa)<Vb,
and hence (2)
(2)holds.
Because
Coxeter diagram
diagram
Because of 29.10 it does little harm to assume the graph of the Coxeter
of (G,
S ) is
is connected.
connected.In that event (G,
(G, S)
S ) is
is said
said to be an irreducible
irreducible Coxeter
Coxeter
(G, S)
system.
Coxeter groups
147
(29.11) Assume
Assume (G,
(G, S)
S )isisan
an irreducible
irreducibleCoxeter
Coxeter system.
system. Then
Then
(1) GGacts
-L.
(1)
actsabsolutely
absolutelyirreducibly
irreduciblyononV/V V/ v'.
(2) IfIfW
(2)
Wisisfinite
finitethen
then (V,
( V ,Q)
Q )isis nondegenerate.
nondegenerate.
Proof. Let
Let U
U be
be aa proper
proper RG-submodule
RG-submodule of
V ,si]
( x i )is
of VV.. For
ForsiSiEESS,, [[V,
si]=
= (xi)
soeither
eitherxixiEE UU or
or UU (
< Cv
(si) =
= xi
of dimension 1 so
Cv(si)
x l..Thus
Thuseither
either there
there exists
==
V1,
iE
E 0A with
withxi
xi EE UUor
orUU <5niEAXi
niEAx;
v',andandI assume
I assumethe
theformer.
former. Claim
Claim
xj EE U
= (xj:
U for
for each jj E 0,
A ,so
so that V
V=
( x j :jj EE0)A <
) _(U,Ucontradicting
, contradictingUUproper.
proper.
As the graph of
of A
0 isis connected
to prove
provexjxj E
E U for mid
> 2.
connected it suffices to
mij >
2. But,
But,
for such
j, (si,
on ((xi,
by 29.7,
29.7, so
so as
asxixi EE U and U is
such j,
(si,sj)
s j )is
is irreducible
irreducible on
x i ,xxj)
j ) by
xj E U.
G-invariant, xj
U.
I've shown
1. FF= =
EndRG
(V) acts
shownG
G isisirreducible
irreducibleon
onV ==V/V
v/v'.
EndRc@)
actsonon[ V
[ v, ,sisi]] =
(xi),
for aa E
E F,
( x i ) ,so, for
F, axi ==bxi
bxifor
forsome
somebb EE R',
R#,and
andhence,
hence, as
as G
G isis irreducible
irreducible
on V and
via bb on V. That is
and centralizes
centralizes aa,, a acts
acts as
as aa scalar
scalar transformation
transformation via
F
F ==R.R.So,
So,by
by25.8,
25.8,GGisisabsolutely
absolutely irreducible
irreducible on V.
Suppose W
W is
is finite.
finite. Then,
Then, by Maschke's
Maschke's Theorem,
Theorem,VV==VV-L
' CB Z for some
Suppose
Ef G-submoduleZZ of
of V
V.. V is Ef
G-isomorphicto
to ZZ and [V,
RG-isomorphic
[ v , si]
si] # 1, so [Z,
[Z, si]
si]
RG-submodule
1.
Hence
(xi)
_
[Z,
si]
<
Z,
so
V
=
(xi:
1
<
i
<
n)
<
Z.
Thus
V1
=
# 1. Hence ( x i ) = [Z, si] 5 Z, so V =
1 ( i ( 5 Z. Thus V' = 00
(2) holds.
and (2)
v
v
v.
v.
v
(29.12) Assume
Assume (G,
(G, S)
S )is
is an
an irreducible
irreducible Coxeter
Coxeter system and W is finite. Then
(V,
under the usual inner
( V , Q)
Q ) is
is isometric
isometric to
to n-dimensional
n-dimensional Euclidean space under
product.
Proof. Let
Let hh be
be the
the bilinear
bilinear form
form on
on V
V which
which makes
makes X
X into
into an
an orthonormal
orthonormal
basis and define g:
g: VV xx V -+
+RRby
by g(u,
g(u,v)v )==F_wEW
CWEw
~h(uw,
( u wvw).
V ,W ) . It is straightforward to check that g is a symmetric bilinear form
form on
on V
V preserved by
by G. As
the quadratic form
form of
of hh is positive
positive definite,
definite,so
soisisthe
theform
formPP of
of g, so (V,
( V , g)
is nondegenerate. But, by 29.11, V
V is
is an
an absolutely
absolutely irreducible
irreducible RG-module,
RG-module,
so, by
by Exercise
Exercise 9.1,
9.1, PP ==aQ
a Qfor
forsome
somea aE Efl8#.
R#.As
AsPPisispositive
positivedefinite
definite
and
1/2
>
0,
a
>
0,
so
Q
is
positive
definite.
By
19.9
there
and Q(xi) =
= 112 > 0, a > 0, so Q is positive definite. By 19.9 there is
is
basisYY== (yi:11 (
< i <5n)n)for
= 00 for
aa basis
for VV with
with (yi,
( y i ,yyj)
j) =
for i # j.
j. As
As Q
Q is
is
positive
definite, Q(yi)
Q(yi) >
positive definite,
> 00so,
so,adjusting
adjustingby
by aasuitable
suitablescalar,
scalar, we
we can
can take
take
Q(yi)
Q(yi)==1,1,since
sinceevery
everypositive
positivemember
memberof
ofERisisaasquare
squarein
inR.
R.Thus
ThusYY isisan
an ororthonormal
( V ,Q),
Q),so
so (V,
( V ,Q)
Q )isisEuclidean
Euclidean space
spaceunder
under the
the usual
usual inner
inner
thonormal basis
basis for (V,
product.
Let0={1,...,n}andforJCOletSj=(SJ:j
E J } and G j = ( S j ) .The
Let A = { I , . . . , n} and for J c A let S j = ( S J :j EJ)andGj=(Sj).The
subgroups
parabolic subgroups
G j and
and their
their conjugates
conjugatesunder G
G are
are the
the parabolic
subgroupsof
of the
the
subgroups GJ
Coxeter system (G,
S).
(G, S).
Presentations of
of groups
148
(29.13) Let
Let J,
J, K C
(29.13)
2A
A and
and g E
E Gj.
Gj. Then
Then
(1)
If
l
(g)
=
m
and
g
=
si,
...
Si,,,
with
i t hE sSi then
, ~ S ik
t hEe Jnfor
i k each
~ J f o1 r<eka <
c hin.
l~k~m
(1) Ifl(g)=mandg=si, ... s i , , , wSi,
(2) (Gj,
Coxetermatrix
matrixMMjj =
= (Mid),
E J.
(Gj,Sj)
SJ)isisaaCoxeter
Coxetersystem
system with Coxeter
(mij), i i,, jj E
J.
(3) (Gd,
Gi fl
= GJnK
(GJ,GK)
GK)==GJUK
GJUKand GJ
n GK
'% =
GJ~K.
(4) If
GK
I f Gj
G J= &
t h ethen
n JJ == K.
K.
Proof. Let
Let gg =
= sat
Proof.
s,, ...
. . .sak
s,, with
with aak
k E
E JJand
andkkminimal
minimalsubject
subjectto
to this
this conconstraint. Claim
Claimkk =
= m. By induction
induction on
onk,k,l(s,,g)
l(sa,g) =
= k --1,1,so
soby
by the
the Exchange
Exchange
Condition
either kk =
= l(g) or
for some
or sa,
s,, ...
. . .sa,_,
s,,-, =
=saZ
s,, ....
. . s,,sar for
some t, in
in which
which
Condition either
case g ==sae
s, .... .sa,_,sa,+,sak,
.s,,-, s,,+,s,,, contrary to minimality of k. Hence
Hence (1)
(1) follows
follows
from 29.5.
29.5.
Next (1) says
says the
the Exchange
ExchangeCondition
Conditionisissatisfied
satisfiedbyby(Gj,
(Gj,SJ),
Sj), so
so (Gj,
(Gj, SJ)
Sj)
Next
system by
by 29.4.
29.4. 29.9
29.9 says
saysMj
Mjisisthe
theCoxeter
Coxetermatrix
matrixofof(Gj,
(Gj, SJ).
Sj).
is a Coxeter system
The first remark in (3) and the inclusion GJnK
GJnK5< G
Gjj flf l GK
GK are trivial. Part
(1) gives the
the inclusion
inclusion GJ
Gj fl
n GK <I GJnK.
GJnK.
If Gj
assumeKK 2
C J.
If
G j ==GK
GKthen
then Gj
G j ==GJnK
G j n by
~by (3),
(3), so
so we may
may assume
J.By
By (2)
(2)
we may
mayassume
assumeJJ =
= A. Hence
G=
= Gj
so,
we
Hence G
G j==GK,
GK,
SO,by
by (1),
(I), SS ==SK.
SK.Hence
Hence
by29.9.1.
29.9.1.
K ==AA ==JJby
30 Root
Root systems
systems
finite dimensional
dimensional Euclidean space over a field
field F
F equal to
In this section V is a finite
rationals. That is
is V is an n-dimensional space over F
F together
together
the reals or the rationals.
with
quadratic form
Q such
such that
that (V,
(V,Q)
Q)possesses
possessesan
anorthonormal
orthonormalbasis.
basis.
with a quadratic
form Q
Hence Q
Let ((,, ))be
be the
thebilinear
bilinear form
form defined
defined by Q.
Q. For
For
Q is
is positive
positive definite. Let
V# there
there exists
exists aa unique
unique reflection
reflection with center (v)
(v)by
by 22.6.2;
22.6.2; denote
denote this
this
v EE V#
reflection
reflection by r,,.
r,.
A root system
systemisisaafinite
finitesubset
subsetCEofofV#
V#invariant
invariant
under
W (E)
E)
under
W(C)
==(r,(r :: vv EE C)
and such that I (v)nflC E
I <2foreachv
2 for each vEEC.Observethatifv
E. Observe that if vEECEthen-v=
then -v =
andsuchthatI(v)
I 5
vr EEE,C,so,
2,2,(v)
{v,{v,
-v).
call
vr,
so,asas((v)
I(v)fln EI
Cl<I
(v)flnEC= =
-v).WeWe
callW(E)
W(C)the
theWeyl
Weyl
group of
of E.
C.
Here's one
one way
way to obtain
obtain root systems:
systems:
(30.1) Let G be aa finite
(30.1)
finite subgroup
subgroup of O(V,
O(V, Q)
Q) generated
generated by aa G-invariant
G-invariant set
reflections.Let
LetCE consist
consistof
ofthose
thosevvEE V#
V#with
withQ(v)
Q(v)=
= 1/2
R of reflections.
112and
and (v)
(v) the
the
center of some member
member of
of R.
R. Then
Then C
E is a root
root system
system and
and G
G=
= W(E).
W(C).
ofsome
somer rEER,
R,so
sor r=
= r,.
r,,.Thus
ThusGG=
= (R) =
=
Proof. For
For v E E,
C,(v)
(v)is
is the center of
W(E),
remains to
to show
show C
E is a root
root system.
system.IfIfuu =
= av EE ECthen,
W(C), so it remains
then,by
by defidefinition
of C,
E, 112
1/2 =
= Q(u)
Thus
Q(u) ==a2Q(v)
a 2 ~ ( v==
)a2/2,
a2/2,sosoaa==±1.
f1.
ThusI E
/ Cfln(v)
(v)II <
5 2.
2.
nition of
149
149
Root systems
then,as
as RR isisG-invariant,
G-invariant, rg EE RR and
and(vg)
(vg)isisthe
thecenter
centerof
of rg
rg
Also if g EE GGthen,
with Q(vg) =
= Q(v)
Q(v) ==1/2,
112,so
sovg
vg EEE.
C.Hence
Hence G
G ==W(E)
W(C)acts
actson
on E.
C.Finally
Finally
< oo.
GO. So
So EC isis aaroot
system.
I1x1
E J==21RI
21R15<21GI
2G I<
root system.
By 29.12 and
and 30.1,
30.1,every
every finite
finite Coxeter
Coxetergroup
groupcan
canbe
be represented
representedas
asthe
theWeyl
Weyl
group
group of some
some root system. We'll see
see later
later in this section that on the one hand
this
this representation
representationisis faithful
faithfuland
andon
on the
theother
otherthat
thatthe
theWeyl
Weyl group
groupof
of each
eachroot
root
system
system is aa finite
finite Coxeter
Coxeter group.
group. Thus
Thus the
the finite
finite Coxeter
Coxeter groups
groupsare
areprecisely
precisely
the Weyl groups
groups of root systems.
systems.
For the remainder of
of this
this section
sectionlet
let CE be
be a root
root system
systemand
and W
W=
= W(E)
W(C) its
its
C will
will be
be called
calledroots.
roots.
Weyl group. The elements of E
(30.2) (1) (E)
El, so
( Cis
) isa anondegenerate
nondegeneratesubspace
subspaceof
of V
V and
and W
W centralizes CL,
soW
W
(C).
is faithful on (E).
(2) The
Thepermutation
permutationrepresentation
representationof
of W
W on
on EC isisfaithful.
faithful.
(3)
(3) W
Wisisfinite.
finite.
Proof.
Proof. Let U =
=(E).
(C).Then
Then H
H ==CW(U)
Cw(U)==CW(E)
Cw(C)and
andW/H
W/Hisisfaithful
faithful on
on
E,
and (3)
(3) itit suffices
sufficestotoshow
showHH=
= 1.
C, so
so W/H
W/Hisisfinite.
finite. Thus
Thus to
to prove (2) and
As Q
definite,UUisisnondegenerate.
nondegenerate.Thus
ThusVV==UU@®u'U'.
. But,
But, for
Q is
is positive
positive definite,
(r,,:
C, rr,centralizes
centralizesvl
vLQ2U1,
u I ,sosoWW==
(r,:vvEEE)C)centralizes
centralizesU1.
u I . Hence
Hence
v E E,
H centralizes
centralizes V,
V, so
so H
H ==1,1,completing
completingthe
theproof
groofof
of the
the lemma.
lemma.
An ordering
ordering of
of V
V isis aatotal
total ordering
ordering of
of VVpreserved
preserved by
by addition
addition and
and multiplimultiplication by
by positive
positivescalars;
scalars;that
thatisisif ifu ,u,v,v,w wEEVVwith
withu u>>v,v,and
and00<< aa E
E F,
F,
then u + w >>vv+ wwand
I leave
andau
a u>2av.
av.
I leavethe
thefollowing
followinglemma
lemmaas
asan
anexercise.
exercise.
+
+
(30.3)
(1) If
(30.3) (1)
If <5isisan
anordering
ordering on V and V+
V+ =
={v
{vEE V:
V: v >>01
0) then
then
V+ isis closed
closedunder
under addition
addition and
and multiplication
multiplicationby
by positive
positive scalars,
scalars, and
and
(i) V+
(i)
-v) nnV+)
(ii) for each
each vv EE v#, I{v,
I{v, -v)
V+I =
=1.
1.
(ii)
V#,
(2) IfIfSSCV#
2 V#satisfies
satisfies(i)(i)and
and(ii)
(ii)ofof(1)
(1)then
thenthere
thereexists
existsaaunique
unique ordering
ordering
of V with SS=V+.
= V+.
ofVwith
(3) IfIf TTC2V#
and I{v,
I {v,-v)
-v) n
nT
< 11 for each v EE V#,
V#satisfies
satisfies (1.i) and
TII 5
v#, then
then
5 V+
V+ for
forsome
someordering
ordering of
of V.
V.
TC
A subset P of E
positivesystem
systemififPP== C+
E+ =
C is aapositive
=ECnnV+
V+for
forsome
someordering
ordering
of V.
linearly independent
independent and each
n of E
C is
is aa simple
simple system
system if Tr
n isis linearly
V. A
A subset
subsetTr
vE
axx such that either
either 005< a,
ax E
EF
EE
C can
can be
be written
written v ==ExE7r
Ex,,a,x
F for
for all x EE 7r
n
orO>a,
or 0 > ax EE F
F forallx
for all x eEnr..
150
150
Presentations of groups
(30.4) Each simple
simple system
system is contained
contained in a unique positive system and each
(30.4)
positive system
system contains
contains aa unique
unique simple
simplesystem.
system.
Proof.
axx E V#
Proof. IfIfnn isisaasimple
simplesystem
systemlet
let TT consist
consistof
of those
those elements FxER
Ex,,a,x
V#
with a,ax >>0.0.By
with
By30.3.3,
30.3.3, TTCEV+
V+for
forsome
someordering
orderingof
ofV.
V.By
Bydefinition
definition of
systems and
and30.3.1,
30.3.1, PP =
simple systems
=TTnnEXisisthe
theunique
unique positive
positive system
system containing
containing
n.
Cyy
for some
someYYE
CP
n . Let
Let SS be
be the set of those vv EE P
P such
such that, if v =
=EyEY
EYE
cYy for
and 00 < cy E F,
then
Y
=
{v}.
Evidently
S
c
n.
I'll
show
n
c
S
which
will
F,
Y = {v}.Evidently S E n .
show n E S which will
prove n ==SSisisthe
theunique
uniquesimple
simple system
system in P.
P.
C,, cYy for
So assume
assumevv EE nn and v ==EyEYCyy
for some
someYY C P
P and 0 << Cy
cy E
E F.
F.
Now,
for YEE Y,y
Y, Y==~,,,b,,xforsomeO
ExEK byxx for some 05 <by,
byxE EF F,
Now,fory
, s osov v== ~Ey
, , ,XcCybyxx
yb,,x
and hence,
hence, by
by the
thelinear
linearindependence
independenceofofn,n,00== Cy
Ey Cybyx
forx XE En n-- (v).
c,b,, for
(v}.
byx==00for
forxx## v,
v, so
soyy ==by
Then, as
asPP ffll (v)
(v) =
_
Therefore by,
b,,vv for
for each
each y E Y.
Y. Then,
(v), Y
= (v),
(v},
Y=
(v},completing
completingthe
theproof
proof of
of the
the claim.
claim.
to show
show that
that ifif P is aa positive system then
then P contains a simple
It remains
remains to
show:
system. In the process I'll show:
(30.5) If
simple system
system and x and
and y are
are distinct
distinct members
members of
(30.5)
If n isis aa simple
of nn,, then
(x, Y) 5 0.
(x,y)<0.
Indeed, returning
returning to
to the
theproof
proof of
of 30.4,
30.4,let
letnn be a subset of
of P minimal subject
to P
P being
being contained in the nonnegative
. I'll
nonnegativelinear
linearspan
spanofofnn.
I'll show
show nn satisfies
30.5 and then
then use
use this
this fact
fact to
to show
show nn is linearly
linearly independent
independent and hence a simple
complete the proof of 30.4.
system. This will complete
Suppose(x,
(x,y)y)>>0.0.By
By22.6.2,
22.6.2,
z=
y-2(x,y)x/(x,
y)x/(x,x),
x),so
soz z==yy-cx
Suppose
z=
yr,yrx
= =y -2(x,
-cx
with cc >
> 0. Now
Now zz =
= C,,,
with
att
with
a,
>
0
z
E
at
5
0
if
z
E
-P.
at
with
at
>
0
if
z
c
P
and
at
<
0
if
z
c
-P.
If
ten
z cE P,
P ,consider
considerthe
theequation:
equation:
(1
- ay)y
(1 aY)y=
=
C att
att ++(ax
(a, +
+c)x.
c)x.
t#x,y
t54x,y
a, >>0.0.But
Butthen
then yy isis in
in the
the
The right
right hand
handside
sideisisininV+,
V+,so,
so,asasyyEc V+, 1 - ay
(y}, contradicting
contradicting the
nonnegative span
nonnegative
spanofofn n-- (y),
theminimality
minimalityofofn .n.IfIfz zEc -P
-P
consider:
consider:
+
(a, + C)X
(ax
Ox =
=
C (-at)t
(-at) ++(1
(1 --ay)y
aY)y
t54x,y
t#x,y
and apply the same
same argument
argument for aa contradiction.
contradiction.
So n satisfies
satisfies 30.5 and it remains
remains to
to show
show nn is linearly
linearly independent.
independent. Suppose
Suppose
O
X E I Taxx
a , xand
a n d let
l e t a==( (x
x E
, >
>0}and/3
y cEn:
n :ay
a , <<O}.Then
0 = CEXEn
c nn:: aax
0) and,8 = ({y
0}. Then
EyE,s(-ay)y)5
<
EXEaaxx
axx== zz =
= EyEf(-ay)y
z) ==(F_xE«
E y E D ( - a , )so
~so00<5(z,(z,z)
(Ex,, axx, CYEB(-ay)y)
0 by 30.5.
for some
some x
30.5. Hence
Henceas
as Q
Q isis positive
positivedefinite
definitez z== 0.
0. But
Butififa,ax #
# 00 for
Exes
Root systems
systems
151
V + a, acontradiction.
contradiction.So
So nnisislinearly
linearlyindependent
independentand
and the
the proof
proof is
is
then z EE V+,
complete.
complete.
be aa simple
simple system
system and PP the
thepositive
positive
For the remainder of this section let n be
system containing n
n..
with (v,
( v ,xx)) >> 0.
0.
(30.6) For each vv EE P there is xx E n
n with
ax 2
> 0 and
andobserve
observe00<<( (v,
= Ex
>x ax
Proof. Write
Write v =
Ex,,
a,x with a,
v , vv)
)=
a, (v,
( v , x).
x).
Proof.
= EX
E, axx
(30.7)
For each
eachxxEEn,
n,xr,
xrx== -x
-x and
acts on
onPP - (x).
and rx
r, acts
(x}.
(30.7) For
Proof.
ofr,r,xrxr =
= -x.
{x},
Proof. Let r ==rx.
r,. By
By definition
definition of
-x. Let
LetVv EE PP- { x }so
,sothat
that
vv ==EYE,
ayy
with
ay
>
0.
As
(x)=
(x)
fl
P,
aZ
>
0
for
some
z
E
n
(x).
EYE,a y y ay 2 As ( x } = ( x )f l P , a, > 0 some z E n - ( x } .
Now
yr = > ayyr
Y
ay(y - 2(y, x)x)
Y
ayy + bx
Y54x
F.As
As aZ
a, >
> 00 itit follows
follows that vr
fromthe
thedefinition
definition of simple
for some b E F.
yr EE PPfrom
system.
system.
W isistransitive
transitiveon
onsimple
simplesystems
systemsand
andon
onpositive
positive systems.
systems.
(30.8) W
Proof. By
By 30.3,
30.3,W
Wpermutes
permutes positive
positive systems,
systems, while it is evident
evident from the
Proof.
, definitions that W
W permutes
permutes simple
simple systems.
systems. By
By 30.4
30.4ititsuffices
sufficesto
to show
show W
W isis
transitive
andRR be
bepositive
positivesystems
systems
transitive on
on positive systems.
systems. Assume
Assume not and let PP and
in different orbits of
of W,
W, and
and subject
subject to
to this
thisconstraint
constraintwith
withIP
( Pf lfl(-R)I
(-R)J =
= nn
minimal.
As RR #
# P,
minimal. As
P ,nn>>0.0.Let
Letnnbebethe
thesimple
simplesystem
systemin
in P.
P .By
By30.4
30.4there
there
Prxnfl(-R)I
(-R)j =
1,1,so,so,bybyminimality
(-R).By
By30.7,
30.7, IIPr,
=nn- minimality of
is x EE nnflf l(-R).
R EE (Prx)W
(Pr,) W ==PW,
P Wa, acontradiction.
contradiction.
n, R
For V
axx for
for unique
unique a,
ax EE F;
v E E,
X ,Vv ==EZE7[
E,,, a,x
F;define
define the height of v to be
h(v)
h ( v )==ExEn
Ex,,ax.a,. Evidently
Evidentlythe
theheight
height function
function h depends
depends on n.
n.Notice
Notice for
for
vE
> 0 and
andh(-v)
h(-v) =
= -h(v)
E P that hh(v)
(v)>
-h(v) <<0.0.Also
Alsoh(x)
h ( x )==11for
for x E
E n.
n.
(30.9)(1)
(1) hh(v)
> 1l for each
eachvvEEPP-- n.
n.
(30.9)
(v)>
x EE 7r).
(2) W
W ==(rx:
(r,:x
n).
(3) Each
Eachmember
member of
of EX isisconjugate
conjugate to an element of
of n
n under W.
Proof. Let G =
=(rx:
(r,: xx E
E n)
n )and
and vv EE P.
P.Pick
Pick U
u EP
E PflflvG
vGsuch
such that
that h(u)
h(u) is
minimal.
By30.6
30.6there
thereisisy yEEnn with
with((u,
0. Then
Thenur,
ury =
= u --cy
u , y)y)>> 0.
cywith
with
minimal. By
Presentations of
of groups
152
152
= 2(u,
y) > Oandifu
by30.7,
30.7,ury
andh(ury)
h(u)c=
2(u, y)
0 andif u 0
# yy then,
then, by
ur, EEPPnvG
n v G and
h(ur,) ==h(u)So uu =
= yy EE jr.
0 n then h(v)
c << h(u),
h(u),aa contradiction.
contradiction. So
n.In
In particular
particular ifif vv 6
h ( v )>
>
h(y)
of Z
E is
h ( y )==1,1 ,so
so(1)
( 1 )holds.
holds.Further
Furtheritit follows
follows that each member of
is conjugate
conjugate
to an element of
of nTrunder
underGG,
sothat
that( (3)
holds.Finally
Finallyv v==yg
yg for
forsome
somegg EE G
G,,
, so
3 )holds.
rU== r,,
ryg== (r,),
(ry)gEEGGand
andhence
henceWW==(r,:
(r,,:vvEEPP)
= G.
so r,
)=
For w
WEEW
Wlet
letNN(w)
I Pwfl n(-P)I.
(-P)I. Then
ThenNN(I)=O
N(rx) =
= 11
( w ) == IPw
( l ) = 0and,
and,by
by 30.7,
30.7, N(r,)
for x EE Tr.
Let R =
n. Let
=Jr,,:
{r, :xx EETr}
n ]and 1l ==lR
1~ the
the length
lengthfunction
functionwith
with respect
respectto
to
R; that is, for w E
E W,
W ,l(w)
l ( w )is
is the
the minimal
minimal length
length of aa word u in
in the
the members
members
of R such that u ==w.
N agree
w .ItItwill
willdevelop
developshortly
shortly that the functions 1l and N
W.
on W.
(30.10) Let w
W EE W
W and
andxXEEn.
r. Then
(1) N(rxw)
=
N(w)
+
1
if
N(r,w) = N ( w ) 1 if xw >>0,
0,and
and
((2)
2 ) N(rw)
N(r,w) ==N(w)
N ( w )--11ifif xw
xw << 0.
0.
+
{x}))w
{xw})
Pr,w ==({-x}
((-XI UU (P
( P- (x}))w=={-xw}
(-xw}UU(Pw
( P w- { x w }by
)by30.7,
30.7,so
so
Proof. Prxw
Prxwnn((-P)
=
Pr,w
-P) =
((Pw
P w n ((-P))
- P ) )UU {-xw}
(-xw]
if xw
if
xw >>0,
0,
- {xw)
((Pw
P w fln ((-P))
-P)) (xw]
xw << 0.
if xw
0.
(30.11)Let
LetririEER,O
R, 05< ii <
(30.11)
5 nn and
and w ==r1
rl ...
. . .r,,r , EE WW with
with N(w)
N(w)=
=nn and
and
N(row)
existsk,k, 11 5< kk 5
< n,
that rorl
rorl ..... .rk_1
= r1
N(row)<5 n.
n. Then
Then there exists
n, such that
rk-1 =
rl ...
. . .rk.
rk.
By 30.10,
30.10,xw
xw < 0,
Proof. By
0,where
where x EE Tr
n with
with ro ==rx.
r,. Let
Letkkbe
beminimal
minimal
= ry.
subject
subject to
to xr1
xrl ...
. . .rk
rk <
< 0. Then, by 30.7, y =
=xr1
xrl ...
. . .rk_1
rk-1 EETr
n and rk
rk =
r,. Thus
Thus
r o ...rk =
Y1
...
r,,
,...
,,-,
=
r,
=
rk,
which
can
be
rewritten
as
rorl
.
.
.
rk-1
=
rl
.
. .Yk.
rk.
Yxrl
=
Yy
=
rk,
which
can
be
rewritten
as
r0r1
rk_
p
...
rA_
I
r;l".rk-l
1
(30.12) 1(w)
= N(w)
l ( w )=
N ( w )for
for each
each w EE W.
W.
(1)=
=0 =
(1). Then 30.10 and induction
Proof. I've
I'vealready
alreadyobserved
observedthat
that 1l(1)
=N
N(1).
on 1(w)
impliesl1(w)
that the
the lemma
lemmaisisfalse
falseand
andpick
pickuu =
=
l ( w ) implies
( w ) 2> NN(w).
( w ) . Suppose that
rorl ...
and
. . .rn
r, EE W
W with
with 1(u)
l(u) =
=nn+ 11>>N(u),
N(u),
andsubject
subjecttotothis
thisconstraint
constraintwith
with
w=
= r1
= nn + 1,1 ,1(w)
n minimal. Let w
rl ...
. . .rnr,.. As 1(u)
l(u) =
l ( w )==n,n,and,
and,by
byminimality
minimality
of
1(u),NN(w)
AsN(row)
N(row)== N(u)
( w ) == 1(w).
l(w). As
N ( u ) <<nn+ 1,1,there
thereexists
existskk with
with
of l(u),
ror1....
rk_1=
= rl
r1 ....
rk by
by 30.11.
30.11.Thus
Thusuu== ro..
ro ....rk-irk..
rk_1rk ....rn
rn=
= rlr1....
r rk+l
rorl
. .rk-1
. .rk
. .rzrk+l
- 1,1 ,contrary
rn is of length at most nn . . .rn
r, =
=r1
rl .. .. .. rk-lrk+l
contraryto
to the
the choice
choice
rk-lrk+1 .. .. .. rn
of U.
u.
+
(30.13) (W,
( W ,R)
R )is
is aa Coxeter
Coxetersystem.
system.
+
+
Root systems
153
Proof. This is immediate from 29.4, 30.11, and 30.12.
(30.14) W is regular on the positive systems and on the simple systems.
Proof. By 30.8 and 30.4 it suffices to show NW(P) = 1. But NW(P) = {w E
W: N(w) = 0) so 30.12 completes the proof.
Let D consist of those v E V such that (v, x) > 0 for all x E Jr.
(30.15) (1) vW n D is nonempty for each v E V.
(2) (d, u) > O for each d E D and u E P.
Proof. Pick Z E v W maximal with respect to the ordering defining P. Then,
for x E Jr, z > zrx = z - 2(z, x)x, so (z, x) > 0 as x > 0. That is z E
vW n D. Let d E D and u E P. Then u = XEn axx with a, > 0 so
(d, u) = >x ax(d, x) > 0.
(30.16) Let d E D. Then Cw(d) is the Weyl group of the root systems E n dl
and has simple system 7r n dl.
Proof. U = (ry: x E Tr n d') < Cw(d), so assume w E Cw(d) - U and subject to this constraint with n = 1(w) minimal. w : 1 so n > 0. Then N(w) =
n > 0 so there is x E zr with xw < 0. Now, by 30.15.2, 0 > (xw, d) =
(x, dw-1) = (x, d), so x E dl by another application of 30.15.2. Thus rx E U
and, by 30.10, l(rxw) < n, so, by minimality of n, rxw E U. But then w E U.
So U = Cw(d). Evidently E n dl is a root system and an easy calculation
using 30.15.2 shows Ir n dl is a simple system.
(30.17) Let S C V. Then Cw(S) is the Weyl group of the root system E n S'.
Proof. Let U = (rn: v EE n Sl). It suffices to show U = Cw(S). Cw(S) =
Cw((S)) = Cw(So) where So is a basis for (S), so without loss S is finite.
Replacing S by a suitable conjugate under W and appealing to 30.15, we may
take d E Dn S. Let G = Cw(d) and E0 = E n dl. Then, by 30.16, G = W(Eo)
and Cw(S) = CG(S), while, by induction on the order of E, CG(S) = U(30.18) Let (G, S) be a Coxeter system with G finite. Then the representation
a of 29.8 is faithful and G is the Weyl group of a root system.
Presentations
Presentations of
of groups
154
Proof.
Proof. By
By 29.12
29.12 and
and 30.1,
30.1, Ga
G aisisthe
theWeyl
Weylgroup
groupof
ofaaroot
rootsystem.
system.By
By 29.7
29.7
and
Gal _=I G
1. Hence a is
and 30.13, Ga
G a isisaaCoxeter
Coxetergroup
group of
of type
type M,
M,so
soI /Gal
IGI.
is an
an
isomorphism.
isomorphism.
(30.19)
= {1,
52 =
(1, ...
. . .,,n), (e) =
Z 12,
Z2,and
and
(30.19) Let H be
be the
the symmetric
symmetric group
group on
on SZ
G
thethewreath
G ==(e)wroH
(e)wrnH
wreathproduct
productofof(e)
(e)by
byH.
H Choose
. Choosenotation
notationso
sothat
thatCH(e)
CH(e) '
isis the
in H
H of n EE Q.
2), (2,
3), ...
the stabilizer
stabilizer in
52. Let
Let G1
G1 ==H,
H,S1
S1=={(1,
((1,2),
(2,3),
. . .,,(n
(n 1,
n)),
G2
=
G,
S2
=
S1
U
(e),
S3
=
S1
U
{(n
1,
n)e),
and
G3
=
(S3).
Then
1, n)), G2 = G, S2 = S1 U (e), S3 = S1 U
- 1, n)e), and G3 = (S3).Then
(Gi,
2, 3, respectively.
(Gi,Si)
Si)isis aa Coxeter
Coxeter system
system of type AnA,-1,1, Cn,
C,, Dn,
D,, for i ==1,1,2,3,
respectively.
Further
product of
of E2"-1
E2'-, by S.
Further G3
G3 is of index 2 in G2
G2 and is the semidirect product
S,.
Proof.
Proof. Let
LetVVbe
ben-dimensional
n-dimensionalEuclidean
Euclideanspace
spacewith
with orthonormal
orthonormalbasic X
X ==
(xi:
with center
center(xi),
(xi),and
andEE== (ei:
(ei: 115< i <(n).
(xi: 1 <
(ii <5n),
n),eieithe
thereflection
reflection with
n).
Represent
xih
for hh E
EH
H.. Then G =
= (H,
RepresentH on
on V
V via xi
h=
=Xih
xih for
(H,E)
E )<5O(V)
O(V)and
andGGisis
the
rn H
H,, where
where ee =
= en.
product
thewreath
wreath product
product (e)
(e)wrn
en.Hence
Hence G
G is the semidirect
semidirect product
of
product of
of D
D=
= EE flnG3
of EE -ZE2,1,
E2",by
by H
H =ZSn,
S,, and
andG3
G3isisthe
the semidirect
semidirect product
G3by
H
D=
= (ei
ej:22(<i i_(< jj <(n)
. The
H where
where D
(eiej:
n) =Z E2-,
E2"-1.
The transposition (i, jj)) isis the
the
reflection
(xih),where
wherexij
xi j==xixi-xi.
- xj. Also Gi ==(Si)
reflection with center (xij),
(Si)and
andSi
Siisisaa set
set
of
by 30.1,
30.1, Gi
Gi =
= W(Ei)
of reflections,
reflections, so, by
W(Ci)where
whereEi
Ciisisthe
theroot
rootsystem
systemconsisting
consisting
of
xn) ifif i ==2.2.Thus
Gi-conjugatesof
ofx12
xl2 if i ==11oror3,3,and
and{x12,
1x12, x,)
Thus
of the
the Gi-conjugates
E1={±xxij:1<i<j<n},
E3 = E1 U (±(xi +x3):1 < i < j < n),
and
and
.
E2=E3U{±xi:2<i <n}.
Next
7r2
Next ni
ni isisaasimple
simplesystem
systemfor
for Ei,
Xi,where
wherenini=={xi,i+1:
(xi,i+l:11<5i i<<n),n),
n2=71
=nl UU
(xn),
as Si
Si =
( ~ n ) rand 73
n3 ==nlnlUU{xn_1
(x,- 1 + xn}.
x,]. Hence,
Hence, as
={rx:
{r, :x EE ni),
xi),(Gi,
(Gi,Si)
Si)isis
aa Coxeter
Coxeter system
system by 30.13.
30.13. Finally
Finally it is
is evident
evident that the
the Coxeter
Coxeter diagram of
(Gi,
C,, Dn,
(Gi,Si)
Si)isis AnA,-l,1, C,,,
D,, for
for ii ==1,1,2,2,3,3,respectively.
respectively.
+
For
(J)(J)
and
WjWj
= (rj:
j EjJ).
thethe
subgroups
and
= (rj:
E JRecall
) . Recall
subgroupsWJ
Wjand
and
For JJCJrnletletVjVj_ =
their
theirconjugates
conjugatesunder
under W
Ware
arecalled
calledparabolic
parabolicsubgroups
subgroupsofofW.
W.
(30.20)
Let00 #
0 JJ CJ r.n .Then
Then
(30.20) Let
(1)
E fl
system
with
simple
(1) Ej
CJ= =
C vi
f l is
VJ aisroot
a root
system
with
simplesystem
systemJ Jand
andWeyl
Weylgroup
group
Wr
.
WJ.
(2)
).
(2) Wi
WJ==Cw(VJ
cW(vJ').
Proof.
Cw(VjL)
UU
and,
byby30.17,
Proof.Wi
Wj< (
cw(vJ') ==
and,
30.17,UUis isthe
theWeyl
Weylgroup
groupofofthe
theroot
root
C J.n isislinearly
linearly independent
independent and
spans Vj,
Vj, so
so JJisisaabasis
basisofofVj.
Vj.
system Ej.Tr
system
and J spans
155
155
Root systems
Hence
member of Ej
C Jisisaalinear
linearcombination
combinationof
ofthe
themembers
membersof
of J,J,so,
so,asas7r
n
Hence each member
simplesystem,
system,JJisisaasimple
simplesystem
system
ThusU U==( r(rj:
= Wj.
is a simple
forfor
E lEj.
. Thus
j :j j Ec-JJ)) =
Wl.
Remarks. The
29 follows
that of
of
The discussion
discussion of Coxeter
Coxeter systems in section
section 29
follows that
presentation of
of root systems given
given here
here
Bourbaki [Bo] and Suzuki [Su]. The presentation
draws heavily on the appendix
appendix of Steinberg
Steinberg[St].
[St].
Coxeter groups
in branches
branches of
of mathmathCoxeter
groups and root systems play an important role in
particularly in
in the study of Lie
ematics other than finite group theory, most particularly
Lie
algebras, Lie groups, and algebraic groups.
groups. We will find in chapters 14 and 16
that they are crucial
crucial to the study
study of the
the finite
finite groups
groups of
of Lie
Lie type.
type.
Exercises for chapter
chapter 10
10
1. Prove
1).
Prove D2n
D2, =Grp (x,
( x ,y:
y: x2
x2 =
=y2
Y 2==(xy)n
( ~ =y=
) 1).
~Prove
Proveevery
everygroup
groupgenerated
generated
by a pair of distinct
distinct involutions
involutions is aa dihedral
dihedral group.
group.
2. Prove
Prove lemma
lemma 30.3.
3. Let
E
be
E,, PP the
Let C beaaroot
rootsystem,
system,7r
n a simple system for C
thepositive
positive system
system of
of
7r, RR =
= (ra:
n,
(r,: a EE7r),
n),and W =
=(R)
( R the
)theWeyl
Weyl group
group of E.
C .Prove
Prove
(1) There
uniquewo
woEEW
Wwith
withPwo
Pwo== -P.
-P.
There exists a unique
(2)
(2) wo
wo is
is the
the unique
unique element
element of W
W of
of maximal
maximal length in the alphabet R.
Further
Further 1(wo)
1(wo)==I I PP II and wo is an involution.
involution.
(3)
(3) nwo
-n and
andRw0
RwO =
rwo =
= -7t
= R.
4. Let
for C
E,, PP the
LetECbe
bean
anirreducible
irreducibleroot
rootsystem,
system,7r
n a simple system for
the positive
positive
system for
forn,
,r,JJ C
subsetofofC,
E,spanned
spannedbybyJ,J,and
and$lU
Ej.
system
c nn,, EC j,j , the subset
==PP-- C
j.
Prove( (E)
(i).
Prove
C ) =_ ($).
5.
5 . Let
Let ECbe
beaaroot
rootsystem,
system,PPaapositive
positivesystem
system for
for E,
C ,and
and w
w EEW.
W.Prove
Provefor
for
exactlyone
oneofofaaand
and-a.
-a.
E P that
that Pw
Pw contains
contains exactly
each aa E
6. Assume
Assumethe
thehypothesis
hypothesis of
of Exercise
Exercise10.3
10.3 and
and let
let wo
wo be the element
element of W
W of
of
maximal length
lengthin
inthe
thealphabet
alphabetR.
R.Define
Definea arelation
relation5<on
onWWby
byuu5< w
w ifif
w
=xu
x uwith
with1(w)
l ( w )==1(x)
l ( x )++1(u).
l(u).Prove
Prove
w=
(1)
(1) <5isisaapartial
partialorder
orderon
on W.
W.
(2)
element of W. That is w
w5
< wo
wE
(2)wo
wo isis the
the unique
unique maximal element
wo for all w
E W.
W.
(3) For
For rr ==rar,E ERRand
andWwEEWWthe
thefollowing
followingare
areequivalent:
equivalent:
(3)
(a) rw <5 w.
w.
(b) l(rw)
Qrw)<51(w).
Qw).
(c) aw
a w <<0.
0.
rn with
with rr =
= r1
= n.
(d) w
= rl ...
(d)
w=
.. .r,
rl and ll(w)
(w)=
(4)
thenur
ur 5
< wr.
(4) If
If uu <5ww and
and rr EE RR with
with l(ur)
l ( u r )<51(u)
l ( u )and ll(wr)
( w r )<
5 1(w)
l ( w )then
(Hint: To
Toprove
prove(2)
(2)let
letwo
wo#0 w EE W
W and
anduse
useExercise
Exercise 10.3.2,
10.3.2, 30.10,
30.10, and
to show
show there
thereexists
existsrr EE RR with
withl1(w)
l(rw).)
30.12 to
( w ) << l(rw).)
11
generalized Fitting subgroup
The generalized
We've seen that the composition
composition factors
factors of a finite group control the structure
of the group in part,
part, but
but that
that control
control is
is far
far from
from complete.
complete. Section
Section 331
introduces
1 introduces
a tool for studying finite groups via
via composition
composition factors
factors 'near
`nearthe
thebottom'
bottom' of
of
the group. The generalized Fitting
Fitting subgroup
subgroup F*(G)
F*(G) of
of a finite group G is aa
characteristic subgroup of
of G generated by the small
small normal
normal subgroups
subgroups of G
G
and with the property
property that
that CG(F*(G))
CG(F*(G)) 5
< F*(G).
F*(G).This
This last
last property supplies a
of G
G as aasubgroup
subgroupofofAut(F*(G))
Aut(F*(G))with
withkernel
kernelZ(F*(G)).
Z(F*(G)). G can
representation of
be effectively investigated via this representation
representation because F*(G)
F*(G)isisaa relatively
relatively
uncomplicated group
group whose
whose embedding
embedding in
in G
G isisparticularly
particularlywell
wellbehaved.
behaved.
It turns out that F*(G)
F*(G)isisaacentral
centralproduct
productofofthe
thegroups
groupsOP(G),
O,(G),ppr=E7r(G),
n(G),
with a subgroup E
E(G)
A
with
( G ) of
of G.
G .To
To define
define E(G)
E ( G )requires
requires some
some terminology.
terminology. A
central extension of a group X is
is aa group
group YY together
together with
with aa surjective
surjectivehomomorphism of Y
Y onto X whose kernel is in the center
center of Y.
Y. The group
group Y
Y will
will
also be said to be a central extension
of X. A
A group L is quasisimple if L is
is
extension of
of G are
perfect and the central extension of a simple group. The components of
its subnormal quasisimple subgroups,
subgroups, and
and EE(G)
of G genera( G ) is the subgroup of
ted by the components of G.
G . It develops that E(G)
E ( G )isis aacentral
centralproduct
product of
of the
the
G.
components of G.
that ifif p is aa prime
Recall that
prime then a p-local subgroup of G is
is the
the normalizer
normalizer
in G of aa nontrivial p-subgroup of
of G.
G .The
The local
local theory
theory of
of groups
groups investigates
investigates
finite groups from the point
point of
of view
view of
of p-locals.
p-locals. A question of great interest in
this theory is the relationship between the generalized Fitting subgroup
subgroup of G
G
and that of its local subgroups. Section 31 contains various results about such
relationships. In the final chapter
chapter we'll get some idea of how such results are
relationships.
used to classify
classify the finite simple
simple groups.
groups.
If F*(G)
F*(G)is
is aa p-group,
p-group,ititcan
canbe
be particularly
particularly difficult
difficult to analyze
analyze the structure
structure
G. One
One tool
tool for
for dealing
dealing with
with such
such groups
groups isis the
theThompson
Thompson factorization.
factorization.
of G.
Lemma 32.5 shows that, if G is solvable
solvable and the Thompson
Thompson factorization
factorizationfails,
fails,
of G is
then the structure of
is rather
rather restricted.
restricted. This result will be used in later
later
Theorem and
and the
chapters to prove the Thompson Normal p-Complement Theorem
Solvable Signalizer Functor Theorem. The Normal p-Complement
p-Complement Theorem
Theorem
will be used in turn to establish
establish the nilpotence
nilpotence of Frobenius
Frobeniuskernels.
kernels.
Finally the importance
Finally
importance of components
components focuses attention
attention on
on quasisimple
quasisimple
groups. In
In section
groups.
section 33 we
we find
find there
there isis aalargest
largestperfect
perfectcentral
centralextension
extension
The generalized Fitting subgroup
157
n:G(? +G
G of
ofeach
eachperfect
perfect group
group G.
G.G(?isisthe
theuniversal
universalcovering
coveringgroup
group ofof
7r:
ker(n) isis the
the Schur
Schur multiplier
multiplier of
of G.
G. In
In particular
particular ifif G
G is
is simple
simple then
then G
(?is
is
andker(7r)
GGand
thelargest
largestquasisimple
quasisimplegroup
groupwith
withGGasasaahomomorphic
homomorphicimage
imageand
andthe
theSchur
Schur
the
multiplierofofGGisisthe
thecenter
centerofofG.
(?.As
Asan
anillustration
illustrationofofthis
thistheory,
theory,the
thecovering
covering
multiplier
groupsand
and Schur
Schurmultipliers
multipliersof
ofthe
thefinite
finitealternating
alternatinggroups
groupsare
aredetermined.
determined.
groups
TheSchur
Schurmultiplier
multipliercan
canbe
bedefined
definedfor
fornonperfect
nonperfectgroups
groupsusing
usingan
an alternate
alternate
The
definitionrequiring
requiring homological
homologicalalgebra.
algebra.The
Thepresentation
presentation given
given here
here isisgroup
group
definition
theoreticand
andrestricted
restrictedtotoperfect
perfectgroups;
groups;ititfollows
followsSteinberg
Steinberg[St].
[St].
theoretic
31 The
Thegeneralized
generalizedFitting
Fittingsubgroup
subgroup
31
In this
this section
section G
G is
is aa finite
finite group.
and
In
group. AA group
groupXXisisquasisimple
quasisimpleififXX==x(')
V and
X/Z(X) isissimple.
simple.
X/Z(X)
X/Z(X) isisaanonabelian
nonabeliansimple
simplegroup.
group.Then
Then
(31.1) Let
LetXXbe
beaagroup
groupsuch
suchthat
thatX/Z(X)
(31.1)
x(')z(x)
andXM
x(')isisquasisimple.
quasisimple.
X ==X(')
X
Z(X) and
Proof.Let
LetYY==Xx(')
andX*
X*==X/Z(X).
NowY*
Y*<9XX*
and XX*
is simple
simple so
so
Proof.
M and
X/Z(X). Now
* and
* is
Y* =
orX*.
X*.InInthe
thelatter
lattercase
caseXX==YZ(X)
YZ(X)and
andininthe
theformer
formerX*
X*isisabelian,
abelian,
Y*
=11or
contrary to
tohypothesis.
hypothesis.
contrary
YZ(X).Thus
ThusX/
X/Y(')
~ ( ' 1 .Further
X* is
is
So X
X ==YZ(X).
So
Y(1) is abelian so Y =
= YO).
Further Y/Z(Y)
Y/Z(Y) E
= X*
simple, so
soYYisisquasisimple.
quasisimple.
simple,
(31.2)
< X. Then
(31.2) Let X
X be
be aaquasisimple
quasisimple group and H <99
Then either H =
=X
X or
or
H <5Z(X).
Z(X).
H
Proof.
XX
==
HZ(X
), so
Proof. If H
H Z(X),
$ Z(X),
HZ(X),
sothat
that X/H
X/H isisabelian
abelianand
andhence
hence XX ==
x(')<5 H.
H.
XM
The
components of
of aa group
group X
X are
are its
its subnormal
subnormal quasisimple
quasisimple subgroups.
subgroups. Write
Write
The components
Comp(X)
of X.
X. Set E(X) =
Comp(X) for
for the
the set of components of
=(Comp(X)).
(Comp(X)).
(31.3)
(31.3) If
If H
H <9<9XXthen
then Comp(H)
Comp(H) =
=Comp(X)
Comp(X) n H.
H.
(31.4)
Let L EE Comp(G)
and H
H <9<9G.
Comp(G) and
G.Then
Theneither
either LLEEComp(H)
Comp(H) oror
(31.4) Let
[L, H
H]I =
=1.1.
Proof.
=G
Proof. Let
Let GGbe
beaaminimal
minimalcounterexample.
counterexample. If L =
G the
the lemma
lemma holds
holds by
31.2,
7.2 we
we may
maytake
takeXX=
= (Lc)
31.2, so by 7.2
( L ~##)G.G.Similarly
Similarlyifif HH ==GGthe
thelemma
lemma
G.
X
n
Y
<
X,
so,
by
minimality
of
G,
either
=
(
H
~
f
)
G.
X
n
Y
9
X,
so,
by
minimality
of
G,
either
is
trivial,
so
take
Y
is trivial,
= (Ho)
L E Comp(X
Comp(X n
n Y) or [L, xXnnY]
Y]==1.1.InInthe
thefirst
firstcase
caseLLEEComp(Y)
Comp(Y)by
by31.3
31.3
158
The generalized
generalized Fitting subgroup
then, as
as H
H<
and then,
5 YY <<G,
G,the
thelemma
lemmaholds
holdsby
byminimality
minimalityof
of G.
G. In
In the
the second
second
[Y,
<[Y[YnX,L]=1,so,by8.9,
[Y,L,L,LIL]I
n x , LI = 1, SO,by 8.9, [Y,L]=1.
[Y,LI = 1.
(31.5) Distinct
(31.5)
Distinct components
components of G
G commute.
commute.
Proof. This
Thisisisaadirect
direct consequence
consequenceof
of 31.2
31.2 and 31.4.
3 1.4.
(31.6) Let L E Comp(G) and H an
an L-invariant
L-invariant subgroup
subgroup of G. Then
(31.6)
(1) Either L
L EE Comp(H)
Comp(H) or
or [L,
[L, HI
H]=
=1.
(1)
1.
(2) IfIf H
H isissolvable
solvable then [L,
[L, H]
HI ==1.1.
(3) If
= 1.
If R
R <_(GGthen
theneither
eitherLL EE Comp([R,
Comp([R, L]) or [R,
[R, L] =
1.
from 31.4
31.4 applied
appliedtotoLLH
in the
the role
role of
of G. Then (1)
Proof. Part (1) follows from
H in
implies
(2). IfIf R
R <5 GGthen
implies (2).
then[L,
[L,R]
R]isisL-invariant
L-invariant by
by 8.5.6,
8.5.6, so
so by
by (1)
(1) either
either
L E Comp([L, R]) or [R,
[R, L,
L, L]
L]==1.1.In
Inthe
the latter
latter case
case [R,
[R,L]
L]==11by
by 8.9.
8.9.
(31.7) Let
Let EE =
= E(G),
E(G), Z
Z=
= Z(E), and E* =
= E/Z.
E/Z. Then
(31.7)
Then
(1) ZZ ==(Z(L):
(Z(L):LL EE Comp(G)).
Comp(G)).
(2) E*
E*isisthe
thedirect
directproduct
productof
of the
the groups
groups (L*:
(L*: L EE Comp(G)).
Comp(G)).
(3) EEisisaacentral
centralproduct
productof
ofits
itscomponents.
components.
The Fitting
subgroup of
of G, denoted by
by F(G),
Fitting subgroup
F(G), isis the
the largest
largest nilpotent
nilpotent normal
subgroup
subgroup of G. O,,.(G)
O,(G) denotes
denotesthe
the largest
largest solvable
solvable normal
normal subgroup
subgroup of G.
G.
(31.8) Let G be a finite group. Then
Then F(G)
F(G) is
(31.8)
is the
the direct
direct product of the groups
(OP(G):
P
E
n(G)).
(Op(G): P E d G ) ) .
Proof. See
See 9.11.
9.11.
(31.9) O.(CG(F(G))) = Z(F(G)).
Proof.
Let Z =
Proof. Let
= Z(F(G)).
Z(F(G)). G*
G* =
=G/Z,
G/Z,and
andHH==O,,(CG(F(G))).
O,(CG(F(G))). Assume
Assume
H*= 1 and let X* be a minimal
subgroup of
of H*.
H*. Then
Then X*
X* is
is a pH*#
minimal normal
normal subgroup
group for
for some
some prime
prime p,
p, so X =
=PZ,
P Z ,where
wherePPEESyIP(X).
Sylp(X). X
X centralizes
centralizes Z,
2,
so P <X.
==Z,
9 X.Thus
ThusPP<OP(G)
5 O,(G) <5F(G),
F(G),SoSoPP< 5CF(G)(F(G))
CF(G)(F(G))
Z,contradicting
contradicting
P*=X*01.
P*
= X* # 1.
(31.10)
solvable then
thenCG(F(G))
CG(F(G)) 5
< F(G).
(31.10) If G is solvable
F(G).
Define
the socle
socle of
of G to be
Define the
be the
the subgroup
subgroup generated
generated by all
all minimal
minimal normal
subgroups of G, and write Soc(G) for the socle of
of G.
The generalized
generalizedFitting
Fitting subgroup
subgroup
The
159
Z(F(G)),G*
G*==
G/Z,
= Soc(CG(F(G)>*).
Then
(31.11) Let Z ==Z(F(G)),
(31.11)
G/Z,
andand
S*S*
= Soc(CG(F(G))*).
Then
E(G)==SM
S('! and
andSS==E(G)Z.
E(G)Z.
E(G)
Proof.Let
LetHH==CG(F(G)).
By31.9,
31.9, O,0(H*)
O,(H*) =1.
= 1.So,
So,by
by8.2
8.2and
and8.3,
8.3,each
each
Proof.
CG(F(G)). By
minimal normal
normal subgroup
subgroup of
of H*
H*isisthe
thedirect
directproduct
productofofnonabelian
nonabeliansimple
simple
minimal
groupsand
andby
by 31.3
31.3 these
these factors are components of
E(H*).Let
Let
groups
of H*.
H*. Thus
Thus S*
S* 5
< E(H*).
K *be
b eaacomponent
c o m p o n eofn H*.
t o f ~By
* .31.1,
~ ~ 3K
K
1 .==
1 K('
,~ ( ')Z
) ~with
w i t KM
h ~ ( quasisimple.
')~uasisim
e.~y31.3
K*
Byp l31.3,
K('
E(G)Z.
K(')EEComp(G),
Comp(G),sosoSS<I
E(G)Z.By
By31.6.2,
31.6.2,E(G)
E(G)<5H.H.Let
LetLLEEComp(G),
Comp(G),and
and
( LH )Then
. Then
M*isisa aminimal
minimalnormal
normalsubgroup
subgroupof
ofH*
H*by
by31.4,
31.4, so
soM
M <(S.
S.
M=
M=
(LH).
M*
ThusSS==E(G)Z.
E(G)Z.By
By31.1,
3 1.1,E(G)
E(G)==
s(').
Thus
Ski).
Define
Fitting subgroup
subgroupof
of G
G to
tobe
beF*(G)
F*(G)=
= F(G)E(G).
F(G)E(G).
Define the generalized Fitting
(31.12) F*(G)
F*(G)isisaacentral
central product of F(G)
F(G) with
with E(G).
E(G).
(31.12)
Proof.See
See31.6.2.
31.6.2.
Proof.
(31.13) CG(F*(G)) < F*(G).
Proof. Let
Let HH==CG(F*(G)),
CG(F*(G)),K
K=
=CG(F(G)),
CG(F(G)),Z =
and G*
G* =
=G/Z.
G/Z.
Proof.
= Z(F(G)), and
Then H*
H* <9K*,
K*, so
so if
if H*
H*01
# 1then
then101 #
H*
Soc(K*),and
andthen,
then,byby31.11,
31.11,
Then
H*
nn
Soc(K*),
E(G)0#Z,Z,a acontradiction.
contradiction.
H nnE(G)
H
I
Recall
by Op,,E(G)/Op,(G)
Op',E(G)/OP,(G) =
= E(G/OP,(G)).
Recall Op',E(G)
O,J,~(G)is defined by
E(G/O,,(G)).
(31.14)
(31.14) Let
Let pp be
beaap-subgroup
p-subgroupof
of G.
G.Then
Then
(1) Op',E(NG(P))
O ~ ~ , E ( N G (<5
PCG(OP(G)),
>CG(O~(G)),
>
and
(1)
and
(2) ifif PP<5Op(G)
O,(G)then
thenOP(F*(NG(p)))
OP(F*(NG(p)))==OP(F*(G)).
Op(F*(G)).
(2)
Proof.
OP,,E(NG(p)).
Proof.Let
LetXX==
0 , 1 , ~ ( N ~ ( pTo
)To
) prove
. prove(1),
(I),ititsuffices
sufficestotoshow
showXXcentralizes
centralizes
R
=
COp(G)(P)
by
the
A
x
B
Lemma.
But
[R,
OP,(X)]
<
R
n
OP-(X)
R = Co,(G)(P) by the A B Lemma.
[ R ,Op,(X)] ( n Op!(X) =
=11
and
XI <5OP,(X)
Op,(X)by
by31.6.2,
31.6.2, so
so[R,
[ R X]
,XI==1 1bybycoprime
coprimeaction,
action,18.7.
18.7.
and [R,
[ R ,X]
So
So take
take P<OP(G).
P 5 Op(G).Then
Then Yo=OP(F*(G))<Y=OP(F*(NG(P))).
Yo = Op(F*(G)) 5 Y = OP(F*(NG(P))).If
If LL
is aa component
NG(P)then
then [L,
[ LOP(G)]
,Op(G)]==11by
by (1),
(I),and,
and,asasYo
Yo <
(Y,
Y, eieicomponent of
of NG(P)
ther
CG(F*(G))
=
ther [L,
[ L,Yo]
Yo] =
=11or
or LLEE Comp(Yo)
Comp(Yo) by 331.4.
1.4. In the first
first case
case LL <_
( CG
(F*(G)) =
Z(F(G)),
Z(F(G)),aacontradiction.
contradiction. So E(NG(P))
E(NG(P))=
=E(G).
E(G). Let
Let qq 0#ppand
andQQ==Oq(Y).
0, (Y).
We must show
show QQ <5O9
0, (G). Passing to G/O9
G/Oq (G) and appealing to coprime
coprime
action,
= 1,
=1.1.But,
action, 18.7,
18.7, we may take Oq(G)
Oq(G)=
1, and it remains to show Q =
But,by
by
(1)
and
as
Yo
<
Y,
Q
<
CG(O9(F*(G)))=CG(F*(G))=Z(F(G)),
so
indeed
Yo 5 Y, Q 5 CG(Oq(F*(G)))= CG(F*(G))= Z(F(G)), so indeed
(1)
Q = 1.
Q=1.
160
The generalized
generalized Fitting subgroup
(31.15) IfIfGGisissolvable
p-subgroup ofofGGthen
(NG (P)) <5Op,
(G).
solvableand
andPPisisaap-subgroup
thenOp,
O,I(NG(P))
O,!(G).
Proof. Passing
Passing to
to G/Op,(G)
G/O,l(G) and
andappealing
appealingto
tocoprime
coprimeaction,
action, 18.7,
18.7, we
we may
may
take Op,(G)
= 1,1,and
O,f(G) =
and itit remains
remains to
to show
show Op'(NG(P))
O,~(NG(P))==1.1.This
Thisfollows
followsfrom
from
31.10 and 31.14.1.
(31.16)
=Op(G)
O,(G)forsomeprime
(31.16) Let F*(G)
F*(G) =
for some primep.Then
p. ThenF*(NG(P))
F*(NG(P))=
=O,(NG(P))
Op(NG(P))
for each p-subgroup P of G.
G.
Proof. This
Thisfollows
follows from
from 31.13
3 1.13 and
and 31.14.1.
3 1.14.1.
The
Schreier Conjecture
Out(L)isis solvable
solvablefor
for each
each finite
finite simple
simple group
group L.
L.
The Schreier
Conjecturesays
says Out(L)
(31.17) Let O,t(G)
Op,(G) =
= 11 and P
P aa p-subgroup
p-subgroup of
of G.
G. Then
Then
(31.17)
(1) Op',E(NG(P))
fixes
O,J,~(NG(P))
fixes each
each component
component of G.
G.
(2)
(2) If each
each component
component of
of GGsatisfies
satisfiesthe
theSchreier
Schreierconjecture,
conjecture, then
then
Op',E(NG(P))°°
<) "E(G).
O p f , ~ ( N ~ ( P >L
H=
Proof. Let H
=NG(P),
NG(P),X ==Op,(H),
O,!(H), H*
H* ==H/X,
H/X,and
andYY ==Op',E(H)°°.
O,I,~(H)*.Let
K <(XXor
K
<
H
with
K*
E
Comp(H*),
and
subject
to
these
constraints
or K ( H with K* E Comp(H*), and subject to these constraints pick
tomoving
movingaacomponent
componentofofG.
G.Let
LetPP(
< Po E
E Sylp(CG(K)).
Sylp(CG(K)).
K minimal subject to
hypothesis with respect to Po, we may
may take
take P =
As K satisfies
satisfies the same hypothesis
=Po.
Po.In
In
particular,by
by31.14,
31.14,Op(G)
Op(G)5< PP.. Let
Let R
R EE Syl,(H
Sylp(H n
fl E(G)).
particular,
Suppose first
first K
K 5< X.
X. Then
Then K is aa q-group
q-group for
for some
some prime q, and
and by
by cocoSylow q-group
q-group Q of Op,(H).
prime action, 18.7, there exists an R-invariant
R-invariant Sylow
O,f(H).
Replacing K
K by
by a suitable
suitable conjugate,
conjugate,we
wemay
mayassume
assumeKK(<Q.
Q.By
By24.4,
24.4,QQ =
=
Replacing
Now[R,
[R,Q]
Q] 5< [E(G),
[R, Q]CQ(R).Now
[E(G), Q]
Q] <(E(G),
E(G),so
so[R,
[R,Q]
Q]fixes
fixeseach
each comcomponent of
of G.
G. Hence
Hencewe
wemay
maytake
take[K,
[K,R]
R]==1.
Thus,by
bychoice
choiceofofPP,
ponent
1. Thus,
, RR 5< P.
P.
So P
(L) for
P flnE(G)
E(G)EESylp(E(G))
Syl,(E(G)) bybyExercise
Exercise3.2.
3.2.As
AsOp,(G)
O,/(G) =
=1,
1, p cE7rn(L)
for
LE
E Comp(G),
Comp(G), so
so 11## P flflLL EE Sylp(L)
Then PP n
fl L $ Z(L), so
each L
Syl,(L) by 6.4. Then
L=
= [E(G),
[E(G),PPflnL]
L]isisK-invariant.
K-invariant.
= 1. In
In the latter
So K $ X. By
By 31.4,
31.4, either
either K*
K* <5 [K*,
[K*, R*] or [K*,
[K*, R*] =
case by
by coprime action,
action, 18.7, K =O,f(K)CK(R)
=Op-(K)CK(R) so, by minimality
minimality of K,
K,
[K, R]
contradiction.
R] ==1.1.But
Butthen
thenan
anargument
argumentin
inthe
the last
last paragraph supplies
suppliesa contradiction.
[K,
former, K
K(
< X[K,
X [K, R]
R]:<
XE(G), so K fixes
of G.
5 XE(G),
fixes each component of
In the former,
< N(L). If LL satisfies
Let L be
be aa component
component of G.
G. II have
have shown
shown Y
Y5
satisfies the
the
Schreier
conjecture, then Autr(L)
Auty(L) <5Inn(L)
Schreier conjecture,
Inn(L)asasYY ==YO°,
Y*, so
soY
Y <(LC(L).
LC(L). Hence,
Hence,
under the
the hypothesis
of (2),
(2), Y
E(G)C(E(G)) and
<
under
hypothesis of
Y 5< E(G)C(E(G))
and then,
then, by
by 31.14,
31.14, Y
Y(
E(G)C(F*(G))
= E(G).
E(G)C(F*(G))<5F*(G),
F*(G),so
soYY <(F*(G)O°
F*(G)* =
The generalized
generalized Fitting
Fitting subgroup
subgroup
The
161
161
balancedfor
forthe
theprime
primeppififOO,l(CG(X))
Opj(G)
order pp
GGisisbalanced
p, (CG M) <<Op(G) for each X of order
G.
ininG.
(31.18) Let
LetOpOp~(G)
oforder
orderpin
p inG,G,LLEEComp(G),
Comp(G),and
andYY==
Op~(C
G(~)).
(31.18)
(G) ==1,1,xxof
Op'(CG
(x)).
Then
Then
(1)
[L,x]x]then
then[L,
[L,Y]
Y]==1 1and
andeither
eitherLLEEComp(CG(x))
Comp(CG(x))or L ##Lx
Lx
(1) IfIfLL##[L,
and
==KKEEComp(CG(x))
CcL,xl(~)(l)
Comp(CG(x))with
withKKaahomomorphic
homomorphicimage
imageof
of L.
L.
and C[L,x](x)('
(2) IfIfLL==[L,
[L,x]x]andAuty(x)L(L)
a n d A ~ t ~ ( , ) ~isis(balancedfortheprime
balancedfortheprime
L)
then[Y,
[Y,L]
L]==1.1.
(2)
p,p,then
Proof.
= 1 or L ##Lx.
Proof. Assume
Assume L ##[L,
[L,x].
x].Then
Theneither
either [L,
[L ,x] =
Lx.InInthe
thefirst
first
case
= [L,
Comp(C(x)), so
so [L,
[L,Y]
Y] =
=1.
1. In
In the
the second
second let M =
[L,x]
x] and
and M*
M* ==
caseLLEE Comp(C(x)),
M/Z(M).
so,so,byby31.4,
M/Z(M).By
By8.9,
8.9,[M,
[M,L]
L]##1, 1,
31.4,L LEEComp(M),
Comp(M),and
andwe
weconclude
conclude
M=
=(0))
(L("))
centralproduct
productof
ofthe
the groups
groups (Lx': 00 5
from31.5.
31.5.
M
is is
thethecentral
< ii <<p)p)from
Hence,
Hence,by
byExercise
Exercise3.5,
3.5,KK==CM(x)(l)
cM(x)(')isis aa homomorphic
homomorphic image
image of L.
L. As
As LL isis
quasisimple,
M1
< <1G,
quasisimple,so
sois
is its homomorphic image K. Also M
G,so
soK
K <1<1CG
CG(x),
(x),
and
Comp(C(x)). So
So [K,
[K, Y]
Y] ==1.1.YYacts
actson
onL*
L*by
by31.17,
31.17,so,
so,
and hence
hence K EE Comp(C(x)).
as [K*,
[K*,Y]
byExercise
Exercise 3.5.
3.5. Then
Then [L,
[L,Y]
= 11by
by 31.6,
31.6,
as
Y] ==1,l,[L*,Y]
[L*, Y] ==1 1by
Y] =
completingthe
theproof
proof of
of(1).
(1).
completing
So assume
assume L ==[L,
[L,x],
x],letletUU= =
Y (x)L,
AutG(L)be
be the
the
So
Y(x)L,
andand
letlet
rr:nU: U
-->+AutG(L)
conjugation map.
O,(U)(x) so
sothat
thatPPEESylp(ker(7r)(x)).
Syl,(ker(n)(x)). By
By 31.14,
31.14,
conjugation
map. Let
Let P ==Op(U)(x)
[P,
YY<5Op,(Cv(P))
==Op,(Nv(P)).
Then,
[P,Y]
Y] ==1,1,so,
so, as
as Cu(P)
CU(P)<5Cv(x),
CU(x),
OPf(CU(P))
Op~(NU(P)).
Then,
as
as PPEESylp(ker(7r)(x)),
Syl,(ker(n)(x)), NUn((x7r))
Nuk((xn)) =
=Nu(P)7r
Nu(P)nby
byaaFrattini
FrattiniArgument,
Argument, so
so
Yrr
(Cu. (x 7r)).Hence,
Hence,ififAutu(L)
Autu (L)isisbalanced
balancedfor
forthe
the prime
prime p, then
Yn<
<Op'
Op~(Cur(xn)).
then
[Y, L] ==1,1,so
sothat
that(2)
(2)holds.
holds.
[Y,
(31.19)
O,<(G) ==11and
andassume
assumeAutH(L)
AutH(L)isisbalanced
balancedfor
for the
the prime
prime pp for
for
(31.19) Let
LetOp-(G)
each
each L
L EE Comp(G)
Comp(G)and
and each H
H <(GGwith
withLL<1H.
H.Then
ThenGGisisbalanced
balancedfor
forthe
the
prime p.
p.
prime
Proof.
Proof.Let
LetXXbe
beaasubgroup
subgroupof
of GGof
of order
order p and
and Y
Y ==Op'(CG(X)).
Op,(CG(X)).IImust
must show
show
Y
Y ==1.1.By
By31.18
3 1.18and
andthe
thehypothesis
hypothesison
on the
the components
components of G,
G, [Y,
[Y, E(G)]
E(G)]==1.1.
By
Op(G)] =
=1.
< CG(F*(G))
= Z(F(G)),
1. So
So Y 5
CG(F*(G))=
Z(F(G)),so,
so,as
asOp,(G)
O,,(G) ==1,
1,
By 31.14,
31.14, [Y,
[Y, O,(G)]
Y=l.
Y=1.
Here's technical
technicallemma
lemmato
tobe
be used
usedin
inchapter
chapter15.
15.
Here's
(31.20) Let
Let A
A be
be an
an elementary
elementary abelian r-group acting
acting on a solvable r'-group
G
(G), pk
p" =
G and
and let
let aa EE A. Let p E nr cC7r
n(G),
=rr'
n' U
U {p},
{p], and P an
an A-invariant
A-invariant psubgroup of
of G.
G.
subgroup
162
generalized Fitting subgroup
The generalized
(1) Suppose
SupposePP <5Op(K)
O,(K) for
forsome
somea-invariant
a-invariant7r-subgroup
n-subgroup K of
of G such that
CK(a)
CK(a)is aa Hall
Hall 7r-subgroup
n-subgroup of CG(a).
CG(a). Then P 1 OPr(N)for each a-invariant
subgroup N
N of
of G
G with
with N
N=
= (P,
( P ,CN(a)).
CN(a)).
(2)
A is
is noncyclic
and let
let A
A be the
(2) Assume
Assume A
noncyclic and
the set
set of
ofhyperplanes
hyperplanes of
A.
o r each
A. Assume
Assume ffor
each BBEE A that
that Cp(B)
Cp(B)<On((CG(a),
5 O,((CG(a), Cp(B))). Then
Then
P <On((CG(a),
5 O"((CG(~>,
P)).
(3) IfIf BB isisaanoncyclic
((a, b)), P))
noncyclic subgroup
subgroup of CA(P)
CA(P) and P <5O,r((CG
O,((CG((a,
P))
for each bb EEB'B#,
thenPP 5
< On((CG(a),
, then
0, ((CG(a),P)).
P)).
Proof.
ProoJ Let
Let GGbe
beaaminimal
minimalcounterexample
counterexample to
to (1),
(I), (2),
(2), or
or (3).
(3). Without
Without loss,
A=
= (a)
=
(a) in
in (1)
(1) and
and A
A=
=(B,
(B, a)
a )in
in(3).
(3).In
Ineach
eachcase
case itit suffices
suffices to assume G =
P <5Q,
(G), where
((P,
P ,CG(a))
CG(a))and
and prove P
O,(G),
whereaa==p"
p"inin(1)
(1)and
andaa==7rninin(2)
(2)and
and (3).
Let
normal subgroup
subgroupof
ofGGand
andG*
G*==GIH.
G/H. Then
Let H be
be a minimal A-invariant normal
P*
and H
H is
P*<5O,,(G*)
O,(G*) ==S*
S*by
by minimality
minimality of G
G and
and coprime action 18.7.4,
18.7.4, and
a q-group for some prime q. Hence if P
P <5O,,(HP)
O,(HP) we
we are
are done,
done, so
so this
this is
is not
not
the case, and in particular
particular qq V
a.
$ a.
Then CH(a)
CH(a)<]I.
Let II=CG(a).
= CG(a). Then
< I. We
We show
show [P,
[P, CH(a)]=l.
CH(a)] = 1.Then
ThenGG =
=
(I,
of H,
H,CH(a)
CH(a)=(aor
(I,P)
P ) <5N(CH(a)),
N(CH(a)),so,
so,by
by minimality
minimality of
1 or H.
H.InInthe
thelatter
latter
case [P, H]
= 1.
HI ==1,1,contradicting
contradicting P $ O,(PH).
O,(PH). Thus CH(a)
CH(a) =
1.
Now to verify that
that [P,
[P, CH(a)]
=1.1.In
< I,
I, CH(a)
CH(a)]=
In(1),
(I), as
as qq EE7r
n and CH(a) <]
CH(a)
is contained
contained in each
each Hall
Hall 7r-group
n-group of CG(a)
CG(a) by
by Hall's
Hall's Theorem
Theorem 18.5,
18.5,and
and
hence in
in K. So [CH(a),
P] 5
<Op(K)
[CH(a), PI
O,(K) flOq(G)=1.
n Oq(G)= 1.InIn(2),
(21,[Cp(B),
[Cp(B),CH
C H(a)]
( ~ )<
]
O,r(GB)
fl
Oq(GB)
=1,
where
GB
=
(CG(a),
Cp(B))
and
B
E
A.
Hence,
0, (GB)no, (GB)= 1, where GB= ( C G ( ~ CP(B))
),
and B A. Hence,
by Exercise
(Cp(B):BBEE A)
A) <5C(CH(a)).
C(CH(a)). Finally,
Finally, in
in (3),
Exercise 8.1,
8.1, P ==(Cp(B):
[P,
fl Oq(Gb)
Oq(Gb)=
= 1, where
where Gb
Gb =
= (P,
[P, CH((a,
C H ( ( ~bb))]
,) ) l<O,r(Gb)
5 O,(Gb) n
( P ,CG((a,
C G ( ( ~b))).
b))).
, So,
So,
again by Exercise 8.1,
8.1, CH(a)
CH(a) =
_ (CH((a,
(CH((a,b)):
b)):bbEE B#)
B') 5 C(P).
C(P).
We've
= 1.1. In
In (2)
(2) and
We've shown
shown CH(a) =
and (3) let
let RR be
beananA-invariant
A-invariant Hall
Hall
7r-subgroup
of S containing P.
P . By
ByaaFrattini
FrattiniArgument,
Argument,GG==HNG(R).
HNG(R).As
As
n-subgroup of
CH(a)=
CG(a)=2CG*(a)
CG*(a)=2NG(R)
NG(R)fli C(a),
l C(a),so
soCG(a)
CG(a)<,<NG(R).
NG(R).But then
CH(a)
=1,1,CG(a)
G=
RRfl i Hl H= =
1,1,
a acontradiction.
=(CG(a),
(CG(a),P)
P)<5N(R),
N(R),soso[P,
[P,H]
HI<5
contradiction.
This
(1). Here
Herewe
wemay
maytake
takePP=Op(K)
=Op(K)<]<K,
K, so,
so, ifif U is aa Hall
Hall
This leaves
leaves (1).
7r'-group
of II,, then
then RR=
= ((PI)
_ (P')
(PU) and
and R
R <]
< (I, P)
nl-group of
PI) =
=(P(inK)U)
(p(InK)')=
P) =
=G.
G. Thus
Thus
may assume
assume H <
5R.
R. Let
Let XX be
bean
ana-invariant
a-invariant Hall
Hall q'-subgroup
ql-subgroup of
of US.
US.
we may
As CH(a)
q'-group, so X is the unique
CH(a)=
=1,
1, CUS(a)
Cus(a) isis aa ql-group,
unique a-invariant Hall q'q1subgroup of
of US by
by Exercise
Exercise 6.2.
6.2.Hence
Hence (P,
(P, U) 5
< X. But then H <5R
R <5X,
X, aa
contradiction.
contradiction.
Thompsonfactorization
factorization
32 Thompson
In this section p is
Denote by
by M(G)
is aa prime and G is
is a finite group. Denote
d ( G )the
the set
set of
of
elementary abelian
abelian p-subgroups
p-subgroups of
of G
G of
of p-rank
p-rank mp(G).
m p(G).Set
SetJ(G)
J(G)=
_ ((.d((G)).
(G)).
Thompson
factorization
Thompson factorization
163
J ( G )the
theThompson
Thompson subgroup of
( G ) depends
depends on
on the
the
We call J(G)
of G
G.. Of course JJ(G)
p.
choice of p.
(32.1) (1) J(G)
J ( G )char(G).
char(G).
(32.1)
A is
G ) and
( H ) Gc d,((G).
(G).
(2) If A
isininda( (G)
andAA5<HH5<GGthen
thend ,((H)
Let PPEESylp(G).
Syl,(G). Then JJ(P)
( P )==J(Q)
J ( Q for
) foreach
eachp-subgroup
p-subgroup QQ of
of GG concon(3) Let
taining JJ(P).
(P).
taining
V is
is aa GF(p)G-module
GF(p)G-moduledefine
defineGJ'(G,
P ( G , V)
V)to
toconsist
consistof
ofthe
thenontrivial
nontrivialelemenelemenIf V
A of
of G
G such
such that
that
tary abelian p-subgroups A
+
+
m(A) + m(Cv(A))
m(Cv(A)) >
1m(B)
m(B) + m(Cv(B))
m(Cv(B))
A. Notice
Notice that if B
B is
is aa nontrivial
nontrivial subgroup of A E
( G , V) for
for each B <5 A.
E P°J'(G,
P ( GV).
, V).
subset
P of
this inequality
inequalityisisananequality,
equality,then
thenBBis is
which this
in in
GJ'(G,
AA
subset
_1:1P
PP(G,
( G , V)
V) isisstable
stableifif GG permutes
permutes°J'
Pvia
viaconjugation
conjugationand,
and,whenever
wheneverAA isisin
in GJ'
P
B is
is aa nontrivial.
nontrivial.subgroup
subgroupof
of A
A with
with
and B
m(A) + m(Cv(A)) = m(B) + m(Cv(B)),
is in
in °J'
9As
As aa final
final remark note that
P ( G , V)
V ) then m(A) 2
then B is
that ifif A
A is in °J'(G,
>
m(V/Cv(A));
in the inequality defining membership
m(V
/Cv (A)); to see this
thisjust
just take
take B
B=
=11in
P ( G , V).
V).
in °J'(G,
(32.2)
),
LetVVbe
beaa normal
normal elementary
elementaryabelian
abelianp-subgroup
p-subgroupof
of G,
G ,G*
G*== G/
G I CG(V
CG(V),
(32.2) Let
Then P
GJ'isisaastable
stablesubset
subsetof
of P7)(G*,
and GJ'
P=
( G )and
andA*
A* # 11.
1).Then
( G * , V).
V).
and
= {A*: AAEE d,1(G)
Proof.
Prooj Let
Let A
A be in a(G)
d ( G and
) andCA(V)
CA(V)<5BB<5A.A.A0
A. ==ACv(A)
ACV(A)isisan
anelemenelemen(G),
m(Ao) 5
< m(A).
so, as
as A
A EE d (, G
) , m(Ao)
m(A). Hence
HenceAA ==A0
A. since
since
tary abelian p-group, so,
Ao. Thus
ilV.
V. Similarly
Similarly m(BCv(B))
m(BCv(B)) <5m(A).
m(A). As
As CB(V)
CB(V) =
A5
< A0.
Thus CV(A)
Cv(A) =
= A fl
CA(V), we have
have
CA(V),
m(BCv(B)) = m(B*) + m(CA(V)) + m(Cv(B)) - m(A fl v)
while
m(BCv(B)) < m(A) = m(A*) + m(CA(V))
i lV
V it follows that
so as Cv(A)
Cv (A) =
= A fl
m(A*) + m(Cv(A)) > m(B*) + m(Cv(B))
with equality only ifif m(BCV(B))
m(BCv(B)) =
=m(A).
m(A). Thus
Thus ifif A*
A* # 1 then A* EE P°J'(G,
( G , V)
while if
if m(B*)
m(B*) + m(Cv(B))
m(Cv(B)) =
= m(A*)
m(A) =
= m(BCv(B)),
m(A*) + m(Cv(A)) then m(A)
m(BCv(B)),
so B0
= BCv(B)
Bo =
BCv(B) is in ,1(G)
d ( G )and
andhence
henceB*
B*==Bo
B,*isisininGJ'(G*,
P ( G * , V).
V).
+
+
The generalized Fitting subgroup
164
(32.3)
Let G
G be a solvable
V a faithful
GF(p)G-module,and
and PP a
(32.3) Let
solvable group,
group, V
faithful GF(p)G-module,
GJ'P is
Sylow p-group of G.
G.Assume
AssumeOp(G)
Op(G)==1, 1,
i sa astable
stablesubset
subsetofof-OP(G,
P ( G , V)
V)
andHH=Then
x 1. xx H,,,
®CV(H)
and
= ( P ) # 1 . Tp<3,H=HI
henp(3,H=H
. - . x V=[V,H]
H,, V = [ V , H
]@CV(H)
®[V,
G permutes
permutes(Hi:
(H,:1
<n),
with [[V,
H] ==[V,
with
V ,HI
[ V Hl]
,H I ](D
@ .. .. . @
[ V ,H,,],
H,], G
1 5<ii _(
n ) ,Hl
Hi S
P)) is
m([V,
Hi]) =
= 2,
SL2(p), m
( [ V ,Hi])
2, and (.( Pflf lP
is aa Sylow p-group of H.
Proof. Let
membersof
of P,
J', pick A
A EEr,
I', and
andset
setUU =
=
Proo$
Let rrconsist
consistof
of the
the minimal
minimal members
Cv(A). If
If BB isisaahyperplane
thenmm(A)+m(U)
m(B)+m(Cv(B)) with
Cv(A).
hyperplane ofofAAthen
( A )+m(U) L>m(B)+m(Cy-(B))
with
of A
A either
either /Al
P or
or B
B ==11inincase
caseof
ofequality.
equality. So
So by minimality of
A /I=
= p or
or
B EE °J'
the inequality is strict, and in that event, as
as m
m(A)
=m
m(B)
( A )=
( B ) + 1,
1, it follows
follows that
U=
=CV(B).
Cv(B).
andlet
letKK=
= [F(G),
[F(G), A].
A]. As
As Op(G)
Op(G)=
= 11,, F(G)
F(G) is a pf-group
p'-group
Suppose ]Al
A /I>
> pp and
1. By Maschke's
Maschke's Theorem, V
=
and A
on F
F(G)
A is faithful on
( G )by
by 31.10,
3 1.10, so
so K
K # 1.
V=
K]] @
®CV(K),
as K
K#
0 1,
0 # [V,
hence on [[V,
[[V,
V ,K
Cv(K),and, as
1,0
[ V ,K].
K ] .A
A acts
acts on K and
andhence
V ,K],
K],
Thus U $ C
Cv(K).
CIV,Kl(A)
# 0. Thus
v ( K ) .On
On the
the other
other hand,
hand, by
by Exercise
Exercise 8.1,
so C[V,K](A)
K ==(CK(B):
by the
last paragraph,UU=CV(B),soCK(B)
<
K
( C K ( B )IA:BI
/:A: BI ==p)p )and
andby
thelastparagraph,
= Cv(B),so C K ( B )5
N(U).
K=
= [K,
N(U). Thus
Thus K <
5 N(U),
N ( U ) ,so,
so, by
by Exercise
Exercise 3.6, K
[ K ,A] <5 CG(U),
CG(U),contracontradicting U
Cv(K).
U $C
v(K).
So JAI
=pp for
for each
each A
AEI'.
/ A /=
E r.As
As AAEE P(G,
P ( G ,V),
V ) m(V/CV(A))
,m ( V / C v ( A ) )(<m(A)=1,
m ( A ) = 1,
so A
A isis generated
generated by
by aatransvection.
transvection. Let
Let S2
52 consist of the
the subgroups
subgroups L =
=
(A1,
withAi
AiEErFand
and1 10
OP(L) (
< F(G).
thereisisLo
LoEEC20
( A l ,AA2)
z ) with
# OP(L)
F(G).For
For each
each Ao
A. EE rF there
= (A,
and W
W=
= [[V,
L]. W
W=
=
with Ao
< Lo as [Ao,
F(G)] # 1. Let L =
A. (
[Ao,F(G)]
( A ,B)
B ) EE S2
52 and
V ,L].
A] + [V,
B] and
and Cv(L)=CV(A)
fl CV(B),
so, as
as 1=m([V,
[[V,
V ,A]
[ V ,B]
C v ( L )= C v ( A ) fl
Cy-(B), so,
1 = m ( [ V ,A])=
A ] )=
m(V/CV(A)),
m(V/CV(L)). IfIf 00 #
o CW(L)
m(V/Cy-(A)),mm(W)
( W )( <2>
2 m(V/CV(L)).
C w ( L )then A and B centralize
W/CW(L),
CW(L),
W/CW(L),
Cw(L),W
/ C w ( L ) and
, V/W,
V / W SO
, so LL centralizes C
w ( L ) ,W
/ C w ( L ) ,and
tralize CW(L),
V// W,
which is impossible by
by Exercise
Exercise 3.1
3.1 since
since LL is
is not
not a p-group. ThereV
W , which
fore W =
= [V,
® [[V,
B]] is of rank
rank 22 and
and V
V=
=W
W@
® CV(L).
Cv(L). In particular LL is
[ V ,A] @
V ,B
< GL(W)
= GL2(p).
the
faithful on W so L 5
GL(W)2
GL2(p).Now A fixes only [[V,
V ,A] amongst the
set 08 of p + 11points
pointsof
ofW
Wand
andhence
henceisistransitive
transitive on
on the
the remaining
remaining pp points.
points.
So as B moves
moves [V,
[ V ,A],
A ] , L is
is transitive
transitive on 0.
8. Thus
Thus as
as LL contains
containsthe
the group
group A
A
of transvections with center [[V,
A], L contains all transvections
transvections in GL(W),
GL(W), so,
V ,A],
= SL(W)
by 13.7, LL =
SL(W) =
ESL2(p).
SL2(p).In
In particular,
particular, as G is solvable, p <
(3 by 13.8.
13.8.
Next
let
Y
=
O"(L)
and
M
=
NG(Y).
Notice
M
acts
on
W
and
CV(L)
Next let Y = Op(L) and M = NG(Y).Notice M acts on W and C V ( L )
as W
(Y). Suppose
Loo €
E 52
S2 with
with Yo
=
Yo=
W ==[V,
[ VY]
, Y ]and
and Cv(L)
C v ( L )==CvCv(Y).
Suppose L # L
OP(L0)
<M
M.. Then either
< Cv(Y),
CV(Y), since Yo is
is irreOP(Lo)5
either W
W==[V,
[ V Yo]
, Yo]or
or [V,
[ V Yo]
,Yo]5
ducible on [V,
W=
= [[V,
thenCCv(Y)
= CV(Yo),
soLL=
= SL(W)
SL(W) =
= Lo,
ducible
[ V ,Yo].
Yo].IfIf W
V , Yo]then
V ( Y )=
Cv(Yo),so
Lo,
contrary
to
the
choice
of
Lo.
Thus
[V,
Vol
<
CV(Y),
so
W
<
Cv(L)
and,
by
contrary to the choice of Lo. Thus [ V ,Yo] 5 C v ( Y ) ,so W 5 Cy-(L)
< CG(W)
fl CG(CV(L))
= CG(V)
= 11
Exercise 3.6, [[L,
L ,Lo] _(
CG(W)i'l
CG(Cv(L))=
C G ( V )=
maximal set of commuting
commuting members
andDD=
= (A).
Let A be a maximal
members of S2
52 and
( A ) .By
By the last
direct product
productof
of((I:
paragraph, D is the direct
I :II EE A)
A ) and
and V
V=
=[V,
[ V ,D]
Dl ®
@ Cv
C v (D)
( D )with
[V,
D] =
_ ®IEO[V,
=A
A and
and JJ =
= (-opfl
[ V ,Dl
@,,,[V, I].
I ] So
. Soititremains
remainsto
toshow
show S2
52 =
( P nPP)) <5 D.
D.
+
+
+
factorization
Thompsonfactorization
165
If S2
11.4,there
thereisisLLEE52c-- 0A with
=OP(L)
<
52 # 0
A then,
then, by
by Exercise
Exercise 11.4,
with Y
Y=
OJ'(L)(
NF(G)(Op(D)).
ThusYYpermutes
permutesZE== (([V,
m([V,
= 2,
NF(G)(OP(D)).
Thus
[ V ,II]:
] :II EE A0)) and,
and, as m
( [ V ,YY])
] )=
it follows that Y
f acts
of
actson
on each
each member
member of E,
Z, and
and hence
hence also on each member of
0. But
next-to-lastparagraph
paragraphand
andmaximality
maximalityofofAA,
contrary
A.
But now by the next-to-last
, I,LEEA0,, contrary
to the choice of L.
L.
So S2
A.Finally
Finallyassume
assumeJ J$DDand
52 ==A.
andlet
letEEEE°J'Pwith
with EEminimal
minimalsubject
subject
to END.
As
E $ D.Let
LetZ=(WE).
Z = (wE).
AsEEisisabelian,
abelian,CE(W)=CE(Z)
C E ( W )= C E ( Z and
)and E/NE(L)
E/NE(L)
is regular on WE.
W E .Let SS be
be aa set
setof
ofcoset
cosetrepresentatives
representatives for NE(L)
N E ( L )in E.
E.
rank Es =
=
Then Z ==®SES Ws
Ws so Cz(E)
C Z ( E =)=(USES
(C,,, ws: ww EECCW(NE(L))]
w ( N ~ ( L ) is
)is) ofofrank
m(CW/NE(L)).
Therefore
2pa - s,
=
m ( C w / N E(L)).Therefore
m (m(Z/CZ(E))
Z / C z ( E ) )==mm(Z)
( Z ) --E s== 2pa
E , where
where aa =
m(E/NE(L)).
m ( E / N E ( L ) )Also
Also
. m(NE(W)/CE(W))
m ( N E ( W ) / C E ( W==
) 3) 6<(11with
withsE==11inincase
caseof
of equality.
equality.
Further
< m(W)
= 2.
2pa - sE >>a a+ S.6.Finally,
Further Es (
m(W) =
2. Thus
Thus ififaa > 00 then
then 2pa
Finally, if
NE(L),
E #N
E ( L ) ,we conclude
conclude
esEs
+
m(Cv(CE(Z))) - m(Cv(E)) > m(Z) - m(Cz(E))
= 2pa - s > a + S = m(E) - m(CE(Z)),
contradicting E EE ?7
P ((G,
G , V).
V).
Therefore
E acts
OP'(GL(W)),
Therefore E
acts on
on each
each member
member of
of A.
A.Hence,
Hence,asasLL==o~'(GL(w)),
E5
<DCG([D,
the action
of D
D on
E
DCG([D,VV])=X.
] ) =X . From
From the
action of
on [D,
[ D ,V]
V ]itit follows
follows that
that
m(E/CE([D,
°J' is
stable either
m ( E / C E ( [ DV])
V
, ] )<(m([D,
m ( [ DV]/C[D,v](E)),
,V]/CID,vl(E)),so
so as
as P
is stable
GJ'.InIneither
either case,
case, by
by minimality
of
CE([D,
or CE([D,
is in P.
CE([DV
,V])
] ) = 11 or
C E ( [ DV
,V])
] )is
rninimality of
E,
CE([D,
V])
<
CD([D,
V])
=
1.
Now
m(E)
<m([D,
VIIC[D,v]
(E)),
so
as
E , CE([DV
, I)5 ~ D ( [ D
v l,> = 1. Now m ( E )5 m ( [ D ,V]/C[D,v](E)),
(Cv (D))=
= D.
E E °J'(G,
P(G, V),
V ) ,E <5CX
C,y(Cv(D))
G S2
, 52 is
is the
the
This completes the proof of the lemma. Notice that, as D =
=H
H < G,
set of all subgroups
subgroups of G generated by a pair of noncommuting members of F.
r.
(32.4) Let F*(G)
and G* =
_
F*(G)=
=Op(G),
Op(G),PP EE Sylp(G),
Sylp(G),ZZ ==c21(Z(P)),
Q 1 ( Z ( P ) )V
V, ==(ZG)
(zG)
G/CG(V
V is elementary abelian and
and Op(G*)
Op(G*) =
= 1.
G
/ C G ( V).) .Then V
1.
= Op(G) <(PP<I
C(Z),
S21(Z(Op(G)))
= Vo.
Proof. F*(G)
F*(G)=Op(G)
C ( Z )so,
, so,byby31.13,
31.13,Z Z< 5
521(Z(Op(G)))=
VO.
So, as Vo
<G
G,, V
V=
= (ZG)
<(Vo
Vo L]
(zG)
Voand
and hence,
hence, as
asVo
Vo is elementary abelian, so is
V.
=Op(G*).
= PP flf lKKEESylp(K),
V . Let K* =
Op(G*).By 6.4, Q =
Sylp(K),so,
so, as
as K*
K* is
is aa p-group,
K=
= CG(V)Q
< CG(Z).
CG(Z).Hence,
Hence,asas
(ZG)
K
C G ( V ) Q5
KK
L] <
G G,
,VV
==
(zG)
5 <Z(K),soK<CG(V).
Z ( K ) ,so K ( CG(V).
is K*
K* =
= 1.
That is
1.
Lemma
32.4 supplies
supplies aa tool
toolfor
for analyzing
analyzinggroups
groupsGGwith
withF*(G)
F*(G) =
= Op(G).
Lemma 32.4
Op(G).
Namely, as V is elementary
elementary abelian, we can regard V as
as aa vector
vector space
space over
over
GF(p)
of G
G on
GF(p) and
and the
the representation
representation of
on V
V by
by conjugation
conjugation makes
makes V into
into aa
faithful GF(p)G*-module. These observations are used in
in conjunction
conjunction with
with
The generalized
generalized Fitting subgroup
166
166
lemmas 32.2 and 32.3 in the proof of the next two lemmas. Both are
are versions
versions
of Thompson Factorization.
(32.5) (Thompson
(ThompsonFactorization)
Factorization)Let
LetGGbebeaasolvable
solvablegroup
groupwith
withF(G)
F(G) =
=
(32.5)
Op(G),
P E Sylp(G),
Sylp(G), Z =
= 01(Z(P)),
(ZG),
CG(V ). Then
Op(G),let P
nl(Z(P)),VV==
( z Gand
) ,andG*
G*==GI
G/CG(V).
either
either
(1)
=NG(J(P))CG(Z),
NG(J(P))CG(Z),or
(1) G =
(2) pp<53,3,J(G)*
J(G)*isisthe
thedirect
directproduct
productofofcopies
copiesof
of SL2(p)
SL2(p)permuted
permuted by G,
and J(P)*
J(P)* EE Sylp(J(G)*).
Sylp(J(G)*).
Proof. IfIf J(P)*
J(P)* # 1 then,
then, by
by 32.2 and
and 32.4,
32.4, G*,
G*, V,
V, dsl(G)*
Proof.
( G ) * satisfies
satisfies the hy32.3, and
and hence
hence (2)
(2) holds
holdsby
by32.3.
32.3.So
Soassume
assumeJJ(P)
pothesis of 32.3,
(P) <
5D
D ==CG(V)CG(V).
By 32.1
32.1 and
andaaFrattini
FrattiniArgument,
Argument,GG==NG(J(P))D
NG(J(P))D 5
< NG(J(P))CG(Z).
NG(J(P))CG(Z).
(32.6) (Thompson
(ThompsonFactorization)
Factorization)Let
LetGGbe
beaa solvable
solvable group
groupwith
with F(G)
F(G) =
_
(32.6)
Op(G).
Let
p
be
odd
and
if
p
=
3
assume
G
has
abelian
Sylow
2-subgroups.
Op(G).
be
= 3 assume G has abelian Sylow 2-subgroups.
= NG(J(P))CG(S21(Z(P)))
NG(J(P))CG(Q1(Z(P))) for
Then G =
for PPEESylp(G).
Syl,(G).
has nonabelian
nonabelian
Proof. This
Thisfollows
follows from
from 32.5
32.5 and
and the
the observation that SL2(3) has
Sylow 2-groups.
33 Central
Centralextensions
extensions
group GG isis a pair
pair (H, nir)) where
A central extension
extension ofofaa group
where H is
is aa group
group and
and
homomorphismwith
withker(n)
ker(ir)5< Z(H).
Z(H). H
nir:: H + G is a surjective homomorphism
H isisalso
alsosaid
said
extension of G. Notice that the quasisimple
quasisimple groups are precisely
precisely
to be a central extension
the perfect central
central extensions
extensions of the simple
simple groups.
groups.
a: (GI,
(G1, nl)
r-+
Ti) (G2,
ir2)n2)
of of
central
A morphism a:
+ (G2,
centralextensions
extensionsof
of GG isis aa group
group
with nl
nt =
=a7r2.
homomorphism a:
a:G1
G1 + G2 with
an2. A central extension
extension (G, nlr)) of G
extension (H,
(H, aa)) of G
is universal ifif for each central extension
G there
there exists
exists aa unique
unique
morphism aa:: (G, nJr)) +-*
= (H,
(H, a)
a )of
ofcentral
centralextensions.
extensions.
(e,
(e,
(33.1)
(33.1) Up
Up to
to isomorphism
isomorphismthere
there is
is at
at most
most one
one universal
universal central
central extension of a
group G.
Proof. If
2, are
If (Gi,
(Gi,iri),
xi), ii ==1,1,2,
areuniversal
universalcentral
centralextensions
extensions of
of G
G then
then there
there
of central extensions ai: (G;,
exist morphisms of
(Gi, ,r,)
ni) + (G3_1,
(G34, n3-i).
As
aia3-i
7r3-0. As aia3-i
and 1 are morphisms
morphisms of (Gi
(Gi,,7ri)
(Gi, ni
), the uniqueness
uniqueness of such
morphism
ni) to (Gi,
ni),
such a morphism
says ala2
ala2 ==11==a2a1.
~ ~ 2 Thus
~Thus
x 1 . aiaiisisananisomorphism.
isomorphism.
'
Central
Central extensions
extensions
167
167
(33.2) If (G,
(6,ir)
n )isisaauniversal
universal central
central extension
extension of
(33.2)
of G
G then
then both
both 6
G and G are
perfect.
perfect.
Proof. Let H
H=
and
=G
6 xx (G/&))
(6/6('))
anddefine
define a:
a : H + G by ((x,
x , y)a
=xir.
x n . Then
Then
y)a =
( H ,a)
a )isisaacentral
centralextension
extensionof
of GGand
anda1:
ai:(G,
(6,ir)
n )+ (H,
( H ,a),
a ) ,ii ==1,1,2,
aremormor(H,
2, are
phisms, where
= ((x,
x , 11)) and
( x ,x&)).
~ 6 "So,
)So,
) by
.bythe
theuniqueness
uniqueness of
phisms,
wherexal
xa1=
andxa2
xa2 =
= (x,
a2 and
and hence
hence G
6 ==0(1).
6").
ThusG6isisperfect,
perfect,so,
so,by
by 8.8.2,
8.8.2,
such a morphism, aaI, ==a2
Thus
G ==Gir
6 nisisperfect.
perfect.
G
be perfect
perfect and
and (H,
( H ,.7r)
n ) a central extension
(33.3) Let
Let G be
extension of
of G
G.. Then H ==
ker(n)HC1)with H
( ' )perfect.
perfect.
ker(ir)HO)
HO
Proof.
Proof.By
By8.8,
8.8,H(1)7r
H ( ' ) n==(Hir)(1)
( H n ) ( '=)=G(1),
G('),so,
so,as
asGGisisperfect,
perfect,H(1)7r
H(')n=
=G.
G .Hence
Hence
H ==ker(.7r)H(1).
ker(n)HC').As
H , nir))isisacentralextension,
( H ) ,so
/ H ( ~=
H
As((H,
a central extension,ker(n)
ker(,r)5<ZZ(H),
so HH/H(2)
=)
Z(H/H(2))(H(')/H(2))
and
z ( H / H ( ~ ) ) ( H ( ' ) / H is
(is~abelian,
)abelian,
)
and hence
hence HO
H(')==H(2)
HC2)by
by 8.8.4.
8.8.4. Thus
Thus HO
H(')
is perfect.
perfect.
possessesaauniversal
universalcentral
centralextension
extensionifif and
and only
onlyifif GGisisperfect.
perfect.
(33.4) GG possesses
Proof.
Proof.By
By33.2,
33.2,ififGGpossesses
possessesaauniversal
universalcentral
centralextension
extensionthen
thenGGisisperfect.
perfect.
Conversely assume G
G is
is perfect.
perfect.Let
Letgg F+
H ggbe
a
be aa bijection
bijection of G with a set (?
and let F be the free group
group on
on(?.
G.Let
Letrr be the
the set
setof
ofwords
wordsX xy(xy)-1,
jilxy)-', xx,, y E
E G,
G,
and let M be the normal subgroup of F generated by F.
r.Next let A be the set ofof
words
subgroup of F generated
words [w,
[w,i],
Z ] ,wwEEP,r,Zz EE G,
G ,and
and let
let N be
be the normal subgroup
generated by
A. As
As M
M 4 F, N
N=
8.5.2.
Then
=[M,
[ M ,flF ]:4 FFand
andM/N
M I N<5Z(F/N)
Z ( F / Nby) by
8.5.2.
Thenby
by28.6
28.6
=xx for
and 28.7 there is a unique
unique homomorphism
homomorphismn:
ir:FF/N
I N + G,
G , with ((xN)7r
XN)n=
for
all xx EE G
G,, and
andindeed
indeedM
M/N
= ker(ir).
IN =
ker(n).Therefore
Therefore ((F/N,
F I Nir)
, n is
)isaacentral
centralextension
extension
G.
of G.
Let (H,
( H ,a)
a )be
bea acentral
centralextension
extension of
of G.
G .For
ForXxEE G,
G , let
let h(x)
h ( x )EE H
H with
with
ker(a)
<
h(x)or
= xx.. Then, for x,
= h ( ~ ) h ( ~ ) h ( x ~E )ker(a)
h(x)a=
x , y,
y, zz EE G,
G ,w =
-'
5 CH(h(z)),
C~(h(z)),
so [w,
= 1. Hence
[ w ,h(z)]
h(z)]=
Hence by
by 28.6
28.6there
thereexists
existsa aunique
uniquehomomorphism
homomorphism
(xN)a =
= h(x)
for each xEEGG.
Notice a:
a : FIN+
h ( x )foreachx
. Noticea:
( F(F/N,
I N , nir)
)+- (H,
( H ,a)
a)
a:
F/N H with (XN)a
morphism.
is aa morphism.
Now
let 6
G=
= (FIN)(').
(FIN)('). By
ker(7r)O
Now let
By 33.3,
33.3, FIN=
FIN =
k e r ( n ) 6and
and G6isisperfect.
perfect. Hence
Hence
(G,
( 6 ,ir)
n )isis also
also aa central
central extension
extension of G,
G , and
and a:
a :(G,
(6,ir)
n )+ (H,
( H ,a)
a )aamorphism.
morphism.
p: ((G,
6 ,n
( H ,a)
a )isisaasecond
secondmorphism,
morphism,and
and define
define y:
y: G + H by
by
Suppose,B:
Suppose
ir))+ (H,
uy=
=ua(uf)-1,
ua(up)-',for
foruuEE G.
6.Then
a=
n ==for,
p a , so GY
Z ( H ) .Thus ((uv)y
uv)y=
uy
ThenaaQ
=7r
Gy c Z(H).
=
= u a v y (up)-' =
( u v ) a ( ( u v ) ~ ) - '=
= uava(v,8)-1(u$)-1
uava(v,!?)-' (up)-' =
= ua(u,8)-lvy
ua(up)-'v y =
=
uyvy,
u y v y , so
soyyisisa ahomomorphism.
homomorphism.Moreover
Moreover Gy
6 isisabelian,
abelian, so,
so, as
as G
6isisperperfect, y is trivial by
by 8.8.4. Thus
Thusaa =
= P.
p.
a
h(x)h(y)h(xy)-1
(uv)a((uv),B)-1
uavy(u,8)-1
The generalized Fitting subgroup
168
been shown to be a universal
((G,
e , 7r)
n ) has been
universal central extension,
extension, so
so the
the proof
proof isis
complete.
(e,
If G is
is aa perfect
perfect group
group and
and (G, 7r)
n ) its
its universal
universal central extension, then G isis
called
called the universal
universal covering
coveringgroup
group of
of G
G and
and ker(7r)
ker(n) the Schur
Schur multiplier
multiplier of G.
G.
Notice that, by 33.2, G is
is perfect.
perfect.
A perfect
perfect central extension
or covering
covering of
of aa perfect
perfect group
group G
G is a central
extension or
extension (H,
(H, aa)) of G with
with H perfect.
perfect.
(33.5) Let
Let (H, aa)) be
(33.5)
be aa central
central extension
extension of a group G,
G , and
and (K,
(K, ,8)
p) a perfect
central extension of
of H. Then (K,
(K, Pa)
p a )isisaaperfect
perfectcentral
centralextension
extensionof
of G.
G.
the composition
composition of surjective homomorphisms
homomorphisms and
Proof. ,8a:
pa: K + G is the
and hence
hence
a surjective
ker(,da) and
and yyEEK.
K. x,d
xpEE ker(a) <
5
surjective homomorphism.
homomorphism. Let
Let Xx E
E ker(pa)
Z(H), so
= [x,d,
yd] =
= 1. Thus
Thus [[x,
y] EEker(p)
ker(b)5< Z(K).
Z(K). Thus [ker(pa),
[ker(da),
Z(H),
so [x,
[x, y],8
y]p =
[xp, yp]
x , y]
by 8.9,
8.9,ker(pa)
ker(8a) 5< Z(K).
K, K]
K l = 1, so, by
(33.6) Let
perfect,
(33.6)
Let (H,
(H,a)
a )and
and(K,
(K,,8)
p)be
becentral
centralextensions
extensions of
of a group G with K perfect,
and y: (H, aa)) -+(K,
Then (H,
(H, y) is a
(K,,8)
p)a amorphism
morphismof
ofcentral
central extensions.
extensions. Then
central extension of K.
Proof. y: H 4 K isis aahomomorphism
homomorphism with aa ==y,8.
yp. The
The latter
latter fact
fact implies
implies
ker(y) <<_Z(H),
ker(y)
Z(H),sosoitremains
it remainstotoshow
showyyisisaasurjection.
surjection.As
As aa ==y,8
ypisis aa surjection,
surjection,
K=
= (Hy)ker(,8),
asker(p)
ker(b)5<Z(K),
Z(K), H
Hyy 49K
( H y)ker(p), so, as
K and
and K/Hy
K / H yisisabelian.
abelian.Hence
Hence
= Hy
H y as
as K
K isis perfect.
perfect.
K=
e
(33.7)
of aa perfect
perfectgroup
groupGG and
andlet
let (H,
(H, aa)) be a
(33.7) Let G be the covering group of
a isis an
perfect central extension of G. Then a
an isomorphism.
isomorphism.
e.
e
Proof.
covering.By
By 33.5,
33.5,(H,
(H,aan)
Proof. Let
Let ir:
n : G + G be the
the universal
universal covering.
n ) is aa perperso by
by the
the universal
universal property there
there is
is aa morphism
morphism
fect central extension of G, so
8: (G,
r =7r
p:
(G, 7r)
n ) + (H, an).
a n ) .Then
Then,8a
,!?an
= nsosoby
bythe
theuniqueness
uniquenessproperty
property of
of univerH
is
an
injection,
while
,8
is
a
surjection
=1.
1.Hence
Hence,8:
p: G + is an injection, while ,3 a surjection
sal extensions,
extensions, pa
,8a =
Thus pfi isis an
an isomorphism
isomorphismand
andasaspa
,8a==1,1,aa=
=,8-1
by 33.6. Thus
p-' is too.
e
(33.8) Let
universal central
central extension
extension of
ofGG,, and
and(H,
(H, aa))
G be
be perfect, (G,
( e ,7r)
n ) the universal
LetG
a perfect central extension of
of G. Then
(1) There
H with
withnr =
= aa.
aa.
(? + H
There exists
exists a covering aa:: G
(2) (G,
(6,a)a )isisthe
theuniversal
universal central
central extension
extension of H.
H.
(3) The
The Schur
Schur multiplier
multiplier of
of H
H isisaasubgroup
subgroup of
of the
the Schur
Schur multiplier of G.
G.
Central
Central extensions
extensions
169
Schur multiplier of
ofG,
G,and
andZ(H)
Z(H)=
= ker(a)
ker(a) S
(4) IfIf Z(G)
Z(G)=1
= 1then
thenZ(G)
z((?) isisthe
the~churmulti~lier
ker(7r)/ker(a) is
of G by the Schur multimultiker(n)/ker(a)
is the
the quotient
quotient of the Schur multiplier of
of H.
plier of
(e,
Proof.
Proof. By
Bythe
theuniversal
universalproperty
property there
there exists
exists aa morphism aa:: (G, 7r)
n ) + (H, aa).
).
= aa
as and,
Then n7r=
and,by
by33.6,
33.6,aaisisaacovering.
covering.Let
Let(H,
(H,,B)
B)be
bethe
theuniversal
universal covercovera).
ing of H.
H. By
By the
theuniversal
universal property
property there
there isis aamorphism
morphism y:
y:(H,
(fi,,B)
,!?)+ (G,
((?, a).
By 33.6
By
33.6 and 33.7, y isis an
anisomorphism
isomorphism so
so (2)
(2) holds.
holds. Now (3) and
and (4)
(4) are
are
straightforward.
straightforward.
(33.9) Let G
G/Z(G)finite.
finite.Then
Then GM
G(')isis finite.
finite.
(33.9)
G be
be aa group with G/Z(G)
Proof.
For Z E
E Z(G)
and g, h EE G,[g,hz]
G, [g, hz] =
= [g,
h] =
Proof. Let
Letnn =
=IG/Z(G)I.
IG/Z(G)I.Forz
Z(G)andg,h
[g,h]
=
[gz,
[gz, h],
h], so
so the
the set
set 0Aofofcommutators
commutatorsisisofoforder
orderatatmost
mostn2.
n2.II claim
claim each
each g EE G(1)
G(')
beexpressed
expressedasasaaword
wordg g==XIxl....
0 of
in 5
< n3.
can be
. .x,x ininthe
themembers
members of A
of length m
The claim together
together with the
the finiteness
finiteness of 0Ashow
showG(1)
G(') isis finite.
finite.
expression for
for gg of minimal
minimal length
It remains to establish the claim. Pick an expression
f i{i:
: x1
= dd}] of
of
m. If m
m.
m>
> n3 then, as 1IA01 <
5 n2,
n2, there is
is some
some dd EE 0Awith
withI'r==
x, =
orderk >>n.n .Asxix1+l
withxx`+'
0 wecanassumer
we can assume F=
= {[1,
orderk
A s ~ , x ,=xa+1xx
+= ~x ~ + ~`+'xwithx?+'
~+'
EEA
I , ...
.. ., ,k}.
k].
can be
be written
written as a product of n commudn+' can
Hence it remains to show that do+1
minimality of
of m
m will
willbe
becontradicted.
contradicted.Let
Letdd =
= [x, y]. As
tators, since then the minimality
(dx)n-'[x2,
IG/Z(G)I ==n,n,do
dnEE Z(G),
Z(G), so dn+'
(dn)xd =
[x2, y]
d"+1 =
= (dn)Xd
= (dn-')xdxd
(dn-')Xdxd =
= (dx)n-'
by 8.5.4. In particular
do+1
is
a
product
of
n
commutators.
particular dn+'
commutators.
n3.
,
(33.10)
(33.10) Let
Let G
G be
be aaperfect
perfectfinite
finite group.
group. Then
Then the
the universal
universal covering group of
G and the Schur
multiplier
of
G
are
finite.
Schur
G are finite.
Proof.
Proof. This
Thisisisaadirect
directconsequence
consequenceof
of 33.9.
33.9.
(33.11)
(33.11) Let
Let (H,
(H,a)a )bebeaaperfect
perfectcentral
centralextension
extensionof
of aa finite group G, p a prime,
and
P
E
Sylp(H).
Then
P
n
ker(a)
<
((P).
and P E Syl,(H). Then P n ker(a) 5 @(P).
Proof. Passing
to H/(@(P)
H/((D(P)nnker(a))
we may
mayassume
assume@(P)
1(P)nnker(a)
= 11 and
Passing to
ker(a)) we
ker(a) =
it remains
remains to
to show
showXx =
= PP n ker(a)
=
1.
But
as
P/(D
(P)
=
P*
is
elementary
ker(a) = 1. But as P/@(P) = P* is elementary
abelian
there isis aa complement
complementY*
Y*totoX*
X*ininP*.
P*.Then
ThenPP=
=X
XxY
abelian there
Y so P splits
splits
over X. Hence by Gaschiitz' Theorem, 10.4, H splits over X.
X . Hence, as H is
is
perfect and
and XX5<Z(H),
Z(H), X
X=
= 1.
perfect
1.
(33.12)
(33.12) Let
Let G
Gbe
beaa perfect
perfect finite
finite group and M the Schur multiplier of G. Then
.7r(M)
c
7r(G).
n(M) S: n(G).
The generalized Fitting subgroup
170
Proof.
Proof. This
Thisisisaaconsequence
consequenceof
of 33.11.
33.11.
ker(a))
(33.13) Let
Let(H,
(H,or)
a )be
be aa perfect
perfectcentral
centralextension
extensionof
of aa finite
finite group
group G
G with
with ker(a
p-group, let
perfect subgroup of G containing a Sylow
Sylow p-group
p-group of G,
a p-group,
let Go
Go be a perfect
ao) is a perfect
let Ho
or,aH0:
Ho ==Q-1(Go),
a - ' ( ~ o )and
,andaoa=0 =
l ~ ,Ho
Ho
: + Go.
Go. Then (Ho,
(Ho,00)
perfect central
central
of Go with ker(ao) =
= ker(a).
ker(o ). Hence
Hence aa Sylow
Sylowp-group
p-group of
of the
the Schur
extension of
multiplier of
of G is aa homomorphic
multiplier
homomorphic image of a Sylow
Sylow p-group of
of the
the Schur
Schur
Go.
multiplier of Go.
Proof.
extension of
of Go
Go with
withker(a)
ker(a) =
= ker(ao),
Proof. Evidently
Evidently (Ho,
(Ho,ao)
00) is a central extension
ker(ao),
so, by 33.3, Ho
=
ker(a)Ho"
with
Hot)
perfect.
As
Go
contains
a
SylowHo = k e r ( a ) ~ i ' )with H:') perfect. As Go contains a Sylow pof G,
G, Ho contains
containsaaSylow
Sylowp-group
p-groupPP of
ofHH and,
and, as
asker(a)
ker(a) is
is aa p-group,
p-group,
group of
Thus
-ker(or)i<PP.. Then,
Then, by
by 33.11,
= @(P)H:').
ker(a)
33.11, ker(a) <
((D(P),
@(P), so Ho
Ho =
Thus
P=
0
P
= t(P)(P
fl n
Ho1))
so P
P=
= P flnHot)
<
P fli lHo
Ho =
@(P)(P
H,'") by
bythe
themodular
modular property,
property, 1.14, so
H:') 5
PHo1)
=
Hot>,
so
Ho
is
perfect,
completing
the
proof.
H'
) by
H~'')
by 23.1.
23.1. Thus
ThusHo
Ho = P H:') = H:'), so is perfect, completing the proof.
(33.14) IfIf G
perfect finite
p-groups then
Gisisaaperfect
finitegroup
group with cyclic
cyclic Sylow p-groups
thenthe
the Schur
Schur
multiplier of
of G
G is
is aa pl-group.
p'-group.
multiplier
Proof. Let
Let (H,
(H,a)
a )be
beaaperfect
perfectcentral
centralextension
extension of G with ker(a) aa p-group;
p-group;
show ker(a)
ker(or)==1.1.Let
LetP PE ESyl,(H),
Sylp(H),
that
=ker(a)(<PPand
andPP/Z
I must
must show
soso
that
ZZ
=ker(a)
/Z
is cyclic. By
By 33.1
33.11,
< @(P),
4)(P), so PI
4)(P) is
hence, by
by 23.1,
23.1, P
1, ZZ _(
P/@(P)
is cyclic
cyclic and hence,
At this point I appeal
H induces
is cyclic.
cyclic. At
appeal to 39.1,
39.1, which
which says some h E
EH
induces a
nontrivial pl-automorphism
p'-automorphism on
on P.
P. But then, by 23.3, [S21(P),
h]#0 1,
1, so,
so, as
[Q1(P), h]
nontrivial
Z
< Z(H), zZni S21(P)
cyclic and Z (< PP this
Z(Z(H),
l Q1(P=
) =1.l . As
As PP isiscyclicand
this says
says Z = 1.
1.
generalized Fitting
The section on the generalized
Fitting group
group focused
focused attention on
on quasisimple
quasisimple
groups. Observe
Observe that
that the finite quasisimple groups are precisely the perfect
groups.
central extensions of
of the finite simple groups.
groups. Hence,
Hence, for each finite simple
central
group G, the universal covering group of G is
is the
the largest
largest quasisimple
quasisimple group
group
with G as
as its
its simple
simple factor, and the center
center of any
any such
such quasisimple
quasisimple group is
a homomorphic
image of
of the
the Schur multiplier
multiplier of
of G.
G. Thus it is of particular
homomorphic image
interest to determine the covering groups and Schur multipliers
multipliers of the
the finite
finite
with a description of the covering groups
simple groups. This section closes with
multipliers of the alternating
and Schur multipliers
alternatinggroups.
groups.
e
e.
e,
(33.15)
Let G=
= A,,
A, n >
coveringgroup
groupofof G,and
and22 =
=
(33.15) Let
>5, G the universal
universal covering
Z(O)
= (1, ..... .,,n).
n). Then
Then
~ ( ethe
the
) Schur multiplier of G. Represent G on X =
(1) 22=G716
z6if n ==66oror7,7,while
while 22=712
E Z2otherwise.
otherwise.
Central extensions
171
171
(2) Let
Letiibe
beaa2-element
2-elementin
in G6such
suchthat
that the
the image
image it of
of i in
in G
6 isis an
an involution.
has 2k
2k cycles
cyclesof
of length
length 2,
2, if isis an
an involution
involution ifif kk is
is even,
even, and i1 is
is of order
order
Then tf has
4 if k is odd.
odd.
(3) IfIf nn ==6 6oror7 7then
then31+2
31+2isis aa Sylow
Sylow 3-group
3-group of G.
6.
proved using
using homological
homological algebra,
algebra,but
butI'll
I'll take a group
This result is usually proved
theoretical approach
theoretical
approach here.
here.
multiplier
There are two parts to the proof. First show the order of the Schur multiplier
= 66 or
of An
A, is at least 2 (or 6 ifif nn =
or 7).
7). Second
Second show
show the multiplier
multiplier has order
order at
33.15.2 and 33.15.3. Exercise
Exercise 11.5
handles part one
most 2 (or 6) and establish 33.15.2
11.5 handles
(unless n ==66or
or77where
wherethe
theproof
proof that
that 33 divides
divides the order
order of the multiplier is
omitted). The second
omitted).
second part
part is
is more
more difficult
difficult and
and appears
appears below.
Assume for
for the
the remainder
remainder of
of the
the section
section that
that (G,
(G, nn)) is aa perfect
Assume
perfect central
central
extension of
of G ==AAn
andwrite
write Sfor
forthe
theimage
imageSSn
G.Let
LetZZ=
= Z(G)
extension
, and
n ofofSSECG.
Z is a nontrivial
nontrivialp-group
p-group for
for some
someprime
primep.p.Let
LetPP E Syl,(G).
Sylp(G). It
and assume Z
suffice to
to show
show pp <53,3,IZI
will suffice
IZI == p,
p , and
and33.15.2
33.15.2 and
and33.15.3
33.15.3 hold.
hold. Assume
Assume
otherwise and choose G to be
be aa counter
counter example
example to
to one
one of
of these
these statements
statements
with n minimal. The idea of the proof is simple: exhibit a perfect subgroup H
multiplier of
of H
of G containing P,
p, use
use the
the induction
induction assumption
assumption to show the multiplier
= 2, and
is a p'-group, and
and hence
hence obtain
obtain a contradiction
contradiction from 33.13. When p =
and
sometimes when p ==3,
sometimes
3,the
the situation
situationis
is more
more complicated
complicatedbut the same general
idea works.
works.
I begin a series
series of reductions.
e
(33.16) n > 2p.
Proof. By
By 33.14,
33.14,Pp isisnot
notcyclic.
cyclic.
(33.17) n > 5.
Proof. Assume nn =
= 5.
33.16,pp =
= 2.
I Z=
I =2.P
2.P(
<H
H<
5. By 33.16,
2. Assume
Assume IZI
(G
6 with
with
H =EA4,
so
P
E4
and
H
is
transitive
on
P#.
By
33.11,
(D(P)
=
Z,
so
A4, so p E E4 and H is transitive on P # . By 33.1 1, @(P) = Z, so by
by
1.13 there is an element of
of order
order 44 in
in PP.. Hence, as H
H isis transitive
transitive on
on P#,
p#,every
every
in P --ZZisisofoforder
element in
order4.
4.So,
So,by
by Exercise
Exercise 8.4, P
P is
is quaternion
quaternion of order 8.
Notice this
Z I(<2.2.Now
NowpP =
=
this gives
gives 33.15.2
33.15.2 in this case,
case, so
so it remains
remains to show IIZI
(g,
h)
so
PM
=
(z)
where
z
=
[g,
h].
If
U
is
of
index
2
in
Z
we've
seen
that
(8,A ) P(') = (z) where z = [g, h ] .If U is of index 2 in Z we've seen that
P/
Q8,
is is
cyclic
P / U -E
Q gso
, soz zV4U.U.Thus
ThusZ/d)(Z)
Z/@(Z)
cyclicsosoZZisiscyclic.
cyclic.So,
So,as
asZ/
Z /U
U isisthe
the
of P/
P/ U
Hence, by
by
unique subgroup of
U of
of order
order 2, P also
also has
has aa unique involution. Hence,
Exercise 8.4, P is quaternion. So,
So,as
asI P
IP:
Z(P) I=
= 4,
4, PP -EQ8,
ZI =
: Z(P)I
Q g , That
ThatisisIIZI
=2.
2.
172
The generalized Fitting subgroup
For S c
C G or G write
write Fix(S) and
and M(S) for
for the
the set
set of
of points
points in X
X fixed
fixed by SS
and moved by S,
S, respectively.
respectively.
(33.18) Let
Let Z3
713Z= Y 5< G
G with
with IM(Y)I
ThenYY==(y)
(y)xx Z
Z for some
I M ( ~=
)=3.
I3.Then
some y
(33.18)
3 3then
(CG (Y)) =-Y
8 5 n # 9oror1010and
andp p= =
then03'03'(cc(E))
=.Y x L with
with
of order 3, and if 8<n=,49
= An-3
LE
An-3.
loss pp =
= 3,
33.17, HH =
=
Proof. Y
E <5 H
H -ZA5
A5with
with IM(H)I
I M ( H )==
I 5.
5. Without loss
3, so, by 33.17,
= (y) x Z where
H(l). Also, if
if 8 5
< n # 9 or 10,
Z x HO).
~ ( ' 1 .Thus Y
Y=
where yy E
E H(').
10, then
then
O3'(CG(Y))=_(7)
(y)xxK,
k, kK-ZAn_3,
of n,
n, K
K=
= Z x K(l).
03'(cG(y))
An-3,and,
and, by
by minimality
rninimality of
K(').
(33.19) IfIf pp =
= 3 then
> 6.
(33.19)
then nn >
6.
E9 Z
- PP==
(y,(9,
g)gwith
Proof. Assume
Assume pp =
=33 and
and n =
=6.
6. Then E9
) withy yand
andggmoving
moving
points. By
By 33.18
33.18 we
we may
maytake
takeyy and
andgg of
of order
order 3.3.Let
Let RR=
= (y,
(y, g).
g). Then
Then R
R
3 points.
is of class at most 2 so, by 23.11,
R is of exponent
3. By
By 33.1
33.11,
= PP and
23.1 1, R
exponent 3.
1, RR=
< @(P).
t(P). PM
3,ZZ=
= P(')
P(l) =
= Z(P).
P(')==([g,
([g,y])
y])so,
so,as
as PP isis of
of exponent
exponent 3,
Z(P). Thus
Thus
Z5
P isis extraspecial
by 23.13,
23.13, P -Z31+2
extraspecial of order 27 and exponent 3, so, by
31+2..Hence
all parts of 33.15 hold in this case.
case.
I now consider two cases: Case I: n is not a power
power of
of p; Case II:
11: n is a power
of p.
onon
XX
so so
there
is aispartition
171,P==[X(XI,
1, X2)
p. In
InCase
CaseIIPPisisnot
nottransitive
transitive
there
a partition
X2)
of X with
1 and
1. Then
with PPacting
actingon
onXXI
and X2.
X2.Let
Let H
H be
bethe
thesubgroup
subgroupacting
actingon
onXXI.
Then
H ==(a)Ho
-ZAn;
(ii)Aowith
with Ho
Ho==H1
H I Xx H2,
Hz,where
where Hi
Hi ==Ox,_,
Gx,-<
A ,isisthe
thesubgroup
subgroup
of G fixing
point of
of X3+,
X3_;,nini =
= IXi1,
and (ii)
(a) Hi
Hi Z
- Sn;
fixing each point
/Xi1, and
S,, (unless
(unless nI
n 1or
orn2
n2
is 1).
II, P is
is a partition
partition P
°J' of
of X
X into p
1). In case 11,
is transitive
transitive on X and there
there is
subset Xi,
Xi, 1 5
< i <5 p,
p,of
oforder
order n/p,
n l p ,with
withPPtransitive
transitive on
on °J'.
P. The
Thesubgroup
subgroup of
preserving P
?Pcontains
containsa asubgroup
subgroupHHwith
withP P5<HH and
and H
HZ
- (An/P)wr
G preserving
(AnIp)wrAp
A,
wrZ2
Z2ififpp =
= 2. Notice that, by 33.16,
if p # 2, while H is
is of
of index
index 22 in
in Sn12
Sn12wr
n/p
nip >
2 p.
P.
Observe
Observe next that:
that:
(33.20) One of the following
(33.20)
following holds:
(1) K
K ==Op'(H)
OP'(H)isisperfect.
perfect.
(2) pp ==22and,
and,in
inCase
Case I, n1
nl >
> 11<<n2n2.
(3) pp==3 3and
andn1
nlororn2
n2isis33or
or 44 in
in Case
Case I.
Moreover, by
by minimality
minimalityofofn,n, Exercises
Exercises11.2
11.2and
and11.3,
11.3,and
and33.13,
33.13,ififKk is
Moreover,
perfect then
then p 5
< 3, and if p ==33 and
n2 is
is 6 or 7. In
and Case
Case II holds,
holds, then
then nnlI or n2
perfect
particular:
particular:
Central
Central extensions
extensions
173
(33.21)
and ifif pp ==33and
andCase
CaseIIholds
holdsthen
thenn1
nl or
or n2
n2 is 3,
3,4,6,
(33.21) p <(33and
4, 6, or 7.
(33.22) p = 2.
ProoJ
not pp ==3.3.Suppose
SupposeCase
CaseI Iholds.
holds.Then
Thenn1
nl or
orn2
n2 is
is 3,
3,4,6,
Proof. IfIf not
4, 6, or 7; say
nl. Then P acts
acts on
on aa subset
subset Xi
X/,of
ofX1
X1of
of order
order 3,
3, so
soreplacing
replacing X1
X1 by Xi
X/,we
we
n1.
may take nl
=
3.
But
now
33.18
implies
K
=
(y)
x
Z
x
L
if
8
_(
n
#
9
or
10,
n1 = 3. But now 33.18 implies K = (y) x Z
< 0 9 10,
7, 9, or
9,
contradicting 33.1
1. Hence
contradicting
33.11.
Hencenn==7,
or 10. As
As nn is
is not
not aa power
powerof
of 3,
3,nn #
0 9,
while if
10 we could have
n =
where we
we
while
if n =
= 10
have chosen
chosen nl
n1==1,1,n2n2==9.9.SoSon
= 7,7,where
choosenl
1, so that K 2 A6 and,
Z l== 33 and
and
choose
n1 =
= 1,
and, by
by 33.13
33.13 and
andminimality
minimalityof
ofn,n, /IZ
2 3l+2.
P
P=31+2
I1 holds.
holds. Here
(ii)KO where
where Ko
KO isis the
the direct
direct product
product of 33
So Case II
Here K
k ==(a)Ko
n. Then
Jgfor
for some
some
copies of
copies
of An/3r
An13,and
andii aisisofoforder
order33with
withI M(ii)l=
IM(a)I =n.
Theniia =
= bg
bE
and Exercise
Exercise11.2,
11.2,K0
KO==ZZxxK0(1)
K;') ifif n >
> 9.
9.
E KO,and,
and,by
by minimality
minimality of
of n and
If
then, by
by33.18,
33.18, K0
KO==(Z,
(Z,gi:
g,:11 <
(i <
(3)
with IM(gi)l
Further
If nn ==99 then,
3) with
I M(gi) =
= 3. Further
gi EE z(03'(cG(cG(gi))))
Z(O3'(CG(CG(gi))))SOso KO
K0 is
abelian. Finally
g,
is abelian.
Finally there
there is
is a 2-group
2-group
T
T],
7 ==[T,
[T,a]
ii]<_(G with
withKo
KO ==[K0,
[KO,
71,so,
so,asasK0
KOisisabelian,
abelian,24.6
24.6 implies
implies K0T
KoT =
=
x O~(KOT)
and
ZX
O3(K0T) and
andaa of
of course
courseacts
actson
on03(KoT).
O3(K0T).Let
LetLL== K;')
Kol)ififnn > 99 and
O ~ ( K ~ififTnn) ==9.9.We
Wecan
canchoose
choosebbEEL,
L,so
sothat
thataa isisof
oforder
order3.3.But
Butnow
now
L ==O3(K0T)
P ==ZZxx((P
L)(a))
contradicting
((Pfl n
L)(a))
contradicting33.11.
33.1 1.
leaves pp =
case pp =
=2.
2.The
Theproof
proof here
here is similar to the case
=3.
3.IfIf nn is
is odd
odd then
This leaves
holds with
withnl
n j=
= 1,
H =2An-1
Case I holds
1, so H
A,-1 isisperfect
perfectand
and 33.13
33.13 and minimality
minimality of n
proof. Thus
Thusnnisiseven
evenand
andHH=
= Ho U with
withH0
Ho =
= HO
1X
H2,
complete the proof.
H0 ==HH1
x Hz,
where, in
in Case
CaseI,I, Hi
Hi 2
= Ani,
and U =
= (u)
where,
A , , ni is even, and
(ii)=2Z2,
Z2,while,
while,ininCase
CaseII,
11,
of 2. Hi 2 An/2
and u
U=
= (u,
n is a power of
An12and
(ii,v)
rir) =2E4.
E4.We
Wecan
cantake
takeUg
Ug<(Ho
Ho
HI(1)
some gg EEG.
G.Let
Letnini=
= n/2 in
By minimality
minimalityofofn,n,Hi
Hi=
= Z ** H,(')
for some
in Case II.
11. By
HH(1)
with Zi
Zi =
= H:'
fl Z
Z of
most2,2,unless
unlessni ni(<4.4.IfIfnini =
= 44 then
with
) n
of order
order atat most
then
Hi I_(<KA5with
Hi
K 2 A5 withIM(K)l=5,andK=Z*KU),soHi=Z*O2(Hi)with
IM(K)I = 5, and K = Z * ~ ( ' 1so
, Hi = Z * O'(H,) with
ni >
>4
Zi ==02(Hi)
0 2 ( ~flin)ZZofoforder
orderat
at most
most 2. Let Li =
=02
0 2(Hi)
( ~ iin) either case. IfIf ni
Hi(1)
set L~
Li =
= H:').
set
If
then there isyiyi€ELI
with IM(yi)l=4and
I M(yi) =4 and yf
yi ==Y2
for some
I f n lj >>44(<n zn2 thenthereis
L i with
y2for
some
aE
Zi =
= (y?)
so, as
asZZ(
< Z(G),
Z(G), Z2 =
= Zf
Zl ==Z1.
Z0 =
= ZZ fl
L0 =
=
E G. Notice Zi
(y:) SO,
Z1.Thus Zo
nLo
L0 =
1 L2.
L2.This
Thisof
ofcourse
coursealso
alsoholds
holdsififnini <
< 4 for some
Z1 where Lo
=LL1
some i,i, since
sincethen
then
ni ==22and
andHi
Hi==1.1.L0Lochar
charHo
Hoso
soUUacts
actson
onL0,
Lo,and
andwe
we may
may take Ug <
(L0,
Lo, so
< Z1.
= Z1, so, by 33.11,
33.11, Z =
n U9
Ug (
Z1. Thus H = Z ** ULo
U b with Z fl
n UL0
ULo=Z1,
=Z1
Z1
Z fl
is of order 2. Thus 33.15.1 is established
established and it remains to establish
establish 33.15.2.
A, consists of
2-element in G with It an involution in
Let t be aa 2-element
in G. As G =
= An
of
even permutations on
on X, t has 2k cycles
cycles of length
length 22 on
on X
X for
forsome
somepositive
positive
integer k. Then
of X with
=
Yo U
U Yl
Yl U .. .. .. U Yk
Yk of
with Y0
Yo =
Then there
there isis aapartition
partition XX==Yo
Fix(t),
YiI I== 44 for
for 1 (
< ii (
< k, and t acting
We've seen that for i >
Fix(t), IIYi
acting on
on Yi.
Yi . We've
>00 there
there
174
The generalized
generalized Fitting subgroup
is Ki fixing
fixing X
X- Y;
pointwise with
withKi
Ki S
- SL2(3)
Gy.. Thus
Yipointwise
SL2(3)and Ki centralizing Gyi
is aa central
central product
product with identified centers and
(Ki: 1 5< ii 5< k)
k) =
= Kl
K1** . - a* Kk is
oforder
order44with
with(ti2)
(t?)== 2,
Z, SO
sott is
t=
= tl
tl . . .. tk with ti E Ki. As Ki
Ki S SL2(3), titi isisof
of order 4 if and only if k is odd.
odd. Hence
Hence the
the proof of
of 33.15
33.15 is
is at
at last
last complete.
complete.
Remarks. I'm
I'mnot
notsure
surewho
whoshares
sharescredit
creditfor
for the
the notion
notion of `components'.
'components'. CerBender [Be
should
tainly Bender
[Be 1],
11, Gorenstein
Gorensteinand
and Walter
Walter [GW],
[GW], and
and Wielandt
Wielandt [Wi
[Wi 1]
11 should
be included. I use Bender's
Bender's notation of E(G)
for
the
subgroup
generated
E(G) for the subgroup generated by
by
L(G) is
the components of a group G. The
The Gorenstein and Walter notation of L(G)
is
also used in the literature
literature to denote
denote this
this subgroup.
subgroup. II believe
believe Bender
Bender[Be
[Be1]
11was
was
generalized Fitting subgroup.
the first to formally define the generalized
subgroup.
Thompson factorization
was
introduced
by
Thompson
[Th 2],
21,
factorization was introduced by Thompsonin
in[Th
[Th111] and [Th
although
32.5 was
was presumably
first proved
proved by
by Glauberman
Glauberman [Gl
[G1 21.
2]. There
although 32.5
presumably first
are analogues
factorization for
for arbitrary
arbitrary finite
finite groups
groups G
G with
analogues of Thompson factorization
F*(G)
GF(p)-represenF*(G)==Op(G).
Op(G).Such
Suchresults
resultsrequire
requiredeep
deepknowledge
knowledge of the GF(p)-represensimple groups.
groups. In particular
particular one
pairs (G, V)
tations of nearly simple
one needs to know the pairs
V)
finite group
groupand
andV
Van
anirreducible
irreducibleGF(p)G-module
GF(p)G-module such
such
with G a nearly simple finite
factorization plays an imthat GJ'(G,
P ( G , V) is nonempty. Generalized Thompson factorization
portant role in the classification;
classification; see in particular the concluding
concluding remarks in
section 48. Thompson
Thompson factorization
factorization for
for solvable
solvable groups
groups isis used
used in
in this
this book
to establish the Thompson Normal p-Complement Theorem, 39.5, and the
the
nilpotence of Frobenius kernels (cf.
nilpotence
(cf. 35.24
35.24 and
and 40.8).
40.8).
Exercises for chapter
chapter 11
11
prime, G
r'-group, and
r-group acting
1. Let
Letrr be
be aaprime,
G a solvable rl-group,
andA
A an r-group
actingon
on G.
G. Prove
Prove
(1) IfIf rnCE7rn(G)
(G) and
-subgroup of G, then
then Op(H)
Op(H) <
andHHisisaHall
a Hall7rn-subgroup
5 Oph(G)
Oph(G)
eachpp EEnr,, where pp"n ==r'n U
p).
for each
1U{{p}.
(2) If
= 1.
If [A,
[A, F(G)] ==11then
then [A,
[A, G]
GI =
1.
(3) IfIf A
p-group of
of G then
then [A, GI
G] I
< Op,(G).
A centralizes
centralizes a Sylow p-group
Opt(G).
2. Let
Let(Gi
(Gi::i EE I)
I )be
be perfect
perfectgroups
groupsand
and Gi
Gi the
the universal
universal covering
covering group
group of Gi.
Gi .
Prove the universal covering group
group of
of the
the direct
direct product
product D of the groups
Gi, i E II,, is the direct product of the
the covering
coveringgroups
groupsG6j,~i ,E
Gi,
E II,, and
and hence
hence
of the
the Schur multipliers
multipliers of
of
the Schur multiplier of G is the direct product of
the groups
groups Gi
Gi,, ii EE I.
I.
coveringgroups
groupsHh and G,
3. Let
Let H
H and
andGGbe
beperfect
perfect groups
groups with universal covering
H.. Prove the universal
let H act
act transitively
transitively on
on aa set
set I,I ,and
andlet
letW
W ==GGwr1
wrl H
I1
covering group of W is the
the semidirect
semidirectproduct
product of
of aa central
centralproduct
productof
of I1 1
copies of G with H,
I?,and
and the
theSchur
Schurmultiplier
multiplier of
of W
W is
is the
the direct
direct product
product
of the multipliers of G and H.
H.
4. Let G
of p! aaG-invariant
G-invariant collection
collection of
4.
G be
be aafinite
finite group,
group, pp aaprime,
prime, Q2
subgroups
of G,
G, P
P a p-subgroup of G, and A E
C P nnQ
subgroups of
2
! with
with A
AC
E Np(O).
Np(A).
e,
175
Central extensions
extensions
Central
that either
eitherAA==PP n
fl Q
Q or
or there
thereexists
existsXXEE( P(PnflQ)
Q)-- A with
Prove that
with X <
(
Np(O).
NP(A).
over CC with
withbasis
basisXX =
= (x, y) and
5. Let
Let U
U be
beaa2-dimensional
2-dimensional vector space over
and
define aa,, $,
B,and
andyytoto1be
be the elements of GL(V) such that
(
Mx(a) = t 0
f ,
Mx($) =
(0 ) , Mx(Y) _ (0 -Ol)
,
= -1.
Let V = Ul ® .. ®U be the tensor product
where
i is in C `with i2 =
whereiisinCwithi2
[email protected]@U,bethetensorproduct
of n copies
copies UJ
skin
inEndc(V)
Endc(V) be defined
Uiof U and let sk
defined (cf. 25.3.5) as follows:
s2J
=(1®...®1®((1®$)+(a(9 Y))®y(9 ...(, Y)l',
1<j<n,
S2J-1
=(1®...®1®(a+$)®y(9 ...®Y)/I,
+
1 <j Vin,
+
where (1
(1 @
(9B)
$) + (a
(a @
(9 y)
y) is
is in
in the
the (j,
(j, jj ++1)-position
of szj
s2j and
andaa + B
$ is
1)-position of
in the j-position of s211.
s2j-1. Prove
Prove
(1) sk
-I)for
S: =
=IIand
and(sksk+l)3
(s~s~=
+= ~-I
~for11<(kk <<2n.
2n.
(2) (sksj)2=
= -I,
-I, for
- jj l1 >> 1.
for k,
k , j with
with lk
(k1.
(3) Let
by (sk:
(sk:115< kk <
LetGGbe
bethe
the subgroup
subgroupof GL(V) generated by
<2n).
2n).Prove
Prove
G/(-I)
=
Stn
with
-I
E
GM,
and
GM
is
quasisimple
if
n
>
2.
G/ (- I) E S2, with -I E G('), and G(') is quasisimple if n > 2.
theuniversal
universalcovering
covering group
group
6. Let
Let GGbe
beaanonabelian
nonabeliansimple
simplegroup
groupand
andG the
of G. Prove
Prove Aut(G)
Aut(G) S
- Aut(G).
~ut(c).
7. Let
with F(G)
F(G) =
LetGGbe
beaasolvable
solvable group with
=Op(G)
Op(G)==T,
T,let
letRRbe
beaa p-subgroup
p-subgroup
C
char
R)
<
H
<
G,
=
of G
T,let
let(NG(C):
(NG(C):11 # C char R) ( H 5 G , let
let X
X=
G containing
containing T,
J(ROp,F(G)), V
J(ROp,F(G)),
V ==(Q1(Z(R))x),
(Q~(Z(R))'),andXR*
and XR*=XR/CXR(V).
= XR/CxR(V).Prove
Prove either
either
(1) Op(H)
Op(H) =
=T,
T, or
or
X =ESL2(p),
(2)
X*=
= X;
Xi x .. .. x X*
(3,3, X*
X,* with
with Xi*
SL2(p),RRflnXXEESylp(X),
Sylp(X),
(2) p <
permutes {Xi:
{X1:11(<ii (
< n).
and NG(R)
NG(R)permutes
n].
8. Let G
G be
be aafinite
finite group,
group, r aa prime,
prime, nr aa set
set of
of primes
primes with
with r V
4 nr,, and
and
E,.3-A<G.For
Er.2=B<Gdefine
A
(
G.
For
Er2
E
B
(
G
define
E,J
n
a(B) ==aG(B)
B]:bbEEB'B#)
a(B)
~ G ( B_=)({On(CG(b)),
([On(CG(b)). Bl:
) n On(CG(b))
on(c~(b)).
bEB#
b€B#
< NG(L)
Assume for each component L of G that A (
NG(L) and for each H <
(G
G
with LL a9HHthat
QAUt
(L)(D)
=
1
for
each
E,.2
=
D
<
AutH(L).
Prove
that CY~,,,,(~)(D)
=
each Er2 E D (AutH(L). Prove
a(B)
-B
= 1.
a(B) =1
= 1for
foreach
each E,.2
Er2 E
B<
(A if On(G)
O,(G) =
1.
9. Let
Let G
G be
be aafinite
finite group,
group, pp aaprime,
prime, and
and HH<I!<I!G.
G.Prove
Prove Op(G)
O,(G) <
5
NG(OP(H)).
NG(OP(H)).
10. Let
Let FI'be
beaageometry
geometryofofrank
rank22and
andGGaaflag
flagtransitive
transitivegroup
groupof
ofautomorautomorphism of rr(cf.
(cf.section
section3).
3).For
ForXx EE FT let
letZX
2, be the
the pointwise
pointwise stabilizer
stabilizer
in G
Fy.
G of
ofUyErx
UYErx
I',. Assume
Assume for
foreach
eachxxEEFrand
andYyEEFX
r, that GX
G, is finite,
ZX## 1, G,
GX## G,,
Gy, and
and NG,
NG,(H)
Z,
(H) <5 GX
G, for each nontrivial normal subgroup
generalized Fitting subgroup
The generalized
176
H of
of Gx
G, contained
contained in Gy.
G,. Prove there exists a prime p such that for each
x cE Fr and
and yy cEFx,
r,, F*(Gx)
F*(G,) and
andF*(Gxy)
F*(G,,) are
are p-groups
p-groups and
and either
either Zx
Z, or Zy
Z,
is a p-group.
(Hint: Let x cE F,
Fx, zz ECr,,
Fy,and
andqq aa prime
prime such
such that
that Z,
Zx is
is not
not a qr , yy C
E r,,
group. For
For H <5 G
(H)E(H) and
group.
G set
setB(H)
0(H)==Oq
O,(H)E(H)
andlet
letQx
Q,==Gx,rz
G,J~.. Prove
Prove
Zx gl
<4 Gy
Z,
G, and
and GyZ.
G,,. Then use 31.4
31.4 and
and Exercise
Exercise 11.9
11.9 to
to show
show O(Gy),
O(Gy),
9(GyZ) 5
< Gx.
O(Gy,)
G,. Conclude:
Conclude:
(*)
B(Gy) < 9(Gxy) < 9(Qy) < 9(Gy)
Then (interchanging the
the roles
roles of
of xx and
and y if necessary) conclude Z,
Zx is a
p-group for some
p-group
some prime
prime p.
p. Using
Using(*)
(*) show
show
9(Zy) < 0(Qx) < O(Gx,y) = 9(Qy)
conclude F*(Z,)
F*(Zy) is a p-group.
O(Gx) =
_
p-group. Finally
Finally use
use (*) to show 0(G,)
and hence conclude
B(Gxy)=
= O(Gy) for
for each
each prime
prime q 54
p. This proof is due to P. Fan.)
0(G,,)
# p.
3]) Prove
11. (Thompson
(Thompson [Th 31)
Provethere
there exists
exists a function ff from
from the
the positive integers
into
the
positive
integers
such
that,
for
each
finite
set
X,
each
primitive
gers into
integers such that, for each finite
each primitive
permutation group G on
on X, each
each x cEX,
X, and
and each
each nontrivial
nontrivialorbit
orbit Y
Y of Gx
G,
on X, either
(a)
< f(lYI19
f(IYI), or
(a) IGxI
lGxl 5
or
(b) F*(Gx)
F*(G,) is
is aa p-group
p-group for
for some
some prime p.
Remark:
Y, 9= (Gx,
(G, ,Gy),
G,), and
and apply
apply Exercise
Exercise11.10
11.10to the action
action
Remark: Let
Let y EE Y,
function
of
of G on the
the geometry
geometry r(G,
F(G, 9 )).. The Sims Conjecture says that a function
ff exists
exists with
with IGxI<
lGxl 5f(IYI)
f (IYI)even
evenwhen
whenF*(Gx)
F*(G,) isisaap-group.
p-group.The
TheSims
Sims
Conjecture
is
established
in
[CPSS]
using
the
classification.
Conjecture established [CPSS]
12
representations of
of finite
Linear representations
finite groups
Chapter
Chapter 12 considers
considers FG-representations
FG-representationswhere
whereGGisisaafinite
finitegroup,
group,FF is
is a
splitting field
characteristic of
does not
not divide
divide the
the order
order
splitting
field for
for G, and the characteristic
of F does
of G.
G. Under
Under these
thesehypotheses,
hypotheses,FG-representation
FG-representation theory
theory goes
goes particularly
particularly
smoothly. For
For example Maschke's
Maschke's Theorem says each FG-representation is
while, as
as F is
the sum of irreducibles, while,
is aa splitting
splitting field for G, each irreducible
FG-representation is absolutely
absolutely irreducible.
Section 34 begins the analysis of the characters of such representations. We
find that, ifif m is the number of
of conjugacy
conjugacy classes
classes of
of G,
G, then
then G has exactly m
find
characters (x,
(Xi:
irreducible characters
: 115< ii <
_( m),
m),and
andthat
that these
these characters
characters form
form aa basis
basis
of class functions
functions from
from G
G into
into FF..
for the space of
A result of
of Brauer
Brauer (which
(which is beyond
beyond the
the scope
scope of
of this
this book)
book) shows
shows that,
that,
under the hypothesis of the first paragraph,
paragraph, the representation
representation theory of G over
F isis equivalent
F
equivalent to
to the
the theory
theory of
of GGover
overC,
C, so
sosection
section 35
35 specializes
specializes to
to the
the
case F
F=
is defined;
this is the m by
case
=C.
C. The
The character
character table
table of
of GGover
overC
Cis
defined; this
by m
m
complex
matrix (x,(g,)),
(Xi(gj)),where
where(gj:
(gj:115< j <5m)
complex matrix
m)isisa aset
setofofrepresentatives
representatives
for the conjugacy
conjugacy classes of G.
G. Various
Various numerical
numerical relations on the
the character
character
table are established;
established; among
among these the
the orthogonality
orthogonality relations
relations of
of lemma
lemma 35.5
35.5
and induced
are most fundamental. The concepts of induced representations and
characters are also discussed.
discussed. These concepts relate the representations
representations and
characters of subgroups of G to those of G.
G. Induced
Induced characters and relations
among characters on the one
one hand
hand facilitate
facilitate the
the calculation
calculation of
of the
the character
character
table,
table, and on
on the
the other
other make
make possible
possible the
the proof
proof of
of deep
deep group
group theoretical
theoretical
results. Specifically chapter 35 contains a proof
proof of
of Burnside's
Burnside's paqb-Theorem,
pagb-Theorem,
which says a group
group whose
whose order
order is
is divisible
divisible by
by just
just two
two primes
primes isissolvable,
solvable,
and of Frobenius' Theorem
Theoremon
on the
the existence
existence of Frobenius
Frobenius kernels in Frobenius
groups.
groups.
Section
Section 36 applies some of the results in section
section 35
35 and
and previous
previous chapters to analyze certain minimal groups and their representations. This section
is in the
the spirit
spirit of
of the
the fundamental
fundamental paper of
of Hall
Hall and
and Higman
Higman [HH],
[HH], which
showed how many group theoretic questions could be reduced
reduced to questions
questions
about the FG-representations
FG-representations of certain minimal
minimal groups,
groups, particularly
particularly extensions of elementary
elementary abelian or extraspecial
extraspecial p-groups
p-groups by groups
groups of
of prime
prime
order.
order.
finite groups
Linear representations
representations of
of Jinite
178
178
34 Characters
Charactersin
incoprime
coprime characteristic
characteristic
In this section G is a finite group and F
F is a splitting
charactersplitting field for G with characterdividing the
theorder
orderof
ofG.
G.Let
LetRR=
=F
F[G]
ring of
of G
G over
over F.
F.
istic not dividing
[GI be the group ring
(34.1) R
R isis aa semisimple
semisimplering and
(1) R is the direct product
product of
ofideals
ideals(Ri:
(Ri:115< ii <5 m)
m)which
which are
are simple
simple as
particular Ri Rj
Rj =
= 0 for
for ii ## j jand
1 e;ei,
, where
rings. In particular
and1 1==Em
CyZl
whereeiei=1
= Ri
1R.~ .
(2) Ri is isomorphic to the ring Fni
Fn' Xnl
"n' of
of all
all ni
ni by
by ni
ni matrices over F.
F.
(3) (Ri: 1 5
< ii <5m)
m)isisthe
theset
setofofhomogeneous
homogeneouscomponents
components of
of R,
R, regarded
regarded
as a right module over itself.
(4) Let Si consist of the matrices
matrices in Ri with 00 everywhere
everywhere except
except in the first
Then (Si:
(Si: 11 5
< i <5m)
classes
row. Then
m)isisaaset
setof
of representatives
representatives for the equivalence classes
of simple right R-modules.
R-modules.
(5) F
F=EndR(Si)
= EndR(Si)and
and dimF(S()=ni.
dimF(Si)= ni.
Proof.
is semisimple
semisimple by Maschke's
Maschke's
Proof.As
Aschar(F)
char(F)does
doesnot
notdivide
dividethe
theorder
orderof
ofG,
G,R
Ris
Theorem. Hence we can appeal to the standard theorems on semisimple rings
xni
(e.g. Lang [La], chapter 17)
D!'
17) which yield 34.1, except that Ri is the ring DF
of matrices
matrices over the division ring Di =
=EndR(Ti)
EndR(z)where Ti is a simple submodule of the ith homogeneous
homogeneous component Ri. But by hypothesis F
F isis aa splitting
splitting
field for G, so F =
=EndR(T)
EndR(T)for
for each
each simple
simple R-module
R-module TT by
by 25.8,
25.8, completing
completing
the proof.
Throughout this
this section
section m,
m, Ri,
Ri,Si,
Si,ni,
ni, and
and eiei will
will be
be as in 34.1.
Notice that
Throughout
34.1. Notice
IGI =
= dimF(R) = EYE1
dimF(Ri)
IG/
dimF(Ri)= ELln?, which I record as:
(34.2) IGI=Ymin?.
I have already implicitly used the equivalence
equivalence between FG-modules and RRmodules discussed in
in section 12, and will continue to do so
so without
without further
comment.
comment.
Let (Ci: 11 5
< ii<5m')
thethe
conjugacy
m')bebe
conjugacyclasses
classesofofG,
G,gig;EECi,
Ci,and
and define
define
zi
by zi
zi =
= F-g,c.g.
= m'.
m'.
zi E
E R by
EgEc,g.ItItdevelops
developsin
in the
the next
next lemma that m =
(34.3)
(34.3) (1)
(1) The
The number
number m
m of
of equivalence
equivalence classes
classes of irreducible
irreducible FG-representaFG-representaof G.
tions is equal to the number of conjugacy classes of
< i <5m)
F..
(2) (zi: 11 5< i <
5 m)
m) and
and (ei: 1 5
m)are
arebases
bases for Z(R) over F
Proof.
<i 5
< m),
Proof. Recall
Recalleiei=1
= R,.
1R ~As
As
. RRisisthe
thedirect
directproduct
productof
of the
the ideals
ideals (Ri
(Ri:: 1 5
Z(R)
=
®m
1
Z(Ri)
=
®m
Fei,
with
the
last
equality
following
from
Z(R) = @Y="=,(Ri) = @y="=,ei,
with the last equality following from the
the
i
isomorphism Ri
Ri =
if I show (zi:
< ii 5
< m')
m') =
= B is
Z Fn'
FniXni
xni and 13.4.1. Thus if
(zi : 1 5
is
isomorphism
'
Characters
Characters in
in coprime
coprime characteristic
characteristic
179
also a basis for Z(R), the proof will be complete.
complete. As
As G
G is a basis for
for R, B is
linearly independent, so it remains
remains to
to show
show B
B spans
spans Z(R)
Z(R) over F.
F.
Let z =
precisely when zh =
for
L-gEG agg
agg E R, a,
ag EE F.
F. Then Zz EE Z(R)
Z(R) precisely
= z for
= CgEG
each h E G,
G, which
which holds
holds in
in turn
turn precisely
preciselywhen
whenag
a, ==a(gh)
a(,h, for
for all
all g,
g, h EE G.
G. Thus
Thus
z EEZ(R)
independentof
of the
thechoice
choice of
of ggEE Ci, or
or
Z(R) ifif and
and only
only ifif a,
ag ==aiaiisisindependent
equivalently,
equivalently,when
whenzz=
=C
F_ aizi.
aizi. So
So indeed
indeed BB spans
spans Z(R)
Z(R) over
over FF..
A class function
function on
on G
G (over
(over FF)) is a function from
from G into F which is constant
on conjugacy classes.
F )the
theset
setof
of all
all class functions
classes. Denote
Denoteby
bycl(G)
cl(G) =
= cl(G, F)
on G and make cl(G) into an F-algebra
F-algebra by defining:
g(a -1- 0) = ga + g,8
a, ,B E cl(G),
g(aa) = a(ga) a E F, g E G,
g(ao) = ga go.
Evidently dimF(cl(G)) =
= m.
m. By
By 14.8, ifif nr isisan
anFG-representation
FG-representationand
and X
x its
its
character, then
then xX isis aa class
class function.
function. Let
Let n,
ni be the representation of
of G
G on Si
character,
set of
of
by right multiplication and
and xi
Xi its
its character.
character.Then
Then(ni:
(Tri:1 15<i i 5
< m) is a set
representatives
representatives for the equivalence
equivalence classes
classes of
of irreducible
irreducibleFG-representations.
FG-representations.
14.8,equivalent
have the same character,
By 14.8,
equivalent representations
representations have
character, so
so {xi:
f Xi 1: 15<i i5< m)
m)
F ; that
that is
is the set of all characters
is the set of
of irreducible characters of G over F;
of irreducible
irreducible FG-representations.
FG-representations.
The degree of a representation
representation 7r
n is just the dimension
dimension of its representation
representation
module. As this is also the trace of
of the d by d identity matrix which is in turn
follows that:
~X ((1),
l )itit, follows
(34.4) Xi
= ni isis the
xi (1) =
thedegree
degreeofofyri.
ni .
Each group G possesses
possesses a so-called
so-called principal representation
representation of degree
degree 11 in
in
which each element
element of G acts
acts as
as the
the identity
identity on
on the
the representation
representation module
which
V. As dim(V)
dim(V) =
=1,
1, the
the principal
principal representation
representation is certainly
certainly irreducible. By
convention 7rl
is
taken
to
be
the
principal
representation.
nl
representation. Hence:
(34.5) Subject
theprincipal
principalrepresentation,
representation, xl
Xi(g)
(g)=
_
nl isisthe
Subjecttotothe
theconvention
conventionthat
that7r1
nn1=1forallgEG.
l = 1 for a l l g ~ G .
Observe that there is a faithful representation
representation of
of the
the F-algebra
F-algebra cl(G)
cl(G) on
on R defined
by(C
(F agg)a
ag(ga),
fined by
a,g)a =
=FC
a,(ga),for
fora aEEcl(G)
cl(G)and
andFCagg
a,gEER.
R.Recall
Recallalso
also that
restriction to G
representation of
(r) =
=Tr(r7ri
ni is
is just the restriction
G of the representation
of RR on
on Si
Si and
and Xi
xi (r)
Tr(rni))
(r) is
is also the
the value
value of
of xi
Xiatatrr obtained
obtained from
from the
the
for r EE R. Finally
Finally notice
notice that
that Xi
xi (r)
representation of cl(G) on R.
180
180
finite groups
Linear representations
representations of
ofjinite
RJ7i=0=Xi(Rj)for#
i 0j.
j.
(34.6) (1)
(34.6)
(1) Rini=0=xi(Rj)fori
(2) Xi
each i.
x i (Ri)
( R i )==FF for
foreachi.
(3) Xi
(ei r) ==Xi
(r) for
(ei) =
(1) =
= ni.
xi(eir)
xi(r)
for each
eachii and
and each
eachrr EER.
R.In
Inparticular
particularXi
xi(ei)
=Xi
xi(1)
ni.
Proof. By 34.1 Ri
RiRj
soso
(1)(1)
holds.
1R;,
Prooj
Ri =0
= 0for
forii0#j, j,
holds.As
Aseiei==
l R i(1)
,(1)implies
implies (3).
(3).
Part (2) is an easy exercise
exercise given the description of Si
Si in 34.1.4.
34.1.4.
(34.7) The irreducible characters form
form aa basis
basis for
for cl(G)
cl(G) over
over F
F.
(34.7)
Proof. As
number
Prooj
As dimF(cl(G)) =m
=m==
numberofofirreducible
irreduciblecharacters,
characters,itit suffices
suffices to
show
(Xi:115< i <5m)
show (xi:
m)isisa alinearly
linearlyindependent
independentsubset
subset of
of cl(G).
cl(G). But
But this
this is
is
immediate
immediate from 34.6.1
34.6.1 and 34.6.3.
34.6.3.
FG-representations was
The sum of FG-representations
was defined
defined in section
section 12.
12. An immediate
immediate conconsequence
= xa+g
Xa+pfor
forFG-representations
FG-representations a
a and
X, + xXp
g=
and
sequence of that definition
definition is
is that Xa
$B and
and their
their sum aa +t-P,B,where
whereXy
X, denotes
denotes the
the character
character of the
the representation
y. Denote by char(G) the
the Z-submodule
Z-submodule of cl(G)
cl(G) spanned
spanned by
by the
the irreducible
irreducible
characters
of G.
G. char(G)
char(G) is the set of generalized characters
characters of
of G. As each
characters of
is the
the sum
sum of
ofthe
theirreducible
irreduciblerepresentations
representationsnini,
< m,
FG-representation is
, 115< ii 5
m,
(cf. 12.10), itit follows
follows from
from the
the preceding
preceding remarks
remarks that
that each
each character
character of
of G
G is
is aa
nonnegative Z-linear combination
combination of the irreducible
irreducible characters.
characters. Thus
Thus characcharacters are generalized characters,
characters, although by 34.7 not all generalized characters
are characters.
characters.
Further if (mi :: 115< ii 5
< m) are nonnegative
nonnegative integers not all zero then >
Cimi
miXi
xi
Further
is the character of the
the representation
representation Ci
>i mi
ni. Finally,
=
mini.
Finally,by
byExercise
Exercise9.3,
9.3,Xa®p
xa@g =
X«Xp,SOsothe
theproduct
productof
of characters
charactersisis aa character.
character. Hence
Hence char(G)
char(G) isis a Zxaxg,
Zsubalgebra
subalgebra of cl(G). These
These remarks
remarks are
aresummarized
summarizedin:
in:
+
(34.8) The Z-submodule char(G) of cl(G) spanned by the irreducible charac(34.8)
ters is a Z-subalgebra
Z-subalgebra of G.
G. The
The members
members of char(G) are
are called
called generalized
generalized
characters. The characters are precisely the nonnegative Z-linear span of
of the
the
irreducible
irreducible characters,
characters, and hence
hence aa subset
subsetof
of the
the generalized
generalizedcharacters.
characters.
Representations and characters
characters of
of degree
degree 11 are
are said
said to
to be
be linear.
lineal:
(34.9) Let G be
be an
an extraspecial
extraspecialp-group
p-groupofoforder
order pl+2nand
andZ Z==Z(G)
Z(G) =
_ (z).
(z).
(34.9)
Then
(1) GGhas
hasper
p2n linear
linearrepresentations.
representations.
(2) GGhas
11
faithful
faithfulirreducible
irreduciblerepresentations
representations(P1,
41, ..... .,,(pp-1. Notation
haspp- can be chosen so that zOi
wikon
zq5i acts via the scalar w
n the representation module
.
Characters in characteristic 0
181
181
Vi of
of q5i,
O , where
where ww is
is some
some fixed
fixedprimitive
primitivepth
pthroot
rootofof 11 in
in FF.. The
The Ot
q5i are
quasiequivalentfor
for1 15<i i5< pp - 1.
1.
quasiequivalent
(3)
(3) O;
q5i is
is of
of degree p".
pn.
(4)
(4) GGhas
hasexactly
exactly pen
p2n +
t-pp --1 1irreducible
irreduciblerepresentations:
representations:those
those described
described
in (1)
( 1 ) and (2).
(2).
(5)
representatives
(5) Let
LetEEbe
bethe
theenveloping
envelopingalgebra
algebraof
of0,
q5i and
andYY aa set of coset representatives
for E
E over FF..
for Z in G. Then E =EFP""p"
F P " ~ P "and
and Y
Y is a basis for
and
Proof. By
By Exercise
Exercise 12.1,
12.1, G
G has
hasexactly
exactlyIG/G(1)I
1 GIG(') I linear representations and
each is
is irreducible.
irreducible.As
AsGGisisextraspecial
extraspecialofof
order
pl+2n
G('
=
Z
and
IG/ZI
_
order
G(') = and I G/Z I =
pp2",
en , so
SO (1)
( 1 ) holds.
For xx EEGG-- Z,
G=
= xZ, so
Z, xxG
SO G has m
m=
=p2n
p2n --11+ ppconjugacy
conjugacyclasses.
classes.
For
representations,this
thisleaves
leavesexactly
exactlypp - 11
has p2n
p2n irreducible
irreducible linear representations,
As G has
representations, by 34.3. Let q5
0 be such a representation.
nonlinear irreducible representations,
representation.
ker((P).But
But ZZ is the unique
12.1, ZZ==G(1)
G(') $ ker(q5).
unique minimal
minimal normal
normal
By Exercise 12.1,
subgroup of G, so ker((P)
= 1.
1.That
Thatisis q50 is faithful.
faithful. By
By 27.16,
27.16, zq5
zO =
= a((P)I
ker(q5)=
a(q5)I for
some primitive
primitive pth
= w.
some
pthroot
rootof
ofunity
unitya((P),
a(@),say
say a((P)
a(@)=
w . By
By Exercise
Exercise8.5,
8.5,there
thereisis an
an
automorphism aa of
of G
Gof
oforder
orderpp--1
regularon
onz'Z#.
LetcPiOj
ai-1 0 1, 15<i i<< p.
p.
1regular
. Let
==a'-'@,
Then
+
-1)
POq5 =
(i-1)
mj(8-')
II
=w
.(,-I)
-1)
zo; =
= zJz'
zq5i
I
whereza
za==z'
f j('-'): 115< ii <
<p}=[i:1
i <i p}
where
z j and
and {j('-'):
p}= {i:<
15
< pmod
} modp.p.So,
So,renumberrenumberwe may
may take
take zq5i
z0i =
= wi
w`I ,I,115< ii <
< p.
p. Hence,
Hence, for
for ii ##j,j,Ot
q5i is
isnot
not equivalent
equivalent
ing, we
Off,SOsowe
wehave
havefound
foundour
ourremaining
remainingpp-- 11 irreducibles,
irreducibles, and established
established (2)
(2)
to q5j,
and (4).
(4).
(P.As
Aseach
eachq5i
qt is
isquasiequivalent
quasiequivalent to
to q5,
0, it too has degree
Let d be the degree
degree of q5.
now appeal to
to 34.2
34.2 to
to conclude
conclude p1+2n==IGI
IGI==P2n
p2n +((p
- 1)d2,
d. We now
pl ) d 2 ,keeping
p2n irreducibles
irreducibles are
are linear.
linear. Of course
course (3)
(3)follows
follows from
from
in mind that our
our first
first pen
this equality.
Aspn
pn =
= deg((P),
= Fp'
Finally let E be
be the
the enveloping
enveloping algebra
algebra of (P.
q5. As
deg(q5),EE E
FP"XP'
xp"
dimF(E) =
by 12.16. In particular dimF(E)
=pen.
p2". Let Y
Y be a set of coset representatives
= pen
for Z in G.
G. Then
Then(Y
IY II== IIGIZI
G/Z/ =
p2n =
=dimF
dimF(E),
( E ) so
,soititsuffices
sufficesto
to show
show Y
Y spans
spans
ag(go),
for
some
ag
E
F.
Further
E over F.
But
for
e
E
E,
e
=
F . But for e E E , = C
g E G a g ( g 4for
) , some ag E F. Further each
each
L+gEG
y)O =
yO with
with wi
w` EEFF,, so
g E G is of
of the
theform
formz`z iyy for
forsome
someyyEEYY and
and(z`
(ziy)q5
=w`
wiyq5
e=
= Eg
C ,ag(g(k)
ag(gq5)==>y
Cby
ybyy,y,for
forsome
someby
byEE F,
F , as
as desired.
desired.
+
35 Characters
35
Charactersin
in characteristic
characteristic 0
continued in
in this
this section. In
In
The hypothesis and notation of the last section are continued
addition assume
assume F
F=
= C.
@.
Linear representations
finite groups
representationsof
ofJinite
182
182
(35.1) FG-representations
(35.1)
FG-representations7r
x and 0C#I are
are equivalent
equivalent ifif and
and only
only ifif they
they have
have the
the
same character.
Proof.
character. Conversely
Conversely
Proof. By
By14.8,
14.8,equivalent
equivalentrepresentations
representationshave
have the same character.
assume x
7r and
andC#Iq have
havethe
thesame
samecharacter
characterX.X.Now
Nown7r
> mixi
mi7riand
andC#I0=
_ >Cki7ri
==C
kixi
for some
to show
show mi
m, =
= ki for each i. But
some nonnegative integers mi, ki. It suffices to
Xi,SO,
so,by
by34.7,
34.7,mi
m i== ki
ki for
for each
each i.
C mi xiXi== xX == C ki xi,
The regular representation
representationof
of G
G is
is the
the representation
representation of G by right multiplication on R.
(35.2) Let
Let eiei =
= F(ai,g)g,
(35.2)
C(ai,,)g,ai,g
ai,, EE F.
F . Then
Then
ai,, =
~ ( e i g - ' ) l l ~==niXi(g-')/IGI,
ln i ~ i ( g - ' ) l l ~ l ,
ai,g
=X(eig-')/IGI
where X
x is the character of the regular representation ofof G.
ai,hX(hg-1).
by
Proof. Observe
~ ( E ~ ( a i , h ) h g - '=
)= F-h
Ehai,hx(hg-').
But,But,by
Proof.
ObserveX(eig-')
X(eig-1)=
= X(F-h(ai,h)hg-1)
Exercise
=) 00 if x
Exercise12.2,
12.2,X
~ (x)
( x=
=) IGI
(ai,g), yield+ 1 and X~ (1)
( 1=
I G ,Iso
, X
~ (ei
( e ig-1)
g - ' )==IGI
IGI(ai,,),
SO
niXi,
ing the first equality in
in the
the lemma.
lemma. Next,
Next, by
by Exercise
Exercise 12.2,
12.2, xX=
= Ey=l
F-m i ni
Xi, SO
so
m
-1
x(eig-')= Y.
C nnjxj(eig
j x j ( e i g1) )==nixi(eig-1)
nixi(eigF1) =
= niXi(g-')
niXi(gP1)
X(eig-')=
j=1
with the last two equalities
equalities holding
holding by
by 34.6.1
34.6.1 and
and 34.6.3,
34.6.3, respectively.
respectively.
We
now define
define aa hermitian
hermitian symmetric
symmetricsesquilinear
sesquilinearform
form( (,, ) on
We now
on cl(G)
cl(G) with
with
respect to the complex conjugation map on @.
C. (Recall (5.c denotes the complex
conjugate
C.) Namely for X,
conjugate of c in @.)
X ,98 EE cl(G)
cl(G) define
define
(X'0)=
g/IGI.
(x()())
It is straightforward
straightforward to check that ((,,))isishermitian
hermitian symmetric
symmetricand
and sesquilinear.
sesquilinear.
Indeed the next lemma
lemma shows
shows the
the form
form is
is nondegenerate.
nondegenerate.
(35.3) The irreducible characters form an orthonormal basis for the unitary
(35.3)
unitary
space (cl(G), (( ,, )).
)). That
That is
is (Xi,
(xi, Xj)
xi) =
=Sip
6ij.
Proof. By
(g) =
=X
and each g EE G.
G.
By Exercise
Exercise9.4,
9.4,Xji(g)
~ ((g-1)
g - ' )for each character xX and
By 35.2:
Xi(ej)lnj =
(xi(g)xi(g1))
I
gEG
Hence
= Jij
Si jby
by 34.6.1
34.6.1 and 34.6.3.
Hence (Xi,
(xi, X
xi)j) =
, Xj)
Characters
Characters in
in characteristic
characteristic 00
183
(Ci:: 1 5
< ii <5m)
Recall (Ci
m)are
arethe
theconjugacy
conjugacyclasses
classesof
of G,
G,gi
gi EE Ci,
Ci, and
and by conven= 1.
(Xi
(g j )).
tion g1
gl =
1. The character
charactertable
tableofofGG(over
(overC)C)isisthe
themmbybymmmatrix
matrix
(xi(gj)).
Thus the rows of the
the character
character table
table are
are indexed
indexed by
by the
the irreducible
irreducible characcharacters of G
G and
and the
the columns
columns by
by the
the conjugacy
conjugacy classes
classes of G.
G. In
In particular
particular the
the
character table is defined only up to aa permutation
permutation of
of the
the rows
rows and
and columns,
columns,
except that by convention
except
convention Xl
~1 is always
always the
the principal
principal character
character and
and g1
gl =
=1.
1.
table has each entry in the first row
Subject to this convention, the character table
the first
first column
column are
are the
thedegrees
degrees of
ofthe
theirreducible
irreducible
equal to 1, while the entries in the
CG-representations.
CG-representations.
LetAAbebethe
thecharacter
character
table
G and
B matrix
the matrix
I).
(35.4) Let
table
of of
G and
B the
(I Ci (ICi
I g j1 Xj
(gi)(gi)/l
/ I GG1).
= A-1.
Then B =
A-' .
xk
Proof. (AB)ij
(AB)ij ==Ek Xi(gk)ICk1Xj(gk)/IGI
xi(gk)lCklXj(gk)/lGI =
xi(g)Xj(g))/lGI =
=(Xi,
(xi,
Proof.
= (CgEG
XI(g)Xj(g))/IGI
Xj)
xi) =
=8ij.
Jij.Therefore
Therefore AB
AB ==I Iisisthe
theidentity
identitymatrix.
matrix.So
SoAAisisnonsingular
nonsingular and
and
B=
= A-1.
A-l.
Observe
that, by
by 5.12,
5.12,I ICil
ICil/lGI
l~G(gi)l-' in lemma
Observe that,
Ci I=
= IG: CG(gi)l,
CG(gi)I, SO
so ICi
I/IGI =
= ICG(gi)I
35.4.
;
(35.5) (Orthogonality
= ICkI
be the order of the kth conjuhk =
lCk1 be
(35.5)
(Orthogonality Relations) Let hk
gacy class of G.
G. Then
Then the
the character
character table
table of
of G
G satisfies
satisfies the
the following
followingorthogoorthogorelations:
nality relations:
n
(1)
0
if is j,
IGI
if i = j,
hkXi(gk)XJ(gk) _
k=1
0
n
Xi(gk)Xi(gi) =
(2)
i=1
ICG(gk)I =IGI/hk
n
(3)
IGI =
n2,
i=1
niXi(gk)=0 if k > 1,
(4)
i=1
n
(5)
>hkXi(gk)=0 if i > 1,
k=1
n
(6)
> hk I Xi (gk) 12 = IGI
k=1
if
if
k=l,
Linear representations
finite groups
groups
representations of
ofJinite
184
part ((2)
of
Proof. Part (1)
( 1 ) is aa restatement
restatement of 35.3, while
while part
2 ) is a restatement
restatement of
Parts (3) and
and ((4)
cases of
of ((2)
35.4. Parts
4 ) are the special cases
2 ) with
with kk == ll==11 and
and11 =
=1,
1,
respectively
=1nni).
Similarly (5)
(5) and
and (6)
(6) are
arejust
just( (1)
withjj =
= 11 and
respectively (using
(using Xi
~ ~(1)(=
i)) .Similarly
1 ) with
j ==i,i ,respectively.
respectively.
The orthogonality relations may
may be interpreted as follows. Part (1)
( 1 ) says
says the
the
inner product of the rows of
hk)
of the
the character
character table
table (weighted
(weighted by the
the factor
factor hk)
is 00 or
whilepart
part(2)
(2) says
says the
the inner
inner product
product of
of the columns (twisted by
or II G 1,J,while
complex conjugation)
conjugation) is 00 or
or IGI/hk.
I G l/hk.
= xi(h)
Xi (h)for
foreach
eachii,, 11 5
<
(35.6) Let h,
h, g EE G. Then hh cEgG
g G if
if and
and only
only ifif Xi
x i ((g)
g )=
i <F n.
n.
Proof. IfIfhhEEgG
= Xi
(h)since
sincecharacters
characters are
are class
class functions.
functions. ConConProof.
gG then
then Xi
x i ((g)
g )=
xi(h)
versely
= Xi
(h) for
for each
each i,
i, but
but h $
0 gG.
Exercise 9.7,
versely assume
assumeXi
x i ((g)
g )=
xi(h)
g C . By Exercise
9.7,Xi
ji is also
also
irreduciblecharacter,
character,so
soXi
j i ((g)
g)=
=Xi
j i ((h).
h ) . Hence,
35.5.2:
an irreducible
Hence, by 35.5.2:
n
n
0= EXi(g)X1(h)= EXj(g)Xj(g) =I CG(9)1
i_i
i_i
which is of course
course a contradiction.
contradiction.
At this point we'll
we'll need a few facts
facts about
about algebraic
algebraic integers.
integers. Recall
Recall an
analgebraic
algebraic
integer is an element of C
The
C which
which is
is aa root
root of
of aa monic
monic polynomial
polynomialin
in 71[x].
Z[x]. The
known and can be found for example in chapter 9 of
following facts are well known
Lang [La].
[La].
(35.7) (1)
( 1 ) The
Thealgebraic
algebraicintegers
integersform
formaasubring
subringof
of C.
C.
(2)
71isisthe
theintersection
intersection of
of the
the algebraic integers
integers with'
Q.
(2) Z
with Q.
(3)
0 0.
(3) JNorm(z)J
]Norm(z)l >2 11 for
for each
each algebraic
algebraic integer z #
(4) An
An element
element cc cE C
(C is
is an
analgebraic
algebraic integer
integer if and
and only
only ifif there
there exists
exists aa
Z[c]-module which
which isis finitely
finitelygenerated
generatedasasaa71-module.
Z-module.
faithful 7/[c]-module
(35.8) ni divides
divides n for each ii..
Proof. Let
Eg1G(ai,g)g
Let a =n/ni.
= n / n i By
. By35.2,
35.2,eiei= =
C g E G ( a i , ,with
) with
g ai,g
ai,,==Xi(g)la.
;Xi(g)/aHence
Hence
aei =
= F-gcG
Xi(g)g.AS
Ase:e?=ei,
aei =
= CgGG
Xi(g)gei.
aei
CgEG
ji(g)g.
= ei, also aei
;Xi(g)gei.By 27.13.1, X(g)
~ ( g )
roots of unity
unity for each character
character X
is the sum of JGJ-th
IGI-th roots
x and each g E G. Let M
M
be the
of R generated by the
the 71-submodule
Z-submodule of
the elements
elements
( { g e , :g E G,
G, { is a IGI-th
/GI-throot of 1).
1).
Characters in characteristic 0
Characters
185
M isisaafinitely
finitelygenerated
generated71-module
Z-module and urgei = {gaei = ChEG
jii(h)
Then M
Xi(h)
{ghei
Mas
, asXZ(h)
jii(h) isisthe
thesum
sumof
ofIGI-th
/GI-throots
roots of
of unity.
unity. Hence
Hence M
M isisaa71[a]Z[a]ghei c EM,
submodule
vector
submoduleof
ofR,
R, and
and certainly
certainlyM
Misisfaithful
faithfulasasZ[a]
7/[a]5< C
C and
and R is a vector
space
space over
over C.
C. Therefore
Therefore aa is an algebraic integer
integer by
by 35.7.4,
35.7.4, and
and then
thenaa is
is an
integer by 35.7.2.
35.7.2.
Define the kernel of X
x to be
ker(X) = {g E G: X(g) = X(1)}.
(35.9)
(35.9) Let
Let X
x be the character of a CG-representation aa and
and let
let gg EE G. Then
(1)
> INorm(x(g)/x(1))1
INorm(X(g)/X(1))Iwith
withequality
equalityifif and
and only
only if
(1) IX(g)/X(1)I
Ix(g)/x(l)I 5< 113
ga
wI for
for some
someroot
root of
of unity
unity to.
w.
g a ==coI
= ker(a) 2
< G.
(2) ker(X)
kerO() =
~ ( g=) Y_'=,
Cy=, wi,
wheren =
=deg(a) =
=X(1)andcoiis
~ ( 1a)n d q is aroot
Proof. By 27.13.1,
27.13.1,X(g)=
wi,wheren
< CyZlIwiI
Ix(g)l I
lwil 5< x(1)
w
of unity. Thus IX(g)I
X(1) with
with equality
equality ifif and only if wi =
=co
is independent of i. Similarly, if Norm(x)
Norm(X) is the norm
norm of
of xX then
then
Norm(X) _ fl X
°EE
C is
is the set of embeddings
w,: so
so IX°I
]xuI 5
where E
embeddingsof
of Q(x)
Q(X)into
intoQ.
Q.But
Butxu
X°== Ci
Yi w°,
(1)and
and hence
hence
xX(1)
INorm(X)I/n < [1 IX°I/n < 1
°
wi =
w for
with equality if and only if coi
= co
for all
all i. Recall in this last case from the
proof of 27.13 that
a=
w I . So
So (1) holds. Also ifif
that as
as C
C is
is algebraically
algebraicallyclosed,
closed,gga
=col.
xX(g)
1 so
a=
Hence (2)
(2) holds.
holds.
(g)==nnthen
thenwco==1
sogga
= II.. Hence
Xj (gi)l nj, and
Let zizi==Y-gECj
: CgEc,
g, aii
aij ==hi
hixj(gi)/nj,
andbilk
bijk==11(g,
I{(g,h):
h):gg EE Ci,
Ci, hh EE
(35.10) Let
g,
Cj,gh=gk)I.Then
Cj,gh=gk)l.Then
Ein=1
(1) zi
Zi =
(1)
= CyZlaijej,
(2) ail
Xj (zi)l nj,
aij==xj(zi)/nj,
ailajl ==Ek
Ck
bijkaklr
(3) aitap
bifkak1,
(4) ail
aijisisan
analgebraic
algebraicinteger.
integer.
Proof.
AsRR==ED' 1Rj,
Rj,zi zi
Proof. As
z i j for suitable z
i j EERj.
34.3, zi
zi E
E
zip
R. By 34.3,
= = Y'
i zij
Z(R),
so
Z(R), so
so zij
zij E
E Z(Rj).
Z(Rj).Now
Now Rj
Rj==Cnj
Cnjxnj
'"j
so Z(Rj)
Z(Rj)==Cep
Cejisis1-dimensional
I-dimensionaland
and
C nxnj
l X n. Thus
~ . zij
cijejej for
consists of the
the scalar
scalarmatrices
matricesininCnj
zit =
= ci1
for some cij in C,
Xj(Zi) =
= xXj(Y-k
cikek)== Ck
Ek cikXj(ek)
= cijnj
and xj(zi)
j ( C k cikek)
cikxj(ek) =
Cijn j by
by34.6.
34.6. Thus
Thusto
toprove
prove (1)
(1)
Linear
finite groups
Linear representations
representationsof
ofjinite
186
(2) itit remains
remainstotoshow
showaiaij
ButciCijnj
g) =
and (2)
j ==
c; jCij.
. But
j n j ==X xj(zi)
j (zi) ==
X xj(CgeCi
j (Y-gEc, g)
_
Y-gEC,xj(g)
Xj(g)== hixj(gi),
hi xj(gi), so
SO the proof of (1) and (2) is complete.
complete.
Next zizj
zizj Ee Z(R), so, as (zi: 11 5< i <5 m)
ziz j=
=
Next
m )is
is basis of Z(R) by 34.3.2, zizj
CijkZk- Also
Y-k CijkZk.
Also
xgECi
xk
zj
= \gE
Eci g/ \gE h
zi
gE
gh
hECj
jk.As
Aszizj
zizj is a linear
and, in the sum on the right,
right, the
the coefficient
coefficient of gk
gk is bi
bijk.
Zk, and
and as
as G is a basis for R,
R, the
thecoefficient
coefficient of
of xx EE Ck
Ck in
combination of the zk,
the sum on the right is
equal
to
that
of
gk
and
that
coefficient
is
Cijk.
That
is
is equal to that of gk and that coefficient is cijk.
bi jk =
= Ci
jk.Next
Next
bijk
cijk.
Zizj = t
\k
Zik
ZilZjl
i
k,I
i
as RkRl
RkRI== 00 for
for kk 0# 1.1.Also
zilzjl
=: ailelajlel
so Zzizj
=
A~SO
Z
ilZjl =
ailelajlel = ailajiei
ailajlel So
iZj =
as
Y-i ailajlel.
ailajlei. On
Y-k bijkZk
On the
the other
otherhand
handzizj
zizj==
bijkzk ==Y-k,l bijkaklel
bijkaklei by the last
last
paragraphand
and(I),
(1),soso Y-kbijkakl=ailaji,
(ei:1
paragraph
bijkakl = ailajl, asas(ei:
1 (<ii <
5 m)
m ) is
is aa basis
basis for
Z(R). Hence
Hence (3) holds.
Fix (i,
= ai,i.
(i, 1) and set a =
ai,l.Then
Then
xi
(*)
xk
xk
as j,l = E bi jkak,i
xk,l
1 < j < n.
k
the following
followingsystem
systemofofnnequations
equationsininn nunknowns
unknownsx x==(XI,
(xi,. ..... ,,x,):
x,):
Consider the
("")
0 = E bijkxk + (bi jj - a)xj 1 < j < n.
k#j
Observe (**)
(**)has
hasthe
thesolution
solutionaa =
= (al,l,
Observe
( a ~ ,. ~. .,,, a,,l)
by (*). xl(g1)
a.,l) by
xi(gi) =nl
= ni #0 0, so
al,i
=
hiXi(gl)/ni
0
0.
So,
as
(0,
...
,
0)
is
also
a
solution
to
(**),
a1.l = hlxl(gl)/nl # 0. So, as (0, . . . , 0) is also a solution to (**),the
the matrix
matrix
0. Consider
Consider the
the matrix
matrixNN with
with entries
entriesininZ[x]
l[x]
M of (**) is of
of determinant
determinant 0.
by replacing
replacing aa by x in
obtained by
in M.
M. (Observe
(Observe that
that by
by definition
definition the bijks
bijks are
nonnegative
integers.)Let
Letf (x)
f (x)==det(N)
det(N)EEZ[x].
l[x]. Then
Then ff (a)
(a) =
= 00 and
andff is a
nonnegative integers.)
monic polynomial,
polynomial, so a is
is an
an algebraic
algebraic integer. Thus (4) holds.
(35.11) Let (ni, hhj)
(35.11)
j) =
=1.
1.Then either
(1) Xi(gj) = 0, or
(2) Ixi(gj)I =ni, so gjker(xi) E Z(G/ker(Xi))
Proof. By
Xixi(gj)
(g j) isisan
is b =
= ai
Proof.
By27.13.1
27.13.1and
and35.7.1,
35.7.1,aa==
analgebraic
algebraicinteger,
integer,as
asis
aijj =
=
h ja/ni by
Assume aa 0# 00and
hja/ni
by 35.10.4.
35.10.4. Assume
and let
let ffbebethe
theminimal
minimalpolynomial
polynomial of a
over Q.
Q. Say
Say ff (x)
(x) =
_ xyd
_k=0
akxk,akakEEQ.Q.Now
Nowthe
thefield
fieldextensions
extensionsQ(a)
0(a) and
f = oakxk,
over
0(b) are
Q(b)
are equal
equal as r =
=hj/ni
hj/niisisininQ.Q.So
Sothe
theminimal
minimalpolynomial
polynomial ff (x) of b
over Q
0 is
of degree
degreed.d.As
Asbbisisa aroot
rootofof
g(x)
is also of
g(x)
==~ k=if akrd-kxk
akrdPkxk,
= ~ ititfollows
follows
Characters in characteristic
Characters
characteristic 0
187
187
that g ==f.fAs
. Asb bisisananalgebraic
algebraicinteger,
integer,gg eE l[x],
Z[x],sosoakrd-k
akrd-kisisan
aninteger.
integer.As
As
r ==hi
/ni
with
(hj,
ni)
=
1,
it
follows
that
nd-k
divides
ak
for
each
k.
Hence
hj/ni with (hj, ni) = 1, it follows that ntVkdivides ak for each k. Hence
d
(akl rid k)xk E 7L[x].
h(x)
k=1
a/ni isisaaroot
root of
of h(x)
h(x)and
and h(x)
h(x)isismonic,
monic,so
soa/ni
a/niisisan
analgebraic
algebraicinteger.
integer.On
On
But a/ni
5 11by
by 35.9.1.
35.9.1. Thus (Norm
INorm (a)[
by 35.7.3,
35.7.3,
the other hand ]Norm
(Norm (a/ni)l
(a/ni)l <
(a)I=ni
=ni by
gjni
l . Therefore
Therefore (2) holds, completing
completing the proof.
35.9.1, gj
so, by 35.9.1,
7ri=
= wwi.
(35.12)
Assume hi
hj =
= ppee is a prime
prime power
powerfor
forsome
somejj >
> 11and
and G
G isis simple.
simple.
(35.12) Assume
Then G is
is of
of prime
prime order.
order.
Proof.
Then, as
as G
Gisissimple,
simple,Z(G)
Z(G) =
=11
Proof.We
Wemay
mayassume
assumeGGisisnot
not of
of prime order. Then,
and ker(xi)
ker(Xi)=
=11for each i >>1.1.So,
each i,i, jj >>11either
(gj) =
=0
So, by 35.11,
35.1 1, for each
eitherXi
xi(gj)
or p divides
dividesni.
ni.Next,
Next,by
by the
theorthogonality
orthogonalityrelations:
relations:
n
0niXi(gj)=1+pc
i=1
not an
an
for some
some algebraic
algebraicinteger
integerc.c.Hence
Hencec c==-pP1,
-p-1, impossible as
as -p-'
-p-1 isisnot
algebraic
algebraic integer,
integer, by 35.7.2.
35.7.2.
(35.13)
GI=
= paqb,
pagb, with
with pp,, q prime.
(35.13) (Burnside's pagb-Theorem).
paqb-Theorem).Let I[GI
prime. Then
Then
G is
is solvable.
solvable.
Proof. Let
counterexample.IfIf11#0 HH 9
a G then G/H
G/H and H
Let G
G be
be aa minimal
minimal counterexample.
are of order less
less then
then G
G and
and both
both are
are{p,
{p,q}-groups,
q}-groups,so,
so, by
by minimality
minimality of
of G,
G,
each is solvable.
solvable. But now 9.3.2 contradicts the choice of G. So
So G
G is
is simple.
simple.
Let QQ EE Sylq(G).
Sylq (G).IfIfQQ==1
then G
G is aa p-group
1 then
p - ~ o u and
pand hence
hence G
G isissolvsolvable. So Q
Q 0# 1,
1, so
so in
in particular
particular there
there is gj
g jEE Z(Q)#.
z(Q)#. Then Q
Q <CG(gj),
5 CG(gj),so
so
hj
G : CG
(gj) I divides JIG:
G :Ql
Q J==p pa
So,by
by35.12,
35.12, G
G is of prime
hi _=I IG:
CG(gj)l
a . .So,
prime order,
order, and
and
hence G is
is solvable.
solvable.
(35.14) Let
@E
E cl(G).
cl(G). Then
Then
Let i/r
(1)
@
=
E
L
I
(
@
,
~
i ) ~and
i,
Xi")`Xi,
and
1r
=
(1)
(2) (Y',
(@,Vf)
@)=
Ey=l(@y
xi)'.
(2)
= yin 1(Y', Xi)2.
irreducible characters
@ *== xy=l
Proof. As the irreducible
charactersare
area abasis
basisforforcl(G),
cl(G),
Em t aiaiXi
Xi
some complex
complex numbers
35.3, (i,
(@, Xj)
xj) = ( x i aiXi,
aiXi, xj)
=:
for some
numbersai.
ai. BY
By 35.3,
Xi) =
xi
ai(xi, xj) =ai,
so (2) holds.
holds.
SO (1) holds. Similarly
Similarly (@,
Eiai(Xi,Xj)=ai,so(1)holds.
xi
Xi
@) = riaiaj(Xi,Xj)=
aiaj()(i, xi) =Y_ia?,
a?,
We
induced representations.
representations.Let
LetHH 5
< G,
G,FFaafield,
field,and
andaaan
anFGFGWe next consider induced
representation. It will be convenient to regard the image ha of h e H under
finite groups
Linear representations
representationsof
ofJinite
188
188
a as
a
as aa matrix
matrix rather
rather than
than aa linear
linear transformation.
transformation. Extend a to
to GGby
bydefining
defining
- H.
ggaa ==00for
for g E
E GH .Let
LetXX==(x1:
(xi : 11 <
5 ii <5n)n)bebeaaset
setof
of coset
coset representatives
= IG
H1.For
ForggEEGGdefine
defineg gaG
((xigx 1)a) to be the
I G : HI.
a G ==((xigxil)a)
for H in G;
G; hence n =
whose (i,
(i, j)-th entry
n by n matrix whose
entry is the deg(a) by deg(a)
deg(a) matrix
matrix (xjgxj-1)a.
(xigxjl)a.
can also
also regard
regardgaG
gaG as aa square
square matrix
matrixof
ofsize
sizedeg(a)n
deg(a)n over
over F.
F. If we take
We can
this point of view
view then:
then:
(35.15) aG
I G : :H1.
aGisisan
anFG-representation
FG-representationofofdegree
degreedeg
deg(a)(a)lG
HI.
Proof. Let u,
G, A
A=
= uaG,
uaG,BB==vaG,
vac, and
u, vv E G,
and C
C=
=(uv)aG.
(uv)aG. We
We must
must show
show
AB =
= C.
from Fdxd
C. Regarding A and B as n by n matrices with entries from
F~~~(where
(where
d=
=deg(a))
deg(a)) we
we have:
have:
(AB)ij =
(xiuxk 1)a(xkvx,j 1)a.
AikBkj =
k
k
Most of
of the
the terms
terms in
in the
the sum on
on the right
right are
are 0, since (xiuxil)a
(xi uxk 1)a== 0 unMost
xiuxk 1EEH.
H.But
Butxiuxkl
xiuxkl E
EH
H precisely
precisely when
when Hxiu
Hxiu =
= Hxk, SO
so (AB)ij =
=0
less xiuxkl
unless there
there exists
exists some k with
= Hxk and
and Hxkv =
= Hxj.
Hxi. This
unless
with Hxi
Hxiuu =
This holds
holds
precisely when
when Hxiuv
Hxi u v==Hxj,
Hxj,and
andininthat
thatevent
eventthere
there isis aa unique
precisely
unique kk with
with
namely kk is defined
= Hxiu. Moreover
in this
this case
defined by Hxk
Hxk =
Moreover in
the property;
property; namely
(xk vxJ 1)a =
vxj 1)a ==C13
=~ C11
u v=
=
(xi
uxk 1)a . (xkvxil)a
(xiuxF1)a
=(xi
(xiuuvxrl)a
cij.. ~Thus
h u (AB
(s A B)ij) =
~cijifif Hxi
H X ~uv
Hx1 then xi uvx
u vxj
1 $VH,
H, so
Hx
= 0 otherwise.
otherwise. Finally
v # Hxj
Hxjj and
and (AB)1
(AB)ijj =
FinallyififHxi
Hxiuuv
j1
Ci
= (xiuvxil)a
(xi uvxi 1)a==00=
= (AB)ij.
(AB)i1.Hence
HenceAB
AB==C,
C, so
so aG
aG is a homomorphism,
Cijj=
homomorphism,
complete.
and the proof is complete.
is called
called the induced representation
aG
Gis
representation of aa to
to G.
(35.16)
(35.16) (1) Up
Uptotoequivalence,
equivalence,aG
aGisisindependent
independentof
of the
the choice
choice of
of coset
coset repfor H
H in G.
resentatives for
G.
(2) IfIfaaand
and,8B are
areequivalent
equivalent FH-representations
FH-representationsthen
then aG
aGand
and pG
BGare
are equivequivalent FG-representations.
FG-representations.
Proof. Let Y
= (yi:
(yi: 11 5
< ii <I:
n)n)bebea asecond
Y=
secondset
setofofcoset
cosetrepresentatives
representatives for
H in
in G.
G. Then
Then yi
yi ==hixi
hixifor
forsome
somehi
hiEE H.
H .Now,
Now, ifif 08 isisthe
theinduced
induced represenrepresentation of
of aa to G defined
with respect
respect to
to Y,
Y,then
then (gO)ij==( h(hixigxj
=
defined with
i x i g x ~'hi
' h j1)a
l ) a=
tation
(hia)(xigxj_1)a(hja)-1.
with
(hia)(xigxi1)a(hja)-'. Let
LetBBbe
bethen
the nby
bynndiagonal
diagonalmatrix
matrixover
over Fdxd
F~~~
with
Bii =
= hi
a. Then (gO)
= (gaG)B, so,
hia.
(go) =
so, by a remark at the beginning of section 13,
13,
aG as
as an
an FG-representation.
FG-representation.
80 is equivalent to aG
Similarly if aa is
Fd Xd
xd with
with (ha)D
(ha)D =
= hO
Similarly
is equivalent
equivalentto
to,B
B there is D E Fd
h8 for each
each
E H.
H. Let E be the n by n diagonal matrix
matrixover
overFdxd
Fdxdwith
withEii
Eii=
= D for each
hE
(gaG)E=
=g1G
gBGfor
foreach
eachgg EE G,
G, so
so (2)
(2) holds.
holds.
i. Then (gaG)E
Characters
Characters in
in characteristic
characteristic 00
189
189
(35.17)
(35.17) Let xx be
bethe
thecharacter
character of aa!and
andextend
extend xx totoGGby
bydefining
definingx(g)
~ ( g==
)00
for
G- H.
for g E
EG
H. Then
Then
(1)
of the
the induced
inducedrepresentation
representationaG
aGisisdefined
definedby
by
(1) The
Thecharacter
characterXXGG of
xG(S) =
Y, x(Sv))
IHI.
(VEG
(2) XG(g) =0 if g
Hv for some v E G.
Proof
X(Sx'
i). Moreover
Proof. XG(g)
xG(g) =
= Tr(aG)
~ r ( a=
=
~ )C:='=, Tr((gX'')a)
~r((g"i-l)a)==Ei=1
C;='=,
x(~";').
Moreover
1
xX(gxi-l)
(gx; ') =
= 0 unless g E H",
Hxi ,so
so(2)
(2)holds.
holds.Finally
FinallyGG== U;='=,
U" 1 Hxi and, as x is
is aa
t)
class
==
x(gx+
') forfor
h Eh H,
so, EyEHX;
t) = I=HI H
I X(Sx'
x(gf ")I")
X(g-l)
EH
so Z,,,, X(Sy
x(gy-I
I~(g";').
classfunction,
function,x(gxi
Hence
Hence (1)
(1) holds.
holds.
X
iscalled
calledthe
theinduced
inducedcharacter
character of
of x to
X GGis
to G.
G.
(35.18)
andx xa character
a characterofofH.H.Then
ThenXG
xG==(xK)G
(XK)G.
(35.18) Let H <5KK<5GGand
Proof.
Proof.Let
~ e08=t=XK.
x K Then,
. hen,for
forgg EE G,
G,
eG(S) =
(9gv) /IKvEG
/IKIIHI.
gEG
UEK
Further
Further the map (v, u) H
Hvu
vuisisaasurjection
surjectionof
of G
G xxKKonto
ontoGGwhose
whosefibres
fibres are
are
of
order
IKI,
so
OG(g)=(E,,,EG
X(Sw))/IHI
=
XG(8)
of order IKI, so OG(g)= (C,,, x(gW))/IHI= xG(g).
For
G -+
-- FFby
= (EvEG
*(gv))/IHI,
For *$r EE cl(H)
cl(H)define
define t/,G:
$rG:G
by1/,G(g)
$rG(g) =
(CVEG
$r(gV))/I
HI,where
where as
as
G
usual
usual *(x)
$r(x)==00for
forxx EE G
G --H.H.IfIf,/r isisa acharacter
characterofofHHthen,
then,by
by35.17
35.17, , $rG
is
the map
map $r
* HH,/rG
is aa character
character of G,
G, and
and evidently
evidently the
$rGpreserves
preservesaddition
addition and
and
scalar
cl(H) we
scalarmultiplication,
multiplication,so
so as
as the
the irreducible
irreducible characters
characters form a basis for cl(H)
we
conclude:
conclude:
+
(35.19)
H ,/1G
of cl(G) into cl(H)
(35.19) The map
map i/r
$r H
$rG isis a linear
linear transformation
transformation of
cl(H)
which
which maps
mapscharacters
charactersto
to characters
charactersand
andgeneralized
generalizedcharacters
charactersto
togeneralized
generalized
characters.
characters.
The
1116
is called
called the
the induction
induction map
map of
of cl(G)
cl(G) into
into cl(H).
cl(H). Notice that
The map
map i/r
$r H
H
$rGis
35.17
$rGshow:
show:
35.17and
andthe
thedefinition
definitionofof,/1G
(35.20)
cl(H).Then
Then
(35.20) Let
Let *$rEEcl(H).
(1)
*G(g)=0forgEG-(UEGH"),and
(1) $rG(g)= O for g E G - (UVEG
HV),and
(2)
(2) *G(1)=IG:HI,/r(1).
$rG(l)= IG: Hl$r(l).
Linear representations
finite groups
representationsof
ofjinite
190
190
Recall that there are hermitian
hermitian symmetric
symmetricforms
forms((,, )H
)H and (,
( ,)G
)Gdefined
defined on
on
cl(H)and
andcl(G),
cl(G),respectively.
respectively.
cl(H)
+
(35.21)
Let 1 EE cl(H)
(35.21) (Frobenius Reciprocity Theorem) Let
cl(H) and
and X
x E cl(G). Then
(+, XIHIH=
(+G, X)G.
(V',XIH)H
=(V'G,X)G
Proof.
('G, X)c = Y *G(g)X(g)
IGI
gEG
= CY- *(g"WO
U
IGIIHI
j(gv)X(gv) /IGHI
_
v,gEG
since jX is
is aa class
class function.
function. As
As the
the map
map gg F+
i-, g"
g" is
is aa permutation
permutationof
of G
G for
for each
each
v EE G,
G, itit follows
followsthat
that($G,
(1G,X)G
X)G== (CgeG
(F-gEG+(g)j(g))/l
*(g)X(g))/IHHI.
fi(g)=
= 0 for
I. AsAs+(g)
H ,this
thissum
sum reduces
reduces to (EhEH
( C h e H+(h)j(h))llHI
(+,XXIHIH.
gE
EG
G- H,
1(h)2(h))/I HI ==(v',
I H)H
A subset T of
of G
G is
is said
said to
tobe
beaaTZ-set
TI-setininGGififTTnnTg
T9Ec{11)
1)for
foreach
eachggEEGGNG
(T
).
NG(T>.
+
+,
(35.22)
Let TT be
be aa TZ-set
TI-set in
in G,
G, H
H=
= NG(T), and
and r, 98 E
cl(H) with
with 1 and
(35.22) Let
E cl(H)
and 08
equal to
to00on
onHH - T.
T. Then
Then
(1) *G(t)
IlrG(t)_=fi(t)
+(t)for
foreach
eachttEE T#.
T'.
(2) IfIf1r(1)
BG)G+(l) ==00 then
then (1,
(+,O)H
19)=
~
=(V"
(+G,
oG)~.
Proof.
G#.
Then+G(g)
lG(g)=
= (CUeG
EVEG+(gU))/I
i(g" ))/IHI
with+(gU)
*(g")=0
ProoJ Let g EEG'
. Then
HI with
= 0unless
unless
g"
==
0 0unless
gUEE T.
T.Thus
ThusfG(g)
IlrG(g)
unlessg gEET"
Tufor
forsome
someuuEEG,
G,ininwhich
whichcase
caseExercise
Exercise
12.3 says Tu
T" is the unique conjugate
conjugate of T containing
containinggg while
while g"
gUEE TT ifif and
and only
only
H.In
In particular
particular ifif ggEETTthen
C h E H+(gh))/l
HHI
I ==+(g).
if vv EEu-'
u`H.
then+G(g)
*G(g)=_((F-hEH
*(gh))/I
f(g)
Thus (1) holds.
holds. Also,
Also, as
as +G
/,.Gisisaaclass
classfunction,
function,+G(g)
fG(g)=
_ +(gu)
*(g") ifif g"
gUEE T for
for
some
G(g) =
= 0 otherwise.
otherwise.
some v E G, and +G(g)
*(1) ==0.
G (l) =
Assume +(l)
0. Then
Then f+G(l)
=00 by
by 35.20.2,
35.20.2, so
(V'G,OG)G =
((g(g))
IGI
gEG
(,(gv)(gv)) /IGI
gEA
Characters in characteristic 0
Characters
191
where A
A consists
consistsof
of those
thosegg EE G#
G' with
with 9'(g)
gu(g) E
E T for
for some
some v(g)
v ( g )EE G.
G. Then,
Then,by
by
Exercise
Exercise 12.3,
12.3,
(*G, OG)G = (*(t)e(t)I HI
e)H
1
ET#
as*is0onH-T#.
A Frobenius group is a transitive permutation
permutation group
group G
G on a finite set
set XX such
that no member
member of
of G#
G' fixes
fixes more than one
one point of X and
and some
somemember
member of
of G#
G'
fixes at least one point
point of X.
X . The
The following
following lemma
lemmaisisleft
leftas
asExercise
Exercise12.4.
12.4.
(35.23) (1) If G is a Frobenius
Frobenius group
group on
on aa set
setXX,, xXEEX X,
andHH==G,,
G, then H
, and
(35.23)
is a proper nontrivial subgroup
subgroupof
of T,
T, H
H is a TI-set
TI-set in
in G,
G, and
and H
H=
= NG(H).
NG(H).Let
K be the subset
K
subset of G
G consisting
consistingof
of 11 together
togetherwith
with the
the elements
elementsof
of G
G fixing
fixing no
no
points
of
X.
Then
IKI
=
1x1.
points of X. Then I K I= I X I.
group G
G such that
(2) Assume H is
is aa proper nontrivial
nontrivial subgroup of a finite group
H is
is aa TI-set
TI-set in
in GG and
andHH==NG(H).
NG(H).Then
ThenGGisisfaithfully
faithfully represented
represented as a
Frobenius
Frobenius group
group by right multiplication
multiplication on the coset space G/H.
Notice that, under the
the hypothesis
hypothesis of Lemma
Lemma 35.23.1,
35.23.1, the representation
representation of
of G
G
on X
X is
is equivalent
equivalent to
to the
the representation
representation of G
G by
by right
right multiplication
multiplication on the
the
coset space G/H
G/H by
by 5.8.
5.8. Hence
Hence the
the permutation
permutation group
group theoretic
theoretic hypotheses
hypotheses
of 35.23.1 are
are equivalent
equivalent to the group theoretic hypothesis
hypothesis of 35.23.2 by that
lemma.
lemma. I'll
I'll refer
referto
to aa group
group G
G satisfying
satisfyingeither
eitherhypothesis
hypothesisas
as aa Frobenius
Frobenius group
group
the subgroup H the Frobenius
Exercise 12.4
and call the
Frobenius complement of G. Exercise
12.4 says H
is determined
K of 35.23.1
determined up to conjugacy in G. The subset K
35.23.1 can be described
group
group theoretically by
K=G-
(UH#).
gEG
K
K will
will be called
called the Frobenius kernel of G.
G. The following important theorem
of G.
of Frobenius shows KK is a normal subgroup of
(35.24) (Frobenius' Theorem)
(35.24)
Theorem) The Frobenius
Frobenius kernel of a Frobenius group G
is a normal subgroup
subgroup of G.
Proof. Let
Let HHand
andKKbebethe
theFrobenius
Frobeniuscomplement
complementand
and kernel
kernel of
of G,
G,respecrespecThe idea
idea is
is to
toproduce
produceaacharacter
characterxXofofGGwith
withKK =
= kerO();
ker(X );then
thenKK 9
aG
tively. The
by applying 35.22 to the TI-set H of G.
by 35.9.2. This will be achieved by
G.
finite groups
Linear representations
representations of
of$nite
192
of H
H,, di
di =
(l), and, for
for
Let Bi
Let
Oi,, 11 5
bethe
theirreducible
irreducible characters
characters of
=Bi
Oi(l),
< i <5k,k ,be
11 <<ii <k,let
5 k , l e tVi=di01-Bi.As
~ , ! r ~ = d ~ O ~ - O ~H#l,k>1.Vfi(1)=di0l(1)-Bi(1)=di.#
A s1H, k > 1. $ i ( l ) =diOl(l)-Oi(l)=di= 0. Hence, by 35.22.2 and the orthogonality relations:
di =
(1/fiG>V`G)G = (Y'i, Vj)H = (diet - ei, dje1 - ej) =didj + 8ij.
E jCy=l
1 CJ cijxj
X j forforsuitable
j . By
Next, by 35.19,
35.19, 1,:if ==
suitableintegers
integerscicij.
By 35.14
35.14 and
and
Next,
Frobenius
1 ==) ($i,
~ ~1X1IH)H
I H ) H ==($i,
SO
Frobeniusreciprocity,
reciprocity,cilci1== ($?,
(iiG, ~X1)G
(Vi,
(fi, 81)
Oi)=
= di. So
2
SO
= C y =1(Cij)2
d? + Ej'_2(cij)2,
c 1i =
d? + 11 =
= (($:, iG, $
~
l= di. Also
Also d;
): 1G) =
l ( ~ i j )=
2= di2
C y = 2 ( ~ i j )so
1lf
;
=
di
X1
+
Ei
Xt(i)
for
some
irreducible
character
Xt(i)
with
t(i)
>
1,
and
some
$? = X I + E i X t ( i ) for some irreducible character xt(i)with t ( i )> and some
Ei=
=±1.
so
~i
5 1 .Next
Next didj
didj+ 8ij
Sij=(riG
= ($:, VfG)G
$,?), =didj
=didj+ EiEj(Xt(i),
E ~ E , ( x ~ ( Xt(j)),
~xt(j)),
),
SO the map
iH
take tt(i)
H t(i)
t ( iis) isananinjection,
injection,and
andwithout
without loss
loss we may take
( i ) ==i.i .Finally,
Finally, by
by
35.20.2,
$i(l)lG : HI
0 , so, as $:(l)
~ i x i ( 1and
)
xi(1)
35.20.2, $:(I)
ii/ (1) ==Yr1(1)IG:
HI =
= 0,
iiiG(1)== didi +EiXi(l)
Xi(1) =
_
is aa positive
positive integer,
integer,ititfollows
followsthat
that~iEi== -11and
= di.
deg ((rri)
x i )is
and Xi
xi ((1)
1) =
di .
= rk_1 di
_ rk=1
Define xX =
d iXi
x i .Then
Then X
x ((1)
1) =
~ f =d?
d;
, ==IHI
1 HIbyby35.5.3.
35.5.3.By
By35.20.1,
35.20.1,
Vf q (g) = 0 for g E
K#, so Xi
-1/''iG( g(g)
Thusx X
Ei1 d?
= di ( g )= 0
E K',
xi ((g)
g )=
) ==didi.
. Thus
( g(g)
) ==EL1
di2 _
=
for
allggEEKK,, so
so KK 5
c ker(X).
hand each
eachmember
memberofofGG-- K
K
~X(1)
( 1 for
) all
kero(). On
On the other hand
to an
an element
element of
of H',
H#, so
so to
to show
show KK =
= ker(X)
is conjugate to
kero() (and
(and complete the
= Y'i
(h) =
=
proof) itit remains
remains to show
show X(h)
~ ( h=
=)00for
forhhEEH#.
H'. By
By35.22,
35.22,V/iG(h)
$?(h) =
$i(h)
di so Xi(h)
xi(h)=
$:(h)
( h=)= >i diOi(h)
- 9Oi(h),
(h), so
= di
di - Vlf
(h) =
= 9Oi(h).
(h). Therefore
Therefore~X(h)
di9i(h) =
=0
by 35.5.4.
+
+
+
_j
+
+
$7
zfz1
xi
complement H
H and kernel
(35.25) Let
Let G
G be
be aa Frobenius
Frobenius group with Frobenius complement
K.
Then
K.
((1)
1 ) KK <9GGand
andHHisisaacomplement
complementtotoKKininG.
G .Thus
ThusGGisisaasemidirect
semidirectproduct
product
ofKbyH.
K by H .
of
(2)
actssemiregularly
semiregularlyon
onK;
K;that
thatisisCH(x)
C H ( x=
)=11for
foreach
eachxx EE K#,
K', or
or equiv(2) HHacts
alently CK
(y)
=
1
for
each
y
E
W.
C K ( y )= 1 for each y E H'.
(3)
Frobenius
(3) KK isisaaregular
regular normal
normalsubgroup
subgroupof
of G
G in
in its representation
representation as a Frobenius
group.
group.
Proof.
K,, H fl
= 1.
Proof. By
By Frobenius'
Frobenius' Theorem,
Theorem, K
K a9G.
G.By
Bydefinition
definition of H and K
nKK =
1.
By 35.23.1,
35.23.1,l KIKI=IG:HI,
by 1.7,
1.7,IKHI
IKHI==IKIIHI=IGI.Thus
By
/ = l G : HI, so,
so, by
lKllHl= IGI. Thus GG=HK.
= HK.
Hence (1)
(3), while ((3)
15.111 imply
imply (2).
(2).
(I) holds. Notice ((1)
1 )and 15.10 imply (3),
3 ) and 15.1
36 Some
Some special actions
representation theory developed in previous
In this section representation
previous sections
sections is used
used to
to
derive various
various results
results on the representations of certain minimal groups, and
used in turn to prove a number of group
these results are used
group theoretic lemmas.
Some of these lemmas will be used in chapter 15 and one is used in 40.7
40.7 to
to
nilpotence of Frobenius
prove the nilpotence
Frobenius kernels.
kernels.
Some
special actions
Some special
193
(36.1)
Letp,
p ,q,
q,and
andrr be
bedistinct
distinctprimes,
primes,X
Xaa group
group of order
order r acting
acting on an extra(36.1) Let
extraspecial q-group
q-groupQQwith
withCQ
CQ(X)=
such
special
(X) = Z(Q),
Z(Q ), and
and V
V aafaithful
faithfulGF(p)XQ-module
GF(p)XQ-module such
that Cv(X)=O.
Cv(X)=0. Then
a Fermat
Then r=2"+1
r =2"+ is
1 is
a Fermatprime,
prime,qq=2,
=2,and
andQQisisofofwidth
widthn.
n.
Proof. Replacing
Replacing V
V by
by an
an irreducible
irreducible XQ-submodule
XQ-submodule of
of [V, Z(Q)],
Z(Q)], we may
is an
an irreducible
irreducible XQ-module.
XQ-module. Let
Let FFbe
beaasplitting
splittingfield
fieldfor
forXQ
XQ over
assume V is
GF(p); by
=V
GF(p);
by 27.13
27.13 we
we may
may take
take FFto
tobe
be finite.
finite.Pass
Pass to VF
vF=
V®
@GF(p)V.
GF(p) V. By 26.2,
F
F isis the
thedirect
directsum
sumof
of ccGalois
Galoisconjugates
conjugatesof
of an
anirreducible
irreducibleFXQ-module
FXQ-moduleW.
W.
CW(X) =
= 0.
0. In particular dimF(W)
dimF(W) ##rr dimF(CW(X)),
dimF(Cw(X)),so
so by 27.17,
27.17,
By 27.12, Cw(X)
is an
an irreducible
irreducible FQ-module.
FQ-module. Therefore
Thereforeby
by34.9,
34.9,dimF(W)
dimF(W)=
= qe, where
where e
W is
of Q, and for any
any set
set YYof
ofcoset
cosetrepresentatives
representativesfor
forZZ== Z(Q)
Z(Q) in
is the width of
is aa basis
basis for
for EE =EndF(W)
=EndF(W) over F.
= Z,
Q, Y
Y is
F. As
As CQ(X)
CQ(X) =
Z, we
we may
may pick Y
Y to
be invariant
Y.
invariant under
under X
X via
via conjugation
conjugationand
andwe
wemay
maypick
pick11EE Y.
Next, XQ is a subgroup of E and
and hence
hence acts
acts on EE via
via conjugation,
conjugation, and
and as
as
Y
Y is a basis for E over
over F,
F ,EEisisthe
thepermutation
permutationmodule
module for
for the
the permutation
permutation
representation
= Z,
- {1},
Y. As CQ(X) =
Z, X
X is
is semiregular
semiregular on Y
Y{ I } ,so, by
representation of X on Y.
Exercise 4.6.1, CE(X)
CE(X) is of dimension
d=1+
JYJ -1 =1+g2e-1
q2e - 1
IYI - 1
d = l r + = l r+ ~
containsaaprimitive
primitiverth
rthroot
rootof
ofunity,
unity,so
soaa
On the other hand we may assume FFcontains
be the
the rth
rth roots
roots
generator x of X can be
be diagonalized
diagonalizedon
onW.
W.Let
Letai,
ai,115< ii 5
< r,r ,be
1, and mi the multiplicity of ai as an eigenvalue
eigenvalue of x. As x is diagonalizable,
of 1,
diagonalizable,
CE(x) is isomorphic
as an
an F-algebra
F-algebratotothe
thedirect
directproduct
productofofalgebras
algebrasFm'
Fml
""I,
isomorphic as
xm;
<i<
5 r,r ,sod
so d==Ei=1
xi=, m?.
m?.Also
Alsoqe
qe ==dimF(W)
d i m ~ ( W=
=
) Ei mi.
mi.Thus
Thus
1-
xi
q2e
=
r
r
i=1
i=1
CEm)2 _ Tm?+T 2mimj.
i>j
Therefore
Therefore
T(mi-mj)2=(r-1)m?)i>j
=r
m2)
i
2mimj
i>
i
- (mi)2 =rd -q2e = r - 1.
i
Pick
I{i : :mi
s=
Pick 1l in
inthe
therange
range115<1l 5< rr,, and
and let
letss=
=s(1)
s(l) =
= I{i
mi=
= ml}l.
mi}I.IfIfs
= rr then
then
=dim(W)
dim(W) ==rmi,
rml,impossible
impossibleasasr rand
andqqare
aredistinct
distinctprimes.
primes.So
Sothere
there are
are
qe =
at least two multiplicities. Now
Now since
since there
there are
ares(r
s(r - s)
differences
of
the
form
s) differences of the form
ml --mj
mi#mj,
mi
m i with
withmi
#mi,
r - 1 = T(mi - M j)2 > s(r - s)
(*)
i>j
Linear
finite groups
Linear representations
representations of
ofjinite
194
194
Mkand
and Iml
Im1-ml and mk
with equality only if there are exactly two
two multiplicities
multiplicitiesm1
Mk
= 1. Then
mkll=
(**)
r(s - 1) < s2 - 1 =(S - 1)(s + 1),
+
soeithers=1
1 1and
r (<s+1.Thenass
s 1. Then a s s <r,itfollowsthats=l
< r , it follows that s = 1orror r and
so
either s = 1 ororr
by the remark
remark after (*), there
(**) is an equality. Thus (*) is also an equality, so by
exactly two
twomultiplicities
multiplicitiesand
and Iml
(mt-- mkl=
mkI = 1. Hence
Hence we
wemay
maytake
takes(1)
s(l)=
=1,
are exactly
1,
m
= ml, s(k) = r --1,l ,and
f 1.1.Therefore
Therefore
m=ml,s(k)=r
a n dMk
m k = m + €E,
, wwhere
h e r e €c== f
+
+
+
m1=ml+(rmi
=mi (r -1)mk=m
1)mk =m +(r(r --1)(m+E).
l)(m c).
dim(W) =
dim(W)
I
AsCv
CV(X)
=0.
0. As
As ZZ==CQ
CQ(X),
scalar
Let a1 =
=1.1.As
(X) ==0,0,mml
1=
(X ), X does not act by scalar
=11 and
Mk==1
fork
Thus s(1)
s(1) =
and mk
1 for
k >
> 1.
1.
multiplication on
on W, so s(1) # r --1.1.Thus
Therefore qe
qe =
= Ft m,
1, 1,
completing
mi==r r- completingthe
theproof
proofof
of the
the lemma.
lemma.
xi
(36.2) Let
Let p,
p,q,
q,and
andrrbe
bedistinct
distinctprimes,
primes,XXaagroup
groupof
of order
order r acting
acting faithfully
faithfully
on a q-group Q,
Q, and
and V
V aa faithful
faithful GF(p)XQ-module.
GF(p)XQ-module.If q =
=22 and
and rr isisaa Fermat
Fermat
prime assume Q
(X) # 0.
Q is
is abelian.
abelian. Then
Then Cv
CV(X)
0.
Proof. Assume
XQI
Proof.
Assumeotherwise
otherwiseand
andchoose
chooseaacounterexample
counterexamplewith
withm(V)
m(V)and
andI /XQl
minimal. Then XQ is irreducible on V by minimality of m(V), and,
and, by miniminimality of IXQJ,
proper subgroup of
of Q. The latter
I XQ 1, XX centralizes every proper
latter remark
remark
and Exercise
8.10 imply
imply Q is
Exercise 8.10
is elementary
elementary abelian or Q
Q isisextraspecial
extraspecial and
and
Z(Q)=CQ(X).
Z(Q)
= CQ(X).Moreover
Moreover in
in either
either case
case X is
is irreducible
irreducible on Q/(D(Q).
Q/@(Q). Now
and 27.18
27.18 supply
supply a
by 36.1, Q
Q is
is elementary
elementary abelian,
abelian, and then Exercise 4.4 and
contradiction.
contradiction.
(36.3)
(36.3) Let a be
be an
an involution
involution acting
acting on
on aa solvable
solvable group G
G of
of odd
odd order,
order, let
let
p EE 7r
{p}
U,r'.
n cg ,r(G),
n(G),and
andp"
p"==
{p}
U n'.Assume
AssumeKKisisanana-invariant
a-invariantsubgroup
subgroup of
contains aa Hall
Hall n-subgroup
ar-subgroupofofCG(a)
CG(a)and
andXX=
= [X, aa]] is aa
G such that CK(a)
CK(a)contains
p-subgroup
(K). Then
pn (G).
OPn(K).
ThenXX<(OOpn
(G).
p-subgroupof
ofOFn
(G) =
= 11and
Proof.
to be
be aaminimal
minimalcounterexample.
counterexample. Then On
OPx(G)
and it
Proof. Take G to
r
remainstoto show
showXX =
= 1.1.Let
of G(a)
remains
LetVVbebea aminimal
minimalnormal
normal subgroup
subgroup of
G(a)
contained in
=G/
G/V.
Byminimality
minimality of
of G,
G,X*
X*<(O OPn(G*),
contained
in G
G and
and G*
G* =
V. By
pn (G*), so
CX(V)
< Opn(G)
Cx(V) (
Opn(G) =
=1.1.Also
AlsoVVisisa aq-group
q-groupfor
forsome
some prime
prime q,
q, and
and as
as
Op-(G)
= 1,
- (P).
Op*(G) =
1, q E n7r{p}.
By coprime action, Exercise 6.2, X isis contained
contained in
inan
ana-invariant
a-invariantHall
Hall7rnsubgroup H
(a)
subgroup
H of
ofK,
K,and
andasasCK
CK(a)
(a)contains
containsaaHall
Hall7r-subgroup
n-subgroup ofofCG(a),
%(a), CH
CH(a)
pr-subgroupof
of &(a).
CG(a).As
AsXX _(< Op.
(K), X <(Op
(H). Thus
is a Hall n-subgroup
Opr(K),
Op(H).
Thus setting
setting
.
Some
Some special actions
195
Ko =
byby31.20.1.
XV)
KO
=(Cxv(a),
(&(a), X),
X),XX<5Opn(Ko)
Opn(KO)
3 1.20.1.Therefore
Therefore(Ko,
(KO,
XV) satisfies
satisfies
XVby
byminimality
minimalityofof(G1.
IGI. In
In particular
particular X
X
so G
G ==XV
the hypotheses of (K, G), so
and G
G is
is aa pr-group.
n-group. As
K and
and
is irreducible on V and
As G
G is
is a n-group,
pr-group,CG(a)
CG(a) 5< K
X <5Op(K),
Therefore,
Op(K),so
so[Cv(a),
[Cv(a),X]
XI <5VVflnOp(K)
Op(K)==1. 1.
Therefore,as
asXXisisfaithful
faithful
and
on V,
V, Cv(a)
and irreducible
irreducible on
Cv(a) ==1.1.But
Butnow
now36.2
36.2supplies
suppliesa acontradiction,
contradiction,
completing
completing the
the proof of
of the
the lemma.
lemma.
Remarks. The
Therepresentation
representationtheory
theoryininsections
sections34
34and
and35
35isis basic
basic and
and belongs
Remarks.
in any introductory
course on
on finite groups. The results
results in section
section 36
36 are
are more
more
introductory course
technical. They are in the spirit of the fundamental paper of Hall and Higman
Higman
Shult [Sh]
[Sh] although
although the
the proof
proof given
given
[HH]. In particular lemma 36.1 is due to Shult
here is from
from Collins
Collins [Co] and
and uses
usestechniques
techniques of
of Hall
Hall and
andHigman
Higman [HH].
[HH].Lemma
Lemma
36.3 will be used in the proof of the
the Solvable
Solvable 2-Signalizer Functor Theorem
in chapter 15.
15. In the
the first
first edition
edition of
of this
this text,
text, section
section 36
36 contained
contained stronger
stronger
results used in
in the
the proof
proof of
of the
the Solvable
SolvableSignalizer
SignalizerFunctor
Functor Theorem
Theorem given
given in
in
results
the first edition. These results have been omitted, since
since they are
are unnecessary
unnecessary
2-signalizers.
for 2-signalizers.
Exercises for chapter
chapter 12
12
1. Let
Let G
G be
beaafinite
finitegroup
group and
and FFaasplitting
splittingfield
field for
for GGwhose
whosecharacteristic
characteristic
does not divide the order of G. Prove
(1) A
<
A character
characterXx of
of G
G isislinear
linearifif and
and only
only ifif Xx is
is irreducible
irreducible and
and G(1)
G(') 5
ker(X
).
ker(x).
(2) If
nth root
root of
of
If G
G ==(g)
(g)isiscyclic
cyclic of
of order
order n then
then FFcontains
contains aa primitive
primitive nth
with
unity w and
and the
the irreducible
irreduciblecharacters
charactersofofGGare
arexi,
Xi,1 15<i i 5
< n, with
xi
wij.
Xi(gj)
(gi) =
_ a)ij
(3) GGhas
hasexactly
exactly IG/GM)
IG/G(')I linear
linearcharacters.
characters.
2. Let
Let7r
n be
be the
the regular
regular representation
representationof aa finite group G over a splitting field
F
does not
not divide
divide the
the order
order of
of G
G and let X
F whose
whose characteristic
characteristic does
x be the
character of 7r.
character
n . Prove
(1) 7r
n is
is the
the representation induced
induced by the
the regular
regular permutation
permutation representarepresentation.
tion.
(2) n7r== Cy=l
Fm l ni7ri,
(nri:115< i <
FG-represen(2)
nixi,where (xi:
5 m)
m )are
are the irreducibleFG-representations and
tations
and ni
ni ==deg(7ri).
deg(ni).
(3) x(g)=Oforg
X(g)=0forgEG#andX(1)=IGI.
(3)
~G#andx(l)=IGI.
3. Let
group,TT gC G, A
A=={tg:
G},and
andHH=
Let G
G be
be aa finite group,
{tg: tt EE T#,
T', g EE G),
=NG(T).
NG(T).
Prove
Prove
(1) TT isis aaTI-set
fl T =
= tH and
(1)
TI-set in
in GGififand
andonly
onlyif,
if,for
foreach
eacht tEE T#,
T', tG n
and
CG(t)
<
H.
C G ( ~5) H.
196
196
Linear
finite groups
Linear representations
representations of
ofJinite
(2) IfIfTTisisaaTI-set
# under
TI-setininGGthen
thenthe
theset
setofofconjugates
conjugatesofofTT#
under G
G partitions
partitions
A,
and,for
foreach
eacht E
t ET T'#,, I ltGl
=IG
IG::H
HI II tHJ I~ ~ I .
0, and,
tG I=
4. (1)
(1)Prove
Provelemma
lemma35.23.
35.23.
(2) Prove
representation
Proveaafinite
finitegroup
grouphas
has at
at most
most one faithful permutation representation
as a Frobenius
Frobenius group,
and
conclude
a
Frobenius
group
has
group, and conclude a Frobenius group has at
at most
most one
one
Frobenius complements.
40.8)
class of Frobenius
complements.(You
(You may use the fact (proved in 40.8)
that Frobenius kernels are solvable.)
groupon
onXX== (1,.
it, .... ., ,5}.
5. Let
Let G
G =ES5
S5be
be the
the symmetric group
5).
(1) Use
Use15.3
15.3to
todetermine
determinethe
theconjugacy
conjugacy classes
classes of G.
G.
(2) As
As in
in Exercise
Exercise 5.1.2,
5.1.2,show
show GGhas
has2-transitive
2-transitive permutation
permutation represenrepresentations of degree 5 and
and 6,
6, and
and find
find their
their permutation
permutation characters
characters (cf.
Exercise 4.5).
(3) Find
Findall
alllinear
linear characters
characters of G.
G.
(4) Determine
Determine the
the character
character table
table of G.
G.
(Hint: Use (2) and
and Exercise
Exercise 12.6
12.6 to
to determine
determine two
two irreducible
irreducible characters
of G.
G. Then
Then use
use the
thenonprincipal
nonprincipallinear
linear character
character from (3)
(3) and
and Exercises
Exercises
9.3 and
9.10
to
produce
two
more
irreducible
characters.
Finally,
and 9.10 to produce two more irreducible characters. Finally, given
given
these characters
orthogonality relations to
characters and the linear
linear characters, use the orthogonality
complete
complete the table.)
Let7r
n be
be aa permutation
permutation representation
representation of
6. Let
of the finite
finite group
group G
G on
on a set X, a
the CG-representation induced
induced by
by n,
r, and
of aa (cf. Exercise
and X
x the character of
is the
the permutation
permutation character
character of
of Jr.
n. Prove
4.5). We say xX is
(1) (X,
(x, Xi)
xl) is
is the
the number
number of orbits
orbits of G on X.
(2) IfIf 7r
n is transitive then (X,
(x, X)
X ) is the permutation rank of
of G on X.
doublytransitive
transitive on
on X if and only ifif G
G
(3) GGisisdoubly
G=
= x1
Xi +xi
+Xi for
for some
some ii >
> 1.1.G
permutation rank
x1 + Xt
X, + Xj
xj for
for some
some
is of permutation
rank 33 ifif and only
only ifif G =
= Xi
i>
>jj>>1.1.(Hint:
(Hint:See
SeeExercise
Exercise 4.5.)
7. Let
r'-group G. Let p be
Let aa be
bean
anelement
element of
of prime
prime order r acting
acting on an r'-group
be
a prime,
with pp =
= 2 if r isis aaFermat
Fermat prime,
prime, let PP==Op(G),
Op(G), and
and assume
assume
prime, with
Cp(a) =
=1.1.Prove
O"(G)] <
Prove [a, OP(G)]
5 CG(P).
Cc(P).
8. Let
prime order
orderrr acting
actingon
onan
anr'-group
r'-groupG.
G.Let
Let 11 #
}
Let aa be
be an
an element
element of prime
X
= [X, a]
a] be a q-subgroup
q-subgroup of
of G
G with
with X
X abelian
abelian ifif qq=
= 2,
2, and
and let
let pp be
be a
X=
(G), X]
prime distinct
distinctfrom
fromq.q.Prove
Prove[Op
[O,(G),
XI==(C[op(G),x1(a)X?
(Cp,cc,,xl(a)x).
+ +
13
Transfer and fusion
Transfer
finite group,
group,H
H(
< G,
G, and
anda:
a: H
H + A is a homomorphism
homomorphism of
of H into an
If G is a finite
abelian group A, then
then itit isis possible
possibleto
toconstruct
constructaahomomorphism
homomorphismV:
V:GG->.
-+AA
a in
in aa canonical
canonicalway. V
V is called the transfer of G
G into
into A
A via a.
a.IfIf we
we can
can
from a
show there
there exists
existsggEEGG-- ker(V),
ker(V),then,
then,as
asG/ker(V)
G/ker(V) isis abelian,
abelian,gg 40 G(')
Gf1 the
show
commutator group of G. In particular
commutator
particular G
G is
is not
not nonabelian
nonabelian simple.
simple.
however in general
general difficult to
to calculate
calculate g V
Vexplicitly
explicitly and
and decide
decide whether
whether
It is however
E ker(V).
ker(V ).To
Todo
doso
sowe
weneed
needinformation
informationabout
aboutthefusion
the fusionof
ofgg in
in H; that is
gE
is
information about
fl H.
H. Hence
gG n
Hence chapter
chapter13
13 investigates
investigates both
both the
the transfer
transfer map
map
information
about gG
determining the fusion of elements in subgroups of G.
and techniques for determining
proof of Alperin's Fusion Theorem, which says that
Section 38 contains a proof
more
p-local subgroups
subgroups control
control the fusion of p-elements. To be somewhat
somewhat more
P is
is aa Sylow
Sylow p-subgroup
p-subgroup of
of G
G then
then we
we can
can determine
determinewhen
when subsets
subsets
precise, if P
fused in
the p-locals
p-locals H of G
P are
arefised
in G
G (i.e.
(i.e. conjugate
conjugate in G) by inspecting the
G
of P
withPflHSylowinH.
with P n H Sylow in H.
Section 39 investigates
investigates normal p-complements. A normal
normal p-complement
p-complement
of G.
G.Various
Various criteria
criteria for
for the
the
for a finite group G is a normal Hall p'-subgroup of
existence of such objects are generated, The most powerful is the Thompson
Thompson
Normal
Normal p-Complement
p-ComplementTheorem,
Theorem,which
whichisisused
usedininthe
thenext
nextsection
sectionto
toestablish
establish
kernels. Section 39
39 also contains
contains a proof of the
the nilpotence of Frobenius kernels.
Baer-Suzuki
Theorem which
whichsays
says aa p-subgroup
p-subgroup XX of
of a finite
finite group G is
is
Baer-Suzuki Theorem
contained in Op(G)
O,(G) if and only if (X,
(X, Xg)
X g ) is a p-group
p-group for
for each
each gg in
in G.
G.
A group
A is
is aa group
groupof
of autoautogroup A
A is said
said to act semiregularly
semiregularlyon
on aa group
group G
G ifif A
morphisms
with &(a)
CG(a) =
=11for each a cE A.
A'. Such
Suchactions
actionsare
areinvestigated
investigated
morphisms of G with
in section
section 40.
40.
37 Transfer
37
Let G be a finite
group, H
H5
< G, and
and a: H + A a homomorphism
of H into
finite group,
homomorphism of
into
an abelian group A. Given a set X of coset representatives for
for H in G,
G, define
define
V:G->.
V : G + AAby
by
fl((xg)(xg)_1)a
gV =
xEX
denotes the unique member of X in the coset Hxg.
We say V
V is the
where xg denotes
Hxg. We
transfer of G into A via a.
a.
Transfer andfision
and fusion
198
(37.1) The
The transfer map V
V is
is independent
independent of the choice of the set X of
of coset
coset
representatives.
representatives.
Proof. Let
for H
H in G. Then there
there is
is
LetYY be
be aa second
second set of coset representatives for
a bijection xx ++
H y(x)
= h(x)x
y(x)of
of X
X with
with Y
Y such
such that y(x)
y(x) =
h(x)xfor
for some
some h(x)
h(x)E
E H.
and gg EEGG,, write
write yg
yg for
for the member
member of
of Y
Y in Hyg.
For y EE YY and
Hyg. Observe
Observe that
that
{y(x)g}==Hy(x)gnY
Hy(x)gfY =
that is y(x)g=
y(x)g = y(xg).
y(xg).
{y(x)g)
=HxgfY
HxgnY==Hxgf1Y
HxgnY=={y(xg)};
{y(xg)};thatis
Also
y(x)g =
=h(x)xg
h(x)xg=
=h(x)xg(xg)-1(xg)
h(x)xg(xg)-l(xg)
=
h(x)xg(xg)-'h(xg)-' y(xg)
= h(x)xg(xg)-1h(xg)-1y(xg)
so that
y(x)g(y(x)g)-1 =
= h(x)xg(xg)-lh(xg)-'.
h(x)xg(xg)-1h(xg)-1
y(x)g(y(x)g)-'
Therefore
Therefore
fl(yg(yg)-1)a = fl(y(x)g(y(x)9)-1)a
xEX
yEY
n
fl(xg(xg)_1)a
=
h ( x ) a ( ~ g ( x g ) - l ) a h ( x g ) -=
= fl h(x)a(xg(xg)-1)ah(xg)-1a
=~ an ( x g ( x g ) - l ) a
xEX
xEX
xex
xex
with the last equality
equality holding
holding as
as A
A is
is abelian
abelianand
and the
the map
map xx i-±
I+ xg is
is aa permupermutation of X. So
So the lemma
lemma holds.
of G into A.
A.
(37.2) The transfer V is a group homomorphism of
Proof. Lets,
Let s,tt EE G.
G.Observe
Observe x(st)
x ( Z )==(xs)1.
(x5)t.Hence
Hence
(st)V = fl((xst)(xst)-1)a = fl(xs(xs)-1(xs)t(xst)-1)a
xEX
XEX
_ fl(xs(xs)-1)a((xs)t(xst)-1)a
xEX
(fl(xs(x)_1)a) (fl(xt(xi)_1)a) =sVtV,
xEX
using the fact that A
A is
is abelian and
and the
the map
mapxxIH
xg is a permutation on X.
+ x5
( ~ x gj:
~ 0g05j<:j j < ni),
the coset
coset
(37.3) Let (Hx1
nl ), 11_(<ii 5< rr,, be the cycles of g E G on the
space G
G/H.
<i<
< j <_(nil.
n1). Then
/ H .Pick
Pick X
X =={x1gi:
(xigj:11 _(
_( r,
r, 00 _(
(1) (g"1)x+`EHfor
( g n i ) x ; ' ~ H f 1o1<i
r5 i s<r.
r.
(2) C;=,ni=lG: HI.
(2)
(3) gV
g~ =
,ccgni)x;')~.
)a.
(3)
= ];=1((gn')x'
Et-1nl=1G:HI.
n;,
Transfer
199
199
Proof. Parts
Proof.
Parts(1)
( 1 and
) and(2)
(2)are
areimmediate
immediatefrom
fromthe
thedefinitions.
definitions.By
By 37.1
37.1 we
we may calcalculate V
V with
with respect
respect to
to this particular
particular choice
choice of coset
coset representatives.
representatives. By deficulate
(xigj)g=xigi-H1
=(xig1)gfor
nition of X, ( x i g j ) g=xigj+'
= (xigj)g
for jj < ni-1.Also
ni - 1. A ~ S O(xig"j-1)g=xig"j
(xigni-')g =xigni
nition
=xi. So
SO(3)
(3)holds.
holds.
and (xigni-')g
(xlg"i-1)g =xi.
(37.4) Let
finitegroup,
group,ppaaprime,
prime,HH<
with(p,
(p, IG
I G:: HI)
H p== 1,
1,KK < H
Let G be a finite
5 G with
with H
H/K
e eEgmK
/ K abelian,
abelian,and
and gg aa p-element
p-element in H --KKsuch
suchthat
thatg"gma
gmK for
for all
all
integers m,
m, and all
all aa EE G,
G ,such
such that
that gma
gmaE H. Then gg V4 G(l).
G(').
integers
A=
= HI
H/K
A the
the natural
naturalsurjection.
surjection.I'll
I'llshow
showggV
Proof. Let A
K and
and a :: H + A
V ## 1.
Hence g V
ker(V), and
and of
of course,
course,as
asGV
GV 5
< A, GV is abelian,
< ker(V).
4 ker(V),
abelian, so GM
G(')5
ker(V).
Choose a set X of coset representatives for
for H
H in G as
as in
in 37.3.
37.3. By 37.3.1,
(g"
g"i K
K.. Hence
Hence (g";x
)a =
= g"
a=
= ((g,)"
EH
H,, so by hypothesis
hypothesis gnix;l
g" xI ' E gni
( g n i ))x
x;l
E
(gnix;l)a
gnia
g c ~. ) ~ ~ .
Hence,
by 37.3.3,
37.3.3, gV
gV =
= (ga)",
(ga)", where
where nn =
= E;-1 ni.
Hence, by
ni.Finally,
Finally, by
by 37.3.2,
37.3.2, n =
=
IG ::HI,
HI, so
so by
by hypothesis
hypothesis (p,
(p, nn)) =
= 1. Hence,
Hence, as g is a p-element
K,,
IG
p-element and
and gg V
4K
also gn
g" 4V K
K.. Thus
Thus 11## g"a
gna =gV,
= g V as
, asdesired.
desired.
EL=,
(37.5) Let G
G: :H1)
G be
be aa finite
finite group,
group, p aaprime,
prime, HH<5GGwith
with(p,
(p,I IG
HI)==1,1,and
and
assume
nH
H=
= ggH
for all
all p-elements
p-elementsgg in
in H.
H. Then
Then (oP(G)G('))
(OP(G)G()) n
nH=
=
assume ggG
G n
H for
OP(H)H(').
OP(H)HO).
K=
= OP(H)HO).
certainly KK 5
< Go fl
nH
H,,
Proof. Let
LetGo
Go==OP(G)G(l)
O ~ ( G ) G (and
') K
O P ( H ) H ( 'Then
) . certainly
and
in Go n
n H isis ininOP(H)
and each
each p'-element
p'-element in
OP(H)<5K,
K, so
soititremains
remains to
to show
show
each p-element
p-element g in Go n
nH
H isis in
in K.
K.But
Butififmmisisan
aninteger
integer and
and aaEE G with
with
= gmh,
for some
somehhEEHH,, so,
so,as
asHH/K
gma
EH
H then
gmaE
then by hypothesis
hypothesisgma
gma=
g m h , for
/ K is
is abelian,
abelian,
gma
= gmh
E gm
gmKK.
followsfrom
from 37.4
37.4 that
that if g 4
VK
K then g V
G(l). Then, as
gma=
gmh E
. ItItfollows
4 G(').
as all
all
p-elements
Go, contrary to the choice of g. So
p-elements in
in Go
Goare
areininG(1),
G('),also gg V
4 Go,
So the
the
lemma
lemma holds.
holds.
Let H <5GGand
fusion
andSSananH-invariant
H-invariantsubset
subsetof
of G.
G.Then
ThenH
H isissaid
saidto
tocontrol
controlfusion
in S if
for
if sG nnS=sH
S = sH
foreach
eachs sEES.S.For
Forexample
exampleone
oneof
of the
the hypotheses
hypotheses in the
last lemma says that H controls
controls fusion
fusion of its p-elements.
p-elements.
Let X C
EHH<5G.
G.Then
ThenXXisissaid
saidtotobe
beweakly
weaklyclosed
closed in H
H with
with respect
respect to
to G
G
ifXGnH={X}.
if xGn H = { X ) .
(37.6) Let
Let pp be a prime,
Sylp(G),W
W5
< TT with
(37.6)
prime, T E Sylp(G),
with W
W weakly
weakly closed in T
with respect to G,
G , and D
D ==CG(W).
CG(W).Then
Then NG(W)
NG(W)controls fusion in D.
<' C(d).
Proof. Let dd EE D and
and ggEE G,
G ,with
with d9
dg EE D.
D. Then
Then W,
W ,W9_'
~ 8 -5
C(d).By
By SySylow's
a p-group.
low's Theorem
Theorem there
there is
is xx EE CG(d)
CG(d)with U
U=
=(W,
( W ,Wg-`x)
wg-lX)a
p-group. Let
200
Transfer
Transfer and fusion
U <iSSEESylp(G).
in T, {W)
(W) =
= wG
We n sS==(Wg-'x).
U
Sylp(G).As
As W
Wisisweakly
weakly closed
closed in
{wg-lx).
= x-lg EENG(W)
Thus h =x-lg
NG(W)and
and dg = ddh.
h.
(37.7)
AssumeTT isis an
an abelian
Sylowp-group
p-groupofof G.
G. Then
Then TT n O'
(G) =
_
(37.7) Assume
abelian Sylow
OP(G)
NG(T)].
[T, Nc(T)I.
Proof. Let
(T). By
Let HH==NG
NG(T).
By the
the Schur-Zassenhaus
Schur-Zassenhaus Theorem there is a comcomplement X
X to
to TT in
in H.
H. Then
Then K
K=
= X[T,
X [T,XI
X] <XT
XT==HH and
and H/K
H/K G
= T/[T, X]
plement
XI is
an abelian
abelian p-group,
p-group,so
so K
K=
=OP(H)HM.
= TOP(G), so G/OP(G) Z
O~(H)H(').Next
Next G =
T/(T nn OP(G))
Op(G)) is
is abelian
abelian and
and hence
hence OP(G)=OP(G)GM.
OP(G) = 0p(G)G('). Certainly
Certainly T is
is
CG(T),and
and hence, as
as
weakly closed in itself, so, by 37.6, H controls
controls fusion in CG(T),
T <(CG
(T ), H
H controls fusion
fusion in
in T.
T. But
But T is the set of p-elements in H so
CG(T),
so
H controls
hence, by
by 37.5,
37.5,Op(G)
OP(G)i? n H
H=
= K. As
controls fusion of its p-elements and hence,
[T, H]
HI are
areSylow
Sylowin
in OP(G)
OP(G)and
and K,
K,respectively,
respectively,itit follows
follows that
T i?n OP(G) and [T,
T nn OP(G)
OP(G)=
=[T,
[T, NA(T)].
NG(T)].
38 Alperin's
Alperin's Fusion
Fusion Theorem
Theorem
G is a finite
finite group,
group, pp is
is a prime,
prime, and
and P
P is some Sylow p-group
p-group
In this section G
of G. A p-subgroup
p-subgroup X
intersection of Sylow p-groups
p-groups
X of
of G
G isis said
said to
to be a tame intersection
Q and R of G if
=Q
Qni? RRand
andNQ(X)
NQ(X)and
andNR(X)
NR(X)are
are Sylow
Sylow p-groups of
if X
X=
NG(X
N G ( ).~ ) .
The main result of this section
section is:
(38.1)
Theorem) Let
Let P E
E Sylp(G),
Sylp(G), gg EEG,
G,and
andA,
A,Ag
AgCC P.
P.
(38.1) (Alperin's Fusion Theorem)
there exists
exists Qi
Qi EESylp(G),
Sylp(G),115< ii 5
< n, and
NG(P inl Qi)
Then there
and xi E Nc(P
Qi) such
such that:
1 . . .x,.
(1) gg == xx1...xn.
(1)
(2) PPnnQiQiisisaatame
of PP and
and Qi for
for each
eachi,i,11(< ii 5
< n.
tameintersection
intersection of
cP
for1 l<
(3) AA CEPPi?
n Q1
Ax' x, E
Qland
andAxl...xl
P inl Qi+i
Qi+l for
I ii<< n.
n.
Alperin's Theorem
Theorem will
will follow
follow from
from Theorem
Theorem 38.2 and
and an
an easy
easy argument.
argument. But
But
some definitions.
definitions.
first some
For R, Q
-+ Q
For
Q E Sylp(G) write R +
Q ifif there
thereexist
existSylow
Sylowp-groups
p-groups (Q1:1
(Qi: 1 5
i<
E NG(P
NG(P i?n Qi)
5n) of G
G and
and elements xi E
Qi) such
such that:
(1) PPni? Qi
of P and Qi for each
each ii with
with 115< ii 5
< n.
Qiisisaatame
tameintersection
intersection of
P
n
Qt+1
for
each
i
with
1
(2) PP n RR( <P nPQnlQ1
a n dand
( P n(P
R )n" l . . . X i i P n Q i + l f o r e a c h i w i t h l ( <i i < nn..
(3) Rx=Q,where
RX=Q,wherex=xl
(3)
x=x1 ... x,.
I'll also
4 QQwhen
also write R 5
whenit's
it'snecessary
necessarytotoemphasize
emphasizethe
therole
role of
of the
the element
element
and say
saythat
that(Qi,
(Qi,xi:
xi:115< ii 5
< n) accomplish R
R+
-+ Q.
x in (3), and
Alperin's Fusion
Fusion Theorem
Theorem
201
38.2. Q +
- PPfor
Theorem 38.2.
foreach
eachQQEESylp(G).
Sylp(G).
reductions. Observe first that:
The proof of 38.2
38.2 involves several reductions.
(38.3) P
P.
Indeed P
:P4 PPisisaccomplished
accomplished by
by P,
P,1.1.
(38.4) -*
(38.4)
+isisa atransitive
transitiverelation.
relation.
Proof. Let (R1,
yi:11 < ii <(m)
x; : 11 (
< ii <(n)n)accomplish
(Ri, yi:
m) and
and (Q;,
(Qi,xi:
accomplish SS -+
+ R
R
and R +
-+ Q,
Then R1,.
R1, ...
Rm,eQ1,
Q,,, and
and yl,
yl,.... .,. ym,
Q, respectively.
respectively. Then
. . ,, R,,
l , . ..... ,, Q,,
, y,,
and
x1,
xl, ..... .,,xn
x, accomplish
accomplish SS -+
+Q.
Q.
(38.5)IfIfSS4P,Qx-+
-+ P.
5 P, Qx +P,P and
, andPP ni lQQ<(PPni S,
l S,then
then Q
Q+
(38.5)
Proof. By
-+ Qx.
(S;, xi:
x; :115< ii 5
< n)
Proof.
By38.4
38.4itit suffices
suffices to show Q +
QX.Let (Si,
n) accomplish
accomplish
< i <(n)
S +-+ P.
P . Then
Then (Si,
(Si, x1:
xi: 11 5
n) also
alsoaccomplish
accomplish Q
Q -+
+Qx.
QX.
(38.6) Assume
AssumeR,
R,QQEESylp(G)
Sylp(G)with
with
andP PnnQQ << R ni? Q.
(38.6)
RR
+AP Pand
Q. Assume
Assume
P..
further,
for all S E Sylp (G) with I S n P I>> IQnPI,thatS+
I Q n P 1, that S -+P.Then
P. ThenQ
further,forallS~Syl,(G)withISnPI
Q -+
+ P
there isisxxEEGGwith
withRR5
4 P.
Proof. By
By hypothesis
hypothesis there
P . Now
Now P ni? Qx
Qx ==Rx
Rx ni? Qx
Qx =
=
so I P n Q
Qxx II=
I>IPPnn Q1.
Hence Q
Qxx +± PPbyhypothesis.
by hypothesis.
((R
R n Q)x,
Q)",soIPn
= IIRRnnQlQ >
Ql.Hence
Now apply
apply 38.5
38.5 to
to complete
completethe
the proof.
proof.
(38.7) Assume P ni? QQisis aa tame
of P and Q such that S +
-+ P for
tame intersection of
(38.7)
E Sylp(G)
Sylp(G)with
withISISi? nPIP)>> lQ
I QnnPI.
PI.Then
ThenQQ+- P.
all SS E
wemay
mayassume
assumeQQ#0PP.. Thus
ThusPP i?n Q
Q < Po
= Np(P i?
n Q).
Proof. By
By 38.3 we
Po=
Q). By
P0 and
and Qo
Qo==NQ(P
NQ(P i?n Q) are Sylow
Sylow in
in M
M=
= NG(P
NG(P i?n Q),
hypothesis Po
Q), so
so there
is x E
= Po.
P0. Notice
Notice Q +
-+ Qx is accomplished
E M With
with Qo
Qg =
accomplished by (Q,
(Q, x).
x). Further
Further
P
n Qx,
± P.P .Therefore,
-+ P.
Pi?n Q <<P0
Po<5PPi?
Qx,so
soby
by hypothesis Qx +
Therefore,by
by 38.4, Q +
P.
We are now
now in
in a position
position to
toprove
prove38.2.
38.2.Pick
Pickaacounterexample
counterexampleQQwith
withPP i?n Q
P,, and
of maximal
maximalorder.
order.By
By38.3,
38.3,PP#0 Q,
Q, so
so P n Q
Q# P
and hence
hence PP i?n Q
Q<
of
Np(P n Q).
Let
S
E
Sylp(G)
with
Np(P
n
Q)
<
Ns(P
n
Q)
E
Sylp(NG
Q). Let E Sylp(G) with Np(P n Q) 5 Ns(P n Q) E Sylp(NG
n i?Q)Q)< 5
P nPS,
((PnQ)).
P n Q)).As
AsPPnnQQ<<Np(P
Np(P
n S,it itfollows
followsthat
thatSS-*
+PPby
bymaximamaximaP..
lity of P
P ni?Q.
Q.Thus
Thusthere
thereisis xx EE G with
with S 5 P
fusion
Transfer and fusion
202
202
Evidently
Q)x 5< QX.
Qx. Also
AlsoPPnn Q
Q5
< S and
and SX
Sx==PP,, so
so((P
Q)x5< PP..
Evidently ((P
P fln Q)X
P fln Q)X
Thus(PnQ)x
Thus(PnQ)x 5< PPnQx.if(PnQ)x
n Q x . I f ( P n Q ) x # PnQxthen
P n Q x t h e nIPnQI
lPnQl<
< IPnQxl,so,
lPnQxl,so,
by maximality of
of II PP nn Q
1,
Qx
-+
P.
But
then
Q
-+
P
by
38.5,
contradicting
Ql, QX+ P. But then + by 38.5, contradicting
the choice of Q.
Q.
So(PnQ)x=PnQx.Next
letT
ESylp(G)with
let T
E Sylp(G) withNQx(PnQx)
N p ( P n QX)<
5 NT(Pn
NT(P n
So ( P n Q)X = P n Qx.Next
Qx) E
E Sylp(NG(P
Sylp(NG(P nn Qx)). Again
AgainPPnn QX
Qx <
< Np(P
NQ=(Pnn QX)
Qx)(
< T, so P n Qx <<
T n Qx.
if T +
- PPthen,
-+
Qx. Hence
Hence if
then,byby38.6,
38.6,Qx
QX
+P,P ,which
whichwe
wehave
have already
already
observed
to be
be false.
false. Thus
Thus we
we do
do not
not have
-+ P
P,, so,
so, by
by maximality
maximality of
observed to
have T +
IPnQI,PnQx=PnT.
I Pn Q I ,P n Q ~ = nP T.
By choiceof
choice of T,
T, andas
and as PP n Qx
have NT
(PnT) EESylp
QX=
=PPnnT,T,we
wehave
NT(PnT)
S y l(NG
p ( N(P
~ (nP n
T)).
By
choice
of
S,
Ns(P
n
Q)
E
Sylp(NG(P
n
Q))
so,
as
(P
n
Q)x
=
T)). By choice of
Ns(P n Q) E
n Q)) so, as ( P Q)X =
P n Qx
= PP nn TT and
n T)
Qx =
andSx
Sx==P,P we
, wehave
haveNp(P
Np(P
n T)E ESylp(NG(P
S y l p ( N ~ (nPnT)).
T)).
Thus
-+ P
P,,
tameintersection
intersection of P and
and T.
T.But
But now,
now, by 38.7, T +
Thus P nn TTisisaatame
paragraph.
contrary to the last paragraph.
This completes
completes the proof of
of 38.2.
38.2.
Now for the proof of Alperin's
Alperin's Fusion
Fusion Theorem.
Theorem. Assume the hypothesis of
Alperin'sTheorem.By38.2,
-+PP.Let(Q1,
x;:115<ii 5< n-1)
n-1)accomplish
Alperin7sTheorem.
By 38.2, pPg-'
g-' +
. Let (Qi, xi:
accomplish
Pg-'-+
P.AsA,AgCP,ACPnPC',so,bydefinition
of-+,A
p g - ' + P . As A, Ag E P , A
P n ~ g - ' SO,
, by definition of +, A c_
E PPn
n
pg-' <
5 P nn Q1
Ql and
and AXl...Xl
- C( P
c
~ . . . ~PI n< PQi+l
for for
1 1 <i i <
Axt...xi
(P n
n P~-')X
Pg-')x,...xi
n Qi+1
<
Pg-'
=P
P,, so
=x-1g
n --1.1.Also,
Also,setting
setting xx =x1
= xl ...
.. xn_1,
.x,-1, Pg-'x
P~-'x=
so xn
x, =
x-lg EE NG(P)
NG(P) and
<
= P and
Ax,...x-' = ~Agxn'
g ==xxn
xx, =x1
=xl ...
. .xn.
.x,. Finally
Finally let
let Qn
Q, =
and observe
observe Axl...xn-l
g"n-'
<
PX;'
=P =
=PP nnQn,
Q,, sosothe
thetheorem
theorem holds.
holds.
39 Normal
Normal p-complements
p-complements
section p is a prime and
In this section
and G
G isis aafinite
finite group.
group. A
A normal
normal p-complement
for G is a normal Hall p'-subgroup of G;
G; that
that is a normal
normal p-complement is aa
normal complement to a Sylow p-subgroup of G.
(39.1) (Burnside
If a Sylow p-subgroup
p-subgroup of
(39.1)
(Burnside Normal p-Complement Theorem) If
p-complement.
G is in the center of its normalizer then G possesses
possesses a normal p-complement.
Proof. This
Thisisisimmediate
immediatefrom
from 37.7.
37.7.
(39.2) If
If p is
of the
the order
order of
of G and G has cyclic
(39.2)
is the
the smallest
smallest prime divisor of
Sylow p-groups, then G has a normal p-complement.
p-complement.
Proof.
Proof. Let
LetPPEESylp(G).
Sylp(G).By
Byhypothesis
hypothesisPP isiscyclic.
cyclic.As
As PPisis abelian
abelian and Sylow
AutG(P) is a p'-group, so, by 23.3, lAut~(P)l
IAutG(P)I divides
dividespp - 1.
in G, AutG(P)
1.Hence,
Hence, as
Normal p-complements
203
AutG(P)
= 1.
p is
is the
the smallest
smallest prime
prime divisor
divisor of (G1,
I G I, AutG
(P) =
1. Equivalently, P is
is in
in the
the
center of its normalizer,
normalizer, so
so 39.1
39.1 completes
completes the
the proof.
proof.
G is metacyclic ifif there
there exists
exists aa normal
normal subgroup
subgroupHH of
of G
G such that
that H and
and
G/H
are
cyclic.
G / H are cyclic.
(39.3) Assume
(39.3)
Assume each Sylow group of G is cyclic. Then G is metacyclic.
Proof. Let
By39.2,
39.2,G
Ghas
has aanormal
normalppProof.
Letppbebethe
thesmallest
smallestprime
primedivisor
divisorofofI IG I.I.By
complement H.
H . By
By induction
induction on the order of G, H is
is metacyclic.
metacyclic. In particular
particular
G/H
/ H isiscyclic,
cyclic,G
G isis solvable.
solvable.
H is solvable, so, as G
F(G).KKisisnilpotent
nilpotentwith
withcyclic
cyclicSylow
Sylow groups,
groups, so
so K
K isiscyclic.
cyclic.
Let K ==F(G).
Thus Aut(K) is abelian. However,
However,by
by31.10,
31.10,KK==CG(K),
CG(K),so
soAu~G(K)
AutG(K)==G/K
G/K
is abelian. As G/K
G/K has
hascyclic
cyclic Sylow
Sylow groups, G/K
G/K isiscyclic,
cyclic,so
soG
G is
is metacyclic.
of G
G isis aap-local
p-local subgroup
subgroup if H
H ==NG(P)
NG(P)for
forsome
somenontrivial
nontrivial
A subgroup H of
p-subgroup P of
of G.
G.
(39.4) (Frobenius
(39.4)
(FrobeniusNormal
Normal p-Complement
p-ComplementTheorem)
Theorem) The
The following
following are equivalent:
alent:
(1) GGhas
hasaanormal
normal p-complement.
p-complement.
(2) Each
Each p-local
p-localsubgroup
subgroupof
of GGhas
hasaanormal
normal p-complement.
p-complement.
(3) AutG(P)
AutG(P)isis aa p-group
p-group for
for each
each p-subgroup P of
of G.
G.
Proof. The
Proof.
The implications
implications (1)
(1) implies
implies (2)
(2) and
and (2)
(2) implies
implies (3)
(3) are
are easy
easy and
and left
left
as exercises.
exercises. Assume G
G satisfies
satisfies (3)
(3) but not (1),
(I), and,
and, subject
subject to this constraint,
constraint,
minimal.
choose G minimal.
Observe first that G =
= OP(G).
OP(G). For if not,
not, by minimality
minimality of G,
G, OP(G)
OP(G) has
has aa
normal p-complement, which
which is also a normal p-complement
p-complement for
for G.
G.
Next let's see that a Sylow p-subgroup
p-subgroup P of G controls
P.. For let
controls fusion in P
g E G, a E P
P,, and
and ag
ag EE P.
P. Apply
Apply Alperin's
Alperin's Theorem
Theorem to
to obtain
obtain Qi
Qi EE Sylp(G)
Sylp(G)
and xi
x, E NG(P
NG (P n
n Q1)
satisfying the
the conditions
conditionsof
ofthat
that theorem
theoremwith
withAA =
= (a}.
Qi)satisfying
(a).
foreach
each 11 <
= X I...
.. .x,,.
x,.ItItwill
willsuffice
sufficetotoshow
showax,... xiEEa aP
P for
5
In particular g =x1
i <5n.n.Assume
Assumeotherwise
otherwiseand
andlet
leti ibebea aminimal
minimalcounterexample.
counterexample. Then b =
=
ax,...X; = LX(
axl...x,-l
ax1...x'-1EEPP n
n Q1
= NG(U),
Q i==U,
U, x,
xiE
EH =
NG(U), P nnH
H EE Sylp(H),
Sylp(H),and
and aX1...Xi
= bxl.
By hypothesis
hypothesisAutG(U)
AutG(U)isisa ap-group,
p-group,sosoHH==CG(U)(P
CG(U)(PflnH),
H),as
as PP fl
nHE
By
ax,...x1=
_
Sylp(H). Thus, as b E U,
bf for some t E
H,so,
so,as
asbb EE aP,
up, aX1...Xz
Sylp(H).
U, bXl
bx'=
=b`
EP
P nnH,
bx' EEa1.
bXr
up.
controls fusion
fusion in P.
P . Hence
Hence by
by 37.5
37.5 (OP(G)G(1))
(oP(G)G(')) nn PP =
OP(P)P(').
So PP controls
=OP(P)PO).
But OP(P) =
=11and,
(OP(G)G(l))nn PP =
=P
P.. Thus P
and, as
as G =
=OP(G), (OP(G)G('))
P ==PM,
P('), so,
so,
P ==1.1.But
But then
then G
G is
is its
its own
own normal p-complement.
are solvable,
solvable, P
as p-groups are
(H/
204
Transfer and
fusion
andfusion
(39.5)
Normal p-Complement
p-Complement Theorem)
Theorem) Let
Let pp be odd and PP EE
(39.5) (Thompson
(Thompson Normal
Sylp(G).
Assume
NG(J(P))
and
CG(Q1(Z(P)))
have
normal
Syl,(G).
NG(J(P))and CG(Q1(Z(P))) have normalp-complements.
p-complements.
Then
Then G
G has
has aa normal
normalp-complement.
p-complement.
Proof.
Proof. Assume
Assume otherwise
otherwise and let G
G be
beaaminimal
minimalcounterexample.
counterexample. By the
the
Frobenius
Frobenius p-Complement
p-Complement Theorem
Theorem there
there is a p-local subgroup
subgroup H of
of G
G which
which
possesses
= PPfln
possesses no normal p-complement. By Sylow's Theorem we take Q =
H
H EE Sylp(H).
Syl,(H). Choose
ChooseHHwith
withQQmaximal
maximalsubject
subjecttotothese
theseconstraints.
constraints.
Claim
Q.IfIfnot
notQQ<<Np(Q),
Np(Q),sosoI NG(C)Ip
ING(C)lp>>I 1QI
Ql for
for each
each C char
char Q.
Q.
Claim PP ==Q.
Hence,
NG(C)has
has aanormal
normalp-complement.
p-complement.So
SoNH(C)
NH(C)
Hence, by
by maximality
maximality of
of Q,
Q,NG(C)
also
this holds
holds for
for C
C=
= J(Q)
J(Q) and
also has
has aa normal
normal p-complement.
p-complement. In particular this
and
S21(Z(Q)),
Ql (Z(Q)), so the hypotheses of the theorem hold in H.
H . Hence,
Hence, by
by minimality
minimality
of
of G,
G, H
H has
hasaanormal
normalp-complement,
p-complement,contrary
contrary to
to the
the choice of H.
H.
So
So PP EE Sylp(H).
Syl,(H). Hence,
Hence, as
as H
H has
has no
no p-complement,
p-complement, H
H ==GGby
byminimality
minimality
of
G.
Moreover
if
Op,
(G)
then, by 18.7
18.7 and minimality of G, G/Op,(G)
G/O,/ (G)
of G. Moreover if O,t(G) # 11 then,
has
has aa p-complement,
p-complement,and
andthen
thenGGdoes
doestoo.
too.So
SoOp-(G)
O,i(G) ==1.1.
As
As G
G=
=H
H isisaap-local,
p-local,Op(G)
O,(G) 0#1.1.Let
LetG*
G*==G/Op(G).
G/O,(G). Then
ThenOp(G*)
O,(G*) ==11
so
so NG(J(P*))
NG(J(P*))and
andCG(Q1(Z(P*)))
CG(Q1(Z(P*)))are
areproper
proper subgroups
subgroups of
of G
G containing
containing P,
P,
and
hence
by
minimality
of
G
have
p-complements.
So,
by
minimality
of
G,
and hence by minimality of G have p-complements. So, by rninimality of G,
G*
p,p,, p(G). Thus E(G)
p(E(G)).
G*has
has aa p-complement.
p-complement.Hence
HenceGG==OO,,,/,,(G).
E(G)==Op,p,,
O,,,~,,(E(G)).
But
But [E(G),
[E(G), F(G)]
F(G)]==11so
so Op(E(G))
O,(E(G)) <(Z(E(G)),
Z(E(G)),and
andthus
thusOp,p,(E(G))
O,,,t(E(G)) =Op
= 0,
(E(G))
(E(G))xxOp,(E(G)).
O,!(E(G)). Now
Now Op,(E(G))
O,t(E(G)) <(Op,(G)
O,t(G) ==1,1,so
soE(G)
E(G)isisaap-group.
p-group.
Hence,
OO,(G).
p(G).
E(G) isis perfect,
perfect, E(G)
E(G)==1.1.So
So F*(G)
F*(G)==F(G)E(G)
F(G)E(G)==
Hence, as
as E(G)
Let
Op,p,(G),
Let pp 0#r cr Eir(G)
n(G)and
andR RE ESyl,(G).
Syl,(G).RR< (
O,,,I(G), so,
so,by
bya aFrattini
FrattiniArguArgument,
ment, Op(G)NG(Z(R))
Op(G)NG(Z(R))contains
contains aa Sylow
Sylow p-group
p-group of
of GGwhich
whichwe
wemay
maytake
take
to
to be
be P.
P .InInparticular
particular PP acts
actson
onOp(G)Z(R)
O,(G)Z(R) so
so PZ(R)
PZ(R)==KKisisa asubgroup
subgroup
of
of G.
G. Now
Now if K 0#GGthen,
then,by
byminimality
minimality of G, Z(R) =
=Op,(K)
O,!(K) a5K.
K. But
But
then
and hence
henceZ(R)
Z(R) =
= 1,
then Z(R) <(CG(Op(G))
CG(Op(G)) <(Op(G)
O,(G) by
by 31.10,
31.10, and
1, aa
contradiction.
contradiction.
So
SoG
G ==PZ(R).
PZ(R).InInparticular
particularGGisissolvable
solvableand
andRR==Z(R)
Z(R)isisabelian.
abelian.But
Butnow
now
byThompson
ThompsonFactorization,
Factorization,32.6,
32.6,GG= =
G2where
whereGGI
=NG
NG(J(P))
andGG2
by
GG
1 1G2
1=
(J (P)) and
2 ==
CG(S21(Z(P)).
By hypothesis
hypothesisGi
G, =
= Op,(G,)P.
CG(Ql(Z(P)). By
Opf(Gi)P.As
AsOp,(G,)
O,t(Gi) <(CG(Op(G))
CG(Op(G)) (
Op
(G), Op(G,) = 1,
so G, =
= P.Thus
G = G 1 G2 =
= P,
contradicting the choice
O,(G),
OPf(Gi)=
l,soGi
P.ThusG=G1G2
P,contradictingthechoice
of
of GGas
asaacounterexample.
counterexample.
The
The next
next result
result does
does not
not involve
involve normal
normal p-complements,
p-complements, but
but itit has
has the
thesame
same
flavor.
flavor.
(39.6)
(39.6) (Baer-Suzuki
(Baer-SuzukiTheorem)
Theorem)Let
Let X
X be
be aa p-subgroup
p-subgroupof
of G.
G.Then
Theneither
eitherXX <(
Op(G)
O,(G) or
orthere
thereexists
existsgg cEGGwith
with(X,
(X,Xg)
X g )not
notaap-group.
p-group.
Semiregular action
205
Proof. Assume
Assume the
xGfln
Proof.
the theorem
theoremisis false
falseand
andlet
let PP EESyl,(G)
Sylp(G)and
andAA =
= XG
P. (Xc)
(xG)
G,so,
so, as
as X
X $O,(G),
(xG)
is
not
a
p-group.
Thus
A
#
xG.
For
P.
<r!G,
Op(G), (Xc) is not a p-group. Thus A 0 XG. For
YE
A,(Y,
(Y,A)
A) isis not
not aa p-group,
p-group, by
by Sylow's
Sylow's Theorem.
Theorem.Let
Let Fr be
beof
of maximaxiY
E xG
XG- A,
sAAand
and (F,
( r ,Y)
Y) aa p-group
p-group for some YY EExG
A. Then
mal order subject
subject to
torr C
Xc - A.
r 0#A,A,butbutononthetheother
otherhand
handasasGGisisa acounter
counterexample
exampleto
tothe
thetheorem,
theorem, at
at
F
least r
nonempty. Let
r , Y).
, rr =
Q. Hence
Hence
r isis nonempty.
Let Q
Q=
= ((F,
Y). By
Bymaximality
maximalityofofr r,
= AA fln Q.
Exercise 11.4
11.4 we
( r ) and
A(r)r . As
As
by Exercise
we may
may take
take YY5<NN(F)
and there
thereisis ZZ EENNo(F)
- F.
(xG
X $ Op(G),
O,(G), NG(F)
NG(r)0#G,G,so,
so,by
byinduction
inductionon
onthe
theorder
order of
of G,
G, (XG fl
fl NG(r))
NG(F)) is
{Z}=
r1and
and (I",
(rl,Y)
Y) isis aa p-group,
p-group, contradicting
contradicting
a p-group. But now
now rr CcFrUU {Z}
= F'
r.
the maximality of F.
40 Semiregular
Semiregularaction
action
section G
G is
is aa nontrivial
nontrivial finite
finite group and A is a nontrivial group of autoIn this section
morphisms of
of G
G which
whichacts
actssemiregularly
semiregularlyon
onG.
G.Recall
Recallthis
thismeans
meansCG
CG(a)
morphisms
(a) ==11
for
for each
each aaEEA#.
A'.
(40.1)
IAl. InInparticular
1.
G I = 11 mod
mod CAI.
particular (IAl,
(IAA,IGI)=
1GI)=1.
(40.1) 1IGI
Proof. As
AsAAisissemiregular,
semiregular,each
eachorbit
orbitofofAAon
onG#
G' isisofoforder
orderI jAl.
A 1.
Proof.
We will wish to apply coprime action, 18.6,
18.6, to the representation of A on
on G.
G.
Thus, until
until 40.7,
assume either
either G
G or A is solvable.
In 40.7
40.7 we
we prove
prove G is
40.7, assume
solvable. In
nilpotent, at
at which
which point
point we
we see
see the
the assumption
assumption was
was unnecessary.
unnecessary.
(40.2) AAisissemiregular
semiregular on
oneach
each A-invariant
A-invariant subgroup
subgroup and
and factor
factor group
group of
of G.
G.
(40.2)
Proof.
Proof. The
Thefirst
firstremark
remarkisistrivial
trivial and
andthe
the second
second follows
follows from
from coprime
coprime action,
action,
18.7.
18.7.
(40.3)
(40.3) For
For each
each pp EEir(G),
n(G),there
thereisisa aunique
uniqueA-invariant
A-invariantSylow
Sylowp-subgroup
p-subgroup
of G.
G.
Proof. By
p-groups is
By coprime
coprime action,
action, 18.7,
18.7, the set A
A of
of A-invariant
A-invariant Sylow
Sylow p-groups
nonempty and
= 1,
and CG(A)
CG(A)is transitive on A.
A. So
So as CG(A)
CG(A)=
1, the lemma follows.
(40.4)
map gg H
H [g,
(40.4) For
For each
each a EE A, the map
[ g ,a]
a ]isis aa permutation
permutation of G.
G.
Proof.
[h,[ ha], aif
Proof. [g,
[ g ,a]
a ] =g-lga,
= ,g-lga,SoSo[g,[ ga]
, a _] =
] ifand
andonly
onlyififhg-1
hg-' EECG(a).
CG(a).So,
So, as
as
CG(a)
= 1,
CG(a)=
1,the
the commutator
commutator map is an
an injection,
injection, and
and hence also aa bijection
bijection as
as
G
G isis finite.
finite.
206
Transfer and
fusion
andfision
(40.5) If
If A is of
of even
even order
orderthere
thereisisaaunique
uniqueinvolution
involutiont in
t inA,A,gtg`==g-l
g-' for
(40.5)
for
E G, and G is abelian.
gE
Proof.
A Iisiseven
eventhere
thereisisan
aninvolution
involutiontt EEA.
A.Let
LetggEEG.
G.By
By40.4,
40.4,gg=
= [h, tt]]
Proof.As
AsI IAl
Therefore,for
for each
each xx EE G,
for some h EE G.
G. Thus
Thus gt
g`=
_ (h-'ht)*
(h-1h`)` =
= h-'h
h-`h ==g-l.
g-1. Therefore,
(g-lx_1)t
xr
= X-l,
x-1, so
= (g-lx-l)t
= g-`x-1
= gx. So
xt =
SO xg
xg =
= (xg)-`
(xg)-' =
=
g-tx-t =
So G
G isisabelian.
abelian.
Finally, if
ifss isisany
any involution
involution in
in A,
A, then
then I've shown
shown ss inverts
inverts G,
G, so
so stst EECA(G)
CA(G)
=1.1.Thus
=
Thus tt isis unique.
unique.
(40.6) Let p,
p, q EE 7r(A).
(40.6)
n(A). Then
(1) IfIfppisisodd
oddthen
thenSylow
Sylowp-groups
p-groupsof
ofAAare
arecyclic.
cyclic.
(2) Sylow
2-groups
of
A
are
cyclic
or
quaternion.
Sylow 2-groups of A are cyclic quaternion.
(3) Subgroups
Subgroupsof
of A
A of
of order pq
pq are
are cyclic.
cyclic.
(4) Subgroups
Subgroupsof
of AAof
ofodd
oddorder
orderare
are metacyclic.
metacyclic.
Proof. By
By 40.2 and 40.3 we may assume G is an r-group for some prime rr,,
and indeed replacing G by G/c1(G)
G/@(G)we
we may
may assume
assumeGGisiselementary
elementaryabelian.
abelian.
Observe that, for B <(A,
= 11 for
A,(CG(b):
(CG(b):bbEE B#)
B') # G as CG(b)
CG(b)=
for each
each bb EE B#.
B'.
So, by Exercise 8.1, m,(A)
mp(A)==1.1.Hence
Hence Exercise
Exercise 8.4
8.4 implies (1) and (2). Part
(3) follows from 27.18, while (1)
(1) and 39.3
39.3 imply
imply (4).
(4).
(40.7)
is nilpotent.
nilpotent.
(40.7) G
G is
Proof.
Proof. Let
Let GG be
beaaminimal
minimalcounterexample
counterexample and a an
an element of A of prime
order. By
By induction
induction on
on the
the order
order of
of A
A we
we may
maytake
take AA=
= (a).
(a). In particular A
A is
order.
solvable, so all lemmas in this section
solvable,
section apply.
apply.
Suppose q EE 7r(G)
Suppose
n(G) and
and 11 # Q is an A-invariant normal elementary
elementary abelian
abelian
of G. By 40.2 and minimality
minimality of
of G,
G, G/Q
G/Q isis nilpotent
q-subgroup of
nilpotent so
so RQ <
9G
G
for each
each Sylow
Sylow r-group
r-group RR of
of G. In particular
= qq then
for
particular ifif rr =
then R < G. But as
G is aa counter
7r(G) with
with Sylow
Sylow r-groups
r-groups of
of G not
counter example there exists r EE n(G)
normal. By
By symmetry
symmetrybetween
betweenr r and
and q,
q, O,(G)
Or(G) =
= 1.
1. As
As RQ
RQ9a G,
G, CR(Q) <
5
Or(G)=
= 1,
1, so
so Z(R)
Z(R) is faithful
on Q.
Q. By
By 40.3
40.3 we
we may
may choose
chooseRR to
to be
be aO,(G)
faithful on
invariant;then
thenaa isis faithful
faithful on
on Z(R).
Z(R). But now,
now,as
as CQ(a)
CQ(a)=
=1,
invariant;
1, 36.2 supplies aa
contradiction.
contradiction.
F(G)
(G)==1.1.Now
Now ifif HHisisaaproper
properA-invariant
A-invariant normal
normal subgroup
subgroupof G
G then,
then,
So F
by minimality
minimality of
of G,
G, H
H is nilpotent, so,
so, as
as F(G)
F(G) =
= 1, H ==1.
1.
As G is not nilpotent,
nilpotent, G
G is
is not
not aa 2-group
2-group so
so there
there is an odd prime pp EE 7r(G).
n(G).
Let PP be
p-group of
of G, G1
G1=
NG(J(P)) and G2
=
bean
anA-invariant
A-invariant Sylow
Sylow p-group
= NG(J(P))
G2 =
CG(S21(Z(P))). As
As F(G)
F(G) ==1,1,G;
CG(n1(Z(P))).
Giisisaaproper
proper A-invariant
A-invariant subgroup
subgroupof G,
G, so
so G,
Gi
207
Semiregular action
Semiregular
minimality of
of G.
G. In
In particular
particular Gi
G; has a normal
normal p-complement,
p-complement,
is nilpotent by minimality
so, by the Thompson
Normal p-Complement
p-Complement Theorem,
Theorem, GG has
has a normal
normal pThompson Normal
complement. This
impossible as G possesses no nontrivial proper
proper A-invariant
complement.
This is impossible
normal subgroups.
subgroups.
that a Frobenius
Frobenius group
group isis aa finite
finite group
group H
H which is
Recall from section
section 35 that
the semidirect
product
of
a
nontrivial
group
K
by
a
nontrivial
group
semidirect
group B with B
semiregular
Frobenius kernel and Frobenius
Frobenius complement
complement
semiregularon
on K.
K. K
K and
and B
B are
are the Frobenius
of H,
H , respectively.
respectively. Notice that,
that, by 40.7:
(40.8)
(40.8) Frobenius
Frobeniuskernels
kernels are
are nilpotent.
nilpotent.
Notice that the lemmas
Notice
lemmas in this
this section
section also
also give
give lots
lots of
of information
information about
about
Frobenius complements.
Frobenius
complements.
Remarks. Transfer
Remarks.
Transferand
andfusion
fusionare
arebasic
basictools
toolsin
in the
the study
study of finite groups. One
can begin to see
see the power of these tools
tools in some
some of the
the lemmas
lemmas and
and exercises
exercises
in this chapter,
chapter, but
but just barely.
barely.
The proof of
of the
the Baer-Suzuki
Baer-Suzuki Theorem
Theorem essentially
essentially comes
comes from
from Alperin
Alperin
and Lyons [AL]. Alperin's Fusion Theorem is (surprise)
(surprise) due to Alperin [Al].
[All.
was the first to prove the nilpotence of Frobenius kernels in his
Thompson was
thesis.
for chapter
chapter 13
Exercises for
group,pp aprime,
a prime, HH <
H1,pp)) ==1,
1. Let G be a finite group,
5G
G with
with (IG
(IG:: HI,
l , and g a
= ICG(g)lp.
ICc(g)1p
p-element in H.
H. gg isis extremal
extrernal in H ifif ICH(g)I
ICH(g)lpp=
(1) Represent
G on G/H
and let
let Hx be
(1)
Represent G
G / H by
byright
rightmultiplication
multiplication and
be aa fixed
fixed
of g on G/H.
G/H. Prove
on G
G/H
point of
Prove the orbit HxCG(g)
HxCG(g)of CG(g)on
/ H is of
in H
H..
order prime to p if and only if
if gX-'
gx ' is extremal in
(2) Assume
is of
of order
orderpp,, KK <HH with
withHH/K
/ K abelian,
abelian, ggEEHH-- K, g is
(2)
Assume gg is
extremal in H,
H, and
and each H-conjugate
H-conjugate of
of gg extremal
extremal in H
H isiscontained
contained
in gK. Prove g 4 G(').
Gel).
2. Let
LetGGbe
beaafinite
finitegroup
group with
with G
G ==02
o ~(G)
( Gand
)andlet
letTTEESy12(G).
Sy12(G).Assume T is
dihedral, semidihedral,
l2n wrZ2,
wr12, and
semidihedral, or Z2"
and prove
prove G
G has
has one
one conjugacy
conjugacy class
class
of involutions.
involutions.
3. Let
=02
Prove
thatififm2(G)
m2(G)>> 2 then
Let G
G be
be aa finite
finite group
group with
with G =
0 2(G).
( ~ )Prove
.
that
m2(CG(x))
m2(CG(x)) >
> 22for
foreach
eachinvolution
involutionxxininG.
G.(Hint:
(Hint:Let
LetTTEESy12(G).
Sy12(G).Prove
Prove
there is E4
=
U
<
T
and
x
G
fl
CT(U)
is
nonempty
for
each
involution
E4 Z U T and xG f l CT(U) is nonempty for each involution xx
of G.)
208
Transfer
Transfer and fusion
4. Let
Let GGbe
beaafinite
finite group,
group, pp aaprime,
prime, and
and assume
assume aa Sylow
Sylow p-subgroup of G
G
modular group
group MpJZ
Mp.,ofoforder
orderpn
p" #
# 8.
is the modular
8. Prove
(1) OP(G)
Op(G) has
has cyclic
cyclic Sylow
Sylow p-subgroups.
p-subgroups.
(2) IfIf pp ==22then
thenGGhas
hasaanormal
normal2-complement.
2-complement.
5. Let
LetGGbe
beaafinite
finite group
group with
with quaternion
quaternionSylow
Sylow 2-subgroups.
2-subgroups.Prove
Prove that either
either
G has a normal
2-complementororthere
thereexists
existsKK9<HH5<GGwith
withHH/K
normal 2-complement
/K 2
SL2(3).
sL2(3).
6. Let
LetGGbe
beaagroup
groupof
oforder
order60.
60.Prove
Prove either
either G
G isissolvable
solvableor
or G
G isis isomorphic
isomorphic
to the alternating
alternating group
(Hint: Let
Let T E
E Syl,(G).
Sy12(G).Show
ShowTT 9
aG
Ag of degree 5. (Hint:
G
group A5
or G
=)1.)
G has
has aa normal
normal2-complement
2-complementororJG
(G: :NG(T)I
Nc(T)(==55and
andkerNG(T)(G)
k e r ~ , ( ~ , ( G=
1.)
14
geometry of groups of Lie type
The geometry
Chapters 4 and 77 introduced
introduced geometries
geometries preserved
preserved by the
the classical
classical groups.
groups.
preserved by
by
Chapter 14 considers these geometries (and related geometries preserved
Coxeter groups) in detail, and uses the representations
representations of the classical
classical groups
groups
geometries to establish
on their geometries
establish various
various group
group theoretical
theoreticalresults.
results.
For example
classical groups L,(q),
Ln(q), U,(q),
Un(q), PSp,(q),
PSpn(q),
example we'll see
see that
that the
the finite
finite classical
PS2n(q)are
are simple,
simple, with
with aa few
few exceptions
exceptions when
when nn and q are small.
and PQE(q)
small. Also
Also
Ln(F)
fields FF,, as are U,(F)
Un(F) and PQ,(F)
PQn(F)
L,(F) and
and PSpn(F)
PSp,(F) are
are simple
simple for infinite fields
on FF.. If F
F is
develop
under suitable restrictions on
is finite of characteristic p, it will develop
geometry of
of aa classical
classical group
group G
G
that the stabilizer B of a maximal flag of the geometry
over F is
of G
over
is the
the normalizer
normalizer of a Sylow p-group of G, and
and the subgroups of
containing B are
are precisely the stabilizers
stabilizers of flags fixed by
by B. These subgroups
theparabolic
parabolic subgroups of G.
G . We say B is the Borel
Bore1
and their conjugates are the
group of G.
G.
It also turns out that to each classical
classical group G there
there is
is associated
associated aa Coxeter
Coxeter
group called the Weyl
group
of
G.
The
Weyl
groups
of
the
classical
groups
Weyl group of G . The Weyl groups of the classical groups are
A, Cn,
C, or
orDn.
D,.The
Thestructure
structureof
of G
G isis controlled
controlledto
to aa large extent by that
of type A,,
Exercise 14.6.
of its Weyl group: see for example lemma 43.7 and Exercise
14.6.
41 Complexes
Complexes
beginning this
this section
section the reader may wish to review
Before beginning
review the
the discussion
discussion
geometries in section 3. Section
of geometries
Section 41
41 is
is devoted
devoted to aa related
related class
class of
of objects:
objects:
A complex
complexisisaapair
pair6'-6'==((F,
-') where r
F is a geometry over some
complexes. A
r , 6')
index set
set II and -6'
collectionof
offlags
flagsofofrFofoftype
typeI.
I. The
The members
membersof
of
finite index
6' isisaacollection
6'
67 are
are called
called chambers.
chambers. Subflags
Subflags of chambers
chambers are
are called
called simplices.
simplices. Simplices
of corank 1 are called walls. A complex is thin if each
each wall
wall is
is contained
contained in
in
of
least
exactly two chambers. A complex is thick if each wall is contained in at least
three chambers.
chambers.
Define the chamber
chamber graph
graph of
of 6'
67 to be
be the
the graph
graph on 6'
6' obtained
obtained by
by joining
chambers which have a common wall. A path in the chamber graph is called a
gallery.
connected. A
gallery. A
A complex
complex is
is said
said to be connected if its chamber graph is connected.
in which
which every
every flag
flag of
of rF of
of rank
rank 1 or
or 2 is
is aa simplex
simplex
connected complex (F,
( r , -6')
6') in
chamber complex.
complex.
is called a chamber
morphismaa:
f))-+
-* (A,
(A,!%)
9) of
morphismrF+- A of
A morphism
: ( (F,
r , 6'
of complexes
complexes isisaa morphism
geometries with
with -6'a
contained in
in !%.
9. AAsubcomplex
geometries
6'a contained
subcomplex of (F,
( r , -2)
6')
is a complex
210
The geometry of groups of Lie type
(0, 5)
_ t'6'flr loAthe
(A,
GB)with
with0Aa asubgeometry
subgeometry of
of Fr and
and _T
GB =
theset
setofofchambers
chambers
contained in A.
The notions of complex, building, and Tits system (which are the
the subject
subject
of a complex
of this chapter) come from Tits [Ti].
[Ti]. However
However the definition
definition of
I've just given
given is somewhat
somewhat less general than that of Tits [Ti]
[Ti] in that under
my definition there is a type function defined on simplices
simplices inherited from the
geometry. However
However Tits
Tits shows
shows how
howto
to associate
associate type
type functions
functions to
to
associated geometry.
and buildings,
buildings, so
so in
in the
the end
end the
the class
class of
of object
object considered
Coxeter complexes and
is the same.
same.
The treatment of complexes,
complexes, buildings, and Tits
Tits systems
systems given
given here
here is
is exextracted from Tits [Ti] and Bourbaki [Bo], modulo remarks above.
Let G
of G. The
G be
be aagroup
groupand
and-IF=
F=(G;:
(Gi:ii E
E I1)
) a family
family of subgroups
subgroups of
The
coset geometry
geometry r(G,
F(G, g)
T) determined
determinedby
by G
G and
and 9 is defined in section 3. Let
-t'(G,
9) bebethe
6'(G, 9)
thecomplex
complexon
onF(G,
r(G,_,F)
F ) whose
whose chamber
chamber set
set (also
(also denoted by
-&(G,
'(G, ST))
consistsof
ofthe
theflags
flagsSI,,,
SI,,,xXEEG,
G,where
whereSI,,
SI,x==(Gix:
(G,x:i i EE I).
fl.
F ) ) consists
(41.1) Let G be a group,
E II)) a family
family of
of subgroups
subgroupsofofG,
G,rF =
=
(41.1)
group, F =
=(G1:
(Gi: i E
F(G, F
ST),
andf3
-' =
= t'(G,
r(G,
) , and
6'(G,ST).
F ) . Then
Then
-6 by right
right multiplicamultiplica(1) GGisisrepresented
represented as
as aa group
group of
of automorphisms of B
tion on the cosets in F.
r.
on the
the simplices
simplicesofof6'-t'of
oftype
typeJJ for
for each
eachsubset
subsetJJ of
of II..
(2) G
G is
is transitive
transitive on
In particular
particular G
G is
is transitive
transitiveon
on chambers.
chambers.
(3) G,I
of the
the simplex
simplexSSjj of
of type
type JJ..
G is the stabilizer of
-6 are the conjugates of
of Si,,
Si', ii E 1,
(4) The
The walls
walls of chambers of 6'
I , under G.
G.
Si, is contained in exactly
exactly IGp:
Sit
IGi,: GI
GI (I chambers.
r of
-t'.
(5) Every
Every flag in r
ofrank
rank11or
or 22 is
is aa simplex
simplex of 6'.
The proof is straightforward;
straightforward;the
the notation
notation is
is explained
explained in
in section
section3.
3.
(41.2) Let C be a chamber in a chamber complex 6'
-0 and
and aa a morphism of B
-6
(41.2)
fixing C.
C. Then
Then
C..
fixes each
each simplex
simplex contained in C
(1) aa fixes
(2) If -0
thin and
andaa is
is aa bijection
bijectionon
on6',
e, then
thenaa =
= 1.
(2)
6' is thin
1.
Proof. aaisisaamorphism
morphismofofgeometries
geometriesso
soititpreserves
preservestype.
type. Hence
Henceas
as C
C contains
contains
a unique simplex of each type,
type, (1)
(1) holds.
holds.
To prove
prove (2)
(2) itit suffices
suffices to
to show
show aa fixes each chamber adjacent
adjacent to
to C
C,, since
ic
Let D
D be such aa chamber;
chamber;then
thenDDnflCc =
=W
f3 is connected.
connected. Let
W is
is aa wall
wall of
of C,
C,
so a fixes
W
by
(1).
As
-t'
is
thin,
D
and
C
are
the
only
chambers
containing
fixes W
6'
and C are the only chambers
W,, so,
so,as
asW
W==WWet
Dar fl
= (C,
asCC =
=C
Ca,
W
a ==Da
l CCa,
a , ( (Ca,
C a , DDa)
a) =
( C ,D).
D). Now,
Now, as
a , also
D=
= Da.
Da.
211
211
Complexes
folding of 6'
-t' is an idempotent morphism
Let 6'
6' be a thin chamber
chamber complex. A folding
whose fibres on chambers
chambers are all of order 2.
-t' is a Coxeter complex
complex ifif for
for each
each
2.6'
C , D of
of adjacent
adjacent chambers
chambers there exists a folding mapping C to D.
pair C,
In the remainder of this section assume 6' =
(r,-')6')isisaaCoxeter
Coxetercomplex.
complex.
= (F,
(41.3) Let 4
0 be a folding of
of 6'.
t'. Then
induces aabijection
(6'(P--6'4)
(1) 04 induces
bijection4:0:
eo) --+->6'4.
eo.
(2)
o and
'j.
(2) 04fixes
fixeseach
each member
member of rr4
and i&?4.
Proof. Part
Part(2)
(2)isisjust
just aarestatement
restatementofofthe
thehypothesis
hypothesisthat
that4)4 isisidempotent.
idempotent.
-6'0. By
By hypothesis
hypothesis there
there is
is aa unique
unique chamber
chamber D
D distinct from C with
Let C E
E -64.
Do
not in
in6'4.
eo.
D@==C.
C .By
By (2),
(2),D is not
For C
C,, D in -'6'let
For
letd(C,
d(C,D)
D )be
bethe
thedistance
distancebetween
between C
C and
and D
D in
inthe
the chamber
chamber
graph.
graph.
(41.4) Let
Let40 be
beaafolding
foldingofof
eo,and
andDDinin6't'-- 6'4.
eo. Then
(41.4)
6',t',CCinin6'4,
Then
(1) If
= Co,
Co, ..... .,, C,,
gallery, there
there exists
with
If CC =
C ==DDisisa agallery,
exists0 05< ii < nn with
Ci E
E -6'0
6'4 and
and Ci+1
Ci+l $ 6 ' 4 .
(2)
to D
D then
then D@
Do =
= C.
(2) If
If C
C is
is adjacent to
C.
1(Ci) - (Ci
} if Ci is in 6'4.
i' .
(3) Let Ci
= Ci
Ci isisnot
notinin6'4
eo and
andCfC =
= @-'(Ci)
Cf =
Ci if Ci
(Ci}
Then Ci
C; is
is adjacent
adjacenttotoC1+1.
C;+l.
eo
Proof. Part
W=
= CCflrl DDisisaawall.
0 fixes
Part(1)
(1)is
is clear.
clear. Assume W
wall.By
By 41.3,
41.3,4
fixes C and
and
Do =
=C
W.
Do, so,
W . Thus W =
=WO
W @c_
E D4,
so, as
as 6' is thin,
thin, D@
C or
or D.
D. As
As D
D isis not
not in
in
-00, Dip
= C,
6'4,
D@ =
C ,so
so (2)
(2)holds.
clear ifif neither
neitherCi
Ci nor
nor Ci+l
Ci+1isisinin6'4,
-00,so
solet
letCi
Cibe
beinin6'4.
eo. Let
PPart
art (3) isis clear
Let
U ==Ci
Ciflr l Ci+1,
Ci+l, vV ==O-1(U)
4-'(u)flr C,
l C;,and
andEEthe
thechamber
chamberthrough
throughVV distinct
distinct from
Ci'.Then
EO=Ci+1then
Cf. Then UU=VcpcEO,so
= V @E E 4 , so EO=C,or
E@ = Ci orCi+1.If
Ci+1. If E@
= Ci+1 then EE=C+1
= Cf+,
adjacent totoCf.
Cl.IfIfE@
EO== Ci
Ci then,
then,as
as4-'(ci)
0-1(Ci)=
= (Cf,
(C, Ci},
have EE =
= Ci.
is adjacent
C i } ,we have
Ci.
= VO
41.2.1, SO
so Cf
C ==Ci+1.
Ci+10 =
= Ci # Ci+1,
Hence VV =
V @=
=UUby
by41.2.1,
Ci+1.Therefore Ci+1@
Ci+17
Ci+1$!e6'4,
o, and
)' =
Ci+1=
=
so Ci+l
andhence
hence(Ci+i
(Ci+1)'
=Ci+i
Ci+1.Thus
ThusC'
C,!==Ci+1
Ci+1isisadjacent
adjacentto Ci+1
Ci+1
Cf+l.
(41.5) Let
0 and
Let C
C and
and C'
C' be
bedistinct
distinctadjacent
adjacentchambers
chambers and let 4
and O'
4' be foldings
C'O =
= C and Cep'
= C'.
C. Then
with C'4
C4' =
(1) If D is in
in 6'4
eo then
then d(C',
d(C1,D)
D) =
=d(C,
d(C,D)
D ) + 1.
1.
(2) If D is not
notinin6'4
eo then
thend(C1,
d(C', DD)) =
= d(C,
d(C, DD)) - 1.
1.
(3)
eo' _ ee -- eo.
(3) -e4l=
64.
+
Proof. As
toC',
C', Id(C,
Id(C,D)
D)-- d(C',
DE
P.
As C
C is adjacent to
d(C1,D)J
D)I <
_< 11 for each D
E 6.
212
The geometry of groups of Lie type
Assume D is in-00 and C' = Co, ... , C, = D is a galleryoflength d(C', D).
Then (Cio:0 < i < n) is a gallery between C = C'O and Do = D as 0 is
a morphism. By 41.4 there is 0 i < n with CiO = C;+1 = Ci+10. Hence
C = CO, ... , C;_10, Ci+10, ... , CnO = D is a gallery of length at most
n -1 from C to D. Therefore d(C, D) < n = d(C', D), so (1) follows from
the first paragraph of the proof.
Similarly if D 0 -00 let C = Co, ... , Cn = D be a gallery of length d(C, D).
Define C' as in 41.4.3. By 41.4.3, C' = Co, ... , Cn = D is a gallery, while
by 41.4.1 and 41.4.2 there is 0 < i < n with C; in ,°¢, C;+1 not in -6'0 and
C;+1O = C. Hence C= C,+1 = Ci+i, so, as in the preceding paragraph,
d(C', D) < n and then (2) holds.
Finally let D E '. If D is in 6' ', then, by (1) applied to 0' and 0, D is not
in e o. So io' c_ -C - e o. If D is not in 6' then, by (2) applied to 0 and 0',
D is in i/'. Thus i' - eo c io'. So (3) is established.
Foldings 0 and 0' are defined to be opposite if e o' = .6' - 60.
(41.6) Let 0 and 0' be opposite foldings and define a = a(O, 0'): r -* r by
va = v/ if v E ro' and va = vO' if v E ro. Then a is an automorphism of
- of order 2 whose orbits on -0 are the fibres of 0.
Proof. By hypothesis 6' = 6' U -00. By definition of chamber complex, each
member of r is a simplex, so r = ro U ro'. By 41.3.2, 0 and 0' agree on
r¢ fl ro', so a is well defined. By 41.3.1, a is bijective on '. If u and v are
incident members of r, then as i' is a chamber complex there is C in -' with
u and v in C. Then ua, va c Ca c e, so ua * va. Hence a is morphism. By
41.4.1 there exist adjacent chambers C and D with C in io and D not in eo.
By 41.4.2, DO = C and CO' = D. Then Ca 2 = Cq5'q5 = Do = C. Hence, by
a-1
= a is a morphism, so a is an automorphism of
41.2.2, a2 = 1. Therefore
order 2.
The element a(0, rp') of 41.6 is called a reflection and is said to be a reflection
through c fl C' for each pair C, C' of adjacent chambers with C E eo and
C' 0'q.
Fix a chamber C in i. is defined over some finite index set I ; let m = I
for the moment. Notice each of the m subsets of I of order m - 1 determines
a wall of C, and, as i' is thin, each wall X determines a unique chamber C'
with C' fl c = X. Let A(C) denote the set of these chambers; then A(C) is
the set of chambers distinct from C and adjacent to C in the chamber graph.
Further A(C) is of order m. As -' is a Coxeter complex, for each C' in A(C)
Complexes
213
there exist
4=
C' and
and C'O'
C'4' =
=C.
C.By
By 41.5,
41.5, 0
4
there
existfoldings
foldings40and
and4'0'ofoffi?6 with
with CCo
= C'
and 0'
4' are
areopposite,
opposite,so
soby
by41.6
41.6they
theydetermine
determinean
aninvolutory
involutory automorphism
automorphism
a(0,
a ( 4 ,0')
4 ' )of
of6,fi?,called
calleda areflection
reflection through
through C nnC.
C'.Let
LetSSdenote
denote the
the set
set of all
n C' as
reflections through C n
as C'
C' varies
varies over
over A(C),
A(C),and
andlet
let W
W be
be the
the subgroup
subgroup
of Aut(6)
Aut(-') generated
generatedby
by S.
S.ItIt will
will develop
developthat
that (W,
( W ,S)
S ) is a Coxeter system and
W
Aut(fi?).
W=
= Aut(o).
For W
W let
let l1(w)
bethe
the length
length of
of w
w with respect to the generating
w EE W
( w ) be
generating set
set S.
S.
(41.7) Let
Let ww =
= sl ...
(41.7)
. . .Sn
s, EE W
W with si EE S.
S. Then
(1)
n. ..... ..,Csl
(1) C,
C ,Csn,
Cs,, CSn-Is
CS,-~S,,
Csl...
. . sn
. s, isisaagallery
gallery from
from C
C to
to Cw.
Cw.
(2)
transitive on
on -0.
fi?.
(2)W
W isistransitive
(3)
1(w).
(3)d(C,
d ( C Cw)
, C w =) =
l(w).
S EE S.
Proof. Let
Lets
S .By
Bydefinition
definitionof
of SS there
there exists C' E A(C)
A ( C )and foldings 4,4'
through
c nf' C'
through C
C' with
with s ==a(0,
a ( 44'),
, 40,and,
and,by
by41.6,
41.6,(C,
( C ,C')
C')isisaacycle
cycle of s.
wi =
We've just
just seen C
C is adjacent
adjacent to
to CS,-,+~
Csn_i+l so
so Cw,
Cwi =
=
Let w,
=Sn-i+l
s,-i+l ...
. . .sn.
s,. We've
Csn-i+l wi -1 is
i _ 1 . Hence C
C,, C
Cw
Cwn =
gallery
is adjacent
adjacentto
to Cw
CwiVl.
w l1,, ....
. .,, Cw,
=Cw
Cwisisaagallery
from C to Cw of length
length n,
n, so
so d(C,
d(C, C
Cw)
< n. In particular d(C,
d(C, C
Cw)
w)5
w)<
5 1(w).
l(w).
Conversely
letCC=
= Co, C
C1,
gallery of
= m.
Conversely let
1 ,....
. . ,, Cbe
C,,, bea agallery
oflength
length d(C,
d ( C ,Cm)
C,) =
m. I'll
E SS,, 115<i i5< m,
m, such
=
show
show there
there exist
exist ri
r, E
such that
that Ck
Ck==CUk,
C u k , where
where Uk
uk=
rm_k+i
Noticethat
that this
this establishes
establishesthe
thetransitivity
transitivityofof W
W on
on fi?.
P. Prorm-k+l . ...
. . r,.rm.Notice
Proceed by induction on m; the case m
m=
= 11has
hasbeen
beenhandled
handledin
in the
the first
first paragraph
paragraph
of this proof,
proof, so take
take m
m > 1.
= Cuk fork
1.By induction Ck =
for k <<m.
m.Then
ThenCm
C, is
is
adjacent to Cm-1
=
Cum_1
so
Cm(um_1)-1
is
adjacent
to
C.
Hence
there
is
CmP1= CumPl C,(U,-~)-' is adjacent
Hence there is
r1
Cr, ==Cm(um_1)-1,
= Crium-1
rl E
E S with Crl
C,(U,-~)-', so Cum =
Crlu,,-1 ==Cm.
C,.
Finally let
let uu =
= r1
Cm== CCw.
Then Cu
Cu =
= Cm
= Cw so
Finally
rl ..... .rm
r, and assume
assume C,
w . Then
C, =
so
uw-1
by 41.2,
41.2, uu =
= w. So
= 1(u)
<m=
uw-' fixes
fixes C and
and hence, by
So 1(w)
l ( w )=
l(u) 5
=d(C,
d(C,Cw),
Cw),
completing
completing the
the proof.
proof.
(41.8) ((1)
).
(41.8)
1) W
W=
=Aut(6
Aut(8).
(2)
on fi?.
e.
(2) W
W is
is regular on
(3)
= II I11;
thatisisthere
thereexists
existsaaunique
unique reflection
reflection through each wall of C.
(3) II SISI =
I; that
C.
', so G =
= Aut(o)
By 41.7,
41.7, W
W isis transitive
transitive on 8,
Aut(&) ==WGc
WGc where
where Ge
Gc is
Proof. By
the stabilizer in
in G of
of C. But, by
by 41.2.2,
41.2.2, Gc
Gc =
=1,1,so
so (1)
(1)and
and (2)
(2)hold. For each
C' EE 0(C)
A(C)I =
A ( C )there
thereisisssEE SSwith
withcycle
cycle (C,
( C ,C'),
C'),so
so ISI
IS1 ?2IIA(C)l
=III.
111.Further
Further if
t is a member of
of SS with
with cycle
cycle((C,
C')then
thenstStEEWc
We== 1,
1, so
soss =
= t.
C , C')
(41.9) (W,
( W ,S)
S )isis aaCoxeter
Coxetersystem.
system.
214
geometry of
of groups of Lie type
The geometry
Proof. It suffices
conditionof
of 29.4.
29.4. So
So let
let w =
=
suffices to establish
establish the exchange
exchange condition
S1....
with l (w) =
=nn ands
l (w s) <
(w ).LetCk
Let Ck=
= C~n-k+l
CSn_k+l....
sl
. .sn
s, E W withl(w)
andsEESSwith
withl(ws)
5ll(w).
. .Sn
Sn
forl1 <k
for
5 k <n.By41.7,
5 n. By 41.7,
L
-C7 = (C = Co,C1,
C1,.... ..,Cn=Cw)
, Cn = Cw)
--9=(C=Co,
gallery and
andd(C,
d(C, Cw)
Cw) =
= l(w)
1(w)2> l(ws)
l(ws) =
= d(C, Cws). Let 0@ be
be aa folding
folding
is a gallery
with Cso =
= C.Nowd(Cs,
Cw),
so, by
41.5, Cw $
withCs@
C.Nowd(Cs,Cw)
Cw)==d(C,
d(C,Cws)
Cws)< 5d(C,
d(C,
Cw),so,
by41.5,Cw
Co. Then by
by41.4
41.4applied
appliedtotog
there existsi,i,115< i <
in Co, Ci+1
&@.Then
$? thereexists
5 nn with
with Ci in&@,
Ci+l not
Co, and
in &@,
and Ci+1/.
Ci+i@==Ci.
Ci.By
Bydefinition
definitionof
ofs,s,Cis
Cis==Ci+1,
Ci+l,SO
SO CSn-i+l
CS,-~+~
..... .SnS
S,S =
=
SnS =
= Sn_i
SO
Ci s =Ci+l
= Ci+1 =
= CS,-~.
Csn_i .... .sn,
and hence, by 41.8.2, Sn-i+1 . . S,S
Cis
sn,andhence,by41.8.2,sn-i+l..
s,-i ...
.. Sn,
. S,,SO
the exchange
exchange condition
condition is
is verified.
verified.
(41.10) Let (G,
(G, R)
R) be
beaaCoxeter
Coxetersystem,
system,R R==(ri:
(r1:
Gi== (ri:
(ri :j j#0 i),
i Ei EI),I),Gi
i), and
(41.10)
= (Gi:
(Gi :i iEEI).I).Then
Then&(G,
P(G,9J)) isis aaCoxeter
G=
=
F=
= F ( G , R)
R) =
Coxeter complex,
complex, G
Aut(&(G,1F,
F ))),
) , and R is
is the
the set
set of reflections
reflections through
through the
the walls
walls of
of the
the chamber
chamber
Aut(-'(G,
C=
= (Gi:i
(Gi: i E
E I).
I}.
C
the notation
notation of
of section
section3.3.By
By29.13.3,
29.13.3,Git
Gi,== (ri)
(r1)(recall
(recalli'=
i'=
Proof. Adopt
Adopt the
by 41.1.4,
41.1.4,L3_T==B(G,
-'(G,9J)) is thin.
thin.Also
AlsoGG== (R)
(R) =
=
II --{i})
{i})and
and GI =
=1.
1. So,
So, by
(Gi,:
By 41.1.2, G is
(Gi,:ii EE II),
) , so,
SO,by Exercise 14.3,
14.3, _T
L3 is a chamber complex. By
transitive
ong- and,
transitive on
and,by
by41.1.3,
41.1.3, GI
GIisisthe
thestabilizer
stabilizer of
of the
the chamber
chamber C, so, as
-9 so, by 441.1.1,
GI =
=1,1,GGisisregular
regular on
on -T.
L3. In
In particular
particular G is faithful on g
1.1.1, G
is a subgroup
Aut(_9). Also G is transitive on 5
_q
subgroup of Aut(g).
B so it remains only to show
for each wall Sit
Si, of
of C
C that there is
is a folding
folding @0 through Sit
Si, with
with C
C in
in g@.
-90.
-+ G by
Let r =
=riri and
and define
define 0:
@: G +
go = g if l(gr) = l(g) + 1
go = gr if 1(gr) = 1(g) - 1.
As G,
H Cg
GI ==11we
wecan
canidentify
identifyGGwith
with _T
L3 via gg H
Cgand
andregard
regard 0@as
asaafunction
function
_Tinto
intog.
-9.Claim
Claimthat
thatififD,
D,D'
D' EEg
-9 and
and D
Dn
fl D'
D' is a wall
wall of
of type
typej'j' then
from L3
Do
of type
typej'.j'. Now
D=
= Cg and D'
D' =
D@ fln D'o
Dl@isis also
also a wall of
Now D
=Crjg
Crjgfor
forsome
some
we must show Cg@
Cgo flflCC(r;g)@
(r j g)o is a wall
wallof
oftype
typej',
j', so it suffices
suffices to
to show
show
g E G,
G, so wemust
(rjg)4
=rj(g4)foreachg
E
G.Thisisclearifl(rjgr)-l(rjg)=l(gr)-1(g).
(rig)@ = rj(g@)for eachg E G. This is clear if 2(rjgr) - 2(rjg) = 2(gr) - 2 (g).
So, by
by symmetry
betweenrjg
rjg and
and g, we may
assume l(gr)
l(gr) =
= l(g) - 11 and
symmetry between
may assume
and
l(rjgr)=l(rjg)
go=gr
(rjg)o=rjg.
l(rjgr)
= l(rjg) + 1.
1. Thus
Thus g@
= gr and
and (rjg)@
= rig. Let
Let gg=sn...si
= s, . . .slwith
with
l(g) ==nnand
(gr) <5l l(g),
(g), the
sn ...
l(g)
andsk
sk ==rik.
ri,. As
Asll(gr)
theExchange
ExchangeCondition
Condition says sm
. . .s1=
sl =
without
loss
sii-1 ..... .sir
S, s,... .Sm+lSm-1
sir, so
sm-1
slrfor
forsome
somem.
m.Thus
Thusg g= =
. .sm+lsm-1.. . . slr,
so
r=
Next rrjgr
= rjs,
=si,
s1, Hence
Hence go
g@==gr
gr==Sn
s, ...
. . .S2.
s2. Next
j gr =
r j s, ...
. . .s2
s2 isis of
of length
length at
at
most nn =
= 1(g)
so l(r;g)
l(rjg) =
1<
1(g).
l(g) so
=l(rjgr)
l(rjgr)- 15
l(g).Then,
Then,by
bythe
theExchange
Exchange ConConmost
dition, r rjs,
j Sn .. .. .. sk+l
Sk+i =
= Sn
for some
some k.k. IfIf kk #0 1 we
may take
take s,sn==r;.r.
dition,
s, ...
. . .Sk
sk for
we may
+
215
215
Buildings
But then
then rjgr
r j gr=
= s,-1
sn_ 1. ....
length
n -22<<n-1
j g ). So
. s2s2isisofoflength
nn - =1 l=(rl(rjg).
Sokk =
= land
1 and
desired.
r j Sn. ....
Thus (rig)@
(r j g)o=
= rjg
r j g=
= s,
Sn....
S2=
= g4,
go, as desired.
rjs,
.s2S2==S,Sn. ..... s1Si== g.g. Thus
. . s2
So the claim is established.
established. By 29.13.3,
29.13.3,
Gkl ==(G{k,j}'
(G{k,jY:
k')
Gk'
ij EE k')
each kk EE I,
I, if III
themap
mapGkg
GkgH
H Gk(g0)
for each
IIl >>2.2.Hence
Hence by
by Exercise 14.8, the
Gk(g4)
morphismof
ofthe
thecomplex
complex(r(G,
(F(G,9().
as 4.
0.
is a morphism
) . -T).
g ) .II also
also write
write this map as
Next observe
observe that
thatfor
foreach
eachgg EE G,
G, g42
g02 =
= go.
g4.So
So04isisan
anidempotent
idempotentmormorphism. Also ifif Cg
Cg E
E g-90
0-1(Cg) ==(Cg,
4 then 4-'(Cg)
(Cg,Cgr}
Cgr]soso04isisaafolding.
folding.Finally
Finally
Crr are
through Sit
Si,and
andCCro
C and C
are the chambers through
r4 =
=CCso
so04isisaafolding
foldingthrough
through
Si'. By construction
0 on
Sij.
construction rr isisthe
thereflection
reflectionthrough
throughSip; that is the fibres of 4
on -9
g
of r on -9.
walls of
of C
are the orbits of
g.So
So RR isis the
the set
set of
of reflection
reflection through the walls
and hence
henceAut(g)
Aut(-T)== (R)
(R) =
= G.
and
$1;
(41.11)
-14-(G,
of
H'(G,
&(G,
F ( G , R))
R)) isis aa bijection
bijection between the set of
(41.11) The map (G, R) H
all Coxeter systems
systems and the set of all Coxeter complexes (up to
to isomorphism).
F4-(G,
( G , R)
R) is
is the
the family
familyof
of maximal
maximal parabolics
parabolicsdefined
defined in 41.12 and
and the inverse
correspondenceisis&6 i-+
of the correspondence
H (Aut(o),
(Aut(&),R)
R) where
where R
R is
is the
the set
set of
of reflections
reflections
through the walls
walls of
of some
somefixed
fixed chamber
chamberof
of -C.
8.
Proof.
41.10, (G,
(G, R)p
R)cp=
= 8-6(G,
Proof. IfIf(G,
(G,R)
R)isisaaCoxeter
Coxetersystem
systemthen by 41.10,
( G , F ( G , R))
while ifif &
6 is
is a Coxeter complex, while
is aa Coxeter
Coxeter complex then by 41.8 and 41.9,
6'* _=(Aut(o),
R)R)
is is
a Coxeter
Picp
&@
(Aut(&),
a Coxetersystem.
system.By
ByExercise
Exercise 14.2.3,
14.2.3, &
@ p=Z-6',
8 ,while
while
by 41.10,
41.10, (G, R)p@
R)cpi ==(G,
R),
so
cp
and
1
are
inverses
of
each
other
and
the
(G, R), so p
@ are inverses of each other
lemma holds.
42 Buildings
Buildings
building isis aathick
thickchamber
chambercomplex
complex930 =
_ (F,
together
A building
( r , ')&)
togetherwith
witha aset
setsad
d of
subcomplexes of
of 93,
0, called
subcomplexes
calledapartments,
apartments,such
suchthat
thatthe
thefollowing
followingaxioms
axioms are
are
satisfied:
1) The apartments
(B 1)
apartments are
are thin chamber
chamber complexes.
complexes.
of 93
0 isiscontained
contained in
in an
an apartment.
apartment.
(B2) Each
Each pair of chambers of
(B3) If A and B are simplices
simplicesof
of93
0 contained
(B3)
contained in apartments
apartments EE and E', then
then
isomorphism of E with
there exists
exists an isomorphism
with E'
E' which
whichisisthe
theidentity
identityon
onAAUU B.
B.
In sections 13 and 22 a geometry r was
was associated
associated to
to each
each classical
classical group
group
G (occasionally
(occasionally subject
subject to restrictions
restrictions on
on the
the field),
field), and
and itit was
was shown
shown that
that GG
transitive on r.
F. Now B
-6'(F)
complex and in
in Exercise
Exercise
is flag transitive
( r ) is a thick chamber complex
14.5 a set of apartments
14.5
apartments sa
disis defined
defined which
which admits
admits the
the transitive
transitive action
action of
of
G, and
isisaabuilding
(&(r), s1)
d)
building admitting
admitting G as
as aa group
group of
of
and it is
is shown
shown that (6'(F),
automorphisms.
representation is then used to study G in section 43.
automorphisms. This representation
216
The geometry of groups of
of Lie type
type
But before all that let's take a closer look at buildings. So for the rest of this
section
letB
A=
= (F,
building with
with apartment
apartmentset
setsf.
a.
section let
(r,6')') be a building
(42.1) Each
Each pair
pair of
of apartments
apartments is
is isomorphic.
Proof. Let
1, 2,
bebe
apartments
E. .By
Let E
X ii, =i =
1,2,
apartmentsand
andpick
pickchambers
chambers Ci
Ci E C,
By axiom
axiom
B2, there
C containing
containing C1
C1and
and C2,
C2,and by axiom
axiom B3, E1
C 1%
there is an apartment E
E=E2.
C
% C2.
Given aa pair
pair ((E,
C)) with E
Given
C ,C
C an
an apartment
apartment and C a chamber in E
C define
define a map
p ==p(E,
C):
F
E
by
vp
=
vo,
where
C,
v
are
contained
in
some
p(C, C ) :r 4 C by up = v 4 , where C , v are contained in someapartment
apartment
E'
C' and
and 0:
4:E'-->
C' +ECisisan
anisomorphism
isomorphismwhich
which is
is the identity on C. There are a
number of
of points
points to
to be
be made
made here.
here. First
First by
by B2
B2 there
there does
exist an
an
number
does indeed
indeed exist
apartment C'
E' containing
apartment
containing C and v, and by B3 the map 0
4 exists.
exists. Moreover
Moreover if
contained in
in an
anapartment
apartment00and
and$:
1/f:
vv,, C are contained
0 04- ECisisan
an isomorphism
isomorphism trivial on
there exists
exists an
anisomorphism
isomorphismaa:: C'
E' + 0 trivial on C U
C then by B2 there
U {v).
{ v ) .Then
= 0,
(0-1)uVf =
= ,B
Aut(E) is trivial
= 11.. Thus
Thus a$
a =
(4-')a$
,!? EE Aut(C)
trivial on
on C,
C ,so,
so, by
by 41.2.2,
41.2.2, ,B
,!? =
4,
sov4
v4==va$
vat == vv$.. What all this
this shows
showsisisthat
thatp(C,
p(E, C)
so
C )is independent of the
of C'
E' and 4,
0, so in particular p(C,
p(E, C)
choice of
C )isiswell
welldefined.
defined. As
As a matter of fact
C trivial on C.
C . Let's see
see
it shows there
there exists
existsaaunique
uniqueisomorphism
isomorphism4:
0:C'E'+
-> E
next that:
(42.2) pp =
(42.2)
=p(E,
p(C ,C)
C )isisaamorphism
morphism of B onto
(c)==C.C .Moreover
Moreover
onto C
E with
with p-'
p-1(C)
if E'
E' + E
C' isis an
an apartment
apartment containing C then p: C'
C is
is the
the unique isomorphism
of C'
E' with
trivial on
on C
E..
of
with E
C trivial
trivial on C.
C. In particular
particular p is trivial
I've already
already made
made the last observation
observation of
I've
of 42.2.
42.2. The
The remaining
remaining parts
parts of
of the
consequence of
of this
this observation
observation and
and the
the building
building axioms.
axioms.
lemma are an easy consequence
Given simplices
simplices A
A and
and B of .M
define dd(A,
B)) to be the minimal integer n
93 define
( A ,B
n )of
of A
Bwith
withAAcontained
contained in
in
such that
that there
thereexists
existsaagallery
gallery97=
= (Ci:
(Ci: 00 5
< i <5 n)
and BB contained
containedin
in C,.
Cn.Then
Then9
C is
is said
said to
tobe
be aagallery
gallerybetween
betweenAAand
andBBof
of
Co and
length n.
n.
(Rainy'Day Lemma)
apartment, C
chamber of
C , and
and
(42.3) (Rainy'Day
Lemma) Let
Let C
E be an apartment,
C a chamber
of E,
simplex of
contains every
X a simplex
of C
E.. Then
Then C
E contains
every gallery
galleryofofB
A between
between XX and
and C
C of
of
length dd(X,
( X , C).
C).
Proof. Let
Let 9
(C,:0 <5i i<5n)n )bebesuch
suchaagallery
gallery and assume
not conconProof.
9 ==(Ce:0
assume9
' isisnot
tained
tained in
in C
E.. Then
an i with
in E
Then there
there exists
exists an
with C;
C, not
not contained
contained in
C and
and Ci+i
C,+1
contained in
in C.
E. Then w
contained
W==CiC,f1nCZ+i
C,+' is
is aa wall
wall of
of C,+1
Ci+' so there is a chamber
T 0# CZ+1
withWWEc TT.. Let
Let pp =
= p(E,
C,+, ofofCEwith
p(C,T).
T ) .As
As ppisisaamorphism
morphism trivial
trivial
I
Buildings
217
217
on E,
so, as
as C
E is
=T
T or
C ,Cip
Cip is
is aa chamber
chamber of E
C containing
containing W,
W, so,
is thin,
thin, C;
Cipp =
or
C,+1.As
As pp-1(T)
C,p =
= C,+1.
But 9qpp isisaagallery
Ci+1.
-l(~=
)= T,
T , Cip
Ci+1. But
gallery between
between X =
=Xp
Xp
and Cp =
=C
p ==Ci+1
o, p ,--and
C of
of length
length n ==d(X,
d ( X ,C),
C ) ,so,
so, as
as C,
Cip
Ci+1==Ci+lp,
c i + l c~COP,
...,,
C1_1p,p, Ci+1
Ci+l p,
p, . ...,
gallery between
between XX and
and CC of
oflength
lengthnn-- 1,
contraCi-1
. , C,,p is a gallery
1, contradicting
= d(X,
d ( X ,C).
C).
dicting nn =
(42.4) Let C and D be
(42.4)
be chambers,
chambers, X
X a subset of C,
C , and E
C an
an apartment
apartment containing
Let pp =
= p(E,
p(C,C)
C and
) andq9a agallery
galleryofoflength
length d(X,
d ( X ,D)
D)between
between X and
taining C. Let
A..Then
lengthdd(Xp,
Dp)
D in B
Then dd(Xp,
( X p ,Dp)
Dp) ==d(X,
d ( X D)
, D )and
andqp
9pisisaagallery
gallery of length
( X p , Dp)
between
= Xp
between XX =
Xp and
and Dp.
Proof. By
By B2,
in an
an apartment
apartmentC'
E' and,
and, by
by 42.3,
42.3, 9
C is
ProoJ
B2, C U D is
is contained
contained in
contained
in C'.
V. By 42.2,
42.2, p:
p: C'
E' -+
EC
is is
ananisomorphism
contained in
isomorphismsosothe
thelemma
lemma holds.
(42.5) Each apartment
apartment is
is aa Coxeter
Coxeter complex.
complex.
Proof. Let
ProoJ
Let CCand
andC'
C'be
beadjacent
adjacentchambers
chambersin
in an
an apartment
apartment E.
C.We
We must
must show
show
there exist
exist opposite
oppositefoldings
foldings40and
and4'0'ofofCEthrough
throughBB=
= C f1
withC'4
C'O =
=C
n C'C' with
and CO'
C4' ==C'.
C'.For
Forthe
thefirst
firsttime
time the
the hypothesis
hypothesis that A
Bisisthick
thick isis used.
used. Namely
as 2B
is thick
thick there
there is
is a chamber
chamber C*
C* distinct
distinct from
from CC and
and C'
C' through
B is
through B. Let
E11be
be an
an apartment
apartmentcontaining
containingCCand
andC*,
C*,plp,==p(C1,
p(E1, CC),
P2== p(C,
p(E, C'),
C
) ,p:!
C'),and
and
40 ==pipe:
pl p2: E
C + E.
C . As
As the
the composition of
of morphisms trivial on C,
C ,04 isis aa mormorphism trivial on C.
C . Moreover
Moreover applying
applying 42.4
42.4 to
to p,
pl and
andp2
pz we
weobtain:
obtain:
(42.5.1)
(42.5.1) If 99isisaagallery
gallery of
of length dd(B,
( B , T)
T )==nnbetween
between BB and
and aa chamber
chamber T
of E,
C ,then
then d(B,
d ( B ,TO)
T 4 )==nnand
and9O
9 4isisa agallery
galleryofoflength
lengthnnbetween
between BB and
andTO.
T4.
Next C'4
C'O is
is aachamber
chambercontaining
containingBB,
C't =
= CCor
E + E1
, sosoC'4
or C'.
C'.As
As pl :: C
C 1is an
isomorphism trivial
trivial on
onCC,, C1pl
C'p1 =
= C*.
Then C'4
C'O =
= C*
p2 #
0 C'
C*.Then
C*p2
C' as
as (p2)-1(C')
(,3:!)-' (c')
=
= C.
C'. So C'q5
C14 =
C.
Similarly
thereisisaamorphism
morphism4'0'ofof CE trivial
trivialon
onC',
C', with
with C4'
CO' =
= C',
Similarly there
C', and
and
satisfying 42.5.1.
Let
in E
Let D be
be aa chamber
chamber in
C and
and (C,:0
(Ci:0 <5 ii <5n)
n) ==C9a agallery
galleryofoflength
length
= CC or
dd(B,
( B ,D)
D ) in
in EC from
from BB to
to D.
D. Notice
Notice Co =
or C'.
C'. Claim
Claim
(42.5.2)
=C
(42.5.2) If Co =
C then
then
(i) 04and
and 0'0
4'4are
aretrivial
trivial on
on D,
D, and
and
(ii) D4'
Do' #
0 D.
(ii)
D.
Assume otherwise and choose acounterexample
a counterexamplewithn
with nminimal.
minimal.As
AsC4'
CO'== C',
C',
C'O
follows that
thatnn >
> 0. Now
Now (Ci:
(Ci:005< ii <
< n) is
C'4 =
=C,
C ,and
and 04isistrivial
trivial on C,
C , it follows
218
218
geometry of
of groups
groups of Lie
Lie type
The geometry
a gallery
gallery from
from BB to
to CnV1
Cii_1=-=EEofoflength
lengthd(B,
d(B,EE)
- 1,1,so
) == nn so (i)
(i) and
and (ii) hold
E by
byminimality
minimalityofofn.n.Let
Let
Do,Dq5
Do =
= D or E.
for E
AA
==
DDi l flE.E.AsAs
AA
==AqAq
5 ccDq5,
By 42.5.1,
42.5.1,nn=
= d(B,
d(B, Dq5),
DO), so,
so,as
asd(B,
d(B,EE)
= nn - 1,1, itit follows
that Dq5
DO =
= D.
By
)=
follows that
D.
Similarly, as
asq50 and
and 4'
0' both satisfy 42.5.1,
42.5.1,so
sodoes
does4'4,
0'0, and then
then as
as4'4
0'0 is trivial
shows Dq5'q5
DO'O=
=D
D..
on E the same argument shows
is established.
established.Notice
Noticeg'-9'== Hq5'
6/'0' isis gallery
galleryof
oflength
lengthnn=
= d(B,
d(B, Dq5')
Do')
So (i) is
with Coq5'
Coo' =
= CO'
= C',
C and C', 0'
with
Cq5' =
C',so,
so,by
by symmetry
symmetry between
between C
4' is
is trivial
trivial on
on
Dq5'. Thus, if if
D D=Do',A=Aq'cDflEq',so
= Dd', A = Aq5' c D n Eq5', so Eq5'
= D since Eq5'
# E.
Do'.Thus,
Eq'=Dsince
Eq'#E.
while d(B,
d(B, D)
D ) ==n.n.This
This
This isisimpossible
impossibleasasn n--11==d(B,
d(B, EE)) =
= d(B, Eq5')
E/') while
completes
completes the proof of 42.5.2.
42.5.2.
Now,
by 42.5.2,
42.5.2,either
eitherCo
Co==CCand
andDD=
= Dq5,
Do, or
Now, by
or Co
Co =
=C',
C',in
in which
which case
case
Coo =
= C and,
42.5.2 to
to gq5,
qq5, we
we get
get (Dq5)q5
(DO)o =
= Do.
Coq5
and, applying
applying 42.5.2
Dq5. In
In either
either case
case
Dq5
Dq52, SO
idempotent. Also
Dq5 then
then by
by 42.5.2,
42.5.2, Co
Co #
Dq ==Dq52,
so q50 is idempotent.
Alsoifif DD =
= Do
# C',
so Co
Co== CC,, and
andthen
thenagain
againbyby42.5.2,
42.5.2,
D Dq5'
Do' and
D=
= Dq'q.
if
so
D#
and D
Dq5'4.Moreover
Moreover if
T##DDisisaa chamber
withTq5
To==DDthen
thenT T#0Tq5,
To,sosoTT==Tq5'
To'==Too'.
T
chamber ofofCEwith
Tq5q5'.
=Do'
Dq5'and
andsoso{D,
{ D Do'}
, D@'}isisthe
thefibre
fibreof
of D
D under
under 0.
q5. Hence
Hence 0q5 is a folding
Thus T =
C through
through B with CO
C'q5 ==C,
C ,and
and0'q5'isisthe
theopposite
oppositefolding.
folding. The
The proof
proof of
of E
42.5 is complete.
complete.
43 BN-pairs
BN-pairs and
and Tits systems
A Tits
systemisisaa quadruple
quadruple(G,
(G,B,
B,N,
N,S)
S) such
suchthat
that G
G is
is a group,
group, B and N
Tits system
are subgroups of
of G, S is a finite collection
collection of
of cosets
cosetsof
of BB ifll N in N, and
and the
the
following axioms
axioms are
are satisfied:
satisfied:
(BNl)
(BN1) GG=(B,N)andH=BIN<N.
=(B,N)andH=BnNaN.
W =N/H
=N/Hisisgenerated
generatedby
byS,S,and
andthe
themembers
membersof
ofSSare
areinvolutions
involutions in
in
(BN2) W
W.
W.
(BN3) For
Foreach
eachssEESSand
andwwEEW,W,sBw
sBwc_EBwB
BwBUU BswB.
BswB.
(BN4)
each ss EE S, B #
0 BS.
(BN4) For each
BS.
By
B =
BW
Bn,for
fornnaarepresentative
representative of the coset
By convention,
convention,wwB
= nB and B'
==B'1,
coset
w
W. As
and H <5B,B,nB
nBand
andB"
Bnare
areindependent
independent of the
w=
= Hn EE W.
As H <I N and
the
choice of coset representative n, so the notation is well defined. Axioms BN3
choice
and BN4 should be read subject
subject to this convention.
B and N are also said to
to be
be aa (B,
(B, N)-pair
N) pair for G.
(43.1) Let
(r,
a buildingwith
withapartment
apartmentset
set.1
dand
andassume
assume G
G is
is a
Let 93
3 ==(I',
t)B)
be be
a building
automorphismsof
of -M
93 transitive
transitive on
on
group of automorphisms
cZ={(C,E):CE
6,EC EE d,
W,CCE}.
S2
= { ( C ,C ) :C E 8,
C C C}.
.
BN-pairs and
andTits
Titssystems
systems
BN-pairs
219
Let (C, C)
E) E 0,
52,BB ==Gc,
Gc,and
andN
N ==NG(E).
NG(C).Then
Then
The representation
representation of
Z maps
maps N
N surjectively
surjectively onto Aut(Z)
(1) The
of N on E
Aut(E) with
kernel
H=
=BB fli lN.
N.
kernel H
(W,S)S)isisaaCoxeter
Coxetersystem,
system,where
whereW
W==N/H
N/Hand
andSSisisthe
theset
set of reflections
(2) (W,
in W
W through
throughthe
the walls
walls of
of C.
C.
in
(G,B,
B,N,
N,S)S)isisaaTits
Titssystem.
system.
(3) (G,
Proof.
Proof.By
By42.5,
42.5,ECisisa aCoxeter
Coxetercomplex,
complex,while
whileby
by hypothesis
hypothesis N
N isistransitive
transitive
on
(2)follows
followsfrom
from(1)
(1) and
and
on the
the chambers
chambersof
of E,
C,so
so(1)
(1)follows
followsfrom
from41.8.
41.8.Then
Then(2)
41.9.
41.9.
Lets
adjacent and
andEE =
=C
and w
w EE W. Then C and Cs are adjacent
C fli lCs
Csisisaawall
wall
LetsEESSand
of
B fixes
fixes E,
E, so
so EE E
C_Csb
Csband
andEw
Ew CCsbw.
LetL+96'=
= (Ci: 00 5
<
of C.
C. Each
Each b E B
Csbw. Let
i <5n)n)bebea agallery
galleryofoflength
length d(C,
d ( C ,Ew)
Ew) from
from C to
to Ew,
Ew, and
and E'
C' an
anapartapartment containing C and Csbw.
Csbw.By
By the
theRainy
Rainy Day
Day Lemma,
Lemma, 42.3, L+9 cEE
Z fl
i l V.
C'.
Let a E
G with
with (C,
(C, Z')a
E')a =
= (C, C).
E). Then
= C,
B. Ci
EG
Then Ca =
C, so
so aa E
E B.
C1 and
and C are
are
the
of C
E or E'
Z' containing
containing C
C fl
n CI
C1 so,
so, as
as aa fixes
fixes C,
C ,ititalso
also
the only chambers
chambers of
fixes C1. Proceeding by
by induction
induction on
on k,
k, aa fixes Ck
Ckfor
foreach
each00 5
< k <5 n.
n. Thus
Thusaa
C,. But
fixes Ew
Ew CC,,.
But Ew
Ew C Csbw
CsbwE
C C'.
V. So
SoEw
Ew==Ewa
Ewa CCsbwa
Csbwa CC'a
E'a =
= E,
C,
so Csbwa
BwB or
Csbwa ==Cw
CwororCsw.
Csw.Hence
Hencesbw
sbweEBWB
or BswB,
BswB, so BN3
BN3 is
is estabestablished.
lished.
thick,there
thereexists
existsaachamber
chamber D
D through
through E distinct
distinct from C and Cs.
As 393isisthick,
Let
an apartment
apartmentcontaining
containingCCand
andD,
D,and
andlet
letg gEeGGwith
with(C,
(C,C)g
E)g =
= (C,
(C, 0).
8).
Let 98 be an
Then g E
EB
B and
and C
C and
and Csg
Csg are the
the chambers
chambersthrough
throughEEininCg
Eg =
= 8,
0, so Csg =
=
Cs. Thus sgs
sgs $ B, so BN4 is established
established and the proof is complete.
complete.
D#
# Cs.
c
c
It will develop
develop later
later in
in this section
section that
that the
the converse
converse of 43.1 also
also holds;
holds; that
that is
is
It
each Tits
Tits system
system defines
defines aa building.
building.
each
Notice
Notice that
that by
by 43.1
43.1 and
and Exercise
Exercise14.5
14.5the
the classical
classicalgroups
groupspossess
possessaaBN-pair.
BN-pair.
The
The BN-pair
BN-pair structure
structurewill
will be used to establish
establish various
various facts
facts about
about the classical
classical
groups.
groups.
In the remainder of this section assume (G, B, N, S) is a Tits system and
let H
andW
W==N/H.
NIH.We
Wesay
sayBBisisthe
theBorel
Bore1subgroup
subgroup of G, H isis
H ==BBfli lNNand
the Cartan
of G. Let 1l be the length
Cartan subgroup
subgroupof
of G,
G,and
andW
Wisisthe
theWeyl
Weyl group of
function on
on W
Wdefined
defined by
by the
the generating
generating set
set S.
S.
function
(43.2) IfIf uu,, w
w Ee W with
with BwB
BwB =
= BuB then uu =
= w.
(43.2)
Proof. Let m
and induct
induct on
on m.
m. IfIf mm =
= 0 then
then w =
= 1,
m=
=1(w)
l(w) <
5 1(u)
l(u) and
1, so
so uu eE
B fl
M. So
ilN =
=H;
H ;that
thatisisuu==1 1(remember
(rememberthe
theconvention
convention on W).
So take
take m >>0.
0.
with ss EE SS and
=m
w =sv with
and 1(v)
l(v) =
m --1.1.Now
Now svB
svB c BuB
BuBby
byhypothesis,
hypothesis,
Then w
so vB
vB E
c sBuB
by BN3.
BN3. Then
ThenBvB
BvB =
= BuB or
sBuB c2BuB
BuBUU BsuB
BsuB by
or BsuB,
BsuB, so,
so, by
by
220
The geometry of
of groups of Lie
Lie type
induction onm,
v ) ==mm-- 11 andm
induction
on m,vv== uu orsu.
or su.As
Asl (1(v)
and m 5< l(u),
1(u),vv#$ u, so vv =
= su
and hence
hence ww =
= sv ==uuasasdesired.
desired.
(43.3) Let
LetW
w EW
Wand
S.
and sS E S.
(1)
( 1 ) IfIf l(sw)
l(sw)>21(w)
l(w)then
then sBw
sBw C2 BswB.
BswB.
(2)
(2) IfIf l(sw)
l(sw)<5l(w)
l(w)then
thensBw
sBwflf'BwB
BwB isis nonempty.
nonempty.
Proof.
induction on m =
= 1(w).
m=
= 00
Prooj First
Firstthe
theproof
proofof
of(1),
( I ) , which
whichisis again
again by induction
l(w).If m
then w
w=
= 11and
trivial,so
sotake
takemm>>0.0.Then
Thenww== ur,
ur, with
withrr E S and
and the result isistrivial,
1(u)== mm-- 1.
If l(su)
l(su) < m - 11 then
l(u)
1. If
then l(sw)
l ( s w )=
= l(sur)
l(sur)<5 l(su)
l(su)+ 11 <<m,
m, conconhypothesis,so
soZ(su)
l(su) > m --11==1(u).
l(u).Then,
Then,by
byinduction
inductionon
onm,
m,sBu
sBu C
trary to hypothesis,
BsuB. Hence
HencesBw
sBw =
= sBur
BsuB.
sBur C2BsuBr
BsuBrC_
2 BsuB
BsuB U
U BsurB
BsurB by BN3. Also, by
BN3, sBw
sBw C
BswB.IfIf BwB
BwB==BsuB
BsuBthen,
then,by
by43.2,
43.2,ww=
= su.
su. Hence
BN3,
2 BwB
BwB U BswB.
u=
= sw
sw so
so m
m --1 1==1(u)
l(u=
) =l(sw)
l(sw)>21(w)
l ( w )==in,
m,a acontradiction.
contradiction.Similarly
Similarly if
BwB
= BsurB
BsurB then
then ww =
= Sur
BwB =
sur =
=sw,
s w ,aacontradiction.
contradiction. It
It follows
follows that sBw
sBw C5
BswB, as desired.
desired.
and it's
it's on to
( 1 ) is established
established and
to (2).
(2).By
By BN3
BN3 and
and BN4,
BN4, BS
BS fl
f' BsB
BsB is
So (1)
nonempty, so
so sBsu
sBsunflBsBu
BsBuisisnonempty
nonemptyfor
foreach
eachu uEEW.
W.Take
Takeuu== sw.
sw. Then
nonempty,
l(su) =
l(su)
=1(w)
l ( w )>
2 l(sw)
l ( s w )==1(u),
l(u),so,
so,by
by (1),
( I ) , sBu
sBuC2BsuB
BsuB==BwB.
BwB. Therefore,
Therefore, as
as
sBsu n
fl BsBu
BsBu is
is nonempty
nonempty and
and w
w=
= su,
su,(2)
(2)isisestablished.
established.
+
s
(43.4) Let
Let w
w=
= si
(43.4)
sl ...
. . .Sn
s, EE W
W with
with Si
si E S. Then
then siB
siB 2
C (B,
BwBfor
for115< ii 5
< n.
((1)
1 ) If n ==1(w)
l ( w ) then
( B , B"-')) D
2 BwB
(2) For each u E W, BwBuB
(2)
Bw BuB c5UiEO
Ui,, Bsi,
Bsil ...
. . .SiruB,
siruB,where A consists of
the sequences i ==i ill,, ...
. . ., ,i i,t with
with ii j EE {( 11 , ..... .,,nn}j and il
< .. . <
< i,..
i,.
i t < i2 <
(3) If
(sow)5<l1(w)
then there exists
< kk 5
< n with
If So
so E SS with 1l(s0w)
( w ) ==nn then
exists 11 5
with
B'_1
SOS1
Sk-1 =
= S1
S0Sl . ....
. Sk-1
S1 ....
. .Sk.
Sk.
Proof. Part
Part ((2)
from BN3
BN3 by
by induction
on n. Next
Prooj
2 ) follows
follows from
induction on
Next part
part (1).
Let
=si ...
s1w for
< i <5 n,
Let wi
wi=si
.. .slw
for 115
n ,nn == 1(w),
l ( w ) , and
and wo
wo =
=w.
w. By
By hypothesis
hypothesis
l(wi+i)
l(wi+l)<<1(wi)
l(wi)so,
so, by
by 43.3.2, si+iBwi
si+lBwiflf' BwiB
BwiBisisnonempty.
nonempty. Hence
si+,B C BwiBwi-1B C (B, B"''.
-1
inductionon
oni,i,sjsjEE( (B,
B") ==XX for
<X
X..
By induction
B , B"-')
foreach
each jj<5i,i ,so
sosi+i
si+l cE(B,
( B ,B-1_')
B W ~) 5
w=
= sl ...
Hence w
. . .sns, cEXX and
andthen
thenBwB
Bw BC5X.
X .So
So(1)
( 1 )holds.
holds.
l ( w ) then, by 43.3.2, soB
w ~ w - ' Bso, by (2),
(2),
Similarly if l(s0w)
l(sow) 5
< 1(w)
SOB2CBBwBw-1B
BsOB
= si,
Sil....sir
BsoB == ~Bxw-1B
x w - for
' ~ xX =
. .sir and some
some i cE A.
A. Hence,
Hence, by 43.2, so =
=
xw-1. But
But ll(xw-1)
= 1,
= n --11and
xw-'.
(xwP1>
) l(w-1)
l(w-') --1(x)
l ( x )>
2 n --r,r ,so,
so,as
as l(so)
l(so)=
1 , rr =
and
15
< k <5 n.
x ==S1
sl .... .Sk-1Sk+1
.sk-lsk+l .. .. ..Sn
S , for some 1
n. Therefore
Therefore (3)
(3)holds.
(43.5) (W,
(W, S)
S ) is
is aa Coxeter
Coxeter system.
system.
.
BN-pairs
BN-pairs and
and Tits
Tits systems
systems
221
Proof. This
Thisisisimmediate
immediatefrom
from 43.4.3
43.4.3 and
and 29.4.
29.4.
(43.6) SS is
is the
the set
set of
of w
w EE W*
W' such
suchthat
that BBUU Bw
Bw B is
is aa group.
group.
(43.6)
Proof. By
B isisaagroup
Proof.
ByBN3,
BN3,BBUUBs
BsB
groupfor
forsS E S.
S . Conversely
Conversely if w ==s1
sl ..... .S"
snE W
with
Si EE Sand
S and nn =
= l(w)
withsi
l(w)>>00then,
then,by
by43.4.1,
43.4.1,SisiEE (B,
(B, w),
w),so
so ifif B
BU
U BwB
BwB isis aa
then, by
by 43.2,
43.2, sisi =
= w for each i.
group then,
Let SS =
= (Si:
andfor
forJJ 5C II let Pj
The
of
(si: ii EEII)
) and
P j ==(B,
(B,sj:
sj:jjEEJ).
J).
Theconjugates
conjugates of
the subgroups
subgroups PPj,
of G. Recall
Recall Wj
Wj =
_
j , JJ C5I,I ,are
arecalled
calledparabolic
parabolic subgroups
subgroups of
(sj:
(sj: j EE J)J )isisaaparabolic
parabolicof
of W.
W.
(43.7) (1)
(1) Pi
P j==BWjB.
BWjB.InInparticular
particular G
G ==BNB.
BNB.
(43.7)
a bijection
HPjPis
j is
a bijectionofofthe
thepower
powerset
set of
of IIwith
with the
the set
set of all
(2) The map JJ H
of G containing B.
subgroups of
(3)
PJUK=
(PJ,PK)PK).
(3) PJUK
= (Pr,
(4)
PJnK
=
Pj
n
PK.
(4) P J ~ K
= PJ f' PK
B9 (
< PJ
NG(Pj) =
= Pi
(5) If
If g EE G
G with Bg
P j then
then g EE PJ.
Pj.In
In particular NG(PJ)
P jisisconconjugate in G to PK only
onlyififJJ =
= K.
K.
(6)
= w n Pj.
(6) wj
Wr=WnPj.
to show
show uBw
uBw2CBWjB
BWjBfor
foreach
eachuu,
Proof. To prove (1) it suffices
suffices to
, ww EE Wi.
Wj.
Proceeding by
by induction
inductionon
onl(u)
l(u)ititsuffices
sufficestotoshow
showthis
this
sj,jj EE J.
J. But
forfor
uu
==sj,
this is just BN3. So
So (1) is established.
established.
LetB <I X
= UYEYByBforsomeY C W.Lety ==rr1
... r nEE
X <I.G.By(1),X
G.By(l),X=Uy,yByBforsomeYSW.Lety
l ...
Y with
withriri EE SS and
andl(y)
1(y)==n.n.By
By43.4,
43.4,ririEE(B,
(B,BY-')
By-) (
< XX.. Let
Let J be
of
Y
be the set of
j EE IIsuch
that
sj
=
ri
for
some
y
E
Y
and
some
such
expression
for
y.
I've
such that s j = ri for some y E Y and some such expression for y. I've
shown PPjj <I.X,
definitioncertainly
certainlyXXI.
<PPj.
ThusXX =
= PJ.
shown
X , while
while by definition
j . Thus
Pj.
Suppose P
Pij ==PK;
remainsto
toshow
showKK=
= JJ..
PK;totocomplete
completethe
the proof of (2) itit remains
By (1)and43.2,
(1) and 43.2,Wj
Wj=
=W
W nnPP,J =
Hence, by 29.13.4,JJ=
=K
K..
=W
Wnn PK
P K ==WK.
WK.Hence,by29.13.4,
So (2) and (6) are established.
established.
Evidently
Evidently the
the bijection
bijection of
of (1)
(1) preserves
preserves inclusion,
inclusion,and
and hence
hence also
also least
least upper
upper
bounds and greatest lower
lower bounds. Hence (3) and (4) hold.
Let Bg
B95< pJ.
Pi.g-1
and,byby43.4.1,
43.4.1,w
(B,fIW-')
B") =
=
g-l E
E BwBforsome
Bw B for some wE
w E W,
W, and,
w E c(B,
(B, Bg)
B9) I< Pj.
Pi. So
(B,g)
g)=
= (B,
(B, Ww)
<.Pj.
Pi.
SOgg EE (B,
) I
Exercises 4.9
4.9 and
and 7.8
7.8 show
show that, ifif G is
Exercises
is aa classical
classical group
group over
over aa finite
finite field
field
of characteristic
characteristic p,
p, then
then the
the Cartan
Cartan group
group H
H is a Hall p'-group
p'-group of the Borel
Bore1
of
possesses a normal p-complement
p-complement UUwith
p (G). Further
Further
group B and B possesses
withUUEESyl
Syl,(G).
(U), so
containing the normalizer
normalizer
B ==NG
NG(U),
sothe
the parabolics
parabolics are
are the subgroups of G containing
222
geometry of
of groups of Lie type
The geometry
of a Sylow p-group of G.
G. Indeed
Indeed itit can
can be
be shown
shown that
that the
the maximal
maximal parabolics
parabolics
P;',
Pit, i EE I,I ,are
arethe
themaximal
maximalsubgroups
subgroups of
of G
G containing
containing the Sylow p-group U,
U,
every p-local
p-local is contained in some maximal parabolic (cf. section
and further every
47).
47).
Somemore
morenotation.
notation.For
For
I let
(recall
I that
J' =I I-- JJ))
Some
i Ei IElet
GiG;
==
Pi,Pi,
(recall
forfor
J cJ ICthat
J' =
and let F =
_ (G1:
(Gi: i EE I)
I )be
bethe
theset
setof
of maximal
maximal parabolics containing B. Form
the geometry rF(G,
thesubgeometry
subgeometryZE== {Giw:
{G;w:ii EE II,, w E W},
( G , _1F),
F ) , the
W), and the
complex
A ==-'(G,
OT).
Let
chamber
E I);
1);then
then & =
=
complex G8
&(G,
9).
LetCCbebethethe
chamber{G1:
{Gi:ii E
G} is
is the
the set
set of
of chambers
chambersofof93,
3, and
{Cg:gg EE G)
and we
we can
can also
also regard E
Z as
as the
the
complex
(E, & fl
= {Eg:
a isisa acollection
complex (Z,
n E).
Z). Let szl
d=
{Zg:g EE G},
G), so
so that d
collection of
of
subcomplexes
A. Write
Write B
3 ((G,
subcomplexes ofof93.
G , B,
B, N,
N, S)
S) for
for the
the pair
pair (A,
( a ,sa?).
d ) . Finally
Finally let
Ul =
= W',
6'8==(U1:
-'(W, d').
Ui
W~J,
(Ui :i EE I)I )and
andform
form the
the complex &(W,
8).
UU
isomorphism
complex
(43.8) (1)
(1) The
Themap
mapG1
Giw i-±
H Ui
ww
is is
anan
isomorphism
of of
thethe
complex
(Z(E,
, Z Enme)
8)
with '(W, 8
e).) .
with&(W,
(2) G
automorphismsof
of%
3 by
(2)
G is
is represented
represented as a group of automorphisms
by right
right multiplimultiplication;
cation; the kernel
kernel of
of this
this representation
representationisiskerB
kerB(G).
(G).
(3) A(G,
B ( G ,B,
B,N,
N,S)S)isisaabuilding.
building.
(4) G is transitive
transitiveon
onC2S2=={(D,
{(D,0):9):DDE E',
(4)
&, 90 E
E s1,
d , DC
D 29}
0).
(5) B ==Gc
are
Gcand
and(BE)N
(Bc)N==NE(E)
NG(Z)
arethe
thestabilizers
stabilizers in
in G
G of
of C
C and
and E,
Z,
respectively.
and H =
=NE.
Nc.
respectively.N'NE==Aut(Z)
Aut(E)S- W and
Proof. By 29.13.3,
29.13.3, WJnK
WJnK==Wj
win
forJ,J,KK CC I.
I. Thus UJ
Uj ==Wj,,
Proof.
n WK
WK for
Wj!, where
where
Uj
=
n
Ui.
In
particular,
as
(W,
S)
is
a
Coxeter
system,
UI
=
WO
UJ = j EJ Uj. In particular, (W, S) a Coxeter system, Ul = WD =
=11
Up =
= W,
= (si)
-'(W, 1)
and Uit
Wi =
(q)is
is of
of order
order 2, so &(W,
8 )isisaathin
thinchamber
chambercomplex
complex by
Exercise 14.3.
41.1.4 and Exercise
14.3.
43.7.6 says
Gi w H
i-+ U;
of (1)
(1)isiswell
well defined,
defined, after
after which
which
Next 43.7.6
says the
the map
map 7r:
n : Giw
Uiwwof
7r
is
evidently
a
bijection
and
7r-1
is
a
morphism
of
geometries.
So
to
prove
(1)
n is evidently a bijection and n-' is a morphism of geometries. So
nonempty then
too. But
But
it remains to show
show that
that ifif Gkw
GkwflnG1
G j v is nonempty
then Ukw
Ukw flnU1
Uj vv isis too.
Gku
nonempty if and
Gku fl
nGj vv isis nonempty
and only
only ifif 11EE Gkuv-1Gj,
G ~ u v - ' Ginin
~which
,whichcase
caseUkw
UkwflnUU
Ujvv
which is
is an
an easy consequence of
of
is nonempty by the following observation, which
43.2 and 43.4.2:
njEJ
andeach
eachwwE E
(GJwGK)
w==UjUJWUK.
(43.9) For each J,
J , KK CEI Iand
(43.9)
W,W,
(Gj
WGK) flnW
WUK.
Hence
Hence (1)
(1) holds
holds and therefore
therefore by the
the first
first paragraph of
of this
this proof,
proof, -04
93 satisfies
satisfies
B1.
B1.
Notice
14.3 and 43.7
43.7 show
show -04
93 is
Notice (2) is immediate from 41.1.1. Exercise 14.3
93 is thick, by 41.1.4 and 43.7 we
we must
must show
show
a chamber complex. To show -04
JP,: BI >2foreachi
> 2 for each i cE I.I.But,by43.6,
But, by 43.6, Pi
U Bs, B so I P, : B I ==22i fifaand
IPi:BI
Pi==BBUBsiBsoIPi:BI
nd
only if si
NG(B). Thus BN4 says 93
si E NG(B).
-04 is thick.
.
BN-pairs and
and Tits
Tits systems
systems
223
For xx E
For
E G and J C
cI IletletCj,x
Cj,, =={Gjx:
{Gjx:j jEEJ}J be
) bethe
thesimplex
simplexof
of type J in
in
Cx.
Cx. Let
Let Cj,x
Cj,, and
andCK,y
C K ,be
~ simplices.
simplices. Then there exist aa,, b E B and n E N with
xy-1 ==anb.
Eby, so
xy-'
anb.Now
NowCK,y
C K =, =
~CK,by
CK,byand
andCj,x
C j ,==CJ,Rby
Cj,nbyare in Cby,
so 93 satisfies
satisfies
B2.
Suppose Ci,x,
CK,y
E nfl Cg.
Eg. Then we can choose x, y E
Cj,,, C
K ,C~c C
E N and there exist n, m E
EN
N with Cj,,
Cj,x =
= CJ,ng
andCCK,y
h =mgy-'
mgy-1E EGK.
G.
Cj,,, and
K ,=
~=CCK,mg.
K , ~Notice
~Notice
. h=
nm-1and
and v
v=
= xy-1.
= ((Gj)ngy-1
Define
xy-'. Then (Gj)uh =
~ j ) n ~ y=
-=' ((Gj)xy-1=
G ~ ) X ~=
-'
Define uu =
= nm-'
(Gj)v, so uu cE(Gj)v(GK).
So,
by 43.9,
43.9, uu =
= rvs for
(G~)u(GK).
SO, by
for some
some r E
E Gi
G j nn N
N and
and
E GGKfN.
Setl
shy. ThenGjxl =
= Gjvyl
Gj vyl=
= Gjvshy
Gj vshy== GJr-'uhy
Gtr-1uhy =
=
sE
KnN.S
etl=
=y-1
y-'shy.ThenGJxl
Gang
= Gjx,
= GKhy=
=GKy,
GK y,and
andClEl==Cyl=
Eyl =Cshy
Eshy =
=
GJng =
Gjx,and
and GKyl
GKyl=
=GKShy
GKshy=
Ehy ==Emg
So ElCl= =
Eg,Cg,(CJ,,)l
(Cj,x)l = Cj,x,
and (CK,y)l ==CK,y.
Chy
C m g==Eg.
Cg.So
=CJ,,,and(C~,,)1
C ~ , ~That
. T h is
atis
l1induces an isomorphism
isomorphism of E
Therefore
3
C with
with Eg
Cgtrivial
trivialon
onCj,x
Cj,, and
and CK,y.
C K , ~Therefore
.
93
satisfies
satisfies B3. This completes
completes the
the proof
proof of
of (3).
(3).
), (1) says N is
6(W, 6'
&),
is transitive
transitive on
As W
W is transitive on the chambers
chambers of -6(W,
-'i?fln E.
holds. Also
AlsoNG(C)
NG(E) =
=
f
C.So,
So,asasGGisistransitive
transitive on
on c1
dby
by construction,
construction, (4) holds.
NNB(E)
is thin, NB(C)
NB(E) =
NNB(C)as
asNB(E)
NB(C)isisthe
the stabilizer
stabilizerof C in NG(E).
NG(C).As 6'
&is
=BE
BE
by 41.2, so (5)
(5) holds.
(43.10)
(43.10) Let G*
G* ==G/kerB(G).
G/kerB(G).Then
Then(G*,
(G*,B*,
B*,N*,
N*,S*)
S*)isisaaTits
Tits system.
system.
Proof. This
This is
is clear.
clear.
kerB (G)=
= 1. By 43.8 this has
Because of 43.10 it does little harm to assume kerB(G)
the effect of insuring that G is
is faithful
faithful on
on its building.
Recall that
Coxeter
S)is
is irreducible
irreducibleifif the
the graph
graph of its Coxeter
Recall
that the
the Coxeter
Coxeter system
system(W,
(W, S)
diagram is connected.
(43.11)
is an irreducible
irreducible Coxeter
Coxeter system
systemand
andkerB(G)
kerB(G)=
= 1.
1.
(43.11) Assume (W, S) is
Then
X<
gG
h e n G ==XXB,
B , aand
nd
(1) IfI fX
Gtthen
(2) ifif GGisisperfect
perfect and
and BBisissolvable
solvablethen
then G
G is
is simple.
simple.
XB<< G
G so,
so, by
by43.7,
43.7,XB
XB==PPjj for
forsome
someJJ E
C I.
I. Let
Proof. Let X <
9 G. Then XB
Let
Jo =
= {i
B) nfl Xx :#Q}.
thensisi Ec Bsi
BsiBBEc XB
XB =
=P
Pjj so
Jo
{i E
E I:
I:(Bsi
(Bsi B)
4).IfIf ii EE JO
JO then
so i E
E J
43.7.6. Conversely,
Conversely,asasPPjj =
= XB, X
X intersects
intersects each
eachcoset
cosetof
of BB in
in PPjj nonby 43.7.6.
nontrivially
so JJ CJO.
trivially so
c Jo. Thus J ==JO.
Jo.
Claim
(W, S)
S)isisirreducible
irreduciblethere
there
exists
I -J Jand
andj j EE JJ
Claim JJ ==I.I.IfIfnot
not as
as (W,
exists
i Ei IEsjsi) >>l l(sjsi)
(sjsi) >>l (si
),
[si,sj] # 1. Therefore, as
as (si,
(si,sj)
sj)isisdihedral,
dihedral,l (si
l(sisjsi)
l(si),
with [si,
so, by 43.3.1, Bsi
Bsi Bsi B=
= BsisjsiB.
Bsi sjsi B.As
AsXX < G
G and
and JJ ==JO
BsiBsjBsiB
Jo itit follows
follows that
that
si
sjsi EE W
W nfl PJ
Pj ==Wj.
sjsi EEWj
}=
= Wj
W3=
= (sj),
sisjsi
Wj.But
Butthen
thensi sisjsi
WjflnW{i,J
Wgi,jJ
(sj),contracontradicting
[si,,sj]
sj]#: 1.
dicting [si
1.
224
224
The geometry of
of groups
groups of
of Lie
Lie type
type
Thus II =
PIPI= G.
Hence
(1)(1)isisestablished
=J,J,so
soXB
XB =
=Pj
PJ==
= G.
Hence
establishedand
andof
of course
course
(1)
(1) implies
implies (2).
(2).
(43.12) Let F
F be
be aa field
field and 11 <<nnand
andinteger.
integer. Then
Then
(1) SLn(F)
and L,(F)
Ln(F) is simple unless
unless (n,
(n, IIFI)
_ (2,
2) or
(1)
SL,(F) is quasisimple
quasisimple and
FI) =
(2,2)
or
(2,
3).
(2,3).
(2) Spn(F)
PSpn(F) is simple unless
unless (n, IIFI)
_ (2,
2),
Sp,(F) isis quasisimple
quasisimple and PSp,(F)
FI) =
(2,2),
(2,
3),
or
(4,
2).
(2,3), or (4,2).
(3) If FFisisfinite
(3)
finitethen
then SUn(F)
SU,(F) isisquasisimple
quasisimpleand
and UU(F)
U,(F) is
is simple
simple unless
unless
(n, IFI)
4), (2,
9) or
4).
IF11==(2,
(2,419
(279)
or(3,
(3,4>.
finiteor
or algebraically
algebraicallyclosed,
closed, and n >>6,6,then
thenStn
Qi(F)
(4) IfIf FFisisfinite
(F) is quasisimPQi(F) is simple.
simple.
ple and PS2n(F)
Proof. Let G ==SLn(F),
(F), ininthe
Proof.
SL,(F), Spn(F),
Sp,(F), SUn(F),
SU,(F), or
or01
Qi(F),
therespective
respective case.
possesses aa BN-pair
BN-pair (B,
(B, N).
N).
By Exercise 14.5, parts 1, 2, and 6, and 43.1, G possesses
Observe next that B is solvable.
solvable. This follows from Exercise 4.9 and Exercise
Indeed these
these exercises
exercises show
show B
B is the semidirect product
product of
of a nilpotent
7.8. Indeed
group by
by a solvable
solvablegroup.
group.InInapplying
applyingExercise
Exercise7.8
7.8observe
observen n--2m
2m 5< 2 as
Z1
thus O(Z',
O(Z1, f)
2' has
has no
no singular points; thus
f ) isissolvable
solvableby
by Exercise
Exercise 7.2.
by Exercise 14.5.5, kkerB(G)
er~(G=
)= Z(G). Further,
Further, except in the
the special
special
Next, by
cases listed in the lemma, G is
perfect.
This
follows
from
13.8,
22.3.4,
22.4,
is perfect.
follows
13.8,
and Exercise 7.6.
7.6. Finally,
Finally,to
tocomplete
completethe
theproof,
proof,43.1
43.11.2
saysG/Z(G)
G/Z(G) is sim1.2 says
ple, since by
by Exercise 14.5.3
14.5.3 the
the Coxeter
Coxeter system
system of
of the
the Weyl
Weyl group
group of
of G
G is
ple,
irreducible.
Remarks.
Remarks. The
Thematerial
materialininchapter
chapter14
14comes
comesfrom
from Tits
Tits [Ti] and Bourbaki [Bo].
Already in this chapter we begin to see the power of the Tits system building
approach to
to the
the study
study of
of groups
groups of
of Lie
Lie type. The
The proof
proof of
of the simplicity of
approach
various classical groups in 43.12 probably provides the best example.
example. Further
results are established in section 47, where groups with aa BN-pair
BN-pair generated
weak version
version of
of the
the Chevalley
Chevalleycommutator
commutator
by root subgroups and satisfying a weak
relations are investigated. These extra axioms facilitate the proof of a number
interesting results.
of interesting
Tits classifies
classifies all buildings
buildings of rank
rank at least 3 with
with a finite
finite Weyl
Weyl group,
group,
In [Ti], Tits
and hence also all groups with a Tits system of rank at least 33 and
and finite
finite Weyl
building is 1
I1
I Iwhile
whilethe
therank
rankofofaTits
a Titssystem
system(G,
(G,B,
B,N,
N, S)
S)
group. The rank of a building
1
is ISI.
IS/.
for chapter
chapter 14
Exercises for
cif
==(A,
letletE C
(sxl)
of
1. If s
( A a)
, d is) aiscomplex
a complex
( d )be
bethe
thegeometry
geometry whose
whose objects of
are the simplices
simplices of sa(
d of
simplices
type i are
of type
type ii and whose edges are the simplices
of
sl of
let B6(I')
-'(I'))
of d
of rank
rank 2.
2. If Fr isis aa geometry
geometry let
( r ) be
be the
the complex
complex ((I',
r , &(r))
BN-pairs and
and Tits
Tits systems
225
of rF of type I.
I. Prove
where -6(I')
& ( r ) is
is the set of flags of
Prove
(1) The
of C
E(sa')
Theinclusion
inclusionmap
map 7r
n is an injective morphism of
( d ) into
into A;
A;7r
n is
an isomorphism ifif and
and only
only ifif each
each rank
rank 22 flag
flag of
of A is a simplex of
of
sxl;7r
an isomorphism
isomorphism of
of-6(E(sy1))
d,
n isis an
& ( C ( d ) ) with d i iff and only
only if every
every flag
flag
of d.
c(.
of A is a simplex of
(2) 7r
of C(&(r))
E(6(I')) with
n is an isomorphism
isomorphism of
with Fr ififand
andonly
only ifif every
every rank
rank 2
of r
r' is contained
contained in a flag of type II..
flag of
(3) For
each simplexTTo of
{C-- T
C EE ds-I] and
=
f da.
l elet
t d.(T
, == {C
T::TT5< C
and ET
CT =
Foreachsimplex
E(AT,
C(AT,aT).
d T )Prove
. Provethe
themap
map7rT:
n T : ET
CT-+
+AT
ATisisan
anisomorphism
isomorphismof
of AT
AT
1 ifif and
with ET
CT for each simplex T of
of ,d
and only
only if
if every
every flag of A is aa
ofd.
s..
simplex of
(4) Aut(F)
= Aut(6(F)).
Aut(r) =
Aut(&(r)).
2. Let
1, let
let W
W=
Let i6
B ==(I',( r-6)
, &)bebea aCoxeter
Coxetercomplex
complex over
over I =
=(1,
(1,...
. . ., ,nn],
=
Aut(6), CC in
in B6,, and
for JJ E
C II let
Aut(8),
and for
let J' =
=II--J Jand
andTjTjthe
thesubflag
subflag of
of
C of type J.
ri be
thethe
reflection
through
J.For
For iic EI let
I let
r, be
reflection
throughthe
thewall
wallTip,
T,I, let
let
R=
= {ri,
and R
Rjj =
{ r l ,...
. . .,, r,], and
=(ri:
(r,:ii EE J').
J']. Prove
Prove
(1) IfIfJJCcI then
(ET,
Coxeter complex and
= Aut(ET,,
I then
(Cz, -6T
, g zf)) isis a Coxetercomplex
and WT,
WE =
Aut(CTI,6Tf )
(in the notation of Exercise 14.1).
14.1).
(2) WT,
= (Rj)
then
WE =
( R j )and
and ifif II##J J
then(WT,,
(WE,Rj)
Rj)isisaaCoxeter
Coxetersystem.
system.
= 9,-(W,
R) =
= (Wi:
1, Wi
= WT,
(3) Let
Let Ti
T, ==(vi
(v,],
W, =
Wz, and
and 9 =
9 ( W , R)
(W,: ii E
E I).
I). Prove
Prove
the map
viwI-Wiw iEI,wEW
of 8
-6 with
with86(W,
is an isomorphism of
(W, 9 ) .
andF
97=
3. Let
LetG
G be
be a group and
=(Gi
(Gi:: i cEI)I )a afamily
familyof
ofsubgroups
subgroupsof
of G.
G. Prove the
is
a
chamber
complex
if
and
only
if
G
=
(G1':i iEE II),
complex &(G,
6(G, 9 )
complex
chamber complex if and only if G = (G,!:
),
wherei'i' =
= II --{i{i1.
ProveB(G,
6(G, 9)
)==
6(I'(G,
only
where
1. Prove
B ( r ( 3T))
G , 9 )if) and
if and
onlyififGGisis flag
flag
trasitive on
on rI'(G,
trasitive
( G , 99-).
).
4. Let
= (F,
complex and
and W
W=
= Aut(8).
Aut(6). Prove
Let -6'
&=
( r ,-6)
&) be
be aa Coxeter complex
Prove
(1) Let
LetAAand
andBBbe
besimplices.
simplices.Prove
ProveAAUUBBisisaasimplex
simplexifif and
andonly
onlyifif AAU B
contained in
in r4
F0 or r4'
Fo' for
0,, 0'
is contained
for each
each pair 4
4 ' of
of opposite
opposite foldings.
bebesimplices
simplicessuch
suchthat
thatAi
A,UU Aj
A, isisaasimplex
simplex for
for
(2) Let Ai,
A,, 11 <5 i i<53,3,
each i, j.
j.Then
ThenU3=1 Ai is a simplex.
(3) Every
Every flag
flag of r'r isisaasimplex.
simplex.
r.
(4) W
Wisisflag
flag transitive
transitive on F.
(5) Prove
of Exercise
Exercise 14.1)
14.1)that
that(rfi,
(IT,,B6T)
Prove (in the notation
notation of
E ) is
is aa Coxeter
Coxeter
complexfor
foreach
eachJJ c
C I.
I.
complex
(6) W
W=
=Aut(I').
Aut(r).
5. Let
LetFFbe
beaafield,
field,VVaafinite
finitedimensional
dimensionalvector
vector space over F,
F,and
and assume
assume
one of the
the following
following holds:
holds:
(A) Fr isisthe
Y=
= {(xi):
theprojective
projective geometry
geometry on V, Y
{(xi):11 5< i <
5 n)
n] for
for some
some
basis X
X=
= (xi:
<:i <5m)
(xi: 1 I
m) of
of V,
V, m =
=dim(V),
dim(V), and G =
=GL(V).
GL(V).
basis
uLl
geometry of
of groups
groups of Lie
Lie type
The geometry
226
226
(C) (V,
unitary space
spaceand
andGG== O(V,
O(V,ff)) or (V, Q)
(V, f)
f ) isisaasymplectic
symplectic or unitary
is an orthogonal
space and
and G =
= O(V,
orthogonal space
O(V, Q). If
If (V,
(V, Q)
Q) isis orthogonal
orthogonal
assume it is not hyperbolic. 0 <<m
misisthe
theWitt
Witt index
index of
of the
the space
spaceand
and
X=
= (xi: I1 <5 ii<52m)
rr its
its polar
polar geometry.
geometry. X
2m)isisaahyperbolic
hyperbolic basis
basis for
a maximal hyperbolic
hyperbolic subspace
subspaceof
of V
Vand
andYY=
x EE X}.
=((x):
((x):x
X).
(D) (V,
orthogonal space
space and
and G
G is
(D)
(V, Q)
Q) is
is aa2m-dimensional
2m-dimensional hyperbolic orthogonal
the subgroup
subgroup of O(V, Q)
Q) preserving
preserving the
the equivalence
equivalence relation
relation of
of 22.8,
22.8,
geometryofof(V,
(V,Q),
Q),XX=
= (xi:
(xi: 115< ii 5
< 2m) is aa
rF is the oriflamme
oriflamme geometry
hyperbolic
: xxEEXXI.
}.
hyperbolic basis of V,
V, and
and YY ==((x)
((x):
each case
caselet
let6%A =
= 8-'(F)
In each
( r )_=(F,
( r ,e)
8 be
) bethe
thecomplex
complex on
on rrdefined
defined
in Exercise 14.1, let Ey
Cy be the subgeometry consisting
consisting of the objects
of rrgenerated
generatedby
by members
members of
of Y,
Y, identify Ey
Xy with
with the
the subcomplex
subcomplex
(Ey,
(Cy,-6
6 fl
ilEy),
Xy),and
and let
let .d' =
=((Ey)g:
((Cy)g:gg EE G).
G).Prove
Prove
(1) (A
isisa abuilding.
( B a)
d)
building.
((C, X):
E): C
C EE -6,
E EE sz'}.
(2) G
G is
is transitive
transitive on SZ
!2 == ((C,
6, Z
dl.
(3) The
The Weyl
Weyl group of A
Bisisofoftype
typeA,,,,
A,, C,,,,
C,, orDm
orD, in case A, C, and D,
respectively.
respectively.
(4) Let C be a chamber
chamberin
in Xy,
Ey, B
B=
= Gc, and
(4)
and N ==NG(Ey).
NG(Xy).Prove
Prove G is
semidirectproduct
productofofaanilpotent
nilpotentgroup
groupUUbybyHH== BB ifll N,
N, and
and H
H
the semidirect
is abelian if
if A or D holds, or if (V, ff)) isissymplectic
if
symplectic or unitary;
unitary, or if
F
F isis finite
finite or
or algebraically
algebraicallyclosed.
closed.
(5) kerB(G)
kerB(G)==kerH(G)
kerH(G)isisthe
thegroup
groupof
of scalar
scalartransformations
transformationsof V
V in G.
G.
(6) SL(V)
SL(V)and
andSp(V)
Sp(V)are
aretransitive
transitiveon
on Q
!2and
and if FFisis finite
finite or
or algebraically
algebraically
Q) and SU(V) are transitive
Q(V, Q)
transitive on Q.
Q.
closed then S2(V,
6. Let
(G,
B,
N,
S)
be
a
Tits
system
with
Weyl
group
andSS =
=
Let
system with Weyl group WW
==NN/H
/ H and
(Si:
I). Prove
(si : ii EE I).
(1) For
J, K C
For each
each J,
EI,I ,the
themap
map
(PJ)w(PK)
H) (WJ)w(WK)
( P J ) W ( ~ KH
(WJ)W(WK)
orbits of the parabolic
parabolic WK
is a bijection of the set of orbits
WK on the coset space
space
B/PJ with
B/PJ
withthe
theorbits
orbitsofofthe
theparabolic
parabolicWK
WK on the coset space W/WJ.
W/ Wj.
(2) Assume
Assume the
the Coxeter
Coxeter diagram A of W
W is one of the following:
1
2
An
o--o
Cn
o- o
1
Dn
2
...
n-1 n
p_...p
n-2 n-1 n
p------ cz::=
BN-pairs
BN-pairs and
andTits
Tits systems
systems
227
Prove:
Prove:
(a) IfIfAAisisof
G/PI,, rank
oftype
typeAn
A, then G is 2-transitive on G/Pl!,
rank 33 on
on G/P2,
G/P21
if n >>2,2,and
andrank
rank66on
onP{t,n}'
Ptl,ny
ifif nn > 1.
1.
(c) IfIf AA isisof
of type
type Cn
C, and nn >
> 1,
1,then
then G
G is
is rank n + 11on G/Pn,,
G/ P,!, rank
rank
3 on G/P1!,
G/P1,, and
and rank
rank66on
onG/P2!
G/P2 if
if nn >
> 2.
2.
Dn, nn 2
> 4,
on G/Pl1
G/Pl, and
(d) If
If A
A is
is of
of type D,,
4, then G is rank 3 on
and rank 8
or 7 on G/
G/P2,
P2t for n =
=44 or
or nn >>4,4,respectively.
respectively.
+
7. (1)
(1) Every
Everyflag
flagin
inaabuilding
building is
is aa simplex.
simplex.
(2) IfIf(G,
(G,B,
B ,N,
N,S)S)isisaaTits
Titssystem
systemthen
thenGGisisflag
flag transitive
transitive on its building
9 (G,
98
(G, B,
B , N,
N, S).
S).
(3) If
If A
98==(F,
( r ,e)8 is
) isa abuilding
buildingthen
then Aut(A)
Aut(98) =
=Aut(F).
Aut(r).
F.
complex and
andfor
forxxEErFlet
let8,
ex =
= B ifll r,.
8. Let
Let (I',
(r,-')
8 )be
beaachamber
chamber complex
(1) IfIfI1
I1II >
> 22 assume
assumefor
for each x EE Fr that
that (Fx,
(r, , -ex)
8 , ) is
is a connected complex.
Assume@:0:8
-.' H
H -'Bisisaafunction
that, for
forall
allC,
C,DD EE8
i' and
Assume
function such that,
and all
isisalso
iE
E I,
I , ifif CflDis
C n D isawall
a wallofoftype
typei'then
i' thenCO
C@flDo
i l D@
also awall
a wall of
of
F define xcr
xa to
i'. For xX EE r
tobe
be the
theelement
elementof
of CO
C@of
of the
the same
same type
type
type P.
a: (F,
8.Prove
Prove cr:
( r , 8')) -->
+ (F,
( r , 8 ')) is
is aa well-defined
well-defined
as x, where
where xx E C EE -e.
morphism.
(2) Assume
F(G,
and -'8==
C(G,
-12-)
for
Assume III
1
1I >>2,2,Fr==
r ( G-12-),
,9
),
C(G,
9)
for some
some group
group
family9
9 ==(G;:
G and some family
(Gi:iiEEI)I )ofofsubgroups.
subgroups. Assume
Assume for each
jjEEi').
i E II that
thatG;,
Git==(Gy,j),:
(G{i,,)~:
it). Prove
Prove(I'x,
(r,,-fix)
8 , ) is aaconnected
connected
complex.
complex.
9. Let
Tits system
Weylgroup
groupW
W== N/H.
N/H.
LetTT ==(G,
(G,B,BN,
, N,S)S)bebea aTits
systemwith
with finite Weyl
38_ =
1-213
(T)bebethe
thebuilding
buildingofofT,
T,CC the
the chamber
chamberof
of 98
A fixed
Let 9
98 (T)
fixed by B,
B,
xxEEC,
of 98
A stabilized
and E
X _=(xw:
(XW:
C,W
w EE W) the apartment of
stabilized by N (cf.
(cf. the
for W and n
it aa simple
discussion before 43.8). Let @ be a root system for
simple
system for
for W
W with
with SS =
= (r,,:
a EE 7r)
Let 5
< be the
system
(r,: cr
n )(cf.
(cf. 29.12
29.12 and 30.1).
30.1). Let
partial order on W
W defined in Exercise 10.6. Prove
(1) If
gallery(Ci:
(C,:00I
<i i I
< n) of
of
If u,
u, w
w EE W
W with
with u <5 w
w then
then there exists aagallery
= Cu.
length n =
=1(w)
l(w) from C to Cw with
with C/(u)
C1(,) =
(2) Ifs
If sEESSand
andWwEEW
Wwith
withl(sw)
l(sw)<<1(w)
l(w)then
thenB Bfl iBS"
l B SBW
W BC
W
CBSw.
BSW.
(3)
Ifss ES
andwwEW
withl(w)
1(w)
<l(ws)then
(3) If
E S and
E W with
5 l(ws)
then BSBSflB'<BandB"'fl
i l BW i B and Bw i l
BSB
B
S B C B.
B.
(4)
(4) IIff uu,wE
, w E Wwith
W w i t hu<wthen
u i w t h e nBflBW
B n B W<Bu.
(BU.
(5) Define
Define T to be saturated if H=
H =nwEw Bw.
B wProve
. Provethe
thefollowing
following are
are
nw,,
equivalent:
(a) T is
is saturated.
saturated.
(b) H is
in G of X.
E.
is the pointwise stabilizer in
= H,
(c) B fl
n Bw0
BWO =
H, where
where wo
wo is
is the
the element
element of
of W
W of
of maximal
maximal length
length
10.5).
in the alphabet S (cf. Exercise 10.5).
(6) Let f, = nwEw Bw,
S EE S).
B ~N
IV,==FIN,
A N ,and
andSS==(Ns:
(IVs:s
s). Prove
Prove (G,
(G, B,
B,
N, S) issaasaturated
saturatedTits
Tits system.
system.
228
The geometry of groups of Lie type
(Hints: Use 41.7
observe
fl n
BSS
BW C
41.7 in
in(1).
(1).To
Toprove
prove(2)
(2)
observeB B
BSWBW
E NB
NB
fl C
Cw)
(Csw n
w ) and use
use the
the Rainy
Rainy Day
Day Lemma
Lemma and
and (1)
(1)to
to prove
prove NB
NB
Cw)
Use((2)
to prove
prove ((3)
and (4) and use (4)
(Csw fl
nC
w )<
5 BSW.
BSW.Use
2 ) to
3 ) and
(4)and
and
Exercise
Exercise 10.6
10.6 to
to prove
prove (5).)
(3.)
10. (Richen
[Ri])
Assume
the
hypothesis
14.9 and
and assume
assume further
further
(Richen [Ri])
hypothesis of Exercise 14.9
that T is saturated. Let
Let aa E
jr and s =
=
that
En
=ra
r, EES.
S. For
For W
wE
EW
W define
define B,"
Bw =
Bn
fl BBw0w-'
, wherewo
w0isisthe
themaximal
maximalelement
elementofofW.
W.For
For B,8 EE n
jr define
B
WOW-',
where
define
Bg
B,, .. Prove
Bp =
= Br,
(1)
aw >> 00 then
(1) If
If w E
EW
W with
with aw
then Ba
B, <
5 Bw
BW-I. '
(2)
EW
W with
withaaw
Ba n
fl BBw-'
= H.
(2) If
If W
w E
w < 0 then
then B,
W-I
=
H.
(3)
EW
W with
withl1(w)
l(ws)then
thenBB== ((B
B').
(3) If
If W
w E
( w ) 5<l(ws)
B flflBB'')(B
S ) ( BflflBW).
(4)
B
=
Ba(B
f1
BS)
with
BS
f1
Ba
=
H.
(4) B = B,(B n B" with BW B, = H .
(5)
H..
(5) Ba
B, # H
(6)
Ba 5
< Bw
(6) For
For w E W,
W, B,
B ~ '- ifif
' and
and only ifif aaw
w >
> 0.
0.
(7) Let A
7r,wwEEW).
W).Prove
Provethethemap
map(B,)W
(Ba)'HH aaw
A =={(Ba)w:
{(B,)": aa EE n,
w is a
permutation equivalence
equivalence of
of the
the representations
representations of
of W
W on
well-defined permutation
A and (D.
In particular
particularwe
wemay
maydefine
defineB,By== (Ba)W
(Ba)' for
@. In
for each
each yy EE(D,
@,
where
= aw.
where yy =
aw .
(8)
0 then
B",Bw
= Ba(B,S,,,)S
= H.
(8) IfI f ww EEW
Wwith
withaw
a w< <
0 then
= B,(Bsw)Sand
andBa
B, fln(B,r",)S
(Bsw)S=
H.
(9) Let w0
= rl
r1 ....
r with
wo =
. .r,
withnn==l(wo)
l(wo)_=I(D+I
I @+ 1 and rl
ri =
=ra,,
raL,a;
ai EE jr.
n. Let
Let
wi
= riri_1 ...
... ri.
BaiBa2wt
... B , , , ,,-, with Ba,w,-,fl
wi =riri-1
rl. Prove
ProveBB= =
Ba1Ba,,, ...
= H.
n;=,+,
BaJwJ-,
=
H . Further
Further the map i H
H aiwi_1
ai w,-1 isis aabijection
bijection of
w
j i+1 Ba
( 1 , . . . , nj
n) with
with 4)+.
@+.
In ((1)
use Exercise
Exercise 10.6.4
10.6.4to
toshow
showww-1 <
< wos
(Hints: In
1 ) use
wos and then appeal
appeal
to Exercise 14.9.4.
14.9.4. In (2)
(2) use parts (2)
(2)and
and (5)
( 5 )of
of Exercise
Exercise 14.9
14.9 to
to
conclude BBw-'S
BS~_(< B
B~and B fl
= H.
~ nfl-BS
n Bw0
BWO=
H . In
In (3)
( 3 )use
use 43.3.1
43.3.1 to
to
get B
and then appeal to Exercise 14.9.3. Use (4)
B C BSBw
BSBWand
(4)and BN4
to prove (5).
,8 E n,
Jr, (Ba)W
(Ba)w =
= Bp
(5).In (7)
(7)use (6)
(6)to show for B
Bg ifif and
and only
only
if aw
a w ==P.
B.InIn(8)
(8)use
use(1)
( 1 )totoshow
showBa
B, <5Bw.
B,. Then
Then use
use (4)
( 4 )to
to show
show
Bw =
= Ba(B
< B.
Bw
B,(B flflBS)
B s )and
and use
use Exercise
Exercise 14.9.3
14.9.3 to show Bsw
Biw 5
B. Use
Use (8)
(8)
to prove (9).)
(9).)
-'
15
Signalizerfunctors
Signalizer
functors
Let r be
of G. An
Let
be aa prime,
prime, G
G aafinite
finite group,
group, and
and A
A an
an abelian
abelian r-subgroup
r-subgroup of
A-signalizer
functor
on
G
is
a
map
0
from
A#
into
the
set
of
A-invariant
A-signalizer functor on G a map 8 from A' into the set of A-invariant
r'-subgroups of G
and 8(a)
0(a) fl
rl-subgroups
G such
suchthat,
that,for
foreach
eacha,ab, bEEA#,
A', 0(a):!:8(a) I
:CG(a) and
n
CG(b)I
<9(b).
functor 80 is said to be complete
there is
is an
:8(b). The signalizer
signalizer functor
complete ifif there
A-invariant r'-subgroup
CB(G)(a)
for
rl-subgroup0(G)
8(G)such
suchthat
that0(a)
8(a)==
Ce(G)(a)
for each
each aa EEA#.
A'.
Notice that one way to construct an A-signalizer
A-signalizer functor is to select some
A-invariant
r'-subgroup X of G and
A-invariant rl-subgroup
and define
define 0(a)
8(a) ==Cx(a)
Cx(a)for
foraaEEA#.
A'. By
By conconstruction this signalizer
struction
signalizer functor
functor is
is complete.
complete. If
If m(A)
m(A) >233itit turns
turns out
out that
that this
this isis
the only way to construct signalizer
signalizer functors. That is, if m(A) >2 33 then
then every
every
A-signalizer functor is complete. This result is called the Signalizer Functor
Theorem. It's one
one of the fundamental
fundamental theorems
theorems in
in the classification
classification of the finite
simple groups. Unfortunately the proof of the Signalizer Functor Theorem is
beyond the scope of this book. However, chapter 15 does contain a proof of a
special case: the so-called Solvable 2-Signalizer Functor Theorem. It turns out
that the Solvable
Solvable Signalizer
Signalizer Functor Theorem
Theorem suffices
suffices for many
many applications
applications
of signalizer
signalizer functors.
functors.
An A-signalizer
functor80 on
on G
G is said to be solvable
0(a) is
A-signalizer functor
solvable ifif 8(a)
is solvable
solvable
for each a EE A'
A#.
We
say
0
is
solvably
complete
if
0
is
complete
and
0(G) is
. We say 8 is solvably complete if 8 is complete and 8(G)
solvable.
solvable. The main result
result of chapter
chapter 15
15is:
is:
-
Solvable 2-Signalizer
ZSignalizer Functor
FunctorTheorem.
Theorem.Let
LetAAbe
bean
anabelian
abelian2-subgroup
Zsubgroup of
a finite group G with m(A) >
2 3.
3. Then each
each solvable
solvable A-signalizer
A-signalizer functor
functor on G
G
is solvably
solvably complete.
complete.
Chapter
Chapter 16
16 contains
contains a discussion
discussion of the Classification
Classification Theorem which illustrates how the Signalizer
Signalizer Functor
Functor Theorem
Theorem is used. Observe
Observe that the condition
condition
that m(A) >233 is
is necessary
necessary in
in the
the Signalizer
Signalizer Functor
Functor Theorem
Theoremby
by Exercise
Exercise15.1.
15.1.
44 Solvable
Solvable signalizer functors
In section 44, r is
r-subgroup of
is aa prime,
prime, G is
is aa finite
finite group, A is an abelian r-subgroup
3, and 08 is a solvable A-signalizer functor on G.
G of r-rank
r-rank at
at least
least 3,
Let Q(G)
r'-subgroups X of
Q(G)denote
denote the
the set
set of
of solvable
solvable A-invariant
A-invariant rl-subgroups
of G
G with
with
Cx(a) =
A#.
ForAA5<HH5< G,
G,let
letQ(H)
Q(H) =
= H fl Q(G) and
and
Cx(a)
=X
X fln0(a)
O(a) for each aa Ee A'
. For
0(H) = (H fl 0(a): a E A#).
Signalizerfunctors
Signalizer functors
230
230
If
is aa set
set of
ofprimes
primesthen
then &(H,
Q(H, 7r)
denotes the set
set of
ofn-groups
7r-groupsin
in&(H).
Q(H). Write
If 7r
rr is
n ) denotes
Q*(H)
&*(H)and
and Q*(H,
&*(H,7r)
n ) for
for the
the maximal
maximal members
members of Q(H)
& ( H )and
andQ(H,
Q ( H7r),
,n),respecrespectively,
X EE &(G),
Q(G), write Qx(G,
tively, under the partial order of inclusion. For X
&x(G,7r)
n )for
for
the
the set of H E
E Q(G,
Q(G,n)
n )with
with X
X <(H,
H ,and
anddefine
define
n(0)= U 7r(0(a)).
aEA#
(44.1)
(44.1) IfIfXX EE Q(G)
&(G)and
andYYisisan
anA-invariant
A-invariantsubgroup
subgroupof
of XX then
thenYY EE Q(G).
&(G).
(44.2)
(44.2)IfIfX,
X ,YYEE Q(G)
&(G)with
withXX<ING
N c(Y)
( Y )then
then XYE
XYE Q(G).
&(GI.
Proof.
Proof.This
Thisfollows
followsfrom
fromExercise
Exercise6.1.
6.1.
(44.3)
==B(a)*
(44.3) Let
Let H
H EE Q(G)
&(G)with
with H
H a5G,
G set
, setG*
G*==G/H,
G I Hand
, anddefine
define9*(a*)
O*(a*)
O(a)*
for
fora*
a*EE A*#.
A*'. Then
Then
(1)
(1) 8*
e*isisananA*-signalizer
A*-signalizerfunctor.
functor.
(2)
==
QH(G)*
==
{X{ X*:*X
(2)Q(G*)
&(G*)
QH(G)*
:X EE Q(G)
&(G)and
and HH <5X1.
X}.
(3)
(3)Q(G*)
&(G*)==Q(G)*
&(G)*=={X*:
{X*:XXEE Q(G)}.
Q(G)}.
Proof.
flf l Cc*(b*)59*(b*).
Proof.Let
Leta,ab, bEEA#.
A'. To
Toprove
prove(1)
( 1 )we
wemust
mustshow
show9*(a*)
O*(a*)
O*(b*).
By
(b*) ==CG(b)*,
soSOititsuffices
By coprime
coprimeaction,
action,18.7,
18.7,CGCG*(b*)
CG(b)*,
sufficestotoshow
showCHOW
C ~ s ( ,(b)
) ( b<5)
H9(b).
so
<9(b),
HO(b).But
Butby
byExercise
Exercise6.1,
6.1,CHO(a)(b)=CH(b)CB(a)(b),
CHsca)(b)
= CH(b)Ce(a)(b),
SOas
asCB(a)(b)
Ce(a)(b)
I
O(b),
(1)
( 1 )isis established.
established.
Let
Q H ( G )Then
.Then X
X isisaasolvable
solvableA-invariant
A-invariant r'-group,
r'-group, so
soits
itsimage
image
Let XX EE QH(G).
X*
(a) <i0O(a),
(a), and
X* has
has the
the same
sameproperties.
properties.Further
Furtherfor
foraaEEA#,
A', Cx
Cx(a)
andby
bycoprime
coprime .
action,
<50(a)*
9*(a*). That
action, Cx{(a*)
Cx*(a*)=
=Cx(a)*,
Cx(a)*,so
soCx.(a*)
Cx*(a*)
O(a)*==O*(a*).
Thatis,
is,X*
X*EE Q(G*).
&(G*).
Conversely,
suppose
Y*
E
Q(G*)
and
let
Y
be
the
full
preimage
of
Y*
Conversely, suppose Y* E &(G*) and
Y
full preimage of Y* in
in G;
G;
to
Q(G). As
As YY/H
to complete
complete the
the proof of (2),
(2),we must show Y
Y EE &(G).
IH =
=Y*
Y* and H
are
Cy(a*) <9*(a*),
soso
are solvable
solvable r'-groups, so
so isis YY by
by 9.3.
9.3. As
AsY*
Y*E
E Q(G*),
&(G*),Cr*(a*)
_( O*(a*),
Cy(a)
C y ( a )<9(a)H.
( B ( a ) HBut
. ButHHEEQ(G),
Q(G),so
soCH(a)
C H ( a<0(a),
)5 $(a),and
andthen
then
Cy(a) < CG(a) fl 0(a)H = 0(a)CH(a) < 0(a)
by
1.14.So
So(2)
(2)isisestablished.
established.
by the
themodular
modularproperty
propertyof
ofgroups
groups1.14.
By
44.2,HX
HXEEQH(G)
Q H ( Gfor
)foreach
eachXX EE Q(G),
&(G),so
soQH(G)*
&H(G)*==Q(G)*.
&(G)*.Hence
Hence(2)
(2)
By44.2,
implies
implies(3).
(3).
(44.4)
EfQ(G).
(44.4) For
For each
each 10
1 #BB<5A,A,9(CG(B))
O(Cc(B))
&(GI.
Proof.
_=(CG(B)
flfl9(a):
a aEEA#).
Proof.Recall
Recall9(CG(B))
O(CG(B))
(CG(B)
@(a):
A'). But
Butfor
forbbEEB#,
B', CG(B)
C c ( B )flfl
0(a)
<
CG(b)
fl
0(a)
<
0(b),
so
9
(CG
(B
))
<
0(b),
and
hence
the
lemma
@(a)iCc(b)flO(a) (@(b),so @(CG(B))
5 O(b),and hence the lemmafollows
follows
from
44.1.
from44.1.
signalizer functors
Solvable signalizerfinctors
2311
23
(44.5) For each X
C.7r(9),
Cx(A)
and Cx(A)
(44.5)
XE
E Q(G)
&(G) and
and Tr
n E:
n (0), Cx
(A) 5<0(CG(A))
O (CG(A)) and
Cx (A) is
is
transitive on Q*(X,.7r).
transitive
&*(X,n). Indeed, Q*(X,.7r)
&*(X,n ) is the
the set
setofofA-invariant
A-invariantHall
Hall.7rnsubgroups of X.
Proof. Let
maxiProof.
LetTTbebethe
theset
setof
ofA-invariant
A-invariantyr-subgroups
n-subgroups of
of X
X and
and T*
T* the set of maximembers of
Q(X,
.7r),
* ==Q*
(X, .7r).
mal members
of TTunder
underinclusion.
inclusion.By
By44.1,
44.1,TT==
&(X,
n),sosoT T*
&*(X,
n).
By Exercise
Exercise 6.2, T*
is
the
set
of
A-invariant
Hall
yr-subgroups
of
X
and
Cx
(A)
T*
n-subgroups of
Cx(A)
on T*. Finally,
as X EE &(G),
Q(G), we
we have
have Cx(A)
Cx(A)5<0(a)
is transitive
transitive on
Finally, as
O(a) for each
each
a EE A#,
Cx(A)
A', so Cx
(A) <
(0(CG(A)),
O (CG(A)), completing
completing the proof.
(0). Then
n(O).
(44.6) Let
LetPpEE7r
(1) For
Foreach
eachaaEEA#
A# there
thereisis aa unique
unique maximal
maximal B(CG(A))-invariant
O(CG(A))-invariantmember
0(a)
A(a>of
of Q(6(a),
&(O(a),p').
PI).
(2) AAisisan
anA-signalizer
A-signalizerfunctor.
functor.
Proof. By
=O(a).
9(a). Let
Let R be
By 44.4,
44.4, YY ==B(CG(A))
O(CG(A))5<XX=
be the
the set
setof
ofY-invariant
Y-invariant
members of
of Q(X,
E S,
S, so as
members
&(X, p') and
and SS==Q*(X,
&*(X,p').
p'). Then
Thenfor
forRREER,
R ,RR<_
(S E
Y is transitive on
on S
S by 45.5, R is contained in each member
member of
of S.
S. Therefore
Y
= (R)
M=
(R)<5S,S,so
soM
Misisthe
theunique
uniquemaximal
maximal member
member of R
Rby
by 44.1.
44.1. This
This estabestabNext for
for bb EE B'
B#,
0(a) fl
= CM(b)
, A(a)
f l CG(b)
CG(b)=
CM(b)5<9(b)
O(b) as
as B
O is a signalizer
lishes (1). Next
functor.
is an
an A-invariant
A-invariantpl-group,
p'-group, so CM(b)5< A(b),
0(b), completing
CM(b)is
completing
functor. Also CM(b)
the proof of (2).
,
(44.7) (Transitivity
Theorem) O(CG(A))
B(CG(A))isistransitive
transitiveon
on &*(G,
Q*(G,p)
p) for each
(44.7)
(Transitivity Theorem)
p EE7r
(B).
P
n(@.
Proof. The
Theproof
proofisisleft
leftas
asExercise
Exercise15.2.
15.2.
(44.8) Let X E Q(G).
&(G). Then
(44.8)
(1) IfIfBBisisaanoncyclic
noncyclicelementary
elementaryabelian
abeliansubgroup
subgroupof
of A
A then
then
X = (B(C(D)) f1 X: I B: D) =r).
(2) IfIf1#T<Athen
(2)
1 # T 5 A then
0(CG(T))
= (0(CG(B)):
O(CG(T))=
(O(CG(B)):T <
5 B <5 A
A and
and B is
is noncyclic).
noncyclic).
(3) X =O(X).
=0(X).
(3)
Proof. IfIf 11 # T <I
A
A then
then by
by Exercise
Exercise 6.5,
6.5,
= (Cx(S):
X=
(Cx(S): T/S
T/S isis cyclic).
cyclic).
Signalizerfunctors
Signalizer
functors
232
011then
by
Further ifif S #
then CX(S)
Cx (S)<0(CG(S))
5 8(CG(S))by
by 44.5,
44.5, so
so (1)
(1) follows.
follows. Similarly by
44.4, YY=
=9(CG(T))
Q(G)and
andAA=
A/T is abelian
on Y,
8(CG(T))EEQ(G)
= AIT
abelian and acts on
Y, so by
44.4,
Exercise 6.5,
Y=
= (Cy
(Cy(B):
Alb isis cyclic)
Y
(B): A/B
cyclic)
Cy(B)
with Cy
(B) ==9(CG(B)).
8(CG(B)).IfIf TTisisnoncyclic
noncyclicthen
then(2)
(2) is
is trivial,
trivial, so
so we may assume
T is cyclic.
cyclic. Then as m(A) >
>3,3, BB isis noncyclic,
noncyclic, so
so (2)
(2) holds.
For H
H5
< G, 9(H)
9(H), while X 5
<9(X)
8(H) <jH
H by
by definition
definition of 8(H),
8(X)by
by (1).
(1).
This about exhausts the results on solvable
solvable signalizer functors which can
outside of
of an inductive
To proceed
proceed further
further we
we must
must
be established outside
inductive setting. To
work inside a minimal counterexample to the Solvable 2-Signalizer Functor
Theorem.
Theorem.
So in the remainder of chapter
chapter 15
15 assume
assume (A, G, 9)
8) is
is aa counterexample
counterexample
to the Solvable
Solvable 2-Signalizer
2-Signalizer Functor Theorem with (G(
IG I + In(9)1
In (8)/ minimal.
minimal. In
In
particular now r =
(A)).
=2.
2. Let
Let II ==9(CG
8(CG(A)).
+
(44.9) (1)
(1) If
< H <<GGthen
(44.9)
If A 5
then 9(H)
8(H)EE Q(G).
Q(G).
(2) GG==A(9(b):
A(8(b):bbEEB#)
B') for
for each
each E4-subgroup
E4-subgroupB
B of
of A.
A.
(3) If
If 10
1 #HHEEQ(G)
Q(G)then
then G
G 0#NG(H).
NG(H).
Proof.
Proo$ In
In each
each case
case we
we use
use the
the minimality
minimality of
of the
the counterexample.
counterexample. Under the
hypotheses
of (I),
(1), OH(a)
9H(a)=
= 8(a)
9(a) n
fl H is an
functor on
on H,
H, so (1)
hypotheses of
an A-signalizer
A-signalizer functor
follows
1.
minimalityofofI G(GI.
followsfrom
fromminimality
(8(b):b cE B#).
B'). If G 0#KA
K A then by (1),
(I),
Let B be a 4-subgroup of A and K =
= (9(b):
Y
=
9(KA)
E
Q(G).
But
now
for
a
E
A#,
Y = 8(KA)
But now for a E A',
9(a) = (9(a) fl 9(b): b E B#) <Cy(a) = (9(b) fl Cy(a): b E B#) <6(a)
by 44.8.1, so 89 is solvably
of ((A,
9) as a
A , G, 8)
solvably complete, contrary to the choice of
counterexample.
counterexample. Thus (2)
(2) holds.
holds.
44.3.1,
Suppose 1 #* H EE Q(G)
Q(G) and
and H
H < G.
G.Define
DefineG*
G*and
and0*
8* as
as in 44.3. By 44.3.1,
9*
1, 9 *isissolvably
solvably complete.
complete.
8* is an A*-signalizer
A*-signalizerfunctor,
functor,sosoby
byminimality
minimalityofofJ G
IG1,8*
Thus there is a unique maximal member X* of
of Q(G*).
Q(G*). By 44.3.2, X EE Q(G).
=8*(a*)<X*,
5 X*,9(a)
8(a)<5X,
X, so
so 9(a)=CX(a)
8(a) = Cx(a)asasXXEEQ(G).
Q(G).Therefore
Therefore
As 8(a)*
6(a)*=9*(a*)
98 is solvably complete,
complete, contrary
contrary to the choice of (A, G, 9)
counterexample.
8) as a counterexample.
a
(44.10) Let
Let nlr 2
c n(9),
and 110
with X
X9
< M. Then
Then
(44.10)
n(8), M
M EE Q*(G,
Q*(G, 7r),
n), and
# X E Q(G) with
(1) M
MisisaaHall
Hall n-subgroup
n-subgroup of
of 9(NG(X)).
8(NG(X)).
(2) Op(M)
(2)
0,(M) <Opn(9(NG(X))
jOpn(8(NG(X)) for each p E7r.
En .
(3) If7r=7r(9),sothatMEQ*(G),then9(NG(X))=M.
HenceifX<YE
(3)
If n = n(8), so that M E Q*(G), then 8(NG(X))= M. Hence
if X jY E
Q(G), then
then Ny
Ny(X)
(X) 5<M.
M.
Solvable signalizer
signalizerfunctors
Solvable
functors
233
Proof. As
ASXX<<IM,
M,M
M<5HH==9(NG(X))
8(NG(X))by
by44.8.3.
44.8.3. By
By 44.9.3,
44.9.3, G 0# NG(X),
NG(X), so
SO
H EE Q(G)
n), so
Q(G) by
by 44.9.1. As M E Q*(G, nn),
), M
M E Q*(H,
Q*(H, n),
so (1)
(1)follows
follows from
follows from
from (1)
(1) and
andExercise
Exercise 11.1.
11.1.IfIfM
MEEQ*(G)
Q*(G)then
thenHH=
=M
M
44.5. Part (2) follows
maximality of M, so
by maximality
so (3)
(3) holds.
(44.11)
CA(P)=
=11for
for each
each pp E7r(0)
E n(8) and
and each
each PP EE Q*(G,
Q*(G,p).
p).
(44.11) CA(P)
Proof. Suppose
Suppose T =
=CA(P)
CA(P)0#1.1.Then
Then by
by 44.7,
44.7, Q*(G,
Q*(G, p) ccCG(T),
Cc(T), so,
so,by
by
Exercise 11.1.3, [H,
[H, TI
T] 5
<Op-(H)
Opf(H)for
for each
each H EE Q(G).
Q(G).
Form the signalizer functor A of 44.6 with respect to p. By minimality of
17r(0)1,
solvablycomplete.
complete.Hence
HenceK
K=
= A(G)
O(G) E Q(G).
ln(8)1, AAisissolvably
Q(G). Observe
Observe
[6(a), TI
Op'(9(a)) < O(a) = CK(a)
by paragraph
paragraph one of
of this
this proof.
proof.Recall
Recallthat,
that,by
by24.5,
24.5,[8(a),
[9(a),T,
T,TI
T]=
= [8(a),
[9(a), TI,
T],
so
SO
[CK(a), Ti _ [9(a), T].
by 8.5.6, [CK(a),
[CK(a), TI
T] <I
< 8(a).
9(a).
Then by
Let S
subgroups of A containing T. By Exercise
S be the set of noncyclic subgroups
Exercise 8.9,
[K, Ti
TI ==([CK(b),
([CK(b),Ti:
TI: bb cE B#)
B')
[K,
for each B E S. Therefore
.
0(CG(B)) = n 9(b)
bEB#
b€B#
acts
9 (CG(T )) acts on
on [K, T]
acts on [K,
[K,T].
TI.Hence
HenceB(CG(T))
TI by
by 44.8.2,
44.8.2, so
so by 44.2
44.2 and 44.4,
44.4,
H ==[K,
EE Q(G).
9(a) <9(CG(T))[9(a),
T],
[K,T]9(CG(T))
T]~(CG(T))
Q(G). But
But by 24.4,
24.4,8(a)
58(C~(T))[8(a),
TI,so,
so,by
by
the second
0(a) <
contradicts
second paragraph of this proof, O(a)
5H
H for
foreach
each aa EE A#.
A'. This contradicts
44.9.2.
44.9.2.
(44.12) For
each lr C
7r(6), a cEA',
A#, and M E Q*(G,
Q*(G, n),
7r), [a,
[a, F(M)]#
F(M)] 0 1.1.
En(@,
Foreachn
Proof.
Proof. Assume
Assumeotherwise.
otherwise.Then
Thenby
byExercise
Exercise11.1.2,
11.1.2, aa centralizes
centralizesM.
M.
Let
p c n (F(M)) and Op (M) = X 5< PP EE Q*(G,
p). By 44.11, [a, PI#
P] 0 1,1,so
so
LetpEn(F(M))andO,(M)=X
Q*(G,p).By44.11,[~,
by the Thompson
Thompson A x B
(X), a]
B Lemma,
Lemma, [Np
[Np(X),
a ] 0#1.1.But
ButNp
Np(X)
(X)is
is contained
containedin
in aa
Hall n-subgroup
H ==0(NG(X)),
8(NG(X)),so
soby
by 44.10.1
44.10.1and
and 44.5,
44.5, Np(X)h
N P ( X ) <~
5M
M for
for
n-subgroup of
of H
some
CH(A). This
This is
is impossible,
impossible, as
as aa centralizes M
M but
but [Np(xlh,
[Np(X)h, aa]] #
0 1.
1.
some h E CH(A).
For 7r
A#,
let M
M(a,
( 8 ) and
and a E
E A'
, let
( a , nr))denote
denotethose
those M
M EE Q*(G, nr)) such
such
n cCn r(0)
that CM(a)
Q*(CG(a), n).
n). Recall that for Jr
7r,, pK
p" =
CM(a)E
E Q*(CG(a),
n a set of primes and p E
En
7r'U{p}.
n ' u {PI.
Signalizerfunctors
Signalizer functors
234
A#, M
M EEMM(a,
n), V
(44.13) Let
(44.13)
LetjrnC-E7r
n ( 8 ) ,p cE 7r,
n, a E
E A',
( a , n),
V EE Q(Op(M)),
Q(Op(M)),and
and N EE
Q(G)
with N
N=
= (V,
Then
V5
<Opn(N).
Q ( G )with
( V ,CN(a)).
C N ( a ) )Then
.
V
O,n(N).
Proof. Assume
pn((N).
By 331.20.2,
there isis aa 4-subgroup
4-subgroup BB of A
Proo$
Assume VV ¢$OOPT
N ) . By
1.20.2, there
A with
Cv(B)
so replacing
replacing VV by
by Cv(B),
C v ( B )¢$ Opn((CN(a),
Opn((CN(a),
CCv(B))),
v ( B ) ) ) ,SO
C v ( B ) ,we
we may
may assume
assume
[V,
withVV¢
OpV)),
so
[ V ,B]
B]=
=1.
1.Next
Next by 31.20.3 there
there is
is b c=
E B'B#with
$O
P n((CN((a,
( ( C ~ ( (bb)a) ),
), ,V
) ) ,SO
we may assume
Cm(a)
EE)Q*
(CG (a), n),
7r), CY
Cy(b)
assume N
N <0(b).
5 8(b).However
HoweverasasYY==
CM(U
Q*(Cc(a),
(b) E
Q*(Co(b)(a),
yr).Further
Further C
Cy(b)
Cv(b) <Cop(M)(b),
Q*(Ce(b)(a),n).
Y ( ~acts
acts
) on
on CoP(M)(b)
C O , ( M ) ( and
~and
) Cv(b)
i C o , ( ~ ) ( bso
so
),
31.20.1
3 1.20.1 supplies
supplies aa contradiction,
contradiction,completing
completingthe
the proof.
proof.
For.7r
andaa EEA',
A#,let
letUU(a,
7r))consist
consist of
of the
the those
those U E
n ( 8 ) and
( a ,n
E Q(G,.7r)
Q(G, n )such
such
For n gc .7r(O)
that [U,
[U,a]
a ] # 1 and U
U isisinvariant
invariant under
under some
somemember
member of
of Q*(0(a),
Q*(8(a),7r).
n).Let
Let
U*(a,
U*(a,7r)
n ) be the minimal
minimal members
members of
of U(a,
U ( a 7r)
,n )under
under inclusion.
inclusion.Given
GivenppcE1r,
n,
let V(p,.7r)
of those V
p) such
V ( p ,n ) consist of
VE
E Q(G, p)
such that V
V _ ( O P r ( H )for each
each
HE
Qv(G).
E Qv(G).
(44.14)
(44.14) For
Foreach.7r
each n cE:7rn ((0)
8 )and
and aa cEA#:
A':
(1) U(a,
U ( a ,Ir)
n )# 0.
0.
(2)
(2) For
ForUUEE U*(a,
U*(a,pr),
n),U
U ==[U,
[U,a]
a ]isisaa p-group
p-group for
for some
some prime p.
(3)
7r)) with
with YY5<Nc(U)
NG(U)and
andMMEEQ*,,(G,
QJ (G, n).
ir). Then U <
(3) Let
LetYY EE Q*(O(a),
Q*(O(a),n
5
O,(M) and
and UUcentralizes
centralizesCF(M)(a).
CF(~)(U).
Op(M)
(4) UUEE V(p,
V ( p 7r).
,n).
Proof.
7r)) and
and YY <XX EEQ*(G,
Q*(G,n).
n). By
By 44.12,
44.12, [a,
[a, FF(X)]
Proof. Let
LetYY EEQ*(O(a),
Q*(O(a),n
( X ) ]## 1,1,
so
so X E U(a,
U ( a 7r),
,n),and
and (1)
(1)is
is established.
established.Let
Let UU EE U*(a,.7r);
U*(a,n);without
withoutloss
loss YY acts
acts
on U.
[Op(U),aa]] #* 1I for some prime
prime p, and
U . By
By Exercise
Exercise 11.1.2,
11.1.2, [Op(U),
and by
by 24.5,
24.5,
[Op(U),
[O,(U), aa,, a]
a ]_=[Op(U),
[O,(U),a],
a ] so
,so(2)
(2)follows
followsfrom
from minimality
minimality of U.
U.
Let M E Q
I (G, 7r).
Thus U <5
QTuy(G,
n).Then
Then YY ==CM(a),
C M ( a )so
so
, U
U isis CM(a)-invariant.
CM(~)-invariant.
Op(M)
O,(M) by 36.3.
36.3. Hence
Hence U
U centralizes
centralizes OP(F(M)),
Op(F(M)),while
while by
by the
theThompson
Thompson
Lemma
COP(M)
(a). This establishes
, Ucentralizes
centralizes
Co,(M)(a).
establishes(3).
(3).
Lemma and
andminimality
minimalityofofU,UU
Finally
let H E
Qu(G)
Finally let
E Q
u ( G ) and N=
N =(CH(a),
( C H ( U U).
)U, ) .By
By 44.13,
44.13, U
U <Opn(N),
Opn(N),so
SO
U=
= [U,
[U,a]
a ]<5[Op"
[O,r (N),
( N ) ,a]
a ]<5Op"
Opn(H)
( H )by
by 36.3,
36.3, establishing
establishing(4).
(4).
(44.15)
(G, p),
b EE B#,
(44.15) Let
LetppcE7rnC7r
E n (0),
( 8 ) ,BB aahyperplane
hyperplaneof
ofA,
A,PPEEQ*
Q*(G,
p),and
andfor
forb
B',
let M(b)
M(b,
M(b)E M
( b ,7r).
n).Then
Then
(1)
=Op(M(b)) and
fl n
M(b)
E EQ*(M(b),
(1)IfIfF(M(b))
F(M(b))=O,(M(b))
andP P
M(b)
P ( M ( b )p),p)for
foreach
eachbbEEB#,
B',
then
then
Z(P) < n Op(M(b))
bEB#
functors
Solvable signalizer
signalizerfunctors
235
Wehave
have
(2) We
Q
n Op(M(b))) c V(p, n).
(bEB#
Proof.
of (1)
(1) and
and let
let Q(b)
Q(b)=
= O,(M(b))
Op(M(b))and
andN(b)
N(b) =
_
Proof Assume
Assume the
the hypotheses of
O(NG(Q(b))).
By 44.10.1, M(b)isaHalln-subgroupofN(b),soas
M(b) is a Hall 7r-subgroup of N(b), so asP(b)
P(b)=
= Pn
0(NG(Q(b))).By44.10.1,
M(b)
of (I),
(1), P(b)
P(b) E Syl,(N(b)).
Sylp(N(b)). Thus Q(b) <5
M(b) EE Sylp(M(b))
Syl,(M(b)) by the hypotheses of
P(b)
<5N(b).
<i
P(b),
so so
as as
CM(b)
P(b) <(P,
P ,sosoZ(P)
Z(P)<B(CG(Q(b)))
5 0(c~(Q(b)))
N(b).Hence
HenceZ(P)
Z(P)
P(b),
CM(~)
(F(M(b)))
(F(M(b))) <(F(M(b)),
F(M(b)),(1)
(1)holds.
holds. Let
Let
1#XEQ(n Op(M(b))
bEB#
and
(G). By 44.13,
X <(Op"
((CN (b), X))
so by
by 31.20.3,
31.20.3,
Qx(G).By
44.13,X
Opz((CN(b),
fo for each
each bb EE B#,
so so
and N
N EE Qx
X <(OOpn
(2).
X
pn(N),
(N), establishing
establishing (2).
(44.16)
7r),aaEc,r(F(M)),
< F(M)
=aa and
(44.16) Let M EE PQ*(G,
( G , n),
n(F(M)), and
and X (
F(M) with.7r(X)
with n(X) =
and
Z(O,(F(M))) <(
Thenfor
foreach
eachNNEEQx(G,.7r):
Qx(G, n):
Z(O.(F(M)))
X.X.Then
[OP(X),Op(N)]
O,(N)] =1
= 1for
for each
each pp EE a.
(1) [OP(X),
a.
thena ==7r(F(N))
n(F(N))and
and [Op(X),
[O,(X), OP(F(N))] =
foreach
(2) IfIfn(F(N))
7r(F(N)) cE.aathence
=11foreach
p E a.
pEa.
(3) IfIf Tr(F(N))
F(N).
n(F(N)) cEaaand
and laI
la1>>11then
then X
X<
(F(N).
Proof. For
p(M)). By
Forpp EEa,
a,let
letXP
Xp==Op(X)
Op(X)and
andZp
Z, ==Z(O
Z(O,(M)).
By 44.10.2,
44.10.2,
Xq
<Ogr(O(NG(Zp)))
a,
E a,
X, I
Oqn(O(N~(Zp))) for each pp,, q E
so
Thus
(NN (Zp)),SOsoby
by31.15,
31.15, OP(X)
OP(X) <
SO Xq
X, <Oq(NN(Zp)).
(Oq(N~(Zp)).
Thus OP(X)
OP(X) <(Op,
O,J(NN(Z,)),
(
Op,(N)
establishing
O,l(N) <CN(Op(N)),
( CN(Op(N)),
establishing(1).
(1).
Suppose
7r(F(N)) Ec a.
a.Then
Then by (1),
(I), 11 ## Op(X)
O,(X) <5CN(Op(F(N)),
CN(OP(F(N)), so
so by
Suppose n(F(N))
31.10, p EE7r(F(N)),
n(F(N)),establishing
establishing(2).
(2).Assume
Assume in
in addition
addition that
that loll
la1 >
> 11 but
but
Op(X)
¢ F(N). Pick p ## qq EE aa and
K=
O,(X) $
and let H ==CN(Op(F(N)))
CN(OP(F(N)))and K
=HZq.
HZ,.
By (2),
OO,(H).
p(H). But
(2), X
X<
iH, so
so as
asHH<9N,
N,we
wehave
haveX,X,¢ $
ButififZq
Z, <(Oq(K),
O,(K),
then
p <Op(NH(Zq))
= Op(H),
then by
by 31.14,
31.14,XX,
5 Op(NH(Z,)) =
O,(H), SO
so Zq
Z, ¢Oq(K).
$ O,(K). However,
However,
F(H)
= 1, so [Zq,
H] i
<CH(Op(N))=
F(H) =
=Op(N)Z(F(N)) and
and [Zq,Op(N)]
[Z,, Op(N)l =
[Z,, HI
CH(O,(N>) =
CH(F(H))
= Z(F(H)). As
OP(F(H))] =
= 1, we
C,y(F(H))=Z(F(H)).
As [Zq,
[Z,,OP(F(H))]
we have
have Zq
Z, <Oq(ZgZ
iO,(Z,Z
(F(H))),
(F(H))),so
so[Zq,
[Z,, H]
HI<Oq(H),
5 O,(H),contradicting
contradictingZq
Z, ¢$Oq(K).
O,(K). This
Thiscompletes
completes the
proof of (3).
(G, 7r),
<) X <5
n c.7r(0),
E n(0), HHEEQ*
Q*(G,
n),X, LL EE Q(G,
Q(G, ir),
n),and
andCF(H)
C F ( ~(X)
) ( X(
(44.17) Let
Let 7r
F(H) nn L.
L. Then
Then
F(H)
[OP(X),Op(L)]
O,(L)] ==11for
for each pp E.7r(F(H)).
E n(F(H)).
(1) [Op(X),
(2)
H) E97r(F(H))
(2) rr(F(L)
n(F(L) n H)
n(F(H)).
Signalizerfunctors
Signalizer functors
236
(3) If7r(F(L))Cn(F(H))then
7r(F(L))=n(F(H))andeither
=
(3)
Ifn(F(L)) 2 n(F(H))thenn(F(L))
=n(F(H))andeither In(F(H))I
In(F(H))I =
or X
X5
< F(L).
1 or
(4) Suppose
Q(Op(L) n
fl O,(H))
Suppose 11 # V EE &(Op(L)
Op(H))with
withNH(V)
NH(V)EEQ*(NG(V),
P(NG(V),7r)
n)
and L E
Q*(G,
7r).
Then
7r(F(H))=7r(F(L)),
and
either
7r(F(H))
=
{p}
E Q*(G, n). Then n(F(H)) =n(F(L)),
n(F(H)) = (p} or
X _<(F(L),
F(L),LLEEHI
HI,, and
a n dLL==H
H ifi f7rn = n7r(9).
(0).
Proof. Part
44.16.2 and44.16.3.
and 44.16.3.
Proof.
Part(1)
(1)follows
followsfrom
from44.16.1,
44.16.1, and
and (3) follows from 44.16.2
As
fl H,
H, OP(X)I
OP(X)] =
=11 by (1)
[Op(L) n
(1) and
CF(H)(X)_(<X,
CF(H)(X)
X,
(2) is a consequence of
of the
the A
AxB
B Lemma
Lemma and
and 31.10.
Assume
let YY==NF(W
(V ). By hypothesis,
hypothesis, NH
(V) EE
andlet
NF(~)(V).
NH(V)
Assumethe
thehypotheses
hypothesesofof(4)
(4)and
Q*(NG(V),7r),
soby
by 44.5,
44.5, Y
Y _(
< H`
P(NG(V),n), so
H' for
for some
some iiEE NI(V).
NI(V). Thus
Thus (L,
(L,Y,
Y, H`)
Hi)
satisfies the
byby
Exercise
1 111.1,
. 1 , OOp(X)
p(X) <_( O
p
the hypotheses
hypothesesofof(H,
(H,X,X,L).
L).Also,
Also,
Exercise
OPn
(B
(NG (V ))),SOX
so X5<F(NL(V)),
F (NL(V )),and
andhenceOP(X)
hence OP(X)_(<F(L)
F(L) by
by 31.14.2.
31.14.2. Therefore
Therefore
(O(NG(V))),
7r(F(H))
(3) to
to (L,
(L, Y
Y,
H'), we conclude
conclude n(F(H))
7r(F(H)) =
=
n(F(H)) C
2 7r(F(L)),
n(F(L)), so applying (3)
, H'),
7r(F(L))
Y5
< F(H') oror7r(F(H))
n(F(L)) and either Y
n(F(H))=={p}.
(p}.So
Soto
tocomplete
complete the
the proof
proof
of (4) we
we may
may assume
assumeIn(F(H))I
I7r(F(H))l>> 1 and
and YY_(<F(H').
F(H`). By
By (3),
(3), X
X(
< F(L).
F(L). As
OP(Y)=OP(F(L))
OP(Y)=OP(F(L)) <OP(F(H1)),
iOP(F(Hi)),IOP(F(L))I
IOP(F(L))I<5IOP(F(H))J.
IOP(F(H))I.As
As (Z(OP(L)),
(Z(Op(L)),
L,
L, H`,
H', Y)
Y)satisfy
satisfy the
the hypotheses
hypotheses of (V, H, L,
L, X),
X), by
by symmetry, JOP(F(H))I
IOP(F(H))I <5
JOP(F(L))I,so
soOp(F(L))
OP(F(L))=OP(Y)=O"(F(H`)).
IOp(F(L))I,
= OP(Y) = OP(F(H')).Then
Thenby
by44.10.1
44.10.1and
and 44.5,
44.5,
H' isisconjugate
conjugate to
to LL ininNI(OP(Y)),
NI(O"(Y)), so
SO LLE
E HI.
HI. Finally,
Finally, ifif 7r
n =7r(0)
=n(0) then
then
L=
= H'
H' =
=0(NG(OP(Y)))
O(NG(OP(Y)))=
=H
H by
by 44.10.3.
44.10.3.
(44.18)
C n(0),
7r(9), U E V(p,
V(p, n),
7r), and
and H,
(G, n70) with
n2
H, LL EE Q*
Pu(G,
with NL(U)
NL(U)EE
(44.18) Let 7r
Q*(NG(U),
yr)
P(NG(U),n).
Then
(1) If F(L)
F(L)=OP(L)
F(H)=OP(H).
(1)
= Op(L) then
then F(H)
= Op(H).
(2) IfIfqqEEJr,
V
E
V(q,
Jr),
NH
(V)
Q*(NG(V),n),
jr),and
and [U,
[U, Vl
v] 5
< U fln VV
n , V E V(q, n), NH(V) EE P(NG(V),
then either
(i) pp ==qqand
and F(H)
F(H)and
andF(L)
F(L)are
arep-groups,
p-groups, or
or
(ii) L E
HI and
andififn7r=7r(O)then
E H'
= n(0) then HH=L.
= L.
(ii)
7r), U _(
<OP(L)
EEQ*(NG(U),
Proof. As UU EE V(p, n),
Op(L)flnO,(H).
Op(H).AsAsNL(U)
NL(U)
P(NG(U),7r),
n),
X=
for some
some ii EENI(U),
NI(U), and we may assume i =
=NF(H)
N F ( ~ )(U)
( U(<
) L'L`for
=1.
1. Thus
Thus we
we
of 44.17.
44.17. If
If F*(L)
7r(F(H)),
have the hypotheses
hypotheses of
F*(L) =
=OP(L)
Op(L) then 7r(F(L))
n (F(L)) C
2 n(F(H)),
so (1) holds by
by 44.17.3. Assume
Assume the
the hypotheses
hypothesesofof(2).
(2).Then
ThenVV (
< Oq(H)
Oq(H) because
so as
as [V,
[V, U]
U] _(
< U, we have V 5
<X <
n), so
(L
L and
and then V
V <_( Oq(L)
Oq(L)
cause V E V(p, 7r),
E V(q,
V(q, n).
7r).So
So we
we have
have the
the hypotheses
hypotheses of
of 44.17.4,
44.17.4, so
so that
that lemma
because V E
completes
completes the proof of (2).
(2).
For 7r
C n(0)
7r(0) let
let X(n)
7-1(7r)consist
consistofofthose
thoseHHEE&(G,
Q(G,n7r)
suchthat
thatPQ*(H,
p) 2
C
n2
) such
( H , p)
Q*(G,
&*(G, p)
p) for
for each
each pp EE 7r.
n.
Solvable signalizer
signalizerfunctors
functors
237
(44.19) IfIf B is
(44.19)
is aahyperplane
hyperplane of
of A
A and
andHHEEM(b,
M ( b 7r)
, n )for
foreach
eachbbEE B#
B# then
then
H EE7-l
(7r ).
H
IFt(n).
Proof. Let
Let p E
E n7r,
, Q
P ( G ,p).
p).As
As H
H EE M(b,
M ( b ,7r),
n ) ,we
Q EESylp(H),
Sylp(H),and
andQQ5PPEEQ*(G,
H ( b )EE pQ*(CG(b),
( C ~ ( b )n),
, SOsoCQ(b)
= Cp(b).Thus
Q by
by 44.8.1.
44.8.1.
have C
CH(b)
7r),
CQ(b)=CP(b).
Thus PP =
=Q
c
(0). Then
Then either
(44.20) Let
Let7r
n C 7r
n (0).
(1)
R(7r)
0
0,
or
(1) IFt(n)# 0,
or
(2)
MEE M (a,
(2)There
Thereexists
existsaprime
aprimeppsuch
suchthat
that F*(M)
F*(M)==OOp(M)
p ( M )for each M
( a ,7r)
n)
A'.
and each aa EE A#.
7r), and
and by
by 44.14.2,
44.14.2, U
U is
is a
Proof. Let
Let aaEE A#.
A'. By 44.14.1
44.14.1 there is U E
E U*(a,
U*(a,n),
=B
p-group for some prime p. By
By minimality of U and
and Exercise
Exercise 8.9, CA(U)
C A ( U )=
is a hyperplane
hyperplane of A. Thus U <5 M(b)
M(b)EE M(b,
M ( b 7r)
, n )for
foreach
eachbbEE B#.
B#.By
By 44.14.4,
44.14.4,
V (p, n),so
7r), soU
U5
< Op(M(b)).By
O p(M(b)). By44.12thereis
44.12 there isU
U(b)
U*(b,
withUU(b)
UE
E V(p,
( b )EE U
* ( ~n7r)
,)with
( b )5<
F(M(b)), and
44.14.3, [U,
[U, U(b)]
U(b)] =
=1.
F(M(b)),
and by 44.14.3,
1.
with N
NH(U)
Q*(NG(U), n).
7r).IfIfFF(H)
Let H EE Q*(G,
P ( G , 7r)
n )with
H ( U )EE Q(NG(U),
( H ) is a p-group then
by 44.18.1, so is F*(M)
(G, 7r).
F(M(b)) =
_
F*(M)for
foreach
eachM
MEE Q*
Q*,(G,
n). In particular
particular F(M(b))
Op(M(b)).
and M
M EEMM(c,
from the
the proof
proof of 44.14
O,(M(b)). Further, for C
c EEAA#
' and
( c , n7r),
) , from
44.14
there is U(c)
U(a,
7r)) with
with U(c)
U(c) 5
< M. By symmetry
U ( c )E
EU
( a ,n
symmetry there is aa hyperplane
hyperplane
B(c) of
of A
A centralizing
centralizing U(c),
U(c), and
andas
as BB nn B(c) #
$ 1, we have
have U
U(c)
< ~M(b)`
for
B(c)
( c )5
( b )for
'
some
I. As F(M(b))
for
EEQ*(G,
F(M(b))is
is aap-group,
p-group, so
soisisF(H(c))
F(H(c))
forH(c)
H(c)
P ( G ,7r)
n)
some i EE I.
with NH(c)
(U (0)EEQ*(NG(U(C)),
Q*(NG(U(c)),n7r)
44.18.2.
HenceF*(M)
F*(M)=Op(M),
NHCc)(U(c))
) byby
44.18.2.
Hence
= O,(M), as
U(c)
<M.
we may
may assume
assumeF*(H)
F*(H) is
is not
not aa ppU
( c )5
M . Thus
Thus (2)
(2)holds
holds in this case, so we
group. Hence
Hence by 44.18.2, M(b)
M(b)EE HI
H' for
foreach
eachbbEE B#.
B'. Thus
Thus(1)
(1)holds
holds by
by 44.19.
44.19.
(44.21)
M EE M
M(a,
(44.21) There
Thereisisaaprime
primepp such
suchthat
thatfor
foreach
each aa EE A#
A# and each M
( a 7r
,n (9)),
(O)),
F(M)
p(M).
F
( M )==O
O,(M).
then for
for p EE7r(0)
Proof. IfIf not,
not, by
by 44.20
44.20 there
there is
is H
H E7-l(7r(9)).
E IFt(n(0)).But then
n ( 0 ) and P EE
Sylp(H),
Sylp(9(a)), and hence
hence 09(a)
_
Syl,(H), PP EE Q*(G,
P ( G ,p),
p), so
sofor
for aaEE A#,
A#, Cp(a)
C p ( a )E Sylp(O(a)),
( a )=
(Cp(a):
( C p(a):pp EE 7E
n ( 0(0))
) )5<HH,, contrary to 44.9.2.
In the remainder
of this
pick p as
remainder of
this section
section pick
as in
in 44.21
44.21 and
and let
let 7r
n=
=7r
n ( (9)
0 ) -(p). If
pickqqEE7r(F(H)),
(p).
If H
H EE 7-1(7r),
IFt(n),pick
n ( F ( H ) ) ,Q E Q*(G,
P ( G , q),
q ) ,and B
B aa hyperplane
hyperplaneof
of A
A
with
z ==Z(Q)
0 1. As H E
R(rr), H E
M(b,
with Z
Z ( Q )nnOq(H)
O q ( H )nnCH(B)
C H ( B )#
E IFt(n),
EM
( b , ir)
n )for each
bEB#.
this case
caselet
letM(b)
M(b)=Hfor
bEB#.
b E B'. In this
= H for b
E B'.
If X
7-l(7r)
0 then by 44.20 there is a prime q such that F*(M)
(M) for
( n ) ==0
F*(M)==Oq
O,(M)
for
each M E M
M(a,
( a 7r)
, n )and
andeach
eachaaEE A#.
A'. Pick Q
Q EE Q*(G,
Q*(G,q),
q ) ,BB aahyperplane
hyperplaneof
of AA
with Z =
01,
b EE B#,
M(b,
=Cz(Q)(B)
C Z ( ~ ) (#
B )1,and
andfor
forb
B', M(b)
M(b)E M
( b ,7r)
n )with Q
Q nnM(b)
M(b)EE Q*
P
(M(b),
44.15.1, ZZ 5
<Oq(M(b)).
(M(b),q).
q). By 44.15.1,
Oq(M(b)).
Signalizerfunctors
Signalizer
functors
238
So in any
any case,
case, Z (
< Oq
(M(b)) for each b E B'.
B#.Hence
Hence by
by44.15.2,
44.15.2, Z E V
V(q,
Soin
O,(M(b))foreach
( q ,7r).
n).
For
a
E
A#,
pick
Ma
E
M(a,
7r(9))
and
let
P
E
Q*(G,
p).
As
I
<
Ma
and
For
A#, pick Ma E M ( a , n ( 0 ) ) and let P E Q*(G, p). As I (
I isis transitive
transitive on Q*(G,
Q*(G, p), PP nnMa
Ma EE Q*(Ma,
Q*(Ma,p). Thus Z(P)
Z ( P )<Op(MQ)
5 Op(Ma)by
44.15.1 and 44.2
44.21.
44.15.1
1. Therefore
< PO
Po== n Op(Mb).
zZ(P)
( p )5
op(Mb).
bbEB"
€BU
Next, ((Z,
so ((Z,
Let U =
= (ZI
Next,
2 ,I)
I )<(0(b)
8(b)<5Mb
Mb for
for each b E B#, so
2 ,II)) acts
acts on P0.
PO.Let
( z ' }),,
)/Vt hthe
setof
ofW
WEEQ(G)
Q(G) with
with I <5 NG
(W), W =
= Op
(W) U and P
Po
< Op(W).
Op (W).
and W
e set
Nc(W),
Op(W)U
o5
(44.22) Po EE VV(p,
andFF(M)
=Op(M)
Q* (G).
( p ,n7r(0)),
( 8 ) ) ,and
( M )=
O p ( M )for each M E Pp0
(G).
Proof.
< L.
Proof. First
FirstPo
PoEE V(p,
V ( p ,n-(9))
n ( 8 ) )by
by 44.15.2.
44.15.2. Let
Let LL EE Q*(G)
Q*(G)with
with 0(NG(PO))
6(NG(Po))(
p,
then
As
Po
E
V(P,
7r(9)),
Po
<Op(L).
If
10Or(L)
for
some
prime
r
As
E V ( P ,n(8)),Po 5 Op(L).If 1 # O,(L) for some prime # then as
as
for
some
B#. But X
L, SO
so by
B is
is noncyclic,
noncyclic, 11 # Co,{L}(b)
C O ~ ( ~for
) (some
~ ) b E B'.
X ==CF{M6}(Po)
CF(Mb)(PO)
5< L,
44.17.2, r EE 7r(F(Mb)),
n(F(Mb)),contrary to 44.21.
Thus F(L)
F ( L )=Op(L),
= O p ( L so
) ,sothe
thelemma
lemmafollows
followsfrom
from44.18.1.
44.18.1.
(44.23)
Q* (G) and
(44.23) Let M,
M, L E
E Q>o(G)
and Op(M)
Op(M<
)(RR EE Q(M,
Q(M,p)
p) with
with B(NG(C))
~ ( N G ( C<
(
) )L
R. Then
Then M
M=
= L.
for each 11 # C char R.
Proof.
As
Proof. Let
Let X
X ==ROp,F(M).
ROp,F(M).
As QQ=Op(M)
= O p ( M<)(R,R,RREESylp(X),
Sylp(X),and
and by
by 44.22,
44.22,
= F(M),
F(M), so
Q=
so QQ==F(X).
F ( X )Thus
. Thusby
byThompson
ThompsonFactorization,
Factorization, 32.6, we have
X=
= Nx(J(R))Cx(Z(R)),
Nx(J(R))Cx(Z(R)), so
As Q =
=Op(X),
X
so X
X <5LL by
by hypothesis.
hypothesis. As
O p ( X )Q
,Q==Op
0,
C
char
Q.
In
particular
(L
<M
M for
( L nnM).
M).By
By 44.10.3,
44.10.3, 0(NG(C))
O(NG(C))5
for each
each 11 # char
particular
NoP(L)(Q)
Q, so
so O
Op(L)
Q by
by Exercise
3.2.1. But
But now
N
o , ( ~ ) ( Q5) <Op(L
O p ( LnnMM)
) ==Q,
p ( L )5< Q
Exercise 3.2.1.
now
hypotheses of
of ((L,
symmetry, Q 5
< Op(L).
Op(L).
((M,
M ,Q,
Q ,L)
L )satisfies
satisfies the hypotheses
L , R, M),
M), so by symmetry,
Therefore Q
so M =
= L by
Q ==OOpp (L),
( L ) ,so
by 44.10.3.
44.10.3.
(44.24)Let
LetWEW,S=Op(W),Y=OP(W),LEQ*(G)with
(44.24)
W E W , S = O p ( W ) ,Y = OP(W),L E Q*(G)w i t 0(NG
h B ( N ~(Y))<L,
( Y )5) L ,
and M E Q*W(G).
QT w(G).Then
((1)
1) Y
y <Op,q(M)
5 Op,q(M).
(2) IfIf Op(M)
= L.
(2)
O p ( M )<5 SS then M =
Proof. Recall
Oqn
p,q} (M)by
by definition
definition of
Recall ZZEEV(q,
V ( q 7r),
,n),sosoZ Z< (
0,"(M)
(M=
)=O{
O1p,ql(M)
V(q,
n). By construction
< ZZ(Q)
E Q*(G,
Q*(G,qq),
and as
as II <
M,,
V
( q ,n).
construction ZZ 5
( Q ) for some
some Q E
) , and
5M
(M). Thus
Thus as ZZ <(O{p,q}
(M),ZZ5OOP,,(M)
p,q (M) by 31.10.
Q nnM
M EE Sylq
Syl,(M).
OIp,,)(M),
31.10. Hence
Hence
U=
= (ZI)
But W
W=
= UOp(W),
UOp(W),sosoYy==O'(W)
_ (Uw)
<Op,q(M),
(2')<Op,q(M).
5 Op,,(M). But
OP(W)=
(uW)
(Op,,(M),
establishing (1).
(1).
Suppose RR==Op(M)
and let
let M*
=M
M/R
= F(M*).
Suppose
O,(M) 5< SS and
M* =
/ R and
and X*
X* =
F(M*). By (1),
(I),
Y*
and as RR <(SS<(NG(Y),
=OP(YR)
a Nx(Y*),
_
Y* 5<Oq(X*)
O,(X*) and
NG(Y),Y
Y=
OP(YR)9
Nx(Y*), so
SO Nx(Y*)
Nx(Y*)=
Solvable signalizer
signalizerfunctors
functors
239
< L.
O p(L). Then NT(R)
< M by
Nx(Y) 5
L. Let
Let TT ==Op(L).
NT(R) 5
by 44.10.3, so
[NT(R)*,
Nx,(Y*)]
< NT(R)*n
n Nx*
Nx,(Y*)
= 1.
[NT(R)*, Nx*
(Y*)]5
(Y*) =
1.
Therefore
< CM.(X*)
< X*, SO
so NT(R)
NT(R) 5
< R and
Therefore by the A x B Lemma, NT(R)*
NT(R)*5
CM*(X*)5
hence TT <5R.
(C)) I
<M
M for
char R,
R, so
R. Finally
Finally by
by 44.10.3,
44.10.3, 0(NG
B(Nc(C))
for each
each 11 # C
Cchar
so
M=L
M = Lby
by44.23.
44.23.
If Mb =
= Mb'
b, b' E B',
B#, then the argument establishing
establishing 44.21 supplies
supplies
Mb, for all b,
a contradiction.
UI(G) for
contradiction.Therefore
ThereforeasasMb
MbEEQ*
Q*&,y1(G)
foreach
eachbbEEB#,
B#,IQP0U,(G)l
] Q>ouI(G)I >
>1.1.
Pick W
EW
W with
Pick
WE
with S=Op(W)
S = O,(W) maximal
maximal subject
subject to
to IQ*1(G)I
IQ*,,(G)l> > 1. Set
Set Y
Y=
=
OP(W) and pick M EE Q*(G)
(S)) 5
<M.
Q*(G) with 0(NG
B(NG(S))
M.
(44.25) For each N E
I(G) --{M},
(44.25)
E Q*
Q&,(G)
{MJ,SSisisthe
theunique
uniquemaximal
maximalWI-invariant
WZ-invariant
member of
of Q(N, p).
< NG
(SO). Then
Then S <
< Nso(S), so
Proof. Suppose
Suppose SS<<So
SoEE Q(N,
Q(N, p)
p)with
withWI
WZi
NG(So).
without loss, SS<9So.
So. Therefore
Therefore So
So <5 0(NG(S))
B(NG(S))<5M.
M.Let
LetWO
Wo==WSo.
WSo.Then
Then
WO
andMM,, N
N EE PWOI
Qw0I(G),
contradicting the
the maximal choice of S.
Wo EEWWand
(G), contradicting
S.
(G)contains
contains aa unique
unique member L.
(44.26) (1)
(1)Q*(NG(y))
QZ;(NG(Y))(G)
L.
= {M}.
(2) Qe(NG(s))(G)
QZ;(NG(S))(G)
=
(3) Q*
(G) =I {M,
(3)
PwI(G)=
M ,L}.
LJ.
!
Proof. Let
andNNEEPwI(G)
Q** (G)-- {{M}.
By 44.25,
44.25, 02(N)
02(N) <5S,
Proof.
LetLLEEQB(NG(y))(G)
Q&NG(Y))(G)
and
M I . By
S,
w
(G)I
1
L,so
so(1)
(1) and
and
so by
by 44.24.2,
44.24.2,NN =
= L. Thus (3)
(3)holds
holdsand
andasasI P
(Q*i
(G))>>1,1,MM#L,
(2) hold.
We are now in a position to
to obtain
obtain aa contradiction
contradictionand
and hence
hence establish
establish the
the
Solvable
< M for
Solvable 2-Signalizer
2-Signalizer Functor Theorem.
Theorem. By 44.26.2,
44.26.2, B(NG(C))
B(NG(C))5
for each
each
#C
(L) 5
::SS.S.Hence
HenceMM=
=LLby
by44.23,
44.23,contradicting
contradicting
1#
C char
charS,
S,while
whileby
by 44.25,
44.25, Op
OP(L)
44.26.3 and the choice of
of W with
with IIQ*
(G) I >
> I.1.
Pw1(G)I
This completes
2-Signalizer Functor Theorem.
completes the proof of the Solvable
Solvable 2-Signalizer
Remarks.
Gorensteinintroduced
introducedthe
theconcept
conceptof
of the
the signalizer
signalizerfunctor
functor and iniRemarks. Gorenstein
tiated the study of these
11. He
He was
was motivated
motivated by earlier work
these objects
objects [Gor
[Gor 1].
of Thompson.
?hompson. Goldschmidt [Gol 1,
1, Gol
Go1 2]
21 simplified
simplified some
some of
of Gorenstein's
Gorenstein's
definitions
definitions and proved
proved the Solvable
Solvable 2-Signalizer Functor
Functor Theorem and the
Solvable
(A) >
> 3. Glauberman
Solvable Signalizer
Signalizer Functor Theorem
Theorem for
for rr odd
oddwhen
whenMr
m,(A)
[Gl 1]
11 established the general theorem for solvable functors. Bender gave a
new
Theorem [Be 21;
2]; the
new short proof of
of the
the Solvable
Solvable 2-Signalizer
2-Signalizer Functor Theorem
proof given here is based on his proof,
although
our
proof
is
longer
and
proof, although our
is longer and more
more
Signalizerfunctors
Signalizer
functors
240
complicated than
than Bender's
Bender's for
for reasons
reasons I'll
I'll explain
complicated
explain in
in aamoment.
moment. Finally,
Finally,
the Signalizer
Signalizer Functor
Functor Theorem
Theorem subject to
to the
thehypothesis
hypothesis
McBride [Mc]
[Mc] proved the
that the composition factors of
of 0(a)
9(a) are known simple groups. The Classification Theorem allows this hypothesis
hypothesis to be removed.
removed.
Our proof
proof is longer than Bender's because
because Bender uses the
the ZJ-Theorem,
ZJ-Theorem,
the ZJ-Theorem is not proved
whereas we use Thompson Factorization, since the
There are
are at
atleast
least two
two advantages
advantages to
tothis
thisapproach:
approach: First,
First, the
theincrease
increase
in this text. There
theproof
proof caused
caused by
bynot
notappealing
appealing to
tothe
theZJ-Theorem
ZJ-Theorem is
isprobably
probably
in the length of the
less than the length of the proof of the ZJ-Theorem, although
although of course
course the
the ZJTheorem is of interest in its own right. Second, the change gives some insight
insight
Factorization is
is used
used in the literature,
into how Thompson
Thompson Factorization
literature, and
and Thompson
Thompson
Factorization is used often, while the ZJ-Theorem
ZJ-Theorem is
is not.
not.
In the first edition of the book, we gave a proof of
of the
the Solvable
Solvable Signalizer
Signalizer
proof given
given here
here but
but
Functor Theorem for all primes which was similar to the proof
still more complicated. In this edition we have opted for less generality in the
hope
hope that aa simpler
simpler proof
proof will
will better
better expose
expose the
the underlying
underlying concepts.
concepts.
Exercises for
for chapter
chapter 15
= ((1,
2)(3, 4), (3,4)(5,6)),
(3, 4)(5, 6)),and
and0(a)
9(a)=
= Os(CG(a))
03(CG(a)) for
for aa E
c
1. Let G
G ==A7,
A7, A =
((1,2)(3,4),
A#.
Prove 09 is a solvable A-signalizer functor on G which is not complete.
A'. Prove
complete.
Assume the lemma
lemma is false and choose P,
2. Prove
Prove lemma
lemma 44.7. (Hint:
(Hint: Assume
P , QQEE
Q*(G, p)
p) not conjugate under
under O(CG(A))
9(CG(A))with
withPP n
fl Q maximal subject to
Q(G,
Prove PP n
fl Q # 1 by
1 # CQ(a)
CQ(a)for
these constraints. Prove
by observing
observing Cp(a)
Cp(a) # 1
some aa EE A'
A#,
andconsidering
considering0(a).
9(a).Then
Thenconsider
considerNG(P
NG(PflnQ)/(P
Q)/(P fl
fl Q)
, and
and use 44.3.)
r-subgroup of
of a finite
finite group
group G
G and
and 09 an A3. Let
Let A
A be
be aa noncyclic
noncyclic abelian r-subgroup
signalizer functor
(b): bbEE B#)
letWB
WE==(9(0(b):
B') and
andlet
letW
W==WA.
WA.
signalizer
functoron
on G.
G.For
ForBB<5AAlet
Prove
WE.
(1) IfIf BBisisaanoncyclic
noncyclicsubgroup
subgroupof
of AAthen
thenW
W==WB.
(2) If
If 6(a8)
0(as)==6(a)9
0(a)gfor
foreach
eachg gEEGGand
anda aE EA#,
A', then
thenF2,A(G)
r 2 , ~ ( G<5)NG(W),
NG(W),
where I'2,A(G)
=) (NG(B):
(NG (B):BB5< A,
A, m(B)
m(B) 2
> 2).
r 2 , ~ ( G=
2).
(3) Assume
r') denote the set of AAssume the
the hypothesis
hypothesis of
of (2)
(2) and
and let
letVIG(A,
MG(A,r')
Ainvariant rt-subgroups
r'-subgroups of G.
r')5c NG(
NG(W).
MG(A, rt)
W). IfIf 09 is comG. Prove
Prove VIG(A,
plete, G =
= F2,A(G)
and O,t(G)
Or'(G) =
= 1,
9(a) =
=11for
r2,A(G)and
1, prove 8(a)
forall
allaac-E A#.
A'.
4. Let
LetAA be
be an
an elementary
elementaryabelian
abelianr-subgroup
r -subgroupofofaafinite
finitegroup
groupG
G with
with m
m(A)
(A) >_> 3.
3.
Prove
Prove
(1) IfIf aaEEA#
(a) is solvable then CG(a)
CG (a)isisbalanced
balancedfor
forthe
theprime
primer.
(1)
A' and
and CG
CG(a)
r.
(See section 31
definition of balance.)
3 1 for the definition
(2) If
If CG(a)
CG(a)is balanced for the prime r for
for each
each aa cE A#
A# then 9
0 is an
an AAsignalizer functor on G, where @(a)
9(a) =
=Or,(CG
(a)).
O,!(CG(a)).
2411
24
Solvable
functors
Solvable signalizer
signalizerfunctors
solvable for
for each
A# and
(3) Assume
(3)
Assume CG(a) is solvable
each aaEEA'
and GG==F2,A(G).
I'~,A(G).
Prove Or1(CG(a))
Or'(CG(a)) <
for
and ifif O,!(G)
Or'(G) =
=11then
5 Or'(G)
Or!(G)
foreach
eachaac-E A#, and
then
F*(Cc(a))
=Or(CG(a)).
Or(Cc(a)).
F*(CG(a)) =
5. Let
Letrrbe
beaaprime,
prime,GGaafinite
finitegroup,
group, Era
Ers =EA
A <5G,
G and
, and7rn aaset
set of
of primes
primes with
rV7r.ForEr2=B
<Gdefine
r e n . For E,z Z B 5 G define
(nO,r(CG(b))
bEB#
ac0(,)(B)
for
hyperplane BB of
of
~ , ( ~ ) (<I:
BO,(CG(c))
O,(CG(c))
)
foreach
eachccc-E A# and each hyperplane
Assume U
A. Also either
either assume
assume the
the Signalizer
SignalizerFunctor
FunctorTheorem
Theoremororassume
assumeOn
0,(CG(c))
(CG(C))
solvable for each c Ee A#.
is solvable
A'. Prove aa(B)
( B )is
is aa 7r-group
n-group for each hyperplane B
of A and aa(B)
( B )isis independent
independentof
of the
the choice
choice of B. (Hint:
(Hint: Define
yB(a) = [On(CG(a)), B](O,,(CG(B)) fl C(a))
= BB 5< A
A and aa EE A'
A#.
Prove yB
for EP2
EP2 =N
. Prove
y~ is an
an A-signalizer
A-signalizer functor
functor and
and
aa(B)
( B )<5a(D)
a ( Dfor
)foreach
eachpair
pairB,
B ,DDofofhyperplanes
hyperplanesof
of A.)
A.)
16
Finite simple groups
To my mind the theorem classifying
classifying the finite
finite simple groups is the
the most
most imimportant result
preface, the Classifiresult in finite group theory. As I indicated in the preface,
cation Theorem is the foundation
foundation for
for aa powerful
powerful theory of finite
finite groups
groups which
which
cation
group theoretical questions
questions to questions
questions about
about
proceeds by reducing suitable group
is devoted
devoted
representations of
of simple groups.
groups. The final chapter of this book is
to aa discussion
discussionof
of the
theClassification
ClassificationTheorem
Theoremand
and the
thefinite
finitesimple
simple
primarily to
groups
groups themselves.
themselves.
Sections 45 and 46 introduce two classes of techniques useful in the study
of simple groups. Section 45 investigates consequences of
of the fact that each
of involutions
involutions generates a dihedral
dihedral group. The two principal
principal results of
pair of
the section
section are
are the
the Thompson
Thompson Order
Order Formula
Formula and
and the
the Brauer-Fowler
Brauer-Fowler TheTheorem. The Thompson
Thompson Order
Order Formula
Formula supplies
suppliesaa formula
formulafor
for the
the order
order of
of a
two conjugacy
conjugacy classes of involutions
involutions in terms of
of
finite group with
with at least two
the
involutions. The
The BrauerBrauerthe fusion of those involutions
involutions in the centralizers of involutions.
there are
are at
at most
most aa finite
finitenumber
number of
of finite
finite simple
simple
Fowler Theorem shows that there
groups possessing an involution whose centralizer
centralizer is isomorphic
isomorphic to any given
groups
group.
group.
Section46
46 considers
considersthe
the commuting
commutinggraph
graph on
on the
the set
set of
of elementary
elementary abelian
abelian
Section
p-subgroupsof
of p-rank
p-rank at
at least
least kk in
in a group
group G.
G. The
The determination
determinationof the groups
groups
p-subgroups
which this graph
graph is disconnected
disconnected for small k plays a crucial
crucial role in
in the
the
for which
Classification
Classification Theorem.
Theorem.
Section
finite simple groups.
groups. The groups
groups
Section47
47 contains
containsaa brief description
description of the finite
Lie type
type are
are described
describedas
as groups
groupswith
with aa split
split BN-pair
BN-pair and generated
generated by
by root
root
of Lie
groups
groups satisfying
satisfying a weak
weak version
version of
of the
the Chevalley
Chevalley commutator
commutator relations.
relations. The
The
last portion of section 47 explores
explores consequences
last
consequences of these axioms. In particular
existence of Levi
Levi complements
complements is derived,
derived, it is
is shown
shown that
that the
the maximal
maximal
the existence
p-groups of groups
groups in characparabolics are the maximal overgroups of Sylow
Sylowp-groups
teristic~,
andthe
the Borel-Tits
Borel-Tits Theorem
Theoremis
(essentially)proved
provedfor
forfinite
finitegroups
groupsof
of
teristicp,
and
is (essentially)
Lie
Lie type.
type. This
This last
last result
result says
saysthat
that in
in aa finite
finite group
group of Lie
Lie type and
and characteristic
characteristic
p,
p, each
each p-local
p-local is
is contained
contained in
in aa maximal
maximal parabolic.
Finally section
section 48
48 consists
consistsof a sketchy outline
Finally
outline of the
the Classification
Classification Theorem.
Theorem.
This discussion
discussion provides
provides aa nice
nice illustration
illustrationof
of many of the techniques
developed
This
techniques developed
in earlier
earlier chapters.
chapters.
finite groups
Involutions in
inJinite
243
45 Involutions
Involutions in
in finite groups
Section 45 seeks
following property of involutions established
Section
seeks to exploit the following
established in
Exercise 10.1:
10.1:
(45.1) Let
group G. Then
Then (x,
Letxx and
andyybe
be distinct
distinctinvolutions
involutionsof
of a group
(x,y)
y)isis aa dihedral
dihedral
21xy 1.
group of order
order 21xyl.
Throughout section 45, G will be assumed to be a finite group. To
To begin
begin let's
let's
look more closely
closely at
at 45.1:
(45.2) Let x and y be distinct
distinctinvolutions
involutionsininG,
G,nn== Ixyl,
IxyI,and
andDD== (x, y).
y). Then
Then
(45.2)
(1) Each
Each element
element in D --(xy)
(xy)isisan
aninvolution.
involution.
oddthen
thenDDisistransitive
transitiveon
onits
itsinvolutions,
involutions, so
so in
in particular
particular x is
is
(2) IfIf nnisisodd
to y in D.
conjugate to
one of
of x,
x, y,
eventhen
then each
each involution
involution in D is conjugate
conjugate to exactly one
(3) IfIfnnisiseven
involutionin
in (xy).
(xy).Further
FurtherzzEEZ(D).
Z(D).
or zz,, where zz is the unique involution
xz is
is conjugate
conjugate to
to xx in
in D
D
(4) IfIfnnisiseven
evenand
andzz isisthe
theinvolution
involution in
in (xy)
(xy) then xz
if and
and only
onlyififnn=
- 00 mod
if
mod 4.
= v-1
Proof. Let
Let uu ==xy
xyand
and U
U ==(u).
(u).Then
ThenuX
uX==u-1
u-' sosoVX
vX=
v-' for
for each v E U.
U.
v x )=
~= V
vvX
= vv-'
==1,
1, soso(1)
holds.
Further
forfor
wE
( V(vx)'=
X )=~vx
Then ((vx)2
VX
= vv-1
(1)
holds.
Further
w UE,U,
vxw=
_
VW-'wXx
VW-~X
soxD
, ==
(v2x:Eu U}.
E UIn
] .Inparticularifn
oddthenxD
vw-1wXx =
= vw-2x,soxD
{v2x:v
particular if n isisodd
then XD =
D --UUand
andUUcontains
containsno
noinvolutions,
involutions,so
so(2)
(2)holds.
holds.
So take n even
even and let z be the involution in U.
U . Then
D-U
{V2x: V E U} U {V2ux: V E U}
with ux
= y,
U U= xD
U yD
(3)(3)
holds.
ux =
y,so
soDD- = xD
U y and
and
holds.Finally
FinallyzxzxEEXD
x precisely
precisely
when zz is a square in U
U;; that is when
when nn = 00 mod 4.
Z(G) =
= 1,1,let
involutions
let m
m be the number of involutions
(45.3) Let
LetG
Gbe
be of
of even order with Z(G)
in G, and n =
Then
G possesses
a proper
=Gm.
IGllm.
Then
G possesses
a propersubgroup
subgroupofofindex
indexatatmost
most
2n2.
2n2.
Proof. Let
Proof.
LetIIbe
bethe
theset
setof
ofinvolutions
involutionsof
of G,
G, RR the
the set
set of
of elements
elements of G inverted
by a member of I,
I, and (xi ::00 5< i <5k)
k)aaset
setof
of representatives
representatives for the conjugacy
classes
classes of Gin
G inR.
R.Pick
Pickxo
xo==1 1and
andlet
letmi
mi==IxG I,I,let
letBi
Bibe
bethe
theset
set of pairs (u,
(u, v)
v)
with u, v EE IIand
xi, and
let bibi==
I Bi
I.
anduv
uv==xi,
andlet
IBil.
Observe
firstthat
thatififu,u,vvEEII then
theneither
eitheruu== vv and
and uv
uv =
= 1 or (u,
Observe first
(u, v) is dihedral.
In either
either case
caseuuinverts
invertsuv,
uv,sosouv
uvEE R.
R. Thus
Thuscounting
countingII xx II in
in two
two
hedral. In
Finite simple groups
244
different
different ways,
ways, we
we obtain:
obtain:
k
(a)
m2=II x71 =
k
i-0
mibi.
Moreover
involution ti inverting xi and
and ifif (u,
(u, v) E Bi
Moreover there
there is an involution
Bi then
then uu inverts
inverts
xi and v ==uxi.
u isu an
injinjection
ection ofof
BiBi
into
ti tiCG
(Xi) so
uxi.Hence
Hencethe
themap
map(u,
(u,v)v)-rH
is an
into
CG(X~)
SO
bi <
tiCG(xi))== ICG(xi)l.
I CG(xi)I.
Also
IGCc(xi)l,
: CG(xi)I,
so mibi
< G.IfIfi i=
=0
bi
iIItiCG(xi)l
Also
mimi
==IG:
so mibi
i IGI.
we can be more precise: mo
mo =
= 11 and
and bo =
=m.
m. Combining
Combining these
these remarks with
(a) gives:
+
m2
m klGI.
m25<m+kIGl.
(b)
(b)
Let s be
of a proper
subgroupofofG.
G.IfIf ii > 0 then
be the
the minimal
minimal index
index of
proper subgroup
Ek
by hypothesis xi
xi V
Z(G), so
G ::CG(xi)I
> s.
6 Z(G),
so mi
mi ==I IG
CG(xi)l L
S . Hence
Hence IGI
(GI>
1 ~ of mi
3
ml =
>
11 + ks,
ks, which
which II record
record as:
as:
+
k < (IGI - 1)/s.
(c)
Combining (b) and (c)
(c) gives:
gives:
((d)4
m2
l)/s) + m.
m25<(lGI(IGI
(IGI(IGI--1)/s)+m.
Then, as
as nn =
= IGI
/m,ititfollows
followsfrom
from(d)(d)
that
s <n(n
n(n--mP1)/(l
m-1)/(1-- m-I),
m-'), and,
IGllm,
that
s 5
and,
as m
m2
> 2, s <5 2n2.
2n2.
(45.4) Let G
in G, and
(45.4)
G be
be aa finite
finite simple
simple group of even order, tt an involution in
CG(t)I. Then
Then IG1
IGI (
< (2n2)!.
n ==IlCG(t)l.
(2n2)!.
Proof.
H 9f index
Proof. By 45.3, G possesses
possesses a proper
proper subgroup
subgroup H,gf
index at
at most
most 2no,
24,
I /m and
and rn
in is the number of involutions
tG 1I=
=
where no ==II G l/rn
involutions in
in G.
G.Then
Thenmm>qI ItG
(G
CG(t)1,SO
sono
noi
< IGJIJG:
IGI/IG : CG(t)J
CG(t)1 =
= n.
IG : CG(t)l,
Represent
G as a permutation
group on
on the
the coset
coset space
space G/H.
G/H. Then
Then k =
=
Represent G
permutation group
IIG/HI
G/HI 5
< 2n2,
2n2, and as G is simple
simple the representation
representationis
is faithful.
faithful. So
So G
G is
is isomorisomorphic to a subgroup
and hence
hence JGI
IGI (
< k!
k! <5(2n2)!.
(2n2)!.
subgroup of the symmetric
symmetric group Sk, and
As an immediate
immediate corollary to 45.4 we have the following theorem of Brauer
and Fowler:
Fowler:
(45.5) (Brauer-Fowler
(Brauer-Fowler [BF])
[BF]) Let
Let H be
(45.5)
be aa finite
finite group.
group. Then
Then there
there exists
exists at
at
most a finite number of finite simple groups G with an involution t such
such that
that
H..
CG(t) = H
Recall
and Thompson
Thompson says
says that
that nonabelian
nonabelian
Recall that the Odd Order Theorem of Feit and
finite simple groups
groups are of
of even
even order
order and
and hence
hencepossess
possess involutions.
involutions. The
The
Brauer-Fowler Theorem
Theoremsuggests
suggestsitit is
is possible
possible to classify
classify finite
finite simple
simple groups
groups
~
245
Connected groups
in terms of the
the centralizers
centralizersof
of involutions.
involutions. This approach
approach will
will be
be discussed
discussed in
in
more detail
detail in
in section
section 48.
48.
has more
more than
than one
one class
class of
of involutions
involutions it is possible to make a much
If G has
precise statement
statementthan
than 45.5:
45.5:
more precise
(45.6) (Thompson
Assume G
G has kk 2
> 22 conjugacy
(Thompson Order
Order Formula) Assume
conjugacy classes
classes of
involutions x f , 1 <
5 ii<5k,k,and
anddefine
defineninitotobebethe
thenumber
numberof
ofordered
ordered pairs
pairs
(u,v)with
E x f ,,v
u Ex2,andxi
E xf, and xicE(uv).
(uv).Then
Then
(u, v) with u Ex1
k
IGI = ICG(xi)IICG(x2)I
ni/ICG(xi)I
i=1
Proof.
in G
G and
and Q
Q=
= xG
ProoJ Let
Let IIbe
bethe
theset
setof
of involutions
involutions in
xf x x2
x.: . Then
(*)
ICI =
JXGJ JXGJ
2 = IG : CG(xi)I IG : CG(x2)I
1
For
so by
by 45.2
45.2 there is a unique involution z(u, v) in
E Q,
a,u $ vG, so
For (u, v) E
(uv). Hence
(uv).
Hence Q
Q isis partitioned
partitioned by
by the
the subsets
subsetsQZ
Q, consisting
consisting of those
those pairs (u,
(u, v)
v)
z(u, v).
v). Thus
Thus IQI
1Q1 =
lQzl.
Further
1Q,1
=
1Q,1
for
z
E
xc,
so
IQz1.
Further
IQ,I
=
IQx,
I
for
z
c
xP,
so
with zz =
= z(u,
= C,,,
>-+ZEI
Y IQzl = IxGIIQxiI = IG:CG(xi)Ini
ZExG
Therefore
Therefore
k
k
101
(**)
=
IG:CG(xi)Ini.
i=1
' Of course
course the
the lemma
lemma follows
followsfrom
from(*)
(*) and
and (**).
(**).
fl CG(xi),
CG (xi ),j j=
= 1,
2, is known
known
Notice that the integer ni can be calculated
calculated if xxyg n
1,2,
for each i. Hence
Hence the order of G
G is
is determined
determined by the fusion of xl and
and x2
x2 in
in
CG(xi),
< ii <5k.k.InInparticular
CG(xi),11 5
particularthe
theorder
orderofofGGcan
canbe
bedetermined
determinedfrom
fromthe
the
fusion of involutions
involutions in local subgroups.
subgroups.
46 Connected
Connected groups
In this section G is aa finite group and pp isis aa prime.
prime.
collection of subgroups of
of G permuted
permuted by G via conjugation,
conjugation, define
If Q is a collection
bethe
thegraph
graphon
onQQobtained
obtainedby
byjoining
joiningAAtotoBBifif[A,
[A,B]
B] =
= 1.
g (5(Q)
Q ) totobe
1.Evidently
Evidently
via conjugation.
conjugation.
represented as
as aa group
group of automorphisms
automorphismsof g (5(Q)
Q ) via
G is represented
(46.1) Let A be a G-invariant collection
collection of
ofsubgroups
subgroupsofofGGand
andHH 5< G. Then
the following
following are
are equivalent:
equivalent:
(1) H
Hcontrols
controls fusion
fusion in H fl
n A and NG(X) _(< H
H for
for each X E H
H flnA.
A.
(2)
forgg EEGG-- H.
(2) H
H flr l Hg
Hg flnA
A is
is empty
empty for
H.
246
Finite simple groups
(3) The
the permutation
permutation representation
representation
Themembers
membersof
of HHfli lA
Afix
fix a unique point in the
of G on G/H
G I Hby
byright
rightmultiplication.
multiplication.
(4) H
ofLB(A),
2(0), and
theunion
union of
of aa set
set Fr of
ofconnected
connected components
components of
and
H fli l AAisisthe
is emptyforgEcGG-H.
rrflnFg
rgisemptyforg
- H.
Proof. Assume
Assume (1) and let g cE G
G with
with H
H fli lHg
Hgfli lAAnonempty.
nonempty. Then
Then there is
X E H fl
<H
= X for
n A with Xg
Xg 5
H so,
so, as
as H
H controls
controls fusion
fusion in H fl
i l A,
A, Xgh
xgh=
for
some hh EE H. Then gh E NG(X)
NG(X) <
5 H,
H ,so
sogg EEH.
H .Thus
Thus(1)
(1)implies
implies (2).
(2).
and consider
considerthe
therepresentation
representationofofGGon
onGG/H.
X E H fl
Assume (2) and
I H. Then X
i lA
fixes Hg
Hg if and only
only ifif X <
whengg Ec H by
5 Hg,
Hg, which
which holds precisely
precisely when
by (2).
(2).
fixes
Thus (2) implies (3).
Assume (3).
(3).Then,
Then,for
forAAEEHH i fll A,
A, {{H}
= Fix(A), so
so NG(A)
NG(A)5< H
H,, as H is
H )=
is
Assume
the stabilizer
stabilizer in
in G
G of
of the
the coset
coset H
H.. In particular
if
B
E
A
is
incident
to
A
in
particular if
incident
1(0) then
soso
B BE EH H
fl iA.
Thus
LB(A)
thenBB<5C(A)
C(A)<5H,H,
l A.
ThusHHfli A
l Aisisthe
theunion
union of
of some
of connected
connectedcomponents
componentsofofg(A).
2(0). Further if
if A EE 6'8 E Fr and
set rF of
and 8g
6'g E F
r then
Ag (
< H,
Ag
H, so
so{H}
{ H )==Fix(Ag)
Fix(A8)=={Hg}
{Hg)and
andhence
hence gg EE H.
H .So
So(3)
(3)implies
implies(4).
(4).
Finally assume
assume (4).
(4).IfIfXXEEHH i fll A
A and
andgg EE G
Gwith
withXg
X g<
X EE 6'9 E I'r
5 H, then X
andXg EE 6"
8' EE rI',so8'
andXg
, SOB' ==8g
6'8 E
E I'nFg.Hence,
r n r g . Hence, by(4),g
by (4), g E
E H.So(4)implies(1).
H. So (4)implies (1).
If k is
(G) for the set
is aa positive
positive integer
integer write
write Sk
&l(G)
set of
of all
all elementary
elementaryabelian
abelian ppsubgroups of G of p-rank at least
least k. G
G isis said
said to
to be
be k-connected
k-connectedfor the
the prime
prime
p if 0(61k(G))
g(&j!(G)) is connected.
p-group. Then
(46.2) Let G be aap-group.
(1) G
G isis 1-connected
1-connected for the prime p.
= U<
(G) isis in
m(G) >>22then
then there
there is Ep2
Epz E
U: G
G and
and X E
E S'
&[(G)
in the
the same
(2) If
If m(G)
connected component
componentof
ofg(&,P(G))
0(S' (G)) as
asUuprecisely
precisely when
when m(CG(X))
m(CG(X)) >
> 2.
2.
(3) IfIf pp==2 2and
has
a normal
andG G
has
a normalE8-subgroup
E8-subgroupthen
then GGisis2-connected
2-connected for
the prime 2.
2.
(4) IfIf pp==3 3and
3,3,then
andm(G)
m(G)>>
thenGGisis2-connected
2-connectedfor
forthe
theprime
prime 3.
3.
Assumem(G)
m(G) >
> 2;
Proof. Part
Part (1)
(1) follows
follows from
fromthe
thefact
factthat
thatZ(G)
Z(G) #
,- 1. Assume
Proof.
2;
by Exercise
Exercise 8.4
8.4there
thereisisEp2
Ep2E= U
U<
G/CG(U) ( SL2(p)
9 G. G/CG(U)
SLz(p) and SL2(p) is of
p-rank 1. So if A E 63(G)
> 2,
&;(G) then
then m(CA(U))
m(CA(U)) 2
2, so
so m(CA(U)U)
m(CA(U)U) >
> 2.
2.Hence
Hence
m(CG(U)) >
> 2.
if u
U ,(G) isisininthe
2. Also, if
# EEEES'&,P(G)
thesame
sameconnected
connected component
as U, then there is E # D EE °z
toEE and
andm(CG
m(CG(E))
>
8;(G)
(G) adjacent to
(E)) 2>mm(DE)
(D E) 2
3. Thus
Thus to prove (2) it remains
remains to
(G)isisin
inthe
theconnected
connected
to show
show that
that each
each A
A EE 613
&f(G)
component of U.
U. But
ButA,
A ,CA(U),
CA(U),CA(U)U,
CA(U)U,UUisisa apath
pathinin_T(6'2P(G)),
LB(&;(G)), so
SO the
the
proof of (2) is complete.
complete.
VaVG.
LetLet
E EE 6122(G);
and E8
E8-E
9 G.
E &$(G);to
to prove
prove (3)
(3) we must exAssume p ==22and
a path from EE to
to V.
V.For
Foree cE E,
(CV (e)) >
> 22 by Exercise
we may
may
hibit apathfrom
E,mm(Cv(e))
Exercise 9.8, and we
assume there is
is ee Ec E
CV
(E)), Cv
(e), VVisisaapathin
path in 2(d0z
E --V,V,sosoE,E,(e,(e,
Cv(E)),
Cv(e),
g ( & ;(G)).
(~)).
Connected groups
247
Similarly
hypothesis of (4) there is Esl
E81 = V
aG
Similarly under
under the hypothesis
V9
G by
by Exercise
Exercise8.11,
8.1 1,
after which we can use Exercise 9.8 and the argument of
of the
the last
last paragraph
paragraph to
to
establish
establish (4).
Given a p-group
acting as
as aagroup
groupof
of automorphisms
automorphisms of G,
G, and
and given
given aa
Given
p-group P acting
integer k, define
positive integer
~ P , L ( G=
(NG(X):X
P , m(X) >
>k)k)
I'P,k(G)
=) (NG(X):
X5
< P,
I'P
k(G) =
= (NG(X):
X 5< P,
r:,,(~)
(NG(X):X
P , m(X) >
> k,k, m(XCP(X))
m(XCp(X)) >
> k).
k).
The following
following observation
observation is an
an easy
easy consequence
consequenceof
of Exercise
Exercise3.2.1:
3.2.1:
(46.3) Let H <(G,
and
(46.3)
G,PPE ESylp(H),
Sylp(H),
andk ka apositive
positiveinteger.
integer.Then
Theneither
either of
of the
the
following implies that
that PP EE Sylp(G):
Sylp(G):
(1) m(P) >>kandFP,k(G)
kandrp,k(G) <
5H
(2) m(P)
<) H.
m(P) >>kkand
and I'P
r $ k(G)
. k ( ~5
H.
(46.4)
Let H <5 G,
(46.4) Let
G, P EE Sylp(H),
Sylp(H), and
and m(P)
m(P) >>k.
k. Then
Then the
thefollowing
following are
are
equivalent:
equivalent:
~ P , L ( G5
(1) FP,k(G)
<) H.
(2) HHcontrols
(H) and NG(X) (
<H
(H).
controlsfusion
fusionin
inolk
&i(H)
H for
foreach
each X
X EE 6k
&:(H).
(3) m
p(H ifll Hg)
Hg) <
< kforeachg
k for each gEEGG- - H.
mp(H
H.
(4) Each
(G) fixes a unique point in the permutation
&f(G)
permutation represenrepresenEachmember
memberofof&'k
tation of G by
G/H.
by right multiplication on GIH.
Proof.
(G)
Prooj Parts
Parts (2),
(2), (3),
(3), and
and (4)
(4) are
areequivalent
equivalent by
by 46.1,
46.1, except
except in
in (4),
(4), °k
&:(G)
should
be &:(H).
gk(H). This
implies (1)
(1) and
and (1)
should be
This weaker
weaker version
version of (4)
(4) evidently
evidently implies
and 46.3 show P EE Sylp(G),
Sylp(G),from
fromwhich
which the
the strong
strong version
version of (4)
(4) follows. It
remains to show (1) implies
implies (2).
(2).
FP,k(G) 5
< H and
(H). ItIt suffices
that ifif gg EE G
Assume rP,L(G)
and let X EE °k
&;(H).
suffices to show that
Xg 5
< H,
Xg) 5
<
H ,then
then gg EE H.
H.By
BySylow's
Sylow's Theorem
Theorem we may assume (X, Xg)
with Xg
P . By
By 46.3,
46.3, PP EESylp(G).
Sylp(G).By
ByAlperin's
Alperin'sFusion
FusionTheorem
Theorem there
there exists
exists Pi EE
P.
Sylp(G), 11 5< i <5n,n,and
= g1
Sylp(G),
and g1
gi E NG(P
NG(P fl P,)
Pi) with
with gg =
gl ..... .g,,,
g,, X <
5 P1,
PI, and
and
Xg,°.gi
<
P
f
Pi.
Then
m(P
f
Pi)
>
k,
so
g;
E
NG(P
f
Pi)
<
FP,k(G)
<
H..
Xg'...g' 5 'I fl
Then m(P fl Pi) 2 SO gi E NG(P fl Pi) 5 rP,L(G) 5 H
Hencegg =
= gl ...
Hence
. . .gg,EEH,Hso
, sothat
that(2)
(2)holds.
holds.
p-core of
If P EE Sy1p(G)
Sylp(G)then
then FP,k(G)
rP,L(G)is
is called
called the
the k-generated
k-generatedp-core
of G.
G. By
By Sylow
Sylow
k-generatedp-core of
conjugacy in
in G.
G. The
The hypothesis
hypothesis
the k-generatedp-core
of G
G is determined
determinedup to conjugacy
to the
the assertion
assertion that
that H
H contains the k-generatedp-core
k-generated p-core of
of
of 46.4 are equivalent to
G, and hence, if H is
is aa proper subgroup
subgroup of G, thatp-core
thatp-coreisis aa proper
proper subgroup
subgroup
k-generatedp-core is
of G. By 46.1, if the k-generatedp-core
is proper then G is
is k-disconnected for
248
Finite simple groups
in the following
following
the prime p, although
although the
the converse
converse need not be true. However in
special
special case the converse
converse is
is valid:
valid:
(46.5)
(46.5) Let P EE Sylp(G)
Sylp(G)and
andassume
assume PPisisk-connected.
k-connected.Then
Then
(1) ek
(1, p,k(G)) isis aa connected
(G)) and FP,k(G)
&;(rp,k(G))
connectedcomponent
componentofof_q((ffk
LB(&:(G))
rp,k(G)isis the
the
stabilizer
stabilizer in G of that connected
connected component.
component.
prime pp if and only if G has
(2) GGisisk-disconnected
k-disconnected for the prime
has aa proper
proper kgenerated p-core.
Proof.
(H) is
Proof. Evidently
Evidently(1)
(1)implies
implies(2).
(2). Let H ==i,P,k(G).
~ P , ~ ( By
G
By)46.1
46.1
. and
and 46.4
46.4 ek
&:(H)
is
(G)),while,
while,as
as PP is k-connected,
the union of connected
connected components
componentsofof_T((ffk
LB(&;(G)),
k-connected,
&[(P)
NG(A).Hence
Hence
(9k (P)isiscontained
containedininsome
somecomponent
componentAAand
andofofcourse
courseHH <_
< NG(A).
ek (H) C
= ,gk
(H) and
and H
H=
&:(H)
EA
A by
by Sylow's
Sylow's Theorem. So A =
&:(HI
=NG(A)
NG(A)by
by 46.1.4.
That is (1) holds.
If H is a proper subgroup of
of G satisfying any of the equivalent
equivalent conditions
conditions of
of
If
46.4 with k =
=1,1,then
then we
wesay
sayHHisisaastrongly
stronglyp-embedded
p-embedded subgroup of G. As
consequence of 46.5 and 46.2.1
a direct consequence
46.2.1 we have:
(46.6)
possesses aa strongly
strongly p-embedded
is
(46.6) G
G possesses
p-embeddedsubgroup
subgroupififand
andonly
only ifif G is
1-disconnected for the prime p.
p.
(46.7) Let mp(G) >>2,2,PPE ESy1P(G),
(46.7)
Sylp(G),and
andlet
let(92(G)°
&;(G)' denote
denote the
the set
set of
of subsubX EE &;(G)
o' (G) with
> 2. Then
groups X
with mp(XCG(X))
9(XCG(X)) >
(1) e2
(G)° isisthe
(G) which are not isolated in LB(&:(G)).
(o'2 (G)).
&:(G)'
theset
setof
ofpoints
pointsof
of e2
&:(G)
(2) Sz
(I'P 2(G))° is a connected component
component of
of LB(&;(G))
0(ff2 (G)) and
and rrPi ,2(G)
&;(ri,,(G))'
2 ( ~is)
the stabilizer
stabilizer in G of this connected component.
component.
(3)
1((ffz (G)°)
connectedififand
andonly
onlyififGG== I';,,(G).
F , 2(G).
(3) LB(&;(G)'
) isisconnected
Proof. Part (1) is trivial,
trivial, as
as is
is (3)
(3) given
given (2).
(2). So
So itit remains
remains to
to establish
establish (2).
(2).
By 46.2.2,
46.2.2, &;(P)'
e (P)° isiscontained
(G)).
containedininaaconnected
connected component
component A
A of
of ?(ff2
LB(&;(G)).
Thus H
H=
= rP
2(G) <
LetrF =
= &:(H)'
ez(H)°.
As &;(P)'
ff2(P)° E
cA
Thus
r;,,(G)
5 NG(A).
NG(A). Let
. AS
A and
and H acts
acts
on A, I'r CEAAby
bySylow's
Sylow'sTheorem.
Theorem.If
If A
A 0#17
r there is xx E F,
r , Y E A --17
r with
X and Y
in LB(A).
1(A). Without
loss X
XE
E &;(P)',
ff2(P)°, so
so YY 5
< CG(X) (
<H
H..
Y adjacent
adjacent in
Without loss
Hence, as mp(CH(Y))
mp(CH(Y)) 2
> mp(XY) >
2 3,
3, Y
Y EE F,
r ,aacontradiction.
contradiction.
So A ==F.r.Thus,
Thus,totocomplete
completethe
theproof
proof of
of (2),
(2), ititremains
remains to
to show
show that if
X, Xg
Xg EE rI' then g E H.
Xg EE rr.. Then
<H
H . Suppose
Suppose X, Xg
Then NG(X) 5
H >2NG(X9)
NG(Xg)
so there
there isis Ep3
EP3 = A
Ai
<: CG(X)
<
H
and
Ag
<
H.
By
Sylow
we
may
CG(X) i: H and A8 i: H. By Sylow we may take
take
A, A
Ag
g <
i:P.P.Now
Nowapply
applyAlperin's
Alperin'sFusion
FusionTheorem
Theorem as
as in
in the
the proof
proof of
of 46.4,
46.4,
17P,3(G)
H,, to conclude g E H.
using ~
P , ~ ( 5G<I';,2(~)
)I'P 2(G)5< H
H.
The finite
finite simple groups
249
mpp(G)
Lemma 46.7 says that ifif m
( G )>> 2 and P EE Sylp(G),
Sylp(G),then,
then, neglecting
neglecting the
the
1((g2 (G)),GGisis2-connected
2-connectedfor
forthe
theprime
prime pp precisely
precisely when
isolated points of LB(&:(G)),
G=
G
= ~O,,,(G).
F 2(G)
These observations
observations are very useful
useful in
in conjunction
conjunction with
withthe
theSignalizer
SignalizerFunctor
Functor
These
Theorem, as the next two
two lemmas
lemmas and
and Exercise
Exercise16.1
16.1 indicate.
indicate.
(46.8) Let
be the
the set
set of
of elements
elements aa of G of order
(a)) >
> 22 and
Let17
r be
order p with
with Mp(CG
mp(CG(a))
and
map from
from r17into
intothe
theset
setofofp'-subgroups
p'-subgroupsofofGGsuch
suchthat,
that,for
forall
allaa,
, bbEE rF
let 89 be a map
with [a,
[a, b] =
= 11 and
with
and all
all g EE G,
G ,9(ag)
B(ag) ==9(a)9
8(a)gand
and 9(a)
B(a)flf l CG(b)
CG(b)<(9(b).
8(b).Let
Let
, 2(G), Op-(G)
= 1,
P EE Sylp(G),
Sylp(G), assume G =
=FI';,,(G),
Op!(G)=
1 , and
and either
((1)
1 ) 9(a)
8(a)isissolvable
solvablefor
foreach
eachaaEE17,
r ,or
((2)
2 ) the
the Signalizer
SignalizerFunctor
Functor Theorem
Theorem holds on G.
G.
Then 8(a)
9(a) =
= 11 for each aa EE r.
F.
Then
Proof. For AAEE (P3
(G),8 9 isis an A-signalizer
d':(G),
A-signalizer functor
functor by
by hypothesis.
hypothesis. For
For
B EE (ffz
(G)°define
define W
WB
(9(b): bb E
E B').
B#). Then there
(G) with
&;(G)'
E == (8(b):
there exists
exists A
A EE6'3
&:(G)
with
B <5AAand,
and,by
byExercise
Exercise15.3,
15.3,WB
W E==WA
W Aand
andFA,2(G)
rA,,(G)<(NG(WA).
NG(WA).In particparticular ifif B and
(G) adjacent
adjacent in
in LB(&;(G))
1(ez (G)) then
ular
and D
D are
aredistinct
distinct members
members of S2
&:(G)
then
BD EE 6'3
WB
= WBD
= WD
Thus, as
2(G) =
= G,
46.7 says
=W
& f (G)
( G )so W
B=
W B D=
WD.Thus,
as F0°
I';,,(G)
G,46.7
says WB
WB=
W
of the
thechoice
choiceofofBB EE&;(G)'.
(9z (G)°.Thus
ThusGG==rrP
is independent of
! ,2(G)
, ( ~(<) NG(W)
NG(W)as
NG(B) <
NG(B)
(FA,2(G)
rA,,(G) <(NG(W).
NG(W).But,
But,by
by (1)
(1)and
and the
the Solvable
Solvable Signalizer
Signalizer Functor Theorem
Theorem or
or by
by (2),
(2), W
W isis aa p'-group,
p'-group, so W
W(
< Op,(G).
Op,(G)
tor
Opf(G).As O
p f ( G=
)= 11 by
hypothesis,
9(a) <(W
hypothesis, and as 8(a)
Wfor
foreach
eachaaEE17,
r,the
the lemma
lemma is established.
established.
be the
the set
set of
of elements
elements aa of
of G of order p with
(a)) > 2,
(46.9) Let
Let 17
r be
with mp(CG
mp(CG(a))
2, let
let
P EE Sylp(G),
F 2(G) ==G,GOp,(G)
==1,1,CG(a)
isisbalanced
Sylp(G),and
and assume
assume I';,,(G)
, Op!(G)
CG(a)
balanced for
for
the prime pp for
foreach
eachaaEE17,
r , and either
either
(a)) is solvable for
for each
each aa E
E r,
F, or
((1)
1 ) Op'(CG
OPt(CG(a))
((2)
2 ) the
the Signalizer
SignalizerFunctor
Functor Theorem
Theorem holds on G.
G.
Then Opt(CG(a))
Op'(CG(a)) =
= 11 for each
each aa E 1.
r.
Proof. For
By
9(a) fl
<
Proof.
For aa EEFrlet
let9(a)
8(a)==Opr(CG(a)).
Op~(CG(a)).
ByExercise
Exercise 15.4.2,
15.4.2,8(a)
f l CG(b)
CG(b)(
9(b)
with [a,
[a, b] =
8(b)for each a,
a ,bb EE17
r with
=1.
1.Hence
Hence 46.8
46.8 completes
completes the proof.
47 The
The finite
finite simple groups
Section
Section 47 describes
describesaa list
list .7C
YC of finite simple groups. The Classification Theorem says that any finite simple
simple group
groupisisisomorphic
isomorphictotoaamember
memberofofYC.
X. The
proof of the Classification
Classification Theorem is far beyond the scope
scope of this book,
book, but
but
there is a brief outline
outline of
of that
that proof
proof in
in the
the final
final section
section of
of this
this chapter.
chapter.
Finite simple groups
groups
250
described as the following
XC can be described
following collection of groups:
groups:
.
(1) The
The groups
groupsof
of prime
prime order.
order.
An of
of degree
degreenn >
> 5.
(2) The
The alternating
alternating groups A,
(3) The
finite
simple
groups
of
Lie
type
listed
The finite simple groups of Lie type listed in Table 16.1.
(4) The
The 26
26 sporadic
sporadicsimple
simplegroups
groupslisted
listed in
in Table
Table 16.3.
16.3.
The groups of prime order are
are the
the abelian
abelian simple
simple groups
groups (cf.
(cf. 8.4.1).
8.4.1). They
They
are the simplest of the simple
simple groups.
groups.
By 15.16,
groups A,,
An, nn >
15.16, the alternating groups
2 5,5,are
aresimple.
simple.The
Thepermutation
permutation
representation
of
An
of
degree
n
is
an
excellent
tool
for
investigating
the group,
group,
representation of A,
excellent
and, as in 15.3 and Exercise 5.3, to determine its conjugacy classes. Lemma
and Schur multiplier of A,.
An. Exercise
Exercise 16.2
33.15 describes the covering group and
unless nn =
= 6.
says Aut(A,) = S,,,
S,, unless
6.
The
analogues of the semisimple Lie groups. This
The groups
groups of Lie type are the analogues
class of groups is extremely interesting; for one thing by some measure most
finite simple groups are of Lie type.
type. We've already
already encountered
encountered examples
examples of
of
these
these groups; the
the classical
classicalgroups
groupsare
are of
of Lie
Lietype.
type.
The groups
representations:
groups of Lie type can be described
described in terms
tenns of various representations:
automorphisms of certain Lie
as groups of automorphisms
Lie algebras
algebras or
or fixed
fixed points
points of
of suitable
suitable
automorphisms of such groups; as fixed points of suitable
suitable endomorphisms
endomorphisms of
semisimple algebraic groups;
groups; as
as groups of automorphisms of
of buildings
buildings with
with
semisimple
finite Weyl
Weyl groups;
groups; as
as groups
groups with
with aa BN-pair
BN-pair and
and aa finite
finiteWeyl
Weylgroup.
group.I'll
I'll take
take
complications.
a modified version of this last point of view here to avoid complications.
A finite group of Lie type satisfies the following conditions:
conditions:
Weyl group
group W
W=
=
(Ll) GGpossesses
possessesaaTits
Tits system
system (G,
(G, B,
B, N,
N, S)
S) with
with aa finite
finite Weyl
N/(B
S!.
N/(Bflf lN).
N).The
TheLie
Lierank
rankofofGGisisthe
theinteger
integerI JSJ.
(L2) H
(L2)
H ==BBflnNNpossesses
possessesaanormal
normalcomplement
complement U in B.
(L3) There
There isis aa root
root system
systemZE for
for W
W and
and aa simple
simplesystem
systemnr for
for E
Z with
with
(L3)
S=
= (sa:
througha.a. Let
Let E+ be
(s,: a E njr)) where
where sa
s, is the
the reflection
reflection through
be the
the
of nrr.. There
Thereexists
existsan
aninjection
injectionaaF-+
H Ua
U, of
of EZ into
intothe
the set
set of
positive system of
subgroups of
of G
G such
suchthat
thatfor
foreach
eachaa EE ZE and
and ww EE W,
W, H
H<
5 NG(Ua)
NG(U,) and
(Ua)'
member
of of
UU
cancan
bebewritten
(U,), ==U.U,,Each
. Each
member
writtenuniquely
uniquelyas
as aa product
product
with u,
ua EE Ua,
of Z
E+
respecting height.
l-IaEC+u,,
U,, in
in some
some fixed ordering of
+ respecting
(L4) The root groups (U,:
(Ua: aa E E+)
(LA)
Z +)satisfy
satisfy
[Ua, UU] C (Z(Ua) fl Z(UU)) f Uia+lp,
a,/R E E+,
+
where the product is over all roots
roots iiaa + jB
j,8 with i and jj positive
positive integers.
integers.
We will be most interested
interested in groups
groups which
which also
also satisfy:
satisfy:
(L5) Let
Letw0
wo be
be the
the unique
unique member
member of W
W of
of maximal
maximal length
length in
in W
W (cf.
(cf. Exercise
Exercise
10.3).
ThenGG=
= (U,
fl B'O
=
10.3). Then
( U ,U'O),
UWo),B n
BWO
=H,
H,and
andW
W isis irreducible.
irreducible.
251
25 1
The$nite
The
finite simple groups
Some observations about these axioms are in
in order.
order. Define a Tits
Tits system
system
T ==(G,
( G ,B,
B,N,
N ,S)
S )to
to be
be saturated ifif fl w B'
BW==H.HBy
. ByExercise
Exercise14.9.5,
14.9.5, T
T is
is
saturated ifif and
and only
onlyififBBn
fl BWO
B"'0== HH.. So the
the condition
conditionUu n
fl UWO
U"'0=
= H in
in L5
can be replaced
replaced by the hypothesis that TT isis saturated.
saturated.
Next, from
from Exercise
Exercise 10.3,
10.3,ZE+wo
= E-.
f wo =
Z-.Hence,
Hence,by
by L3,
L3,47.1.1,
47.1.1, and
and 30.7,
30.7,
Ua
on LLl-L4
U, ==UUflf lU"'0,°
U wOSafor a EE r.n Moreover
. Moreoverthe
theproof
proof of
of 47.1 depends on
1-LA but
not L5. By 30.9 each
each yy E E
Z can
can be
be written y ==aw
a wfor
forsome
someaaE ETr,
n,W
w EE W.
W.
U,, _=(Ua)"'
(U,)W==(U
(UflnUwOS°)'.
UwOS~)W.
determine the
So, by L3, U, ==UaN,
Hence L1-LA
L1-L4 determine
EE
Z uniquely.
uniquely.
root groups U,,
Uy, yy E
Conversely if T
T isis aasaturated
saturatedTits
Titssystem
system satisfying
satisfyingL2
L2 then
then by
by Exercise
Exercise
14.10 we
wecan
candefine
defineU,Uy==(U(UnnUWOSa)W,
U'0'°)', where
whereyy =
= aaw,
and
w , a EE r,n,
andww EEW.
W.
Then, by Exercise
parts (7) and (9), L3
Exercise 14.10
14.10 parts
L3 is
is satisfied.
satisfied. Thus L3 can be
dispensed with
with ifif we
we assume
assume Ll
L1and
and L2
L2 with TT saturated.
saturated. Equivalently
Equivalently L1,
L1, L2,
L2,
dispensed
imply L3.
L3.
and L5 imply
essentially characterize the finite groups of
The properties L1 and L2 essentially
of Lie
type (cf.
(cf. [FS],
[FS], [HKS],
[HKS], and
and [Ti]).
[Ti]). There are
are some
some rather
rather trivial
trivial examples
examples of
satisfying L1-L5
whichare
arenot
notof
ofLie
Lietype.
type.The
Thefinite
finite
groups of Lie rank 11 satisfying
Ll-L5 which
groups of Lie type
type satisfying
satisfying L5 and
and possessing
possessing a trivial
trivial center are
are listed
listed in
in
Table 16.1. Column 1 lists G. The parameters qq and n are an arbitrary
arbitrary prime
power and positive integer, respectively,
respectively, unless
unless some
some restriction
restriction is
is listed exlists the order
order of a certain
of G.
G .Usually
Usually
plicitly. Column
Column 2 lists
certain central
central extension G of
G is
of G
G,, and dd(G)
is simple,
simple, G isis the
the universal
universal covering group of
( G )is
is the order of
the Schur multiplier
multiplier of
of G
G.. In
In any
any event
event ]GI
IGI =
= I61/d(G).
~ e l / d (Finally,
Finally,
~ ) . column
column 4
lists the Dynkin diagram of the root system E
of
G.
Table
16.2
explains
Z of G . Table 16.2 explains the
the
notation An,
A,, B,
B,,,etc.
etc.The
Thenodes
nodesofofthe
theDynkin
Dynkindiagram
diagramare
areindexed
indexedby
by the
the set
set
nr of
of simple
simpleroots.
roots.Distinct
Distinct nodes
nodes aaand
and,B
B are
are joined
joined by
by an
an edge
edge of
of weight
weight
2(a,
2(a,,8)(,8,
B)(B, aa)/(a,
) / ( aa)(,8,
,a)(B,,B)
B ) ==map
r n , ~--2.2.ItItturns
turnsout
outIsasp
Is,sSII ==man,
r n 4 ,so,
so,ififwe
weneneglect the arrows in the diagram, the Dynkin diagram of
of Z
E is
just
the
Coxeter
is
the Coxeter
diagram
diagramof
of the
the Weyl group W of E
Z or
or G,
G ,as
as defined
defined in section
section 29. By 30.9, each
member
member of EZ isisconjugate
conjugateto
to aa member
member of rnunder
underW,
W,and
andititturns
turnsout
outthat
thatthere
there
two orbits
orbits of W
W on
on EZdepending
dependingon
onwhether
whether the
the Dynkin
Dynkin diagram
diagram
are one or two
and,B
B
has no multiple
multiple bonds
bonds or
or has
has multiple
multiplebonds,
bonds, respectively.
respectively. Indeed ifif aa and
are
are joined by
by an
an edge
edge of weight
weight 11then
then aa isisconjugate
conjugatetoto,B
B under (sa,
(s,, so).
s S ) If
. the
the
diagram has multiple
multiple bonds, roots in different
different orbits are of different lengths,
and hence are called long or short roots. The arrow
arrow in the diagram points to
short root in the pair joined.
the short
Let q be a power of the prime
prime p.
p. pp is
is called
called the characteristic of
of G.
G .Recall
Recall
that G can
can be
be defined
defined in terms of one of several representations. Consider for
the moment the representation
of G as
representation of
as aa group
group of
of automorphisms
automorphisms of aa Lie
Lie
algebra L.
L . LL isisaaLie
Liealgebra
algebraover
overaafield
field of
of characteristic
characteristic ppobtained
obtained from
from
a simple
simple Lie algebra L' over
over the
the complex
complex numbers. L' has
has an
an associated
associated root
root
ow,,
e
Finite simple groups
groups
252
252
Table 16.1 Finite
Finite groups
groups of Lie type with L5 and trivial center
G
IGld(G)
I ~
IGId(G) =
= IGI
I
An(q)
qn(n+l)/2 rl(gi+l - 1)
d(G)
E
Z
(n+1,q-1)
An
- 1)
(2, q - 1)
Bn
Cn(q)
q"2 fl(g2i - 1)
(2, q - 1)
Cn
Dn(q), n > 2
qn(n-l)(qn - 1)
(4, qn - 1)
Dn
(3, q - 1)
E6
(2, q - 1)
E7
1
E8
1
F4
1
G2
(n + 1, q + 1)
C[n+l/2]
1
Al
(4, q" + 1)
Cn-1
(q2 - 1)
1
C2
g36(g12 - 1)(q9 + 1)(q8 - 1)
(3, q + 1)
F4
1
dihedral 16
16
Jr
i=1
Bn(q), q odd
q"2
fl(g2,
i=I
i=1
n-1
(g2i - 1)
i=1
E6(q)
g3fi(g12 - 1)(q9 - 1)(q8 - 1)
(q6 - 1)(q5 - 1)(q2 - 1)
E7(q)
g63(g18 - 1)(g14 - 1)
(812 - 1)(g10 - 1)
(q8 - 1)(q6 - 1)
(q2-1)
E8(q)
g120(g3° - 1)(g24 - 1)
(q20 - 1)(g18 - 1)
(814 - 1)(g12 - 1)
(q8 - 1)(q2 - 1)
F4(q)
g24(g12 - 1)(q8 - 1)
G2(q)
(q6 - 1)(q2 - 1)
q6(q6 - 1)(q2 - 1)
2An(q)
2B2(q), q = 22,n+1
n
qn(n+1)/2 rl(gi+l -(-1)i+l)
i=1
q2(q2 + 1)(q - 1)
n-1
2Dn(q), n >
qn(n-l)(q" + 1) rl (q2i - 1)
i=1
3D4(q)
2E6(q)
g12(g8 + q4 + 1)(q6 - 1)
(q6 - 1)(q5 + 1)(q2 - 1)
2F4(q), q = 22m+I
gl2(g6 + 1)(q4 - 1)
(q3 + 1)(q - 1)
2G2(q), q = 32m+1
q3(q3 + 1)(q
- 1)
1
system E'
system
C' with
with Dynkin
Dynkin diagram
diagramV.
C'.IfIfGGisisofoftype
typeA,
A,B,
B,C,
C,D,
D,F,
E or
or G then G is
ordinary Chevalley
Chevalleygroup
groupand
andCE =
= V.
called an ordinary
C'.On
Onthe
theother
other hand
hand the groups
'X are
(q) is the fixed
points on
of type "X
arecalled
calledtwisted
twistedChevalley
Chevalleygroups.
groups.mX
"X(q)
fixed points
on an
an
group of
of type X(qe)
X(qe) of a suitable
ordinary Chevalley group
suitable automorphism of order
order
in this
thiscase
case X
E is
is not
not equal
equaltotoXI.
V.
m, and in
that U
U is a Sylow p-subgroup of G, where p is
It turns out also that
is the
the characcharacteristic of G (cf. 47.3).
253
Thejnite
simplegroups
groups
The
finite simple
Table
Table16.2
16.2Some
SomeDynkin
Dynkindiagrams
diagrams
2
1
o---
A.
C
-
n-I n
2
n-2 n- I
1
2
n-2 n-1 n
1
2
1
B
...
n
-<
n-1
n
1
2
3
2
3
n-I n
5
0-0
En
4
4
ate-c
1
F4
o
G2
o<Em
2
1
4
1
Dihedral 16
16
Dihedral
1
2
2
^ 66 ^
The
The classical
classical groups
groups are
are groups
groups of
ofLie
Lietype;
type;namely
namelyAn(q)
An(q)==Ln+1(q),
Ln+l(q),
Bn(q) =
PQ2n+l(q), for qq odd,
odd, Cn(q)
Cn(q) =
=PSP2n(q),
PSp2n(q), Dn(q)
Dn(q)==PQ2n(q),
f'filn(q),
Bn(q)
= P22n+1(q),
2An(q)
2 ~ n ( q==)Un+1(q),
Un+l(q),and
and2Dn(q)
Dn(q)==PS22n(q).
P!2,(q). Also
Alsothe
thegroups
groups2B2(q)
B2(q)were
werefirst
first
discovered
M.Suzuki
Suzukiand
andhence
henceare
arecalled
calledSuzuki
Suzuki
discoveredas
aspermutation
permutation groups
groupsby
byM.
andsometimes
sometimesdenoted
denotedby
by Sz(q).
Sz(q).
groupsand
groups
There
Thereare
aresome
someisomorphisms
isomorphismsamong
amongthe
thegroups
groupsof
of Lie
Lie type,
type, and
and of
of groups
groups
of
of Lie
Lietype
typewith
withalternating
alternatinggroups.
groups. Also
Also some
some centerless
centerlessgroups
groups of
of Lie
Lie type
type
are
are not
not simple.
simple. Here's
Here's the
the list
list of
of such
such exceptions;
exceptions;we've already
alreadyseen
seen aa number
ofthem:
them:
of
A1(q) = B1(q) - C1(q)
B2(q) - C2(q)
2A1(q) - L2(q),
PSP4(q),
D2(q) - PS24 (q) = L2(q) x L2(q),
2D2(q) = PS24 -(q) --- L2(q 2),
D3(q) - A3(q),
2D3(q)
2A3(q).
In
In particular
particularrecall
recallL2(2)
L2(2)and
and L2(3)
L2(3) are
are not
not simple,
simple,and
and of
of course
courseneither
neitherisis
L2(q)
- S6,
L2(q) x L2(q).
L2(q). Also C2(2) %
S6,and
and2A2(2)
2 ~ 2 ( 2-%
)PSU3(2)
Psu3(2)are
arenot
notsimple.
simple.IfIf
Finite simple groups
254
= 22 with
=' )U3(3).
G ==G2(2)
G2(2)or
or2F4(2)
' ~ ~ ( then
then
2 ) IG
IG :: G(I))
G(')I =
with G(I)
G(') simple. (G2(2))(1)
( ~ ~ ( 2 ) ) (E
(2F4(2))(1)isiscalled
calledthe
the Tits
Titsgroup.
group. '~'(2)
2B2(2)isis aa Frobenius
Frobenius group of order
(2~4(2))(1)
order 20,
20,
and hence
2G2(3)(I)I==33 with
with '2G2(3)(1
hence solvable.
solvable.12G2(3)
I2G2(3)::2G2(3)(1)1
~ ~ ( 3 ) ( ="
'=) L2(8).
L2(8). All other
other
centerless groups
groupsof
ofLie
Lietype
typeare
aresimple.
simple.L2(4)
L2(4)E=L2(5)
L2(5)E- A5,
L2(9) E
= A6,
As, L2(9)
Ag,
L4(2)
- A8,
L4(2) E
A8,U4(2)
U4(2) =EPSp4(3),
PSp4(3),and
andL3(2)
L3(2) =="L2(7).
L2(7).This
Thisexhausts
exhausts the
the isoisomorphisms
among groups
groups of
of Lie type and between
morphisms among
between groups of Lie type
type and
and
alternating
alternating groups.
Later in this section we will explore
explore further consequences of the properties
properties
Ll-L5,
El-L5, using
usingthe
thetheory
theoryof
ofBN-pairs
BN-pairs developed
developed in chapter 14. But first let's
take a look at the sporadic
sporadic simple
simple groups.
groups.
groups and
and their orders are listed in Table 16.3. At
The sporadic simple groups
present there
there is no nice
to describe
present
nice class
class of
of representations
representations available
available to
describe the
sporadic
groups in
in a uniform
some recent
recent work
work on the
sporadic groups
uniform manner,
manner, although some
the
Table 16.3 The sporadic
sporadic simple
simple groups
groups
Notation
Name
Order
Order
M11
MII
Mathieu
Mathieu
24
2 4 ..32.5.11
32.5.11
26.33.5.11
26.33.5.11
27
2 7 ..32.5.7.11
32.5.7.11
27.32.5.7.11.23
27 . 3 2 . S . 7 . 1 1 . 2 3
210.33.5.7.11.23
210.33.S.7.11.23
23.3.5.7.11.19
23.3.5.7.11.19
f.
27 33
. 3 3.52
. 5 2.7
.7
2.35.5.17.19
z7 . 35 . S . 17 . 19
221
2". .33
33 . 5.5. 7.7. .l 1113
3 . 2.23-29-31-37-43
3.29.31.37.43
29.32.53.7.11
29.32.~3.7.11
27.36.S3.7.11
213.37.52.7.11.13
213 .37 .s2. 7 . 1 1 . 1 3
28
2 8 ..337
7 . S.56
6 . 7.7-11-31-37-67
.11.31.37.67
21° .33
33 .52
.S2 .73
. 73.,17
17
210
214 . 33 . S3 . 7 . 13 .29
214.33.53.7.13.29
29.34.5.73.11.19.31
29.34.5.73.11.19.31
210.37.53.7.11.23
21° .37 .S3 . 7 . 11 .23
2 1 8 .36
3 6.53
. S 3.7-11-23
. 7 11
. .23
218
221
2" . .39
3 9 ..54
S 4 ..72
7 2 . 1111 ..13.23
13.23
217.39.52.7.11.13
217.39-~2~7.11.13
218.313.52.7.11.13.
218 .313 .s2. 7 . 1 1 . 13. 17.23
17 + 2 3
221.316.52.73.11.13.17.23.29
2" . 316 .S2 . 7 3 . 11. 13. 17.23 .29
215 . 310.53 . 72 . 13. 19.31
215.310.53.72.13.19.31
214.36.56.7.11.19
214. 36 . s 6 . 7 . 1 1 . 1 9
241 .313
.s6 .72 . 11 . 13 . 17
241
.313.56.72.11.13.17
..19
1 9 . .23-31-47
23.31.47
246.. 320
112.133
133.. 17
17
246
320.59
59 .76.
.76 . 112.
.-19-23-29-31-41-47-59-71
1 9 . 2 3 . 2 9 . 3 1 .41 . 4 7 . 5 9 . 7 1
Mil
M12
M22
MZZ
M23
M
7.3
M24
M24
J1
J1
J2 =HJ
=HJ
J2
J3 =HJM
=HJM
J3
J4
J4
HS
Mc
Sz
Sz
Ly =
= LyS
He =
= HHM
Ru
O'N =
= O'NS
Co3
= .3
Cog =
.3
Cot
= .2
Coz =
.2
Col =
=.1
.l
M(22) =
=F22
Fz2
M(23) ==F23
M(23)
FZ3
M(24)' =
M(24)'
=F24
FZ4
F3 =
=E
=D
F5 =
=B
F2=
F1=
=M
F1
M
Janko
Janko
Hall-Janko
Hall-Janko
Higman-Janko-McKay
Higman-Janko-McKay
Janko
Janko
Higman-Sims
Higman-Sims
McLaughlin
McLaughlin
Suzuki
Suzuki
Lyons-Sims
Lyons-Sims
Held-Higman-McKay
Held-Higman-McKay
Rudvalis
Rudvalis
O'Nan-Sims
O'Nan-Sims
Conway
Fischer
Thompson
Thompson
Harada-Norton
Harada-Norton
Baby monster
Monster
The finite simple
simple groups
Thejnite
255
representations of the sporadic groups on geometries may eventually lead to
such a description.
description. Instead
Instead the
the sporadic
sporadic groups
groups were
were discovered
discovered in
in various
various
ways. The Mathieu groups
groups all have multiply transitive permutation representations; specifically M,
M isis3,3,4,
4, or
or5-transitive
5-transitiveof
of degree
degree n.
n. J2,
Jz, Mc,
Mc, HS,
HS, Sz,
Sz, Ru,
Ru,
M(22),
representations. Hence
M(22), M(23),
M(23), and
and M(24)
M(24)all
all have
have rank-3
rank-3 permutation representations.
each is aa group
group of
of automorphisms
automorphisms of aa strongly
strongly regular
regular graph via
via the
the conconstruction
struction of section
section 16.
16. The Conway groups act on the Leech lattice, a certain
discrete
discrete integer
integer submodule
submoduleof 24-dimensional
24-dimensionalEuclidean
Euclidean space.
space.The
Theother
othersposporadic groups were
were discovered
discovered through
through the
the study
study of
of centralizers
centralizersof
ofinvolutions
involutions
(cf. section
section 48).
48).
L1-L5.
In the remainder of this section assume G satisfies
satisfies the conditions L1-L5.
We'll
these conditions
conditions together with the theory in chapter
chapter 14
14 have
have aa
We'll see that these
number of interesting consequences.
consequences.
A subset
C is
is closed
closed if, for
C, we
we have
subset A
0 of E
for each
eachaa,, ,!?,BEEA0with
withaa + ,!P? E E,
a + P,!?EE A.
A.AAsubset
subset of rr of
of 0Aisisan
anideal
idealof
of 0Aif,
if,whenever
whenever a EE F,
r,,B
,!? EE A,
0,
aand
n daa +,B
+p E
E E,
C,thena+,!?
(U,:aa E
then a +,B EE r.DefineUA
r. Define Uo =
= (Ua:
E A).
0).
+
+
(47.1)
of C+
E+ and rran
anideal
idealof
of A.
A. Then
Then
(47.1) Let 0Abe
be aa closed
closed subset of
(1) Uo
UA==11aEAUa,
naeAU,, with
with the
the product
productin
in any
any order,
order, and
and each
each element
element in
in Uo
UA
product lIaEAu,,
11aEoua,u,ua EE A
0 (for
can be written uniquely as a product
(for any
any fixed ordering
of A).
0).
of
(2) Urd
u r Uo.
UA
(3)
(UA).
(3) H
H <iNG
NG
(~A).
H acts
actson
on Ua
U, by
by L3,
L3, so (3) holds.
holds. Property
l] UA.
Recall
Proof. H
PropertyL4
L4says
saysUr
Ur4
U. Recall
the
the definition
definitionof
ofthe
theheight
heightfunction
functionh hfrom
fromsection
section30,
30,and
andletleta aEEA0 with
with
h(a) minimal.
minimal. Observe
{a}
idealofofA.
A.Thus
ThusUo-4
UAl: Uo,
UA,so
so
ObserveA'0' =
= 0A- {a}
isisananideal
Uo ==Ua
UA,, and
and then,
then,by
byinduction
inductionononthetheorder
order
UP
UA
U,UAf,
of of
A,0,UU== r1,,,
nSEAUg
with respect to some ordering of A.
A. Now
Now L3
L3 and
and Exercise
Exercise 16.3
16.3 complete
complete the
the
proof.
(47.2)Let
LetwWEEW,n
W, nE ENNwith
with
and
E-w. Then
0 isis
HnHn
==w,w,and
A 0==E+
C+ ifll C-w.
Then A
(47.2)
a closed subset of E
C and
and each
each element
element of
of BwB
BwB can be written uniquely in the
form bnu,
bnu,bbEEB,
B,uuEEUA.
U.
E+ and EProof. Observe C+
C- are
areclosed
closedsubsets
subsets of
of E,
C,the
theimage
image of
of aa closed
closed
under an element of
of W is closed, and the intersection of closed subsubset under
setsisisclosed.
closed.SoSoA0isisclosed.
closed.
Similarly
sets
Similarly
r =r =C+E+-- A0 ==E+
C+flnE+w
C+wisisclosed.
closed.
By
47.1,
U
=
UrUA
so
BwB
=
BwHUrUA
=
B((Ur)'
')wUo
=
BwUo
By 47.1, U = Ur UA SO BwB BwHUr UA B((u~)w-')wUA =: BwUA
as ((Ur)'"-'
= UrW-l
Urw-[
Suppose
= any,
U. Then
as
u ~ ) w - '=
5 U.< U.
Suppose
bnubnu
= anv,
a, a,
b Eb EB,B,u,u,v vEEUA.
Then
=I:
256
25 6
Finite simple groups
a-'b
E Bn
( U ~ ) W -5'< BBn
C+wo,by
by
E
n (UA)w-'
n UWo
U'OasasAW-'
Aw-1&CCE- ==E+wo,
a-1b =(vu-')"-I
=
Exercise
l BWO
1,
Exercise 10.3.
10.3.But
Butby
byL5,
L5,BBi fl
B'O== H,
H, so B
B ifll UWO
Uw°=
= 1. Hence
Hencea-'b
a-1b =
= 1,
s o a == bbaand
n d uu==vv.
.
so
(vu-1)n-'
characteristicpp satisfying
satisfyingL5
L5then
then U
UE
If G is aa finite
finite group of Lie type in characteristic
Syl,(B).
Sylp(B). Thus the next result shows
shows U
U E Syl,(G).
Sylp(G).
(47.3)
Syl,(B) then U EE Sylp(G).
Syl,(G).
(47.3) If G is finite and U E Sylp(B)
w EE W let
let A(w)
A(w) =
= E+
Proof. For w
C+ fl
i l E-w.
C-w. Then,
Then,by
by 47.2,
IGI = IBI E WA(w)l
WEW
AsU
S yp(B),
l , ( IBB) I, p=
~ BI U~ I, and
= ~IUUA(w)
~ a nl-d0~mod
U ~p(if~UA(w)
) ~ O r# n1.l .oBut
BduptUA(w)
U
i f~U( , ~)==(11~ ~ #
As
U E~Syl
if and only
only ifif A(w)
A(w) =
= 0,
when w =
0,and,
and,by
by30.12,
30.12,this
thishappens
happens precisely
precisely when
=1.1.
IBlmod(plUI), so
so U EE Sylp(G).
Syl,(G).
Hence [GI
IGI =
- IBlmod(pIUI),
j=
J)and
and further that
Recall from
fromsection
section43
43that,
that,for
forJ J&Cn,
rr,SSJ
= (sj:
(s1:jj E J)
that Wj
Wj =
=
(Si)
Wj)
Let C
EJ be
(Sj) and
and Pi
P j ==(B,
(B,
Wj)are
areparabolics
parabolicsofofW
Wand
andG,
G,respectively.
respectively. Let
be
the set of
ofroots
rootsspanned
spannedbybyJ Jand
andletlet$rJ
*J=
= C+
E+ --ECJ.j. Observe
Observe *J
$rJisis an
an ideal of
C+
U$, .Similarly
Similarly Ei
Cfisisaaclosed
closedsubset
subset of C
U,;.
E+ and let
let Vj
Vi =
= U*,.
E and set UJ
UJ =
= UE+.
Finally set
setLLJj =
= (WJ,
Finally
(Wj, UJ).
UJ).
(47.4) (1) Pi
P j==NG(VJ)
NG(Vj).
(47.4)
in PPj.
(2) LJ
L isisaacomplement
complement to Vj in
j.
(3) (Li,
(L j,HUj,
H U jN,
, N,
a Tits
systemsatisfying
satisfyingL1-L4
L1-L4 with
withrespect
respect to
to the
Si)Sj)
is aisTits
system
EJj and the simple system
system J.
J.
root system C
Proof.
root system
simple system JJand
Proof.By
By30.20,
30.20,EJ
C isisaaroot
systemwith
withsimplesystem
andWeyl
Weyl group
group WJ.
Wj.
As Yf J is an ideal of E+, VjGUandU
As$rJisanidealofC+,
Vi :!l U and U==UJVJby47.1.ThenPJ
UJ Vi by 47.1. Then PJ==(U,
(U, Wj)
WJ)=
_
VJ).Finally,
Finally,forfor
and
1frj,
2(a,j )j)j/(j,
((LJ,
L j , Vj).
s j sjE ES Si
j and
a aE E$rJ,
a s jasj
==
a a--2(a,
j / ( j , j)j )EE1/rj,
1+9j,
because
J,and
and hence
hence as
a sjj does
does too.
too. Thus
Thus
because aa has a positive
positive projection
projectionon
onn7r-- J,
U."
= U.,,
U
: =
U,, <5Vj,
VJ,sosoWj
WJ<5NG(Vi).
NG(Vj).Therefore
Therefore PPjj =
=(U,
(U,Wj)
Wj) <_(NN(Vi).
NG(Vj).
So, by
J J&CK.
then
by43.7,
43.7,NG(Vj)
NG(VJ)==PK
PKfor
forsome
someK K&Cn7rwith
with
K.IfIfk kEEKK-- J then
$rJ and
SO U_k
UPk =
U
: (
But then U_k
U-k 5
UWO
nflUu ==1.1.
k EE 1/rj
and ksk
ksk== -k,
-k, so
= Ukk
< Vj.
V. But
< Uw0
K=
= JJand
So K
andhence
hence(1)
(1) holds.
holds.
LetA
HUJ,ClaimLJ
AWJA.ItsufficestoshowwUJwl
& AWjA
Let
A=
= HUJ,Claim
LJ = =
A Wj
A. It suffices to show w Ujw' C_
AWJA =
=
Y
Wj. Then
Then by
by induction
induction on the length
length of w',
wl, itit suffices
suffices to
Y for
for all w, w1
w' E Wj.
show
w Ujs &j CY Y
each
and
J. Let
Let A
A=
=C
EJj - {j}.
forfor
each
wW
E E
WjWj
and
j jEEJ.
{j). By
By
show wUJsj
30.7,
As
j
C
E
,
so
Uo
=
UAsj
C
Uj.
Thus
as
Uj
=
U
j
UA,
it
suffices
30.7, Asj C f , so U: = UA,, & UJ. Thus as UJ = UjUA, it suffices to
show wUJSI
0, so
so
show
wUJs..&C Y.
Y. Similarly,
Similarly,ififl(wsj)
l(wsj)>> l(w),
1(w),then
thenby
by30.10,
30.10,jw-'
jw-1 >
> 0,
c
257
257
finite simple
simple groups
groups
The jinite
-I
U,,-I 5
UJ and
and hence
Y. So
So assume
assumel(wsj)
Uj'-'
==Uj",
< Uj
hencewU;s;
w Ujsj== U;-'WS;
Uj'-'wsj cGY.
l(wsj) <
<
Uw
J
Then by
by induction
induction on l(w),
l(w), ws
ws,U;s,
Y. Now
Now ififAW;A
l(w). Then
j Uj sj C
c Y.
AWjA is a group then
UJ'
U
: cGAWjA,
AW; A, so
wUjsj = wsjUjj c ws1AWWA = wsjA U wsjAsjA C Y.
Thus itit suffices
AW; A is a group.
group.
Thus
sufficesto
toshow
showAWj
and
Let X
X=
=
By Exercise 10.3, ss"y ==sisifor
forsome
somei iEE7t,
n,
andwo
wo is
is an involution. Let
Pi". I'll
AWjA,
PTO.
I'llshow
showXXni Pj
l Pi= =
AW; A,which
whichwill
willshow
showAWjA
AW; Aisis aa group
group and
and hence
Bs;j B
A W,A.
establish the claim. By
By 43.7,
43.7, Pi
Pj ==BBUU Bs
B=
= VVjT,
j T,where
whereTT =
= AWj
A. SimBwO
UBWOs;BWO
==BWO
UsjBWoSj
BWo,
and, ass; assj
E X,
so
ilarly
ilarlyXX==
BWOUBWOsjBWO
BWOUSjBW0SiBW'0,and,
E sjX
X,sjX==X,
X,so
X ==s s;BWOUBWoS~BWO.Let
i-2 =
X
j B'OU BWOSJ BWO. Let
Q =C--{-j).ThenUWO
E - - {- j }. Then UWO=
= U_;UQ
U_ j Uu soUwOSj
so Uw°S;=
=
Uj Uc2by
by30.7,
30.7, and
and hence BWOS;
B'O==UUj
BwO.
NextBBi lnssj BWO
j Bw0==(Bwo
(Bwo i nl
UjUQ
BWoSj
BWO
; BWO.
Next
s jwoB)woand,
and,as
aswo
wo#0 sjwo,
s jwo,BwoBnBsjwoB
Bw0B n Bs jwoB==@by
0 by 43.2. So
So BnsiBWO
B nsj B'O =
=
sjwoB)wo
0. Hence BBnX
= BnU;B"'0
= UjH ==A.
@.Hence
nX =
=BnB'Os;BIO
BnBWOsjBWO
=
BnUjBWO=
= Uj(BnBw0)
U;(BnBWO)=
A.
Assj,ACX,TCX,soXnPj=TVjnX=T(VjnX)c_T(BnX)=
ASS^, A c X, T c X, S O X P;
~ = TV, n x = T(V; n x )
T(B n x ) =
TA
= T.
TA =
T. Thus
Thus the
the claim
claimis
is at
at last
last established.
established.
that (L
(LJ,
j, A, N,
N, Sj)
Sj)isisaaTits
Titssystem.
system. The
The only
only one of the
It's now easy to see that
BN-pair axioms
axioms which
whichisisnot
notevident
evidentisisBN3.
BN3.But,
But,for
forssEESj,
Sj, w E Wj,
Wj,
BN-pair
c
sAw G
C
SAW
(BwB
u BswB)
A
(BWBu
BSWB)nn AWj
AWjA
= AwA
=
AwA U
U AswA
equality following
with the equality
following from
from 47.2 and 43.2. Similarly
Similarly itit is clear
clear that L2-L4
L2-L4
inherit to L j.j. So (3) holds.
Wee ssaw
Vj<
43.2
W
awV
J g Pj
P J==(L
( Lj,J Vj),
, V J )so, sPj
o P=JL= jVj.
L J VBy
J.B
y 4 3and
. 2 a47.
n d1,
4 7Vj
. 1n, VLJj n=L J =
V.
Vj ni l A
A ==1,1,establishing
establishing(2).
(2).
The subgroup
factor of
of the
the parabolic
parabolicPPj,
and Vj
Vj is the
j , and
subgroup LLj is
is called
called a Levi
Levi factor
radical of
of Pj.
Pj.
unipotent radical
==
1, 1,
where
xoxoisisthe
(47.5) (1) HUj
HUjni u,'l Up
where
theelement
elementof
ofWj
Wjof
ofmaximal
maximal length.
(2) CH(UJ)
(2)
CH(UJ) =
=kerHU,(LJ)
~~~Hu~(LJ>.
p(B) for
p, then Vj =
=
(3) IfIfGGisisfinite,
finite, Z(G)
Z ( G )==1,1,and
andUUEESyl
Sylp(B)
for some prime p,
Op(Pj) =
Op(Pj)
=F*(Pj)
F*(Pj)for
foreach
each JJCcTr.
n.
Proof. Let
LetAA =
= HUj and
L ==LLj.j. By
10.3,(Cf)xo
(Ei )xo==CEJ
Proof.
and L
By Exercise
Exercise 10.3,
J, so
so
UJ°
n
B
<
U"'°
n
B
=
1.
Thus
(1)
holds.
Let
D
=
kerA(L).
Then
D
U p i l B 5 UWoi l B = 1. Thus (1) holds. Let
= kerA(L). Then D <
5
AnAxO=Hby(1).So[D,Uj]<HnUj=1,andhence
D<CH(Uj)=
A
i l AXO
= H by (1). So [D, UJ] 5 H n UJ = 1, and hence D
5 CH(Uj) =
E. I'll prove
prove EE <5DDbybyinduction
inductionononthe
theLie
Lierank
rank11of
of L.
L.Without
Without loss
loss of
of
generality,
= I.I.(I'll
not L5.)
L5.)As
AsU<1
U< B,
B, EE =
= CH(U)<
(I'llonly
only use
use (1),
(I), not
CH(U): H,
H , so
so
generality, JJ =
E<a BB==UH.
(Li) for each
each ii EETr.
E
UH.IfIf11>>1 1then,
then,by
byinduction
inductionon
on1,I, EE <5kerHU;
kerHu,(Li)
n.
Finite simple groups
258
25 8
Thus
UiL')
Thus EE <
(C(Uf)
C(U,")for
for each
each x eE Li,
Li,so
soL,
Li==(H,
(H,
u:')<5NG(E).
Nc(E).Therefore
Therefore
G=
(Li,
B:
i
e
n)
<
NG(E),
so
E
<
D.
So
take
l
=
1,
and
consider
G
= (Li, B: E n ) ( NG(E), so E ( D. So take 1 = 1, and considerthe
the
action of G on BG
B' by conjugation. By BN3, G ==BBUUBwB,
BwB, so
so B
B is
is transitive
transitive
on BG
- {B).
B' {B).By
By 47.2,
47.2, BwB =
=BwU,
BwU, so
so BG
B' --{B)
{B)==BW'U.
B ~ ' . Hence, as E <(Bw
Bw
and
U(
< C(E),
ngEG Bg
(2)
and U
C(E), E<
E (
Bg==D.D.SoSO
(2)isisestablished.
established.
Let X =
= CG(Vj).
By 47.4.1, X 5< PJ.LetY=BXandforl
Pj. Let Y=BX and for 1 (<mm E Z
define
LetX
CG(VJ).By47.4.1,X
Zdefine
Q. =={a{aEE*j:
h(a)
> m).
idealininC+,
E+,so
soV,,,
V ==Vim
a,,,
$j:
h(a)
> m).Observe
Observe52,E
a, is an ideal
VQm< B.
Vm+19<BBfor
foreach
eachaa E
L4, U,,
U, V,+l
E Qm.
M
., So
So also
also UaVm+14
U, V,+l <I Y. Now for
for
Indeed by L4,
and hence by 47.1.1, ay =
= aa..
Y
UU, u
==Uay
=l
y EY
Y fl~Wj,
w~
;U , yso
~ ~UcgyVm+1
U , y V m +=U,Vm+landhenceby47.1.1,ay
Thereforeyyfixes
fixeseach
eachmember
member
'/ij.Now
Now
forsome
somej j Ee JJ then
Therefore
of of
$j.
if if
y y
==s jsjfor
then
Vri
c C(E)(y). Hence,
(E), aa contradiction.
$
j G
Hence, by Exercise 10.4, y centralizes
centralizes (C),
contradiction.
However,by
by43.7,
43.7,YY==PK
PKfor
forsome
someKKGC J so
Therefore SSjj fl
i lY
Y is empty. However,
so as
SjnY=l,K=IZIandY=B.
SiflY=1,K=0andY=B.
So X
X5
< B.
complementtotoVj
VjininPj,
Pj,XVj
XVj=
= Vj(XVj
Vj(XVj i fll LLj)
B. Next, as LLj is a complement
J)
and, as
asXVj
XV j5< B,
B,XVj
XVV
l Lj5< kerA(L
ker,l(Lj)
i lL
j) ==CH(UJ).
CH(Uj).Assume
Assumethe
thehypothesis
hypothesisof
(3). Then
Then H
H is a p'-group and
and Ho
Ho =
=XVj
XVj fli lLj
L induces
inducesinner
innerautomorphisms
automorphisms
on the
the p-group
p-groupVJ,
Vi,so
so[Ho,
[Ho,Vj]
Vj]== 1.
1.Thus
ThusHo
Ho== CH(UJVj)
CH(UJVJ)== CH(U)
CH(U) =
_
on
kerB(G).But,
But,asasGG=
= (u'),
(UG), kerB(G)
kerB(G)== Z(G)
Z(G) =
= 1.
1.
CG(VJ) (
< Vi
the
I've shown CG(Vj)
Vj under
under the
the hypothesis
hypothesis of (3). Thus to complete the
proof of
of (3)
(3) itit remains
remains only
onlyto
toobserve
observethat,
that,by
by(2),
(2),Vj
Vj== O,(Pj).
Oy(Pi).
proof
n,,,
(47.6) UG = UwEw(Uw)U
Proof.
so
ProoJ By
By43.7,
43.7,GG==nWEwBwU,
nWEwBwU,
sothe
theremark
remark holds.
holds.
(47.7)
U(
<X=
= (UG
<G
H <(NI(X)
= Pj
(47.7) IfIf U
(u' ifll X) (
G then H
NG(X)and
and HX =
P jfor
for some
some
J cG.r.n.InInparticular
particular(P,-{r}:
(P,+): iiEE.r)n )isisthe
theset
setofofmaximal
maximalsubgroups
subgroups of G
G
containing U.
Proof.
47.6, XX =
ProoJ By 47.6,
=(Ud:
( u d :dd EED)
D)for
forsome
someDDCGW.
W.So,
So,as
asHHnormalizes
normalizes
Uw for
for each
each w
w EE W,
W, H
H normalizes
normalizes X.
X. So
So HX
HX5< G
G and, as B =
=HU
HU <5HX,
HX,
HX
= Pj
HX =
P j for
for some J cGn,n ,by
by43.7.1.
43.7.1.
Let M be a maximal
subgroupofof GG containing
containingUUand
andXX =
= (UG
(u' fli l M).
M).
maximal subgroup
Then
(X) or X 9
< G. In the latter
latter
Then X <
9 M, so by maximality of M either M ==NG
NG(X)
case G
G=
= (U,
= X, aa contradiction.
SoHX
HX(<NG(X)
NI(X) =
case
(U,Uw0)
UwO) =
contradiction. So
=M
M and
and hence
hence
J ==7r
=Pj
P jfor
forsome
some JJcG7r.n.By
Bymaximality
maximality of M, J
n --{i)
{i)for
forsome
some ii EE 7r.
n.
M=
If G
G is
is aa finite
finite group
group of
of Lie
Lie type
type in
in characteristic
characteristic p then,
then, as
as aa consequence
consequence
of a theorem of
of Bore1
Borel and
and Tits
Tits [BT],
[BT], each
eachp-local
p-local subgroup
subgroup of
of G
G is
is contained
contained
L1-L5 together
in a maximal parabolic. This result can be derived using L1-L5
together with
The
finite simple
simple groups
TheJinite
259
extra properties of finite
finite groups of Lie type
type listed
listed as
as hypotheses
hypotheses in
in the
the
three extra
next lemma:
lemma:
(47.8) Assume
Assumethe
the following
following hold:
hold:
(a) GGisisaafinite
finiteand
and U
U EE Sylp(B).
Sylp(B).
(b) Let
LetI'rbe
bethe
theset
setof
ofnontrivial
nontrivial p-subgroups
p-subgroups R
R of
of G
G with
with RR ==Op(NG(R)).
Op(NG(R)).
Assume for
for each
eachRR EE rF that
thatRR =
= P fli lQQfor
for some
some P,
P, Q
QEESylp(NG(R)).
SylP(NG(R)).
(c)
U=(Ui:iE7r).
(c) U
= (Ui: i E n ) .
Then
Then the
the following
followinghold:
hold:
(1) (NG(R):
(NG(R):RR EE F)
r )isisthe
theset
setof
ofproper
proper parabolic
parabolic subgroups of G.
nontrivialp-subgroup
p-subgroupofofGGthen
thenthere
thereexists
existsaaproper
properparabolic
parabolic
(2) IfIf QQisisaanontrivial
M of G
(M).
G with
with NG(Q)
NG(Q)<5M
Mand
andQQ:S5Op
Op(M).
(3) F*(X)
of G, ifif Z(G)
Z(G) =
= 1.
F * ( X )==Op(X)
O,(X) for
foreach
each p-local X of
1.
Proof. Let
LetRREEFrand
andM
M==NG(R).
NG(R).By
Byhypothesis
hypothesis there
there are P,
P,Q
Q EE Sylp(M)
Sylp(M)
=PPflnQ.
Q.By
By47.3,
47.3,UUEESylp(G),
Sylp(G),so
sowithout
without loss P <5U.
U.Similarly,
Similarly,
with R =
by 47.6, Q <
forfor
some
5 U"
Uwu
somewwE EW,
W,UuEEU,
U,and
andreplacing
replacingRRby
by R"-`
R"-' we
we may
may
assume
Observe
R=
=U
For
R<
fl M
M<
assume QQ 5< U'.
UW.
Observe R
U fl
n U'.
UW.
For if
if not, R
<U
U fl
n U"'
UWn
5
P fli lQQ==R,R,a acontradiction.
contradiction.
Z+fln E+w
Z+w and
and S2
2
' =
Z+w .Then Uw
UAUnwith
with Uo
UA 5
Let A =
= E+
= ZE- flf l E+w.
U' ==UoUQ
<
R, so R =
Uo(U
fl
UQ)
c
Uo(U
fl
U"'°)
=
Uo.
That
is
R
=
U.
I'll
show
= UA(U n U,) E UA(U n UWo)= UA.That is R = UA. I'll show
UT
Ulu'<5MMfor
foreach
eachi iEEn nand
andsome
someviv,E EU.U.Then,
Then,byby(c)
(c)and
andExercise
Exercise8.12,
8.12,
U
<5
M,M,
soso
MM
is is
a parabolic
U <5M.
M.Also
AlsoHH<5N(Uo)
N(UA)
a parabolicbyby43.7,
43.7,and
and(1)
(I)holds.
holds.
Let i EE 7r;
Ui <
Hence -i-i EE
n;it remains to show U,
5 M.
M.IfIfnot
notU;
U, -$ R,
R, so
so i & A. Hence
(E+)w
by
Exercise
10.5,
so
U_i
<
U'.
By
47.5,
U
fl
US,
=
U*<
(U,
US!
),
(Z+)w by Exercise
so U-, 5 Uw. By 47.5, U i l US' = U+<1
(U,
US'),
where
{i).
USi
fli USi)
where ,k
I) =
=E+
Z+-{i).SoSoU_i
U-,< 5
USa< 5NG(U
NG(U
l US,)and
andofofcourse
courseU_1
U-, <5
U"'
soU-,
U_iacts
actsononUUnflUw
U"'n flUS'
Usi
U*nflUA
Uo==UA
U. =
= R,
Uw <5 NG(Uw),
NG(Uw),so
= =U+
R,
as A
AG
c ,.I).That
ThatisisU_i
U-, <<M.
M.
Observe next
nextthat,
that,asasPPEESylP(M),
Sylp(M),PP=
= U ifll M. IfIf PP <
even PP =
=
5 U,,
U+ then
then even
U,,
U+ fl
i l M so,
so, as
as U_i
U-, acts
acts on
on U,,,
U+, U_;
U-, acts
acts on P.
P. But
But then,
then, as
as PP EESylp(M),
Sylp(M),
U_i
fl UwO
Ulo =
= 1.
U-, <
5 P,
P ,contradicting
contradicting U n
1. So P -$ U*.
U+.Consider
Consider the
the parabolic
parabolic Pi.
P, .
By 43.7, Bs;
of P,,
Pi, while, as P -$ U*,
Mni lPi
BSl is a maximal subgroup of
U+, M
P, -$ BS1.
BSi.
Hence,
as BSl
Bsi =
= HU-,
HU_iU*
< (M
that PI
Pi =
= (M ni l Pi)U,,.
Hence, as
U+ 5
(M ni lPi)U*,
P,)U+,itit follows
follows that
P,)U+.
Therefore s,v
siv EE M
M for
for some
somevv Ee U, so U
Ui"
= (U-,)SIU
(U-i)'1° 5
<P
: =
P.
This
Thiscompletes
completesthe
theproof
proofof
of (1).
(1). Arguing
Arguingby
by induction
inductionon
on the
the order
order of Q,
Q, there
there
exists R eE Frwith
withQQ<5RRand
andNG(Q)
NG(Q)<5NG(R),
NG(R),soso(1)
(1)implies
implies(2).
(2).Finally (2),
47.5.3,
47.5.3, and
and31.16
31.16imply
imply(3).
(3).
260
Finite simple groups
48 An
An outline
outline of
of the
the Classification
Classification Theorem
. 9 Cbebethe
thelist
listofoffinite
finitesimple
simplegroups
groupsappearing
appearingininsection
section 47.
47. Section
Section 48
Let K
provides a brief outline of the Classification Theorem, which asserts:
asserts:
Classification
ClassificationTheorem.
Theorem.Every
Everyfinite
finitesimple
simplegroup
groupisisisomorphic
isomorphicto
toaamember
member
of K.
of
X.
of objects
objects is
is to associate to
The usual procedure for classifying a collection of
collection some family of invariants, prove
prove that
that each
each object
object
each member of the collection
is uniquely determined
determined by its
its invariants,
invariants, and determine
determinewhich
which sets
sets of invariants
invariants
actually correspond
correspond to objects.
objects. The
The invariants
invariants used
used to
to classify
classify the
the finite
finite simple
simple
groups are certain local subgroups of
of the group, usually the normalizers of
of
suitable
suitable subgroups
subgroups of prime order, particularly centralizers
centralizersof involutions.
involutions.
obtained from the Odd Order Theorem
A rationale for this approach can be obtained
Theorem
Brauer-Fowler Theorem (Theorem 45.5).
of Feit and Thompson and the Brauer-Fowler
of odd
Odd Order
OrderTheorem.
Theorem.(Feit-Thompson
(Feit-Thompson [FT])
[FT]) Groups
Groups of
odd order
order are
are
solvable.
solvable.
group G is
The Odd Order Theorem says that every nonabelians simple
simple group
is of even
involution t.t. The Brauer-Fowler
Brauer-Fowler Theorem says
order and hence possesses an involution
finite number
number of
of finite
finite simple
simple groups
groups Go
Gopossessing
possessing an
an involution
involution
there is only a finite
to with
with CGo(tO)
CG0(to)Z= CG(t).
to
CG(t). In practice, with a small
small number
number of
of exceptions,
exceptions, G
G
is the unique simple group with such a centralizer.
centralizer. Even in the exceptional
L5(2),
cases at most three simple
simple groups
groups possess the same
same centralizer
centralizer (e.g. L5(2),
M24,and
and He
He all
all possess
possess an
an involution
involution with
with centralizer
centralizer LL3(2)/Dg).
MZ4,
~ ( ~ ) / D ; )Exercise
.
16.6
isomorphism type of a simple
16.6 illustrates how the isomorphism
simple group
group can be recovered
recovered
from the isomorphism
centralizersof involutions.
involutions.
isomorphism type of the centralizers
centralizers of involutions provide a set of invariants upon which to base
So centralizers
Theorem.For
Forvarious
variousreasons
reasonsititturns
turnsout
outto
to be
be better
better to
to
the Classification
Classification-Theorem.
enlarge this
this set of invariants to include suitable normalizers
normalizers of
of subgroups of
of
odd prime order.
To be more precise,
precise, define aa standard
standard subgroup of a group
group G
G for the
the prime
prime pp
be aa subgroup
subgroup HH of
of G such that
that H
H=
to be
=CG(x)
CG(x)for
for some
some element
element x of order p,
H has a unique component L, and CH(L)
CH(L) has cyclic
cyclic Sylow
Sylowp-groups.
p-groups. Standard
Standard
H
subgroups provide
provide the
the principal set of invariants for the classification of
of the
subgroups
groups.
finite simple groups.
certain small groups
groups either
either possess
possess no
no standard
standard subgroup
subgroup or
orcannot
cannot
However certain
be effectively characterized
characterized in terms of this invariant. Such groups are characterized
terized by other
other methods.
methods. Hence
Hencewe
we have
haveour
our first
first partition
partition of
of the
the simple
simplegroups
groups
for purposes of the classification:
classification: the partition into
into generic
generic groups
groups and
and small
small
outline of
of the
the Classification
ClassiJicationTheorem
Theorem
An outline
261
26 1
groups.I'll
I'll define
definethe
theappropriate
appropriatemeasure
measureof
ofsize
sizein
inaamoment;
moment;but
butbefore
beforethat
that
groups.
another
partition.
another partition.
When possible, we'd
we'd like to characterize a simple group
group G in terms of a
standard
standard subgroup
subgroup for
for the
the prime
prime 2.
2. But
But often
often GGpossesses
possessesno
nosuch
suchsubgroup.
subgroup.
O,(H) for
for each p-local
G is said to be of
of characteristic
characteristicp-type
p-typeififF*(H)
F*(H) =
= Op(H)
subgroupH
Hof
ofG.
G.For
Forexample,
example,ififGGisisof
ofLie
Lietype
type and
and characteristic
characteristicp,
p,we
wesaw
saw
subgroup
in
characteristic p-type.
2in 47.8 that G
G is of characteristic
p-type. In
Inparticular
particularifif G
Gisis of
of characteristic
characteristic2type then
then itit possesses
possessesno
no standard
standardsubgroup
subgroupfor
forthe
theprime
prime2.
2. Our
Our second
secondpartition
partition
type
the simple
simple groups
groups is
is the
the partition
partition into
into groups
groups of
of even
even and
and odd
odd characteristic,
characteristic,
of the
where by
by definition
definition G
G isis of
of even
even characteristic
characteristicifif GGisisof
ofcharacteristic
characteristic2-type
2-type
where
and G
G isis of
of odd
odd characteristic
characteristicotherwise.
otherwise.
and
Define the 2-local p-rank of G to be
m2,
p(G) =
= max{m
p(H): His aa 2-local
m2,,(G)
max{mp(H):
2-local of G},
G),
and define
define
and
e(G)
e(G) ==max{m2,p(G):
max{m2,,(G): pp odd}.
odd).
group of
of Lie
Lie type
type and
and characteristic
characteristic 2, e(G) is
is aa good
good approximation
approximation of
In aa group
the Lie
Lie rank.
rank.
the
group G
G of even characteristic
5 22and
and generic
genericotherwise.
otherwise.
A group
characteristic is small ifif e(G)
e(G) <
A group
group of odd
odd characteristic
characteristic is small if G
G isis 2-disconnected
z-disconnected for
for the
the prime
prime 22
and generic
generic otherwise.
otherwise. Thus we have a four part partition of the finite
finite simple
groups for
for purposes
purposes of
of the
the classification.
classification.
groups
The
proof
of
the
Classification
The
Classification Theorem
Theorem proceeds by induction on the order
of the
the simple
simple group
group to
to be
beclassified.
classified.Thus
Thus we
we consider
consideraaminimal
minimalcounter
counter
example G to
that is G is aa finite
to the
the Classification
Classification Theorem; that
finite simple
simple group
of minimal
order subject
subject to
to G
G @¢ X.
X. Define
Define aafinite
finite group
group H
H totobe
beaa.7CXminimal order
group
if
every
simple
section
of
H
is
in
.7C
(a
section
of
H
is
a
factor
group
group every simple
is in X
is a factor group
A/B,
A/ B,where
whereBB<5AA<5H).
H).Observe
Observethat
thatevery
everyproper
propersubgroup
subgroupof
of our
our minimal
minimal
counterexample
-group. This
X-group.
Thisproperty
propertywill
willbe
beused
usedrepeatedly.
repeatedly.
counterexampleis
is aa X
If G
< 2,
G is
is aa finite
finite simple
simple group with m2(G) 5
2, aa moment's thought
thought shows G
to be
2-disconnected
for
the
prime
2.
If
m2(G)
>
2
then,
by
> 2 then, by 46.7,
46.7, neglecting
neglecting
be z-disconnected
g(8?j(~)),G
G isis 2-disconnected
2-disconnected for the
the prime
prime 22
isolated points in the graph -T(ez(G)),
isolated
precisely when G $#I'r:,2(G),
, 2(G), where
wherePPEESy12(G).
Sy12(G).As
Asaamatter
matter of
of fact, if G is
simple
m2(G) >
>2,
2,itit can
can be
be shown
shown that G is 2-disconnected
Zdisconnected for the prime
simple and m2(G)
2 if and only if G
G has
has aa proper
proper 2-generated
2-generated 2-core.
2-core. Thus
Thus to
to classify
classify the
the small
small
simple groups
prove the following
following two
results:
simple
groupsof
of odd
odd characteristic
characteristicititsuffices
sufficesto
to prove
two results:
Theorem
with m2(G) 5
< 2 then
Theorem48.1.
48.1. IfIfGGisisaanonabelian
nonabelianfinite
finite simple
simple group with
either:
either:
(1) aaSylow
Sylow2-group
2-groupof
of GGisisdihedral,
dihedral,semidihedral,
semidihedral,or
orZ2»
Z2,,wr
wr Z2,
22, and
and G
G "=
(1)
L2(q),
(q), U3
(q), q odd,
L2(4)7L3
L3(4),
U3(4),
odd, or
or M11,
M11, or
or
(2) G
G 2 U3
U3(4).
(4).
262
Finite simple groups
Theorem 48.2.
group with
with rnz(G)
m2(G) > 22
48.2. Let
Let G
G be
be aanonabelian
nonabelian finite
finite simple group
and assume G has a proper 2-generated
2-core. Then
Then either
either G
G is
is a group of
of
2-generated 2-core.
or Sz(2")) or
Lie type of characteristic 2 and Lie rank 11 (i.e.
(i.e. L2(2"),
L2(2"), U3(2"), or
G=Z J1.
G
J1.
In brief, Theorem 48.1 is proved by using local theory to restrict the subgroup
subgroup
structure
counterexample G.
structure of our minimal counterexample
G. At
At this
this point
point there
there is either
either enough
enough
conclude G E 3%
. C or
or G possesses many TI-sets.
information available
availableto conclude
TI-sets. TI-sets
using character theory (along the lines
lines of
of 35.22
35.22 or
or Frobenius'
Frobenius'
can be exploited using
or to
to restrict
restrict the
the structure
structure of
of G
G further
further and
and
Theorem) to derive a contradiction or
show
G EE3%.
X.
show G
plays an important role
role in
in the
It's interesting to note that character theory plays
rarely used
used in
in the
the remainder
remainder of
of the
the classification.
classification.
proof of Theorem 48.1, but is rarely
essence character
character theory is used to deal with small groups such as the groups
In essence
of Lie rank 1 and some groups of Lie rank 2. The generic groups of higher
dimension can be identified using geometric
geometric or quasigeometric
quasigeometrictechniques.
techniques.
The local
Transfer
local theory
theory used in the proof of Theorem
Theorem 48.1 is of two sorts.
sorts. Transfer
and fusion techniques are used to pin down the structure
structure of a Sylow 2-group
2-group
of G and
and the
the fusion
fusion of
of 2-elements;
2-elements; see
see for
for example
example Exercise
Exercise 13.2.
13.2. One such
sparingly in the proof of Theorem
tool, used sparingly
Theorem 48.1
48.1 but frequently
frequentlyin later
later stages
stages
is Glauberman's
Glauberman's Z*-Theorem:
Z*-Theorem:
of the classification,is
Glauberman Z*-Theorem.
Let
GG
bebea afinite
Glauberman
Z*-Theorem.[G13].
[GI 31.
Let
finitegroup
groupand
andtt an
an involution
involution
Z(G*), where
where G*
G*=
_
CG(t). Then
Then t* EE Z(G*),
in G such
such that t isis weakly
weakly closed
closed in CG(t).
G/OT(G).
G/Oz4G).
Recall a subset S of G is weakly closed in
in a subgroup
subgroup H
H of
of G (with respect
fl H =={S}.
-Theorem uses modular character
to G) if 'SG
S n
{S).The
Theproof
proof of
of the
the Z*
Z*-Theorem
theory and is beyond the scope of this book.
The second kind of local theory used in the proof of Theorem 48.1 involves
an analysis of
of subgroups of
of G of odd order using signalizer functor
functor theory
theory or
or
some variant of that theory. Notice the Odd Order Theorem is one step in the
proof of Theorem 48.1, since groups of odd order are of 2-rank 0. When Feit
proved the Odd Order Theorem signalizer
and Thompson proved
signalizer functor theory did
not exist;
exist; instead they generated their own techniques, which in time
time evolved
evolved
theory. I'll
I'll illustrate the signalizer functor approach in
into signalizer functor theory.
generic case a little
the generic
little later.
later.
The
techniques used to establish
establish Theorem
Theorem 48.2 are rather
rather different.
different. The
Theproof
proof
The techniques
heavily on
on the fact that each pair of involutions
depends heavily
involutions generate a dihedral
makes possible
possible aa number
number of
of combinatorial
combinatorial and
and group
group
group. This observation makes
theoretic arguments of the flavor of sections 45 and
and 46.
46. Exercises
Exercises 16.5
16.5and
and
An outline of the Classification
Class$cation Theorem
Theorem
263
16.6 illustrate some of these
these arguments.
arguments. Indeed
Indeed Exercise
Exercise 16.6
16.6 establishes
establishes aa
very special case of Theorem 48.2. It would be nice to have the analogue
analogue of
Theorem 48.2 for odd primes, but as nothing in particular can be said about
groups
groups generated
generatedby
by aa pair
pair of elements
elementsof
of odd
odd prime
prime order,
order, a different
different approach
approach
is required.
required.
Let's turn
turn next
next to
to the
the generic
generic groups
groups of
of odd
odd characteristic.
characteristic.Observe
Observe that by
Exercise
Exercise 16.1
16.1 aa generic
generic group
group G
G of
of odd
odd characteristic
characteristicpossesses
possesses an
an involution
involution
O2',E(Cc(t)) #
:O2'(Cc(t)).
I encourage
t such that OZ',E(CG(t))
OZ'(CG(t)).
I encourageyou
youtotoretrace
retracethe
the steps
steps in
the proof of
this
exercise;
the
proof
provides
a
good
illustration
of
signalizer
of this exercise; the proof provides a good illustration of signalizer
functor
functor theory. Similar
Similar arguments reappear
reappear in the
the proof
proof of
of Theorem
Theorem48.1
48.1 and
and
analysis of
of the
the generic
genericgroups
groupsof
of even
even characteristic.
characteristic.
in the analysis
This brings
brings us
us to
to the
the following
following fundamental
fundamentalproperty
property of
of finite
finitegroups:
groups:
Bp-Property. Let
with Opf(G)
Op,(G) =
= 1,
Bp-Property.
Let ppbe
beaaprime,
prime, G
G aafinite
finite group with
1, and x an
element of order pp in
in G.
G. Then
Then Op',E(CG(x))
O p f (CG
, ~ (x)) =Op'(CG(x))E(CG(x)).
=Opt (CG(x))E(CG(X)).
The
critical and difficult steps
The verification
verification of the Bp-Property
Bp-Propertyis
is one
one of the most critical
in the classification.
for odd p,
p,
classification. Only the B2-Property
Bz-Property is established directly; for
the Bp-Property
Bp-Propertyfollows
followsonly
only as
as aa corollary
corollaryto
to the
the Classification
Classificationby
by inspection
inspection
of the groups in X.
X. (Notice
(Noticethat
thatby
byExercise
Exercise16.4,
16.4,to
toverify
verify the
the Bp-Property
Bp-Property
it suffices to consider the case where F*(G)
F*(G)isis simple.)
simple.)
Observe next that the B2-Property
Bz-Propertyfollows from the following result:
Unbalanced Group
Group Theorem.
Theorem. Let
Let GGbe
beaafinite
finitegroup
group with
with F*(G)
F*(G) simple
simple
which is unbalanced for the
the prime
prime 2.
2. Then
Then F*(G) is a group of Lie type and
odd characteristic,
L3(4),
characteristic,A2n+i,
AZ*+~,
L3(4),or
or He.
He.
For to verify
verify the B2-Property
to assume
assume F*(G)
F*(G) is simple
Bz-Property it suffices to
simple by Exercise
Exercise
16.5. If
If Ozl(CG(t))
O2'(Cc(t)) ==11for
foreach
eachinvolution
involution ttininG,
G,then
thenthe
theB2-Property
B2-Propertyis
is
trivially satisfied, so we may assume G is
is unbalanced.
unbalanced. Hence,
Hence, by
by the
the UnbalUnbalGroup Theorem,
Theorem, F*(G)
F*(G) EE YC.
X. But,
anced Group
But,by
byinspection
inspectionof
of the
the local structure
of Aut(L) for L E X,
X, ififF*(G)
GG
satisfies
the
F*(G)EEX3then
% then
satisfies
theB2-Property.
Bz-Property.
Suppose for the moment that the Unbalanced
suppose
Unbalanced Group
Group Theorem,
Theorem, and
and hence
hence
established. The B2-conjecture
Bz-conjecture, is established.
Bz-conjecture makes possible maalso the B2-conjecture,
nipulations which prove:
nipulations
[As 11)
1]) Let
Let G be
Component Theorem.
Theorem. (Aschbacher
(Aschbacher [As
be aa finite
finite group
group with
with
F*(G)
simple
satisfying
the
B2-Property
and
possessing
an
involution
t
such
F*(G) simple satisfying the B2-Property
possessing
such
O2' (CG (t)).The
TheGGpossesses
possessesaastandard
standard subgroup
subgroup for the
that 02',
E (CG (t)) # OZ'(CG(t)).
02',E(CG(t))
that
prime
prime 2.
2.
Finite simple groups
264
Actually
Actually the
the definition
definition of
of aa standard
standard subgroup
subgrouphas
has to
to be
be relaxed
relaxedaalittle
littleto
to make
make
Theorem correct
correct as
as stated above,
above, but
but the spirit is accurate.
the Component
Component Theorem
accurate.
Notice
that at this
Notice that
this stage
stage we
we have
have associated
associated to
to each
each generic
generic group
group of
of odd
odd
characteristic the desired set of invariants: its collection of standard
characteristic
standard subgroups
subgroups
for the prime
prime 2.
2. ItIt remains
remains to
to characterize
characterizesimple
simplegroups
groups via
via these
theseinvariants.
invariants.
Thus
Thus we must
must consider:
consider:
Standard
Standardform
formproblem
problemfor
for(L,
(L,r):
r):Determine
Determineallallfinite
finitegroups
groupsG
G possessing
possessing
a standard subgroup
subgroup H
H for
for the
theprime
primerr with
withE(H)
E(H)Z= L.
If G is our minimal counterexample and
and H
H a standard subgroup of
of G,
G, then
then H
H
is a K-group,
.X-group, so
so E(H)/Z(E(H))
E(H)/Z(E(H)) EEX.YC.
AsAs
wewe
know
the
universal
covering
know the universal covering
group of
of each
eachmember
memberofof3%,
X, we know
know E(H),
E(H), and
and to
to prove
proveGG Ee YC
X ititremains
remains
to treat the standard
standard form problem
problem for
for each
each perfect
perfect central
central extension
extension of
of each
each
member
of X.
member of
How does one retrieve
from that of the standard
standard subgroup
subgroup
retrieve the structure
structure of G from
H?? Let L =
=E(H).
E(H).Then
ThenH/CG(H)
H/CG(H)< 5Aut(L),
Aut(L),sosowe
wehave
havegood
goodcontrol
control over
H
HI CG(H), while as CG(H)
control
H/CG(H),
CG(H)has cyclic Sylow p-groups we have good control
of CG(H).
CG(H). Then
p-elements of
Then analysis
analysisof fusion of p-elements
of H
H gives
givesus
us aa conjugate
conjugateHg
Hg
of H such
intersection H
similar to Theorem
Theorem 48.2
such that the intersection
H fl
flHg is large. Results similar
allow us
us to conclude
conclude G
G=
= (H,
(H,Hg).
Hg).This
Thisinformation
information can
can be
be used
used to
to define
define a
construction
representation of G on a subgroup
subgroup geometry
geometry along
along the lines of the construction
in section 3, or to obtain a presentation of
of G.
G. For
For example
example we
we might
might represent
represent
so, the
the machinery
machinery in
in chapter
chapter
G on a building or show it possesses a BN-pair. IfIf so,
14 becomes
becomes available
availableto
toidentify
identifyGGas
asaamember
memberofof3%.
X.
the Unbalanced
Unbalanced Group Theorem. By the Odd Order TheIt remains to prove the
orem
(CG(t)) is solvable
solvable for each involution t in G,
orem O2
O2,(CG(t))
G, so
so by 46.9
46.9 and
and Exercise
Exercise
= CG(t)/O2'(CG(t))
CG(~)*
=
CG(t)/o2l(C~(t))
13.3, ifif G is not balanced for the prime 2, also CG(t)*
prime 2 for some involution
is not balanced
balanced for the prime
involution tt in G. Hence by 31.19
31.19
component L*
L* of
of C(t)*
C(t)* and
and X*
X* 5
< N(L*)
there is a component
N(L*) such
such that Autp(L*) is not
the prime
prime 2.
2. By
By induction
induction on
on the
the order
orderof
ofG,
G, L*/Z(L*)
L*/Z(L*) is one of
balanced for the
the groups listed in the conclusion
conclusion of the
the Unbalanced
Unbalanced Group Theorem. This
information is critical; together
together with some
piece of information
some deep local theory
theory itit can
can be
be
used to produce a standard
standard subgroup
subgroup for
for the
the prime 2, reducing
reducing us to
to aa previous
previous
case.
The groups of odd characteristic have been treated; let us turn next
next to the
the
groups of even characteristic.
characteristic. If
If G is a generic group of even
even characteristic
characteristic we
seek to produce a standard subgroup for some odd prime
prime p.
p. More precisely pp
is a prime in the set
set o-(G)
a(G) where
where
x.
a(G) =
{p:p odd,
odd, m2,p(G)
rn~,,(G) >
Z 31.
31.
or(G)
= {p:
An outline of the Classification
An
Class$cation Theorem
Theorem
265
a(G) isisaalittle
The actual definition of n(G)
littlemore
morecomplicated,
complicated,but
but again
again the
the definidefinition above
above is in the right spirit.
spirit. Using signalizer functor
functor theory and other local
group
group theoretic
theoretic techniques
techniques one
one shows:
shows:
Theorem 48.3.
minimal countercounterTheorem
48.3.(Trichotomy
(TrichotomyTheorem
Theorem[As
[As2],
21,[GL])
[GL])let
letG
G be a minimal
example
Classification Theorem
Theorem and assume
characexample to the Classification
assume G
G is
is generic
generic of even characteristic.
teristic. Then one
one of
of the
the following
following holds:
holds:
(1) GGpossesses
subgroupfor
forsome
somepp Ee n(G).
a(G).
possesses aa standard subgroup
(2) There
There exists
exists an involution
involution tt in G such
such that
that F*(CG(t))
F*(Cc(t)) is aa 2-group
2-group of
of
symplectic
symplectic type.
type.
inthe
theuniqueness
uniqueness case.
case.
(3) GGisisin
case if,
if, for
for each
each p eE a(G),
n(G),GGpossesses
possessesaastrongly
strongly
G is in the
the uniqueness
uniqueness case
p-embedded
p-embedded maximal
maximal2-local
2-local subgroup.
subgroup. Recall
Recall that a p-group
p-group is
is of
of symplectic
symplectic
type if itit possesses
possessesno
no noncyclic
noncyclic characteristic
characteristic abelian
abelian subgroups,
subgroups, and
and that
that
groups of symplectic
type
are
described
completely
in
chapter
8.
symplectic
described completely
I've already
alreadydiscussed
discussedbriefly
briefly how one
one deals with standard subgroups. In case
2 of Theorem 48.3,
48.3, the
the structure
structureofofF*(C(t))
F*(C(t)) =
= QQisisdetermined
determined from chapter
8, as is Aut(Q)
Aut(Q) (cf.
(cf.Exercise
Exercise8.5).
8.5).With
With this
this information
informationand
and some
some work
work one
one
can recover C(t)
C(t) and
and then proceed as though C(t)
C(t) were
were aa standard
standard subgroup.
subgroup.
The arguments used to deal with the uniqueness case and the small groups
of even characteristic
characteristic are
are quite
quite similar.
similar. They involve factoring 2-locals as the
product of normalizers
normalizers of certain subgroups
subgroups of aa Sylow
Sylow 2-group
2-group of
of the
the local.
local.
The Thompson
Factorization,
discussed
in
section
32,
is
the
prototype
of
Thompson Factorization, discussed in section 32, is the prototype of such
such
proof of the Thompson Normal p-Complement Theorem
factorizations. The proof
(39.5) and of
of the
the Solvable
SolvableSignalizer
SignalizerFunctor
Functor Theorem
Theorem give
give some
someindication
indication
of how such
such factorizations
factorizationscan
can be
be used.
used.
Remarks. See
SeeGorenstein's
Gorenstein'sseries
seriesof
of books
books [Gor
[Gor2,
2, Gor
Gor 3]
31 for a more detailed
outline
Theorem and a more complete discusdiscusoutline of the proof of the Classification Theorem
sion of the sporadic
particular [Gor
3] contain explicit
sporadic simple
simple groups.
groups. In particular
[Gor 2, Gor 31
references to the articles
articles which,
which, taken
taken together,
together, supply
supply aa proof
proof of
of Theorems
Theorems
48.1 and 48.2 and the Unbalanced Group Theorem.
Theorem.
Carter [Ca]
[Ca] and
and Steinberg
Steinberg [St]
[St] are good places
places to learn
learn about
about groups
groups of
of Lie
Lie
type.
type.
chapter 16
Exercises for chapter
16
1. Let
with PP EE Sy12(G)
andGG=
= 02(G).
=
Let G
G be
be aa finite
finite group with
Sy12(G) and
o~(G).Assume
Assume G =
T
, 2(G), m;?(G)
m2(G) > 2,
rOp,2(~),
2, and
and 02',E(CG(t))
0 2 1 , ~ ( C ~ (==
t )O2(CG(t))
)O2(CG(t))for
for each
each involution
involution
t in
in G.
G. Assume
Assume O2,(G)
Or(G) =
=1.1.Prove
ProveGGisisofofcharacteristic
characteristic 2-type;
2-type; that is
F*(H)
for
F*(H)==02(H)
02(H)
foreach
each2-local
2-localsubgroup
subgroup H
H of
of G.
G.
Finite simple groups
266
2. Let
group on
on nn L
> 55 letters
Let G
G ==A,,
A, be
bethe
thealternating
alternating group
letters and A =
=Aut(G).
Aut(G).
Prove:
Prove:
(1) AA=S,,
(1)
= S, ififnn#6.
# 6.
(2)AIG
A/G=E4ifn=6.
(2)
E E4 ifn = 6.
3. Let
i i<5n)n)bebea afamily
Let (Gi:
(Gi: 11<I
familyofofsubgroups
subgroupsof
of aagroup
group G
G such
such that
(1) Each
Each element
element of G
G can
can be
be written
written uniquely as a product
91 ...g,,
... gn, with
gi EE Gi, and
gi
withgi
and
(2)
G for
foreach
eachrn,
m,115<rnm5<n.
(2) GmGm+1
GrnGrn+'..
G , s< G
n.
....Gn
Then,
permutation aa of
Then, for each permutation
of 11,
11, ...
. . ., ,n),
n},each
eachelement
elementof
of G
G can
can be written
uniquely as a product gl,
gia ...
. . .gnu,
gnu,gi cEG1.
Gi.
4. Let G
4.
G be
be aafinite
finitegroup
group and
and ppa aprime.
prime.Assume
AssumeOP,(G)
Oe,(G) ==1.1.Assume
Assume
AutH(L) satisfies the
the Be-Property
BP-Propertyfor
foreach
each component
componentLL of
of G
G and each
subgroup H
H of G with L <5 HH<5NG(L).
NG(L).Finally
Finallyassume
assumeeach
eachcomponent
component
satisfies the
the Schreier
Schreier conjecture.
conjecture.Prove
Prove G
G satisfies
satisfiesthe
the BP-Property.
B,-Property.
of G satisfies
5. Let
LetHHbe
bestrongly
strongly2-embedded
2-embeddedin
inGGand
andrepresent
representG
G by
by right
right multiplication
multiplication
X=
= G/H.
ofinvolutions
involutionsofof
I flH H,
GIH.Let
Let IIbe
be the
the set of
G,G,
t Et E
In
, u uE cI I-- H,
on X
D=HfH",m=IIfHI,andJ={dED:du=d-1}.Prove:
D = H n HU,rn= II n HI,and J = {d E D:dU =d-'].Prove:
(1) G
I.
G isis transitive
transitive on I.
(2) H
H isistransitive
transitive on I fl
n H.
H.
(3) DD isisof
of odd
oddorder.
order.
(4)
(4) UuJ=uDfI.
J = U D ~ I .
(5) CG
(j) is of
of odd
odd order
orderfor
foreach
eachjj Ec J#.
CG(j)
J'.
(6) Distinct
involutions
in
uD
are
in
distinct
Distinct involutions in uD are in distinct cosets
cosets of
of CG(t).
CG(t).
(7) For
For each
each x,
x, yy EE XX with
with xx ##yythere
thereare
areexactly
exactlym
rn involutions
involutions in G with
cycle (x, y).
Y).
on II fl
(8) D
D isis transitive
transitive on
n H.
For(1)
(1)observe
observeeach
eachmember
memberofofI Iisisconjugate
conjugate
some
H;;
(Hint: For
to to
some
s Es cI I-- H
use 45.2
45.2 to
to prove
provess isis conjugate
conjugatetotot tin
t). Use
then use
in (s,
(s, t).
Use 46.4 in (2) and (3)
Y=
= CG
and 45.2 in (4). In (5) set Y
CG((j)
j ) ((u)
u )and observe that if t E CG
CG((j)
j ) then
H fln Y
Y isis strongly
strongly embedded
embedded in Y;
Y; now
now appeal
appeal to (1) for aa contradiction.
contradiction.
Derive (6) from
from (5).
(5). To
To prove
prove(7),
(7),let
letS20 be the set of triples (i, x, y) with
i EE II and
and(x,
(x,y)
y)aacycle
cycleininiion
onG/H.
GIH.Count
CountIQ1
1 S21 in
in two
two ways, using (6) to
conclude there are at most rn
m involutions with
with cycle
cycle (x,
(x, y).)
y).)
6. Let
Let GGbe
beaafinite
finitegroup
groupwith
with noncyclic
noncyclic Sylow
Sylow 2-group
2-group T.
T. Assume
Assume CG(t)
CG(t)
is an elementary abelian 2-group
2-group for
for each
each involution
involutiont tof
of G
G and
and T is not
normal
in G.
G. Let
Let qq =
1, HH =
), XX =
= G/H,
normal in
=IT
[TI,
=NG(T
NG(T),
GIH,uuan
aninvolution
involution in
G- H,
H, and
and D
D ==HHflnH".
H ULet
.LetFF==GF(q)
GF(q)and
andYY ==FFUUtoo).
{oo].Regard
Regard
G
L2(q) =
Lz(q)
=G*
G*as
as the
the group
group of
of all
all permutations
permutations of Y of the form
+
+
0(a,b,c,d):y
i-+(ay
(ay+b)/(cy+d)
@(a,
b, c, d): y H
b)/(cy d) a,b,c,d
a , b, c, d cE F,ad
F , a d --be
bc ## 00
An outline of the Classification
An
ClassiJicationTheorem
Theorem
267
as in Exercise 4.10. Let
H*={4(a,b,1,1):aEF#,bEF).
H* = {#(a, b, 1,l): a E F', b E F ) .
Prove:
(1) TTisiselementary
elementaryabelian.
abelian.
(2) HHisisstrongly
strongly2-embedded
2-embedded in G.
(3) DDisinvertedbyu,~H~=q(q-1),andDisacomplementtoTinH.
is inverted by u, I H I = q (q - 1), and D is a complement to T in H.
(4) NG(D)
NG(D)==D(u)
D(u)and
and{H,
{H,Hu}
Hu}isisthe
thefixed
fixed point
point set of D on X.
(5) GGisis3-transitive
3-transitiveon
onXXwith
withonly
onlythe
theidentity
identityfixing
fixing33 or
or more
more points.
points. TT
regular on
on XX - {H}.
{HI.
is regular
(6) There
H -+
a: X -+
There is an isomorphism
isomorphism nn:: H
+H*
H *and
and aa bijection
bijection a:
+YY such
such
that H
Haa ==oo,
that
oo,(Hu)a
(Hu)a ==0,0,and
and (xa)h7r
(xa)hn =
= (xh)a
(xh)a for
for all
all xx EE X and
and
E H.
H.
hE
(7) There
exists vv Ec uD
uD with
with(aa-')v
(act-1)v==a-la-'
a-la-1 for
There exists
for all
all a EE F#.
F'.
(8) nnextends
= L2
(q) with
withvn
vn =
1, 1, 0).
extendstotoisomorphism
isomorphismof
of G and G* =
L2(q)
=4(0,
#(0,1,1,0).
(Hints: Use Exercise
(D) =
_
Exercise 16.5
16.5 in the proof of (3) and (4). In (4) show NG
NG(D)
E(u)
and
E(u) where
where E ==02'(NG(D))
02t(NG(D))
andE(u)/D
E(u)/Disisregular
regularon
onthe
thefixed
fixedpoint
point set
set
0 of
that ii E
c uE
A
of D
D on
on X.
X. Let rrbe
bethe
theset
set of
of triples (i, x, y) such that
uE and (x, y)
of i on A.
r inintwo
0 I=
= 2.
is a cycle of
A. Count
Count r
two ways
ways to
to get
get IlAl
2. Use
Use (4)
(4) and
to get
get TT regular
regular on
onXX - {H}.
Phillip Hall's Theorem, 18.5, to
{HI.Then complete
(5) using 15.11.
that, as D is regular on
15.11. For (6) use (5) and the observation that,
T#,
EndGF(2)D(T)== F.)
F.)
T'
, D ==EndGF(2)D(T)#
~ n d ~ D(T)'
~ ( 2 )and EndGF(2)D(T)
Appendix
Solutions to
to selected
selected exercises
exercises
Solutions
Chapter
Chapter3,
3,Exercise
Exercise 5.
5. First,
First, as
as aaisisof
oforder
order nn and
and (a)
(a)isistransitive
transitive on
on A
A=
=
{Gi
n}ofoforder
ordern,n,(a)
(a) is
is regular
regular on
on A,
0, so
{Gi::11(<i i(<n}
sorenumbering
renumbering ifif necessary
necessary
we may take Gia=Gi+1,
G i a = Gi+1,with
withthe
thesubscripts
subscriptsread
read modulo
modulo n. Thus
Thus a`-1:
a'-':
G1
Gi
G1
Gi is
is an
an isomorphism.
isomorphism.
Next, by definition of "central product" in 11.1,
11.1,
n
G = fi1Yi:YiEGij
ni
with yi yj
y, =
= yy,jyiyifor
forii#$ j,
j, so
so fli yiyiisisindependent
independentofofthe
theorder
orderof
ofthe
thefactors.
factors.
Now as ai-1:
a'-': G1
G1 + Gi
Gi is
is an
an isomorphism,
isomorphism,
n
G = lFIxjai_1:XiEG1I.
(*)
i=1
ni
Define
= fli xac-1.
n: G1
G1 + G by xx7r
n=
xui-'. Then
Then
Define7r:
x7ra =
(flxat_1) a = FI XU'= X7r,
i
i
as the
the product
product is
is independent
independent of the order
order of
of its
its factors
factors and
and indices
indices are
are read
read
modulo n. Thus
< CG(a).
Thus G17r
G 1n (
CG(a). Similarly
Similarly
fxai-1yai-1 = T7xai-1 f yai-1 =x7r
(xy)7r = fl(xy)a1-1=
i
i
i
i
y7r,
ni
so 7r
is aa homomorphism.
homomorphism.Further,
Further,ififx xEEker(n)
ker(7r)
then1 =
1= flixai--',
xai-1, so
n is
then
x=
(flxa')
1
E G1 r1 G2G3 ... Gn < Z(G1),
i>1
so
< Z(G1).
so ker(7r)
ker(n) (
Z(G1).
Claim
< CG(a),
Claim CG(a)
CG(a)==G17rZ,
G l n Z ,where
whereZZ==CZ(G)(a).
Cz(G)(ff).We
We just saw
saw G17r
Gin (
CG(a),so
SO
G17rZ
< CG(a).
CG(a). BY
By (*),
(*), gg=
= fli xiai-1
GlnZ I
CG(ff).Suppose
Suppose g E
E CG(ff).
xiai-'for
forsome
somexixiEEG1.
GI.
Then
Then
ni
flxia`-1 =g =gai+1 = fxiai+' =x-iu.i
x1u-1 =
i
i
270
270
Appendix
ni+-j
for each
eachj,j,where
whereujuj== fi 0 -i xiai+j
xia`+j EE G2G3
for
G2G3. . . Gn,
Gn, so
SO
xlxI = u u-1 E G1 fl G2G3
Gn < Z(G1)
and hence
hencexjxj=xlzj
wherezz=
and
= xlzj for
forsome
some zjE
z j EZ(G1).
Z(G1).Therefore
Therefore gg==xiJrz,
x l n z , where
=
fi zjaj-1
AsAs
g, g,
x17r
zjaj-lE Z(G).
E Z(G).
x l nEECG(a),
Cc(a),also
alsozz=g(xiJr)-1
=g(xln)-lE ECZ(G)(a),
Cz(c)(a),comcompleting the
CG(a)==G1nCZ(G)(01).
GlnC~(~)(a).
pleting
theproof
proofthat
thatCG((X)
Let K ==Glrr
As
G l nand
andZZ==CZ(G)(a).
Cz(G)(a).
As7r:
n:G1
G 1 -+
+KK isisaasurjective
surjectivehomomorhomomorG11 is perfect,
perfect, K is perfect and as ZZ is
= 1.
phism and G
is abelian
abelian Z(1)
z(')=
1. Thus
nj
CG((X)(1) = (KZ)(1) = K(1)Z(1) = K.
Thus
f1
NAut(G)(Gl)
n C A ~ ~ ( G )4( CAut(G)(G1).
C
KA
) ~ ~ ( G ) (Let
G~).
Thus itit remains
remainstotoshow
showthat
thatNAut(G)(G1)
Let
G I and
and M
M ==CG(L).
CG(L).As
ASG2
G 2.... .Gn
. G,,< 5M,MG, G= LM.
= LM.
E NAnqc)(L)f1
n
L=
= G1
LetLet
0 EB NAnt(G)(L)
CAut(G)(K).Then
Thenp ,B
acts
For
=x-1 .x7r
CAut(G)(K).
acts
onon
M.M.
For
x xE EL L,
, y y=x-'
x n EE M,
M ,so
so as
as ,B
p acts on
M,, y-1
y-l . yfi
yp E M.
M .Now
M
y-1 yN=x-17r x x-1,B x7rf=(x .x-1P)x" EL
centralizesxxir
and LL 9
< G.
G. Thus
Thus (x
=) y-l
y-1
as ,B
B centralizes
n EEKK and
( x . x-1,8)x'7
x - ' ~=
~ ~ . y,B
yp EEL
L fl
n
M=
= Z(G),
Z(G), so x . x-'B
x-1,B EEZZ(L),
=1.
M
( L ) , so [,B,
[B,LL,, LL]] =
1. Hence
Hence [,B,
[p, LL]] = 11 by 8.9,
8.9,
so indeed
indeed,B
p E CAut(G)(G1)
CAut(~)(G1).
Chapter 4,
1:jCi,j
'j ai,1xi
y3 EE V
= f [x,
Chapter
4, Exercise
Exercise7.7.(1)
( 1 )Let
Letf (x,
f ( xy)
, y)_=
ai,jxiyj
V=
[ x ,y]
y] and
and
g E G.
G. By definition
definition
.f (g7r) = E ai,i (xg)` (yg)j ,
i,i
so as xg=
= ax+byandyg
ax + by and yg =
= cx
+ dy for some a, bb,, cc,, d E F,
we have f (g7r)
ex+dyforsomea,
F,wehave
( g n E) ~
soasxg
V.. It is easy
V
easy to
to check
check that
that gir
g n preserves
preserves addition
addition and
and scalar
scalar multiplication,
multiplication,
so g7r
EndF(V). For h E G,
g n EE EndF(V).
G,
.f (gh)ir _ .f (x(gh), y(gh)) = .f ((xg)h, (yg)h) _ .f (xg, yg)(h7r) _ (.f (g7r))(h7r),
so (gh)ir
= girhir,
(gh)n =
g n h n ,and
andhence
hence7r:
n:G -+
+GL(V)
GL(V)isisaarepresentation.
representation.
.
.
(2) Observe Vn
has basis
basis B,
Bn=
= {xlyn-"
{xi yn-i::005<ii_(< n).
n}.Further,
Further,(xiyn-')(pr)
(xi yn-i)(gir) =
_
V, has
(ax
(cx + dy),-'
dy)n-i with
(ax + by)i
by)' (ex
+
+
(ax + by)k
(k)
()a1xi(by)k_J E Vk,
=
i
j
+
so (xiyn-i)(gir)
(xiyn-')(gn)EE Vn.
V,. Thus G acts on V, of dimension n + 1.
1.
(3) For
For i EEZ,
Z, let
let 7i be the residue
of i modulo
(3)
residue of
modulo p; that
that is
is 00 <
5 ii <
< p and
and
i i- m
i mod
p < n and
Define
o d pp.. Assume
Assumep<n
a n dnF =
i =rr f pp --1.1.Define
U={f
thatii >
>r}.
U
= { f EVV:aij
E V,: ai,j=0
= O for
forall
all ii such
such that
r].
271
27 1
Appendix
yn_P
yn-P+l U . Claim U is G-invariant,
Then 0 # U # Vn as
X P - l yn-p+' 6 U.
as xpyn-p
xP
E U but xP-1
yn-i
)g7r E
EU
U for all i
G is
is not
not irreducible
irreducibleon
onV,,.
Vn.It
It suffices
suffices to show
show (x'
(xiyn-')gn
so that G
=s<
monomials
with 1i =
5 r.r .To
To do
doso,
so,we
wewill
willshow
showthat
thatall
allmonomialsininx1x jg7r
g n and
andyjy jg7r
gn
withi t5<7.
J. Hence as nn - ii ==rr --s,s,allallmonomials
xtyj-'yi_twith
monomials in
in
are of the form xt
(,i yn-i
) g n =xignyn-'gn
areare
ofof
thethe
form
xtyn-'
with
t 5t <
s s +( r(r-- ss)) =
(x'
yn-' )g7r
= x i g7r yn-i g7r
form
Xt yn-t
with
= r.r.
Let J=
j = v,
As(f(f + h)P
=
Let
v, so
so that
that jj == up
u p + vvand
andxig7r
x j g n==(xng7r)Px"g7r.
(xUgn)Pxugn.As
h)P =
fP
g gEEV,V all
f P+ hP
hPfor
forall
allf,f,
, allmonomials
monomialsinin(xng7r)P
(xugn)Pare
areof
of the
the form
form xptyp(j-'),
can assume
assume jj =
= v. Then
so we can
+
xPty
+
+
+
c (L)
i
xjgn =
x g n ) j=
= (ax
(ax + by)j
(ax)*(by)j-*,
_ ((xgn)'
by)'== k1 (i)()k(by)J_k,
k)
k=O
=O
+
establishing
establishing the
the claim.
claim.
Finally assume
assume nn << p;p;ititremains
hyremainstotoshow
showGGisisirreducible
irreducibleon
on Vn.
Vn. By hypothesis, X =={x,
{ x ,y}
y ] is
is aa basis for U.
U .Identify
Identify g EE G with its matrix
matrix MX(g)
Mx(g) and
define
define
a),
ga =
(0
ha=
(1a
10
/
Then
S={ga:aEF} = F=T={ha:aEF}.
Let
Mi=(yjxn-j:0<
that
= ( y j x n - j : O i j <i)
i i )for
for -1<i
-1 ( <n,
i i nso
, so
thatMn=Vn,
Mn= Vn,Mo=(xn),
Mo= ( x n ) ,
Let Mi
M-1 =O.
= 0.
and MW1
Lemma. TT acts
/Mi_2, TI =
= M,-1
/Mi_2 for ii >
> 1.
Lemma.
actson
onM,
Mi for
for all
all ii and
and [Mi
[Mi/Mi-2,
Mi-l/Mi-2
+
Proof.
= ax + y,
Prooj: xha
xh, ==xx and
and yha
yh, =
y, so
so
[y'xn-j,
'
ha] _ (y' x'-' )ha j-1
n-i
(I)yai_kXn_k
ajyj 1xn-j+1 modMj_2
k=O
andifa0O jthen aj00.
M, - M_1
Mi -1.
Corollary. IfIf ii >> O0 and
and w E Mi
Mi-1 then
then[w,
[ w T]
, TI==Mi-1.
Proof. The proof
w Ew (x
(T),
Prooj:
proof is
is by
by induction
induction on
oni.i If
. Ifi =i 0
= Othen
then
E (n)
x n<)5Cv
C"(T),
[w,, TI
T] =
= 00 =
-1.1.By
so [w
=M-1.
M - l . Assume
Assumethe
the result
result holds at ii Bythe
the Lemma,
Lemma, [w,
[w,h]
h ]EE
Mi-1
Mi-2
for
some
h
E
T,
so
by
the
induction
hypothesis,
Mi_2
=
Mipl - Mi-2 for some h E T , so by the induction hypothesis, Mi-2 =
[w,
< [w,
T] =
= Mi-i
Mi_1 as
asdim(Mi-l/Mi-2)
dim(Mi-1/Mi-2)=
=1.
[ w ,hh,, T]
TI 5
[ w , T],
T I , so
SO [w,
[w, TI
1.
Appendix
Appendix
272
Now
=H=
= (T,
Now SL(U) =
( T ,S)
S ) and
and we
we will
will show
show H
H isisirreducible
irreducibleon
on Vn.
Vn.Let
Let
WWbebeanan
FH-submodule
0#
FH-submoduleofofVn.
Vn.By
Bythe
theCorollary
Corollaryand
andsymmetry
symmetry between
between
= (yn),
W. Hence by the
T and
and S,
S, yn
yn EE [W,
[ W ,SS]] or W
W=
( y n ) so
,so in
in either
either case
case yn
yn E W.
Mn-1 =
= [[W,
T], so Vn
= ((yn,
Mn-1)
Corollary, Mn-1
W , TI,
Vn=
y n, M
n P 1<
)< W.
W.
(4) Suppose
Supposeg
c ker(nn).Thenxn
forsome
(4)
g E
ker(nn).Then xn ==xng7r
x n g n ==(xg7r)n,soxg7r
(xgn)",so xgn =
=Xx
hx for
some
xyn-1
),µn-1
nth root of unity A.
Similarlyygn
yg r =
= µy.
h. Similarly
py. Then
Then xyn-'
= =( (xyn-1)g7r
x ~ - ' ) ~==nhpn-'
xyn 1, so
xYn-l,
SO h = pl-" = It.
p.
Chapter 4,
that ifif II isis the
Chapter
4, Exercise
Exercise 10.
10. (1)
( 1 )and
and (2):
(2):First
First observe
observe that
the identity
identity
matrix
in G
G then
then zq5(I)
z'(I) ==zz for
matrix in
forall
allzzcE17,
r, so 0(I)
@ ( I=1
=
) 1isisthe
theidentity
identity of
of the
the
monoid SS of
of all
all functions
functionsfrom
fromrr into F.
monoid
r.
Next, if A
A=
= (at,1)
and B
B=
= (b;,1)
arein
inGGthen
thenfor
forz zEcFF with
withal,2z+a2,2
al,2z+a2,2 # 0:
(ai,j ) and
(b;,j ) are
0:
Ca1,1z + a2,11
z(P(A)(P(B) =
(P(B)
(a 1,2Z + a2,2 /J)
(a1,1z + a2,1)bi,I + (al,2Z + a2,2)b2,1
(al,iz + a2,1)b1,2 + (al,2Z + a2,2)b2,2
- (a1,ib1,1 +a1,2b2,1)z +a2,ib1,1 +a2,2b2,1
(ai,1bi,2 + a1,2b2,2)z + a2,1b1,2 + a2,2b2,2
+
If a1,2z
a2,2=
= 00 then
then zq5(A)
zo(A) =
= oo,
al,2z + a2,2
m,so
so
zO(A)O(B)= oo0(B)=bi,i/b1,2
+
On the other
otherhand
= 00 and det(A) # 0,
0, a1,2
a1,2 # 0, so
al.2z + a2,2
a2,2 =
==-a2,2/a1,2
hand as
as ai,2z
SOz Z
-a2,2/ai,2
and
and
z.P(AB)
a2,1b1,i
- ai,1a2,2b1,1/a1,2 - bi,i(ai,2a2,i
- a1,1a2,2) =b1,1/b1,2.
b1,2(a1,2a2,1 - a1,ia2,2)
a2,ib1,2 - ai,ia2,2b1,2/a1,2
Finally, ooo(A)O(B)
and
mq5(A)q5(B)_=(a1,1/a1,2)0(B)
(a131/al,2)q5(B)
andififa1,2
al,2 # 0 then
(a1,1/a1,2)(P(B) =
b1,1a1,1/a1,2 + b2,1
_ b1,1a1,1 + b2,1a1,2
b1,2a1,1/a1,2 + b2,2
b12a1,1 + b2,2a1,2
=
oo(P(AB).
On the other hand if a1,2
= 0 then al,l
a1,1#:00and
/ai,2 =
= oo,
al-2 =
anda1,
a l ,i1 /al,2
m,so
so
oo(P(A)(P(B) _ oo(P(B) = b1,1/b1,2 -
bi,iai,i +b2,1a1,2
b1 za1 1 + b2,2a1,2
= oo-P(AB).
verified that
that q5:
0: G
G -+
+SSisisaamonoid
monoidhomomorphism.
homomorphism. Hence
Hence
Thus we have verified
as G is a group
groupand
and@
0(I)
of SS and
andq5:
0: G
G+
-+ G*
(I=
)=1,1,G*
G*_=O(G)
q5(G)is
is aa subgroup of
G* is
a surjective group
group homomorphism.
homomorphism. As
As G*
G*isisaasubgroup
subgroupof
ofS,S ,each
eachO(A)
@ ( AE)EG*
G*
in SS and hence (1)
the proof
proof of (2)
(1) holds. Further to complete the
(2)
is invertible
invertible in
ker(o) is the group
it remains to show ker(q5)
group of scalar
scalar matrices.
matrices. So let A E ker(o).
ker(q5).
273
273
Appendix
Appendix
Thus
a1,1
/ai,2, so
Thus zo(A)
zq5(A)==zzfor
forall
allzzEEF.
r.InInparticular
particularoo
m==ooo(A)
mq5(A)==
a1,1/a1,2r
SOa1,2
al,2 =
=
O0 #a1,1.
Also 00=O
(A)=a2,1/a2,2,
sosoa2,1
al,l. Also
= Oq5(A)
= a2,1/a2,2,
a2.1 =O
= 0 #a2,2.
a2,2.Finally
Finally
a1,1 +a2,1
1 1/a2,2,
a1,2 + a2,2
so
1,1 =
= aa2,2
1,11I is
2,~
and hence
hence AA==aal,l
is scalar.
scalar.
soaal.1
a:Q
C2 -+
+Frisisaabijection.
bijection.
(3) By
By construction
construction a:
(3)
FxlEEQ.C2.By
By13.5,
13.5,GGisis2-transitive
ZtransitiveononS2,
C2,so
so
Let HHbe
bethe
thestabilizer
stabilizerininGGofofFx1
Let
GisprimitiveonC2
by 15.14.
15.14.Hence
ismaximalinG
by5.19,soG
(H,t),
t),
G
is primitive on Q by
Hence H is
maximal in G by
5.19, so G =
= (H,
interchangesx1
xl and
and x2. Thus to
foreach
each
where tt cEGGinterchanges
where
to show
show (wg)a
(t)g)a =
= (wa)$(g) for
g EE G, it suffices to show
(wa)@(g)for each g cE H and
and gg ==t.t.
show (wg)a
(t)g)a =
= (coa)0(g)
H isisthe
theproduct
productof
ofaascalar
scalarmatrix
matrixand
andaamatrix
matrix
Each hh EE H
Each
c,
g
_ (a
0
b
1
aEF#, bEF,
+
so
(t)g)a ==(wa)o(g).
But
z zv-->
(wa)@(g).
But0(g):
q5(g):
H az
az + b,
b,so
soggfixes
fixes Fx1
Fxl
so itit suffices
sufficesto show (wg)a
and
forfor
cow==F(Xx1
F,,
and 0(g)
q5(g)fixes
fixesoo
m==(Fx1)a.
(Fx1)a.Further,
Further,
F(hxl+ x2),
x2),,Xh E F
+
(wg)a = F((a.1. + b)x1 + x2)a = a), + b = )fi(g) _ (wa)o(g)
+
xl and
and x2, so
SO t has
has cycles
cycles (Fx1,
(Fxl, Fx2) and
x2),
Next, tt interchanges
interchangesx1
Next,
and (F(hxl
(F(ax1 +x2),
F(a.-1x1
onC2.
Q.Then
Thenq5(t):
0(t):zzH
H l/z,
1/z, so @(t)
0(t) has
~ ( h - l x l+ x2)),
x.~)),a.
h EEF'F#,
, on
has cycles
cycles (oo,
( m ,0)
0)
and
(X,
A
1)
on
F.
Hence
as
(Fx1)a
=
oo,
(Fx2)a
=
0,
and
F(.lx1
+
x2)a
and (h, h-')
r.
=m ,
=
F(hxl x2)a =A,
the
the proof
proof isis complete.
complete.
+
+
Chapter
Chapter5,5,Exercise
Exercise6.6.(1)
(1)Induct
Inducton
onin.
m.By
ByJordan's
Jordan'sTheorem,
Theorem,15.17,
15.17,GGisis
2-transitive
= 1. Again by Jordan, Gx
G, is
2-transitive on X, so
so the
the result
result holds
holds when
when in
m=
primitive
on X
X- {x}
= X'
X' for X
SoasasY'Y=
Y- {x}
in - 11
primitive on
{x)=
x EEY.Y.So
=Y
{x) is of order
order m
with
=G
Gy primitive
primitiveon
onX'X'-- Y'=
G,,yf =
Y' =XX --Y,Y,Gx
G, isisin-transitive
m-transitive on X' by
with Gx,y'
induction
induction on
on in.
m. Thus
Thus G is
is (m
(m ++1)-transitive
1)-transitiveon
on X
X by
by 15.12.1.
15.12.1.
(2)
Fix(t).
(2) Let
Let tt be
beaatransposition
transposition or
or cycle
cycle of
of length
length 33 in
in G
G and
and set
setYY ==Fix(t).
Then
- Y.
Then t E Gy
G and (t)
Y. Thus
Thus Gy
G isis primitive
primitive on X --YYasas
(t)is
is transitive on X IX - YI
Y I isisprime.
prime. Therefore
Therefore G
G is
is (n
(n --2)-transitive
2)-transitiveon
onXXby
by(1),
(I),so
soGGcontains
contains
the
the alternating
alternatinggroup
groupby
by 15.12.4.
15.12.4.
(3)
= M(a)
M(a) n
rl M(b), y =xu,
=xa, zz=xb,
(3) Let {x}
{x)=
=xb,and
andAA=={x,
{x,y,y,z}.
z).Now
Now
Fix(a)
= Fix((a,
Fix(a) rl
nFix(b) =
Fix((a, b))
b)) C
5 Fix([a,
Fix([a,b]).
b]).
If V
M(a) - {x,
va-1 E M(a)
M(a) --{x}
{x,y}
y) then v, va-'
{x)C5Fix(b),
Fix(b),so
so
v EEM(a)
v[a, b] = va-lb-lab = va-lab = vb = v.
Similarly
Similarly if v cE M(b)
M(b) --{x,
{x,z}
z)then
then v,
v, vb-1
vb-' EEFix(a),
Fix(a), so
so
v[a, b] = va-lb-lab = vb-lab = vb-1b = v.
274
Appendix
X- AACGFix([a,
- {x}
Thus X
Pix([a,b]).
b]).Next
~ e xxb-1
xb-'
t EE Mov(b) {x}C
G Fix(a),
Fix(a), so
so
Y[a, b] =Ya-lb-lab =xb-lab =xb-lb =x.
xa-1 E Fix(b), so
Similarly xu-'
x [a, b] =xa-lb-lab =xa~lab =xb = z.
Finally zz =xb
=xb E Mov(b)
Mov(b) - {x}
c Fix(a)
{x} G
Fix(a) and
and yy =xa
=xa EE Fix(b),
Fix(b), so
z[a, b] = za-lb-lab = zb-lab =xab = yb = y.
Thus (y,
z) is
is aa cycle
cycle of
of length
length 33 in
in [a,
[a,b],
b], completing
completingthe
the proof
proof of
of (3).
(3).
(y, x,
x, z)
Assume G does not contain the alternating
alternating group
n/2; we
(4) Assume
group on X and IIYYI I>>n/2;
contradiction. As
n /2, r
F=
must derive a contradiction.
As II YYI I>> n/2,
=X
X --YYhas
hasorder
orderless
lessthan
thanI Y
I YI,I,
by minimality
minimalityof
of IY
IY1,
Gr## 1.
1.Pick
PickaaEEG#,,
G,, y EE M(a),
M(a), and set A =
=Y- {y}.
so by
I, Gr
{y}.
Then IlAl
A I<< IYI,
IYI,sosoby
byminimality
minimalityofofIYI,
IYI,Ga
Go##1.1.pick
Pickb bE E
G. As
As Gy
Gy =
=1,
G#,.
1,
y E M(b), while
M(a) fl M(b) c (X - F) n (X - A) = X - (I' U A) = {y},
so M(a)
M(a) flflM(b)
M(b)==(y).
{y}.Thus
Thus[a,
[a,b]b]isisaa3-cycle
3-cycleby
by(3),
(3),so
so(2)
(2)supplies
supplies aa concontradiction, establishing (4).
(5) Pick Y asin(4)andletH
as in (4) and let H =Sym(X)y.ThenIHI=rn!,wherern=
= Sym(X)y. Then I H I m!, where m = IIX-YI
(5)PickY
X -Y I
and by (4),
(4),rnm>> [(n
[(n + 1)/2]. Now
Now H fl
andby
n G = GGyy ==1,1,so
so
+
ISym(X)1 > IHGI = IHI IGI = m! I GI ,
establishing
establishing (5).
(5).
Chapter 6,
of H by
Chapter
6, Exercise
Exercise 2.
2. (1)
(1) Let
Let G
G be
be the
the semidirect
semidirect product of
by A.
A. We
We
must show there
there exists
exists an
an A-invariant
A-invariantHall
Halln-subgroup
n-subgroupofofH.
H. Let
Let A be the
of Hall
Hall n-subgroups
n-subgroups of
of H. By
set of
By Hall's
Hall's Theorem,
Theorem, 18.5,
18.5, there is K E
E A and
(K). By
By the
the SchurSchurH is
is transitive
transitive on A, so
so by
by aa Frattini
Frattini argument,
argument, G
G ==HNG
HNG(K).
Zassenhaus Theorem 18.1, there exists a complement
complement B
B to NH(K)
NH(K) in NG(K)
NG(K)
and Bg
Kg is
is an
an A-invariant Hall n-subgroup
B9 =
= A for some
some gg EE G.
G. Hence
Hence JJ =
= K9
n-subgroup
of H.
H.
(2) We must show CH(A) is transitive on
on the set Fix(A) of fixed point of
transitive on
on the
the set
set A'
Ac fl
A on A. But
But by
by Schur-Zaussenhaus,
Schur-Zaussenhaus, NH(K) is transitive
fl
NG(K)
(K), so
NG(K)of complements
complements to NH(K)
NH(K)in
inNG
NG(K),
SO by 5.21,
5.21, NH(A)
NH(A)is
is transitive
transitive
on Fix(A).
Fix(A).
(3) We
We must
must show
show each
each A-invariant
A-invariantn-subgroup
n-subgroup X
X of
of H is
is contained
contained in aa
member
so assume
assume G is a minimal
member of Fix(A).
Fix(A). The
Theproof
proofisisby
by induction
inductionon
onI I G
G 1I,, so
counter example.
example. Let
Let M
M be a minimal normal
normal subgroup
subgroupof
of G
G contained
contained in
in H
H
and
= G/M.
G/M. By
of G there
Hall nn-and G* =
By minimality
minimality of
there exists
exists an
an A-invariant
A-invariant Hall
subgroup Y*
of H*
H* containing X*. Now IYI,
HI,r, so Y contains a Hall
Y* of
IYI, ==IIHI,,
Hall
n-subgroup of H,
H, and
and ifif HH:h#YY then
then XX isiscontained
contained in
in an
an A-invariant
A-invariant Hall
275
Appendix
7r-subgroup of
of YYby
byminimality
minimalityof
ofG.
G. Thus
Thus H
H=
= Y, so H
HIM
n-subgroup
I Misisaa7r-group
n-group and
and
hence H =
= JM, so H* =
= J*.
J*.
By 9.4, M is
n then
then H
H isisaa7r-group,
n-group, so
so
is aa p-group
p-group for
for some
someprime
prime p.
p.IfIf Pp EETr
7r-subgroupof
ofHH containing
containing X.
X. Thus
Thuspp 0$ 7r,
X is aa
H is an A-invariant Hall n-subgroup
n , so X
Hall n-subgroup
7r-subgroupof
ofMX.
MX.Let
LetZZ==JJ fl XM.
XM.As
AsH*
H*=
= J*,
J*, Z*
Z* ==X*,
X*,so
soZZ and
and X
X
hence by (2) there is g E CH(A)
CH(A)
XM, and hence
are A-invariant Hall 7r-subgroups
n-subgroups of XM,
Z9 =
= X. Thus
of H containing
with Zg
Thus Jg
J g is
isan
anA-invariant
A-invariant Hall
Hall 7r-subgroup
n-subgroup of
containing X.
and hence remains valid.
(4) Part (4) does not depend on p and
Chapter 6,
modulefor
forGG==Alt(I)
Alt(I) on
on II
Chapter
6, Exercise
Exercise 3.
3. As
As V
V is
is the
the permutation
permutation module
F=
and F
=GF(2),
GF(2), we
we can
can identify
identify V
V with
with the
the power set of I,
I ,and
andfor
foru,u,vvEE V,
V,
u + vv is
is the
the symmetric
symmetricdifference
difference of u and v;
v ; that is,
+
u+v=uUv-(u fl v).
Define the
ofthe
thesubset
subset vv of
of I.
I.
the weight
weight of
of vv to
to be
be the
theorder
orderIIv 1Iof
W# of
of weight
weight m.
m. By
By
(1) Let 0:A
0 # W be an FG-submodule
FG-submodule of V
V and
and W
wE
E W#
15.12.3,
- 2)-transitive
2)-transitive on I,
I ,so
so as
as n >>2,
2, G
G isis transitive
transitive on m-subsets of
15.12.3, G is (n wG
GG
C W.
II and
and hence
hence on
on vectors
vectors of
of weight
weight m.
m. Thus
Thus for
for each
each v of weight m, v EE w
W.
Assume w ##I,I ,the
thegenerator
generatorofofZ.Z.Then
Thenthere
thereisisvvofofweight
weightmrn such
suchthat
thatw
w flf l vv
of order
order rn
m- 1,1,so
is of
soww+ vvisisofofweight
weight2.2.So
Soasasww+ VvEE W,
W, we
we may
may take
take
m=
= 2. But as w is of
rn
of weight
weight 2, (wG)
(wG) =
=U.
U .Namely,
Namely, if 0 ## uu EE U,
U ,pick
pick ii EE u,
u,
and
observe
u
=
Ei
{i,
j).
Thus
W
=
U
or
V.
and observe u =
+ j,,{i, j}.
W = U or V.
O jEU
and (3):
(3): First
that G =
= 002(G).
2 ( ~ )If. n >>44this
thisfollows
follows because G
(2) and
First observe that
by 15.16. If
If nn =
= 3 or 5 it is
is easy
easy to
to check
check G
G=
= o02(G)
~ ( G )directly.
directly.
is simple by
Observe next
next that
thatififnnisisodd
oddthen
thenI10
so VV== UU @
®ZZ and
andhence
hence0UE
=U
$ UU,, so
is of
of dimension
dimensionnn-- 1. On
On the
theother
otherhand,
hand,ififn nisiseven
eventhen
then
0
<
Z
<
U
<
0 < Z < U <V
V
with dim(U)
dim(U)=
=nn - 2.
with
2.
By Exercise 4.6, Z =
= Cv
(G) and U =
= [V,
CV(G)
[V, G].
GI. Further, if X <
5 G is
is of
of odd
odd
order, then
then by
by coprime
= CV(X), so
so as G
order,
coprime action
action 18.7,
18.7, CV(X)
Cv(X) =
G ==O2(G),
02(~),
CM(G) = CM(G) = 2 =0. By
By 8.5.3,
8.5.3, [V,
[V, G]=[V,
G] _ [V, G]=U.
G] = U.
Cv(G)=CV(G)=Z=O.
As CU(G)
largest FG-module
FG-module W such that
that [W, GI
G] 5
<
Cc(G) =
=0,
0, by 17.11
17.11 there is a largest
= 0. In
In particular as
as V
V is such a module,
module, V
V5
< W. Also
Also by
by
uU 5< W and CW(G)
Cw(G)=
definition, H'(G,
H1(G, U)
U) =
= w/U,
W/U, so
definition,
so itit remains
remains to
to show
show W
W=
=V,
V, or
or equivalently
equivalently
H
1
(G,
U)
=
0
or
F
for
n
odd
or
even,
respectively.
We
prove
this
by induction
induction
H' (G, U) = 0 or F
prove this by
on n.
If n =
= 3 then G =EZ3
'(G, U)
Z3isisofoforder
orderprime
prime to
to 2,
2, so
so by
by 17.10,
17.10, H
H1(G,
0 )=
=0,
desired. Thus
Thus we
we may
maytake
takenn>>3.3.Let
Let HH =
= G1
G 1and
andJJ==II - {{I},
1), so that
as desired.
H =Alt(J)
{1,
=Alt(J)=ZA,i_1.
A,-1. For
Forj jEEJ,J let
, letvjv =
,=
{I,j}.
j}.Then
Then{vj:
{ v j j:jEEJ}
J }isisaabasis
basis for
for
U with vjh
H,, so
v, h ==Vjh
vjh for h E H
SO U is
is the
the permutation module for H of degree
- 1.1.In
by Exercise
Exercise4.6.1,
4.6.1,vv== C,,,
EjEJ vi
and
nn In particular,
particular, by
vj generates
generates Cu(H),
CU(H),
and by
induction on
on n,
n, (v)
= CU(H).
= JJ so
(H),
induction
(3) =
Cu(H). Notice if n is odd then v =
so(J)
( I=)=CC,
Cu(H),
while ifif nn is even then
then vv =
= II,, so
CU
(H)
=
0.
so Cu(H) =
+
xi
+
276
Appendix
Appendix
Assume nn is
isodd,
odd,so
sothat
thatUUZ= 0
U as an
anFG-module.
FG-module.Suppose
SupposewWEECw(H)
Cw(H)-- 0
U
let M
M==(wG).
image of
of V
V as an
and let
(wG).By
ByExercise
Exercise 4.6,
4.6, M
M isis aahomomorphic
homomorphic image
FG-module.
As CM(G)(
<Cw(G)
Cw(G)==0,0,00#0 [M,
[M,GI
G]_(<[W,
[W,GI
G]==0U,
soaa G
G is
, so
FG-module. As
irreducible
, 0U=[M,
= [M, G]
GI <M.
5 M.Then
Then as
as CM(G)=0
CM(G)= 0 and
and by
by (1),
(I), U
0 is the
irreducibleon
on0U,
nontrivial image
image N
N of
of V with CN(G)
= 0, we
we conclude
conclude M
M=
= 0U,, contraCN(G)=
only nontrivial
dicting w
0.Thus
Thus we have shown that
J ) ,so
so also
also
w EE W
W- U.
that Cw(H)
Cw(H) =
= Co(H)
CU (H)=
= ((J),
CW1(j) (H)==00asasHH==0022 (H).
Further
U/(J) isisthe
Cw,,,-,(HI
( ~ )Further
.
U/(J)
the image
image of
of the
the permutation
permutation
0 Z U for H modulo Co(H),
n, n,W/(J)
/(J),
module U
CO(H),so
soby
byinduction
inductiononon
W/(J)==oU/(J),
= U,
so W =
0 , completing
completing the
the proof in this case.
case.
This leaves
leaves the
the case
case nn even.
even. Thus
Thusnn - 11isis odd,
odd, so
so as
as UUisisthe
thepermutapermutamodule for
for H,
H, by induction on
on n,
n, W =
=0
Ue
®
Cw(H). Let K ==G2.
3 Cw(H).
G2.Then
Then
tion module
Cw(Hf1K)=Cw(H)®CU(H2)isofdimension
d+l,where
d=
dim(Cw(H)).
Cw(HnK) = Cw(H) @ C,7(H2)is ofdimensiond+l, whered = dim(Cw(H)).
Thus Cw(H)
Cw(H) and
of Cw(H
Cw(H n
fl K). Further
and Cw(K)
Cw(K) are
are hyperplanes of
0o = Cw(G)
= Cw((H,
Cw((H, K)) =
= Cw(H) n CH(K),
Cw(G>=
CH(K),
dim(Cw(H n
fl K))
so dim(Cw(H
K)) <5 22 and
and d <(1.
1. Therefore
dim(W) = d + dim(U) < n - 1= dim(V ),
= V,
V , completing
completing the proof.
so again W =
+
Chapter 7,
in
Chapter
7,Exercise
Exercise5.5.Let
Letuu==ax2 ++bxy
bxy++cy2
cy2and
and vv ==rx2
rx2 +
+sxy
SXJ + ty2
ty2 be in
W. Then
Q(x+y)=(b+s)2-4(a+r)(c+t)
+
+
+
=
- 4rt)
- 4(at
=(b2
(b2 --4ac)
4ac) + (s2
(s2 4rt) + 2bs
2bs 4(at + rc)
rc)
=
++Q(v)
++f f(u,
= TO
Q(u>
Q<v>
( ~v).
v).
3
Evidently ff isisbilinear
bilinearand
andsymmetric
symmetricand
and Q(,ku)
Q(hu) ==A2Q(u)
h2~ ( ufor
) hE
t F.
F.To
To prove
prove
Q) is
is nondegenerate
nondegenerate itit suffices
suffices to
to observe
observethat
that X
X ==[x2,
{x2,xy,
xy, y2}
y2}is a basis
basis
(W, Q)
(q) =
andxy
xy is orthogonal
orthogonal to
for W with x2,
x2, y2
y2 singular,
singular, Q
Q(xy)
= 1,
1, ff (x2,
(x2, y2)
y2)=
= -4,
-4, and
x2 and y2.
x2
y2.
(2) Let
g=
Ca
c
bl
E
dJG.
To prove
prove (2),
(2),itit suffices
sufficestotocheck
checkQ(wga)
Q(wga)==,k(ga)
Q(w)and
andff(wga,
(wga,zga)
zga) =
_
h(ga)Q(w)
To
A(ga)ff (w,
(w,z)
z)for
forw,
w,zztEX,
X,where
whereh(ga)
k(ga)== det(,g)'
det(g)2==a ad
- bc. For example
h(ga)
d-
+
+
+
Q(x2ga)
~ ( x ~ g==
aQ((ax
)Q((ax+ by)2)
by)2)==Q(a2x2
~ ( a ' x ' + 2abxy
2abxy + b2y2)
b2y2)
= 4a2b2 - 4a2b2 = 0 = X(ga)Q(x2),
+
+
+ +
+
Q(xy(ga)) =
= Q(acx2
bdy)
=Q((ax
Q((ax+ by)(cx
by)(cx + dy))
dy)) =
~(ac+
2 (ad + bc)xy
bc)xy + bdy2)
+
Q(xy),
= (ad
- 4abcd
= (ad - bc)2
= gga)
=
(ad + bc)2
hc)' 4abcd =
bc12 =
h(ga)Q(xy),
Appendix
277
and
and
f (x2ga, y2ga) = f ((ax + by)2, (cx + dy)2)
= f(ax2 + 2abxy + b2y2,c2x2 + 2cdxy + d2y2)
= 8abcd - 4(a2d2 + c2b2)
_ -4(ad - bc)2 = )(ga)f (x2, y2).
that uu is
is singular iff
if b2 =
= 4ac, so the set S
S of singular points
points in
in
(3) Observe that
W is So
where
SoU
U {Fx2},
{FX~},
where
So = {{F(e2x2
+~2exy
e E F}.
F}.
So=
~ ( e+
2exy
x ~+
+y2):
y2): e~
The subgroup
subgroup
T=
{(i ?)F}
n SSandhence
of G is transitive
transitive on
on So,
So,so
soGGisis2-transitive
2-transitiveoon
and henceA(W)
A (W)==GaAF2,,,2.
Ga OF,Fy2.
AFX2,,,2,
then
asasFxy
==(x2,
y 2y2)',
) l , Mx(h)
Now if hh EE OFx2
Fy2,
then
Fxy
(x2,
MX(h)isisdiagonal
diagonal with
with x2h
x2h =
=
px2,
A p - l y2, and
x y ,for
forsome
some µpEE F#,
F', where
where k
h=
On
µx2, y2h =
=.lµ-1
and xyh
xyh==A5xy,
_)h(h).
(h). On
the other
g aEEAFx2,Fy2,
A,,2,,,2,
where
other hand
handga
(1
g
o
0- )EG,
andx2ga =x2,
=x2, y2ga
y2ga=hp-2y2,andxyga=
=µ-2y2, andxyga = Ap-'xy,soh
soh = g a . p I
E Ga.S.
Therefore
Therefore (3)
(3) holds.
holds.
is aa nondefinite
nondefinite 3-dimensional orthogonal
(4) If U is
orthogonalspace
spaceover
overFF then
then U =
=
H
I D where
, whereHHisisa ahyperbolic
hyperbolicline
lineand
andDD==Fd
Fdisisdefinite.
definite.Multiplying
Multiplyingthe
the
HID,
QUon
on U
U by a suitable scalar, we
1. Thus
quadratic form Qu
quadratic
wemay
mayassume
assumeQU(d)
Qu(d) =
= 1.
cw, Q).
Q).
(U, Qu)
Qv) r
- (W,
(5) Part (4) implies (5),
( 3 , since if F
Fisisfinite
finiteor
or algebraically
algebraically closed,
closed, then
then no
no
3-dimensional
3-dimensionalorthogonal
orthogonalspace
space over
over FFisis definite
definite by lemmas 21.3 and 20.10.1.
(6) Let A =
=O(W,
A(W, Q).
Q).By
By Exercise
Exercise4.7.4,
4.7.4,
ker(a)={aI:aEFand
ker(a)
= { a t a E F and aa2=1}=(-I).
2 = 1) = (-I).
Thus G(')CY
G(1)aZ- L2(F),
0 ==SGa
< Z(A),
Z(O), A(')
0M =
= G(')a
GO)a Z
= L2(F).
Thus
L2(F), so as A
SGa and
and S (
L2(F).
0, so
0M. But
As 0/0(1)
A/A(') isisabelian,
abelian,rgrgEErO(1)
rA(') for each r E R and g EE A,
so rrg
rfl E A(').
But
43.12.1, L2(F)
L2(F) is
F I==3,3,so
soexcept
exceptpossibly
possibly in
in that case,
is simple
simple unless IIFI
case,
by 43.12.1,
0(1)
Fl1 =
= 3 then
A(') =
=(rrh:
( r 9 :hh EE 0(1)).
A(')). Finally,
Finally, if II F
r= (-10 0)
1
ER
278
Appendix
Appendix
with center Fxy,
Fny, and r inverts
inverts
g_ (1
1
1) EG
1
G(')m is generated by conjugates of
order 3.
of order
order 3, so again GO)a
of
of r+
rrg =
= g-l
g-1 of order
Chapter
Chapter8,8,Exercise
Exercise11.
11.Let
LetVVbe
beaanormal
normalelementary
elementary abelian
abelian subgroup
subgroup of
G of maximal rank,
rank, H =
= CG(V),
and
G*
=
G/H.
By
23.16,
V
=
CG(V), and
= GIH. By 23.16, V =Q1(H).
C21(H).As
As
m(G)
by 23.17, sosowe
may assume
m (V) =
=3.
Thus G*
),
m(G) >>2,m(V)
2,m(V)>>2 2by23.17,
wemay
assumem(V)
3.Thus
G* <(GL(V
GL(V),
so
so m(G*)
m(G*) <
5m(GL3(p))
m(GL3(p))==2.
2.
LetEp.
A<Gwith
LetE,. E A
s G w i t h nn>
> 3.AsV=Q1(H),AnH=AnVisofrank
3 . A ~ V = C 2 ~ ( H ) , A n H = A n V i s o f r a natk a t
most
most 2.
2. Also
Also m(A*)
m(A*)<(m(G*)
m(G*)==2,
2,so
so
4 < m(A) = m(A*) + m(A n H) < 2 + 2 = 4
and
m(A) = 4 and
andm(A*)
m(A*)=
=
and hence
hence all
all inequalities
inequalities are
are equalities,
equalities, so m(A)=4
m(AnH)=2.
m(A n H) = 2.
Let
LetBB ==AAnnV.
V.Then
ThenAAcentralizes
centralizesthe
thehyperplane
hyperplaneBBof
of VVand
andm(A*)
m(A*)==2,2,so
so
A* is the
Cv(a(a))
A*
the full
fullgroup
groupof
oftransvections
transvections with
with axis
axis B.
B.Therefore
Therefore [A,
[A, V]
VI =
= BB ==Cv
foreach
a EE A-B.
Suppose
DD= Z
Ep4
with
Then JAV:
foreacha
A- B.
Suppose
Ep4
withDD<AV.
(AV.Then
IAV:A(_
A1 =JAV:
IAV: DI
Dl _=
p,
m(A nn D)
= 3, and
hence tthere
- B.
p, so
so ifif A,A # D then
thenm(A
D)=3,
andhence
h e r eisi sa aE~A
AnnDD B. Then
Then
D
C ~ v ( a=
)=ACv(a)
ACv(a) =AB
=AB =
=A,
A, aa contradiction.
contradiction. Thus A is the unique E4Ep4D <5CAV(a)
subgroup
subgroupofofAV,
AV, so
soAAchar
charAV.
AV.
Let
may assume
is ggEENG
(K) - K. As
K ==NG(A);
NG(A);we
wemay
assume K#
K #G,G,sosothere
thereis
NG(K)-K.
AsA*
A*is
is
Let K
ofindexp
G*,A*
A* <<IG*,
G*,so
so (Ag)*
(Ag)* =A*
g (AHn
=AX,
of
index pininG*,
= A* andhence
and henceA#
A #AA8
<AHn K =
AX,
where
NH(A).Now
Now
where X
X ==NH(A).
[A, AX]=[A, X] <A n H = B <Z(AH),
so
AAg is of class
class at
at most
most 2.
2. Hence
Henceby
by23.11,
23.11,AAg
AAg is
is of
of exponent
exponent 3.3.Thus
Thus
so AAg
AAg =AAg
=A(AAg
A(AAg nn X)
with AAg nn X
C21(H)==V.
V.Therefore
ThereforeAAg
AAg <(
AAg
=AAg n X =
x) with
x <5Q1(H)
AV,
AV, so as A is the
the unique
uniqueEp4-subgroup
E$-subgroup of
ofAV,
AV, A
A=
=Ag,
Ag, contradicting
contradictinggg @ K.
K.
Chapter9,9,Exercise
Exercise6.6.(1)(1)Fix
Fixvivic EVZ,
&, ii ==1,1,2,
anddefine
define
Chapter
2, and
fV,,U2: V1 x V2 -* F,
(u1, u2) H f1(v1, u1)f2(v2, u2)
Asf1
f l and
andf2f2 are
arebilinear,
bilinear,sosoisisf,,,,,,2,
f,,,, ,so
soby
by the
the universal
universal property
property of
of the
the tensor
tensor
As
product,
product, f,,,,, extends
extends to
,,: VV-*
-+FFwith
with f,,,,
,(ul @(9I u2)
to aa linear
linearmap
mapff,2:
11I,,,2(u1
u2) =
f,,,,,,2(u1,
u2).
Similarly
fv, ,vz (u 1,
Similarly
V1xV2-*
v1 x vz -+Vv*,
(v1, v2) H f VhV2
279
279
Appendix
Appendix
bilinear,so
sothere
thereisisf fE E
Hom(V,V*)
V *with
) withf f(v1
( v l(9@v2)
v2)== f v , , , , . Therefore
Therefore
isisbilinear,
Hom(V,
F defined
f (x,y)y=) =
f ( ~ )is(isbilinear
~bilinear
)
and
ff::VV xxVV-*-+
F defined
bybyf (x,
f (x)(y)
and
f (V1 ® v2, u1 (9 u2) = 1(V1 (9 v2)(ui (9 u2)
U
U2)-
As
thefundamental
fundamentaltensors
tensorsgenerate
generateVV ==V1
Vl 0@V2,
Vz,ff isisunique
uniquesubject
subjecttotothis
this
A
s the
As
property.Finally
Finally
property.
f (ul ® u2, v1 (9 v2) = f (u1, vi)f (u2, V2)= (-f (VI, ul))(-f (v2, u2))
= f(v1, ul)f(v2, u2) = f(V1 0 v2, u1 0 u2),
Vl®
@V2,
V2,f isissymmetric.
symmetric.
soas
asthe
thefundamental
fundamentaltensors
tensorsgenerate
generateV1
so
(2)
(9 v2)
v2)=
= fl(v1,
fl (vi, vl)f2(v2,
= 0 as fi
(2)First
Firstf f(v1
(vl®@v2,
v2,V1
VI @
v l )f2(v2, v2)
v2)=
fiisissymplectic.
symplectic.
But ififchar(F)
char(F): #2,2,recall
recallfrom
fromChapter
Chapter77that
thatthe
theunique
uniquequadratic
quadraticform
formQQ
But
associatedto f satisfies
satisfiesQ(x)
Q ( x )==ff (x,
( x x)/2.
, x ) / 2Thus
. ThusQ(v10
Q(vl @v2)
v2)==f f(v1(9
(vl @v2,
v2,v1®
vl @
associated
v2)/2==0,
0,as
asdesired.
desired.On
Onthe
theother
otherhand:
hand:
V2)/2
Lemma.IfIfchar(F)
char(F)==
basisfor
foran
anF-space
F-spaceU,
U f, fisisa asymplectic
symplecticform
form
Lemma.
2,2,XXisisa abasis
on
U,
and
a,
E
F
for
each
x
E
X
,
then
there
is
a
unique
quadratic
form
Q
on
on U, and ax E F for each x E X, then there is a unique quadratic form Q on
associatedto
toff such
suchthat
thatQ(x)
Q ( x=)=a,
foreach
eachxxEEX.
X.
UUassociated
ax for
Let Xi
Xi =={vi,
{vi,ui)
ui)be
be aabasis
basis for
for V1;
Vi;then
X2 is a basis for V,
V ,so
so
Let
then X
X=
= X1
X1 @
®X2
V associated
associated to f with
with
by the
the lemma
lemma there
there is
is aa unique
unique quadratic form Q on V
by
Q(xl
x2) =
=00for
foreach
eachx,xiEEXi.
Xi.Now
Nowfor
forwwc V1,
E V lw,w= =avl
and
Q
(x 1@
0 x2)
a v 1 + bubul
1 and
+
Q(w (9 v2) = Q(a(v1 (9 v2) + b(u1 (9 v2))
+
+
2
a Q(vl(9@v2)
~ 2+) b2Q(ul
b 2 e ( u (9
l @v2)
vz)+ abf
abf(v1
(vl®@v2,
vz,u1
u l (9
@V2)
v2)=O.
==a2Q(vi
= 0.
+
SimilarlyQ(w
Q(w(9@u2)
uz)==0,0,sosofor
forzz==cv2
cv2+ due
duzEEV2,
V2,
Similarly
Q(w (9 z) = Q(c(w (9 v2) + d(w (9 u2))
+
+
Q(W(9@v2)
v2)+ d2
d 2Q(w
e ( w(9@u2)
u2)+ cdf
cdf(w
( w®@v2,
vz,ww (9@u2)
u2)==00
==c2c2Q(w
completingthe
theproof
proofof
of(2).
(2).
completing
(3)Pick
Pick Xi
Xi to
tobe
beaahyperbolic
hyperbolicbasis
basisfor
forVi.
V,. Then
Thenx1
xl ®
@ x2
x2 is
is orthogonal
orthogonal to
to
(3)
xl ®
@ Y2
y2 and
xz for all xi,
X i , while
while
x1
and yl
y1@
®x2
xi, yi
yi E
c Xi,
@ v2,
v 2 ,u1
u l(9@u2)
~ 2=1=
) = 1f =(V1
f(v1 ®@ u2,
U z , u1
U1®
@v2),
v2),
ff(v1
(V1 ®
so
so X
X isis aa hyperbolic
hyperbolicbasis
basis for
forV.
V.
(4)
Let
g
=
(g1,
92)
E
A.
Then
(4) Let g = ( g i , g2) E A. Then
f(V1 (9 v2)gn, (ul (9 u2)gn) = f(v1g1 ® v2g2, u1g1 (9 u2g2)
= fi(v1g1, u1g1)f2(v2g2, u2g2)
= A(g1)fl(vl,ul)?(g2)f2(v2,u2)
A(g1)X(g2)f (vl ® V2, U 1 (9 u2),
280
280
Appendix
Appendix
so
tensorsgenerate
generate
_
so as
as the
the fundamental
fundamental tensors
V ,V,
g ngnE EA0(V,
( V , f f)
) with
with h) ((grr)
gn) =
.l(gi))l(g2).
Similarly
h(gl)h(gz).
Similarly
Q((vl (9 v2)(g7)) = Q(v1g1 (9 v2g2) = 0 = Q(v1 (9 v2)
as
u2)=0
as Q(u1
Q(u1 (9
@ uz)
= 0for
forall
alluiuiEVi,
E Vi, so
SO by
by the
the lemma,
lemma, gn
g n also
alsopreserves
preserves Q.
Q.
Of course ggn
n Ec O
(V, Q
1 = h(g)ififf,l($2)
h(gz)=,l(gi)-1.
= A ( ~ ~ ) -Also
'Also
.
ker(n) iff
Of
O(V,
Q)) iff
if 1=,1(g)
gg EE ker(rr)
if
x1
=xlg1 O
for all
all xi
xi EEX,,
xl (&X2=(XI
@ xz = ( X I (&x2)g7r
@ xz)gn =xlgl
@X292
xzgz for
Xi, so
so as
as X
X isisaabasis
basisfor
forV,
V,
1
follows that g cE ker(rr)
ker(n)ififfgigi==µip Ii IforforUiuiEEF#
F' with
withA2
p2 ==µp;i .
itit follows
(5)
(5)The
Themap
map
Ta:
V1@®vzV2+- V,
T,: Vl
v,
(u,
H vv ®ua
( u , vva)
a) H
@ ua
is bilinear,
bilinear, so there exists a unique
a )=
is
unique tt =
= t,t,, Ec End(V)
End(V) with
with t,(u
ta(u @
(9 vva)
=v@
® ua
ua
for all
all u,
u ,vv cEV1.
V l .Evidently
Evidently t2
t 2==1.1.Further,
Further, as
as aaisisan
anisometry,
isometry,
for
f ((vi (9 va)t, (ul (9 ua)t) = f (v ® via, u (9 ula)= fl(v, u) f2(vla, ula)
==flfl(V1,
(VI, u1)fz(va,
ui)f2(va, ua)=
ua)= f(vl
f (vl(9@va,
va,u1
U 1®
@ ua),
ua),
so tisisanisometry.
an isometry.Next,
Next,we
wemay
maychoose
choosenotation
notationsosothat
thatvia
vl a==
andu u11 aa ==u2,
UZ,
sot
v2vzand
SO
so
(v1 (9 v2)t = (v1 (9 vla)t = (v1 (9 via) = v1 ® v2.
Similarly
2 and
reflection
Similarlyttfixes
fixesuu1®u2
1 @ uz and
andinterchanges
interchangesv1®u
vl @uz
andu1
ul ®v2,
@ vz, so
so t is aareflection
or transvection.
transvection.
or
Let
01,
g l , 1)
1 E) A,,
~ so that ggnn EE O1n.
A l n . Then
Let g = ((g1,
(u (9 va)gt = (v (9 ua)gt = (vgi ® ua)t = u ® vgla = u ® (vagia*),
A2,gt
g' E A
z n . Thus
Aln)'=
A z nand
and similarly
similarly (Gin)'
Gzn.
so as Ala* =
= 02,
O2n.
Thus((Din)t
= Len
(Gin)` =
= Gen.
( 6 )and
and (7):
(7):For
For Fv
Fvaapoint
pointininV1
Vllet
letFv
Fv®@V2
V2=={v
{ v®@uu: :u uEEV2}
V2)and
and set
set
(6)
L1=
{ F v®@V2
V2::vv EE V141.
v:).For
V2define
@ Fu
and G2
L z similarly.
similarly.
G1
= {Fv
ForFu
Fu apoint
a point in V2
define Vl
V1®
Fu and
Observe G
L=
L1U
Q(vl(9
@ v2)
v2)==00
Observe
= G1
U LZ
G2 isis aa set
set of
of totally
totally singular
singular lines in V
V,, as Q(vl
for all
all vi
vi cEVi.
Vi.
for
singular vectors
vectors are
are the
the fundmental
fundmentaltensors
tensors v1
vl ®
@ v2
vz and G
L is the
Claim the singular
all totally
totally singular
singular lines. For if x EE V# is totally singular then 00:0
# xx1
' fln
set of all
Fv@V2,
V2,so
SO x E (v@u)'
forsome
someuuEEV20.
V$ Now
Fv®
(v(gu)1 for
Nowas
asVVisis6dimensional
4-dimensional hyperbolic
hyperbolic
= Fy
space and
and yy =
= vv @
® uu is
is totally
totally singular,
singular,we
wehave
have'y y1=
FyI
IH
H,, where H is a
hyperbolic line, and if H1
H1 and
and H2
Hz are the
the two
two totally
totally singular
singular points in H,
H,
hyperbolic
H1 and
and Fy
Fy ++H2
Hz are
are the
the two
two totally
totally singular
singular lines through Fy and
then Fy +
+ H1
contain all totally
totally singular
i , and
and
contain
singular points
pointsininy'.y1. Thus
Thus xx Ec Fy
Fy + Hi
Hi for some i,
as Fv
Fv ®@V2
VZand
and V1
Vl ®
@ Fu are
are totally
totally singular
singular lines through y, we
we may
may take
take
Fv ®@V2.
Vz.This shows
shows each totally singular
singular point is aa fundamental
fundamental tensor,
tensor,
x EE Fv
and each totally singular
singular line l1 is one of the two totally singular
singular lines Fv
Fv ®@V2
Vz
+
28
2811
Appendix
and
Vl @®FFu
u through
on 1,
I , so L is
is the
the set
set of
of all
all totally
totally
and V1
throughaapoint
pointF(v
F(v 8
0 u) on
singular lines.
Observe
the two
two classes
classes of maximal
maximal totally singular
ObserveG1
L1 and G2
Lz are the
singularsubspaces
subspaces
of V
22.13, so
so by
by 22.13,
22.13,the
thesubgroup
subgrouprFofofXE=
= A(V,
0(V, Q) acting
V described in 22.13,
L1 and
(t).
on G1
and L2
G2isisofofindex
indextwo
twoininX,
E,with
withXE=
= rF(t).
ui) isis our
our hyperhyperNext by Exercise 7.1.1,
7.1.1, Gi
Gl =
= SL(&).
SL(V, ).Further
FurtherififXiX,=_(vi,
(v ul)
bolic
gµ,; =
= v;
gµ,, = µu;
F' and
andgµ,,
g,,i EEGL(V;)
GL(&) with
with v;VigLL,i
Vi and
and ui
UigCL,i
p i ,, then
bolic basis, µpEEF#
gµ,?
Ai with
with h(g,,i)
)(gµ,i) =
= µ,
g,,i EEAi
II.9 so
SO
Ai =SL(V,)(gµj: µ E F#) =GL(VV).
Now 01
A lnn <5ro,
r othe
, thestabilizer
stabilizerininFrofofV1®Fv2
Vl @ Fvzand
andV1®
Vl @Fu2.
Fuz.For
For x cEVl
Vl ®
@ Fv2,
Fvz,
define f,
fx cE (V1
®Fuz)*
Fu2)*bybyf,(y)
fx(y)==ff (x,
(x, y).
y). Then
Thenthe
themap
mapxx H
i-* f,
fx is aa To(Vl @
roisomorphism
® Fvz
Fv2 with (Vl
(V1@
(9Fuz)*,
Fu2)*,so
soasasCr(Vl
Cr(Vi @
(9 Fvz)
Fv2) n
n Cr(Vi
isomorphism of
of V1
Vl @
Cr(Vl ®
@
Fu2)
Fuz) =
=1,1,ro
roisisfaithful
faithfulon
onV1®
Vl @ Fv2,
Fvz,and
and thus,
thus, as
as 01
A lnn acts
actsfaithfully
faithfullyas
asGL(V1(9
GL(Vl @
Fv2)
® Fvz,
Fv2, it follows that
that ro
F0 =
= Aln.
Otn. Further
Fvz) on V1
Vl @
Further0271
A2n is
is 2-transitive
Ztransitive on
the points of A2
and
hence
also
on
G2,
so
F
=
F0L
27i
=
O7i.
This
completes
Az and hence also on CZ,SO r = r0A2n = An.
completes
the proof of
of (6).
(6).
Finally
= c2(V,
the derived
derivedgroup
groupof
ofO(V,
O(V, Q),
Q), and
andO(V,
O(V, Q) =
_ (O(V,
Finally c2
Q=
Q(V, Q) is the
(O(V,
Q) n
n r)(t)
r ) ( t )with
with
ro =O(V, Q) n F = ((gl, g2)7r:,1(g1)=A(g2)-'}
1
(6). Thus Fo
ro==(G1G2)2
(G1G2)nTom,
. T n ,where
whereTT==(gµ7r:
{g,n:µpEEF#}
F') and
andgN,
g, =
=
by (4) and (6).
(g,,l, 9,,-1,2)
g,-~,~).Furtherg,m
G ~ G z wherevlhl
)where
~ , v1h1==~µ-1v1,
l . - ~ vu1h1
l , u ~=
(gµ,1,
FurthergN,22r=hlh2n
=h1h22rEE ((G1G2)n,
=h l
g.-I,so
so[Tn,
[Tn, tl
t] i
< (G1
G2)jr and
pul,
V and
Z , u2h2
U Z ~=
Z= µ-1
p-luz,
and g`µ
g; =
(G1Gz)n
µu
1, v2h2=
=~
µv2,
u2, and
= g,-I,
Q) <5(G1G2)n.
(GIGz)n.Conversely,
Conversely,ifif IIFFI
1>
> 33 then
then G;Tr
Gin 2
SLz(F)
hence Q =
= Q(V, Q)
= SL2(F)
is perfect
1 G2)jr . If IIF
perfect by
by 43.12.1,
43.12.1, sosoQQ==(G(GIGz)n.
directly
F1I=
= 33 itit can be checked directly
that (G1Gz)n
(G1G2) r 5
<Q.
Q.
Chapter
f 0fE@ HOMFG
Chapter9,9,Exercise
Exercise9.9.ByByExercise
Exercise9.1,
9.1,a =
a=
E H o m ~W,
~ (V0*)
v'*)
v , and as
ff ##0,0,a:0
a #0.0.Thus
ThusasasGGisisirreducible
irreducibleon
onV,
V,aaisisan
anisomorphism
isomorphismby Schur's
Lemma
= 0*
(F), the group
CJ =
O* regarded
regarded as
as an
an automorphism
automorphismof
of GL,,
GL,(F),
group
Lemma 12.4.
12.4. Let
Let or
of invertible n by n matrices
(F).
matrices over F,
F ,and
andregard
regardGGasasaasubgroup
subgroupofofGL,,
GL,(F).
As
gB*
(F) from
As V
V -2VQ,
V", g°
g"==
gB*for
forallallggEEGGand
andsome
someBBEEGL,,
GL,(F)
fromthe
thediscussion
discussion
in section
section 13.
13. Thus
Thus
gal
= (gB*)° = g°(Bo )* =
gB*(BQ)*
= g(BB°)*
v"'. However,
V 2 VQ2.
so V
However,6'0commutes
commuteswith
with the
the transpose
transpose inverse
inverse map
map ** on
GL,,(F),
V° as (V*)*
(V*)*2- V.
GL,(F), so V°z
v"' ==((V*)*)Bz
((v*)*)" 2=V"
V.
Similarly if V 2
V' then
then V*
V* -2VB*
v'* =2V,V,soSObybyExercise
Exercise9.1.4,
9.1.4, G
G preserves
preserves
Similarly
= Vo
aa nondegenerate
nondegeneratebilinear
bilinear form
form on
on V.
V. That
Thatisis(1)
(1)holds
holdsin
inthis
thiscase,
case,so
sowe
wemay
may
V02,
assume
assumeV
V 7 V0.
v'. Then
rn is
is even,
even, since
sinceotherwise
otherwise06'isisaapower
ofB2.
6''.
Thenas
asVV2- v", m
power of
282
282
Appendix
fieldof
of0'.
02.Then
Then(0')
(02)==Gal(F/K),
Gal(F/K), so by 26.3, V
V=
= F ®KU
Let K be the fixed field
BKU
for some irreducible KG-submodule U of V. As 0 is an automorphism
automorphism of K of
order
order 2, Exercise 9.1.4 says G preserves a nondegenerate
nondegenerate hermitian
hermitian symmetric
symmetric
form on U.
U.
Chapter
Chapter9,9,Exercise
Exercise10.
10.Let
LetG1
G1==G2
G2==G.
G.As
Asora isis1-dimensional,
1-dimensional,or
a is irreducible and by hypothesis
hypothesis 7r
irreducible, so by 27.15,
27.15, aor€3®nn is
is an
an irreducible
irreducible
n is irreducible,
ducible
representationof
of G1
G1 x G2.
G2. Recall that if we
we identify
identify G
G with
with the
the diagonal
diagonal
representation
subgroup
g): g E G)
G} of G1 x G2
g), then the
subgroup {(g,
{(g,g):
G2 via the isomorphism
isomorphism g i-+
H ((g,
g , g),
the
tensor product representation
®nn of G is the restriction of the tensor product
a €3
representationor
r of
of G1
G1 xx G2
G2to
tothe
the diagonal
diagonal subgroup
subgroup G,
G, so it remains to
representation aor€30 n
show the diagonal
diagonal subgroup
subgroup G
G is
is irreducible
irreducible on
on UU €3
® V,
V, where
where U
U and V are
n,,respectively.
the representation modules for or
a and n
respectively.
EU
U#
then UU=
= Cu
Cu and the
the map
map (o:
lo:au
au€3®v vH
H av
av is an isomorphism
But ifif u E
' then
of U
G and
U®
8V
V with V
V such
such that
that x(g,
x ( g ,g)(o
g)(a®€3n)cp
n)(o=X(g)(x(p(g7r))
= h(g)(x(o(gn))for g E G
x EU
U €3
® V,
V,where
whereuu(ga)
=
)
(g)u
and
X(g)
E
C.
Namely
x
=
au
®v,
( g a )= h(g)u
h(g)E c'. Namely x = €3 v,
x(g, g)(a (9 7r)lp = (auger, vgn)1P = (a) (g)u, vgn)co = a) (g)vgn
and xcp(g7r)
= avg7r.Thus
Thus (ocoinduces
inducesaa bijection
bijection (o:
lp:W
WH
H W(o
Wlo of
of the CGx(o(gn)=avgn.
CGof U
U €3
0V
subspaces of
V and
and V,
V, so
so as
as G
G isis irreducible
irreducible on V,
V, itit is
is also
also irreducible
€3 V.
on U ®
Chapter 10,
6. (1)First,
First,ww==1
= 0, so
sow
<
10, Exercise 6.
1 .ww with
with l(1)
l(1)=
w<
5 w.
w . Thus 5
Second, ifif uu (
< w then w =xu
= xu with 1(w)
= l1(x)
+l (u), so as l1(x)
> 00
is reflexive. Second,
l ( w )=
( x )+l(u),
( x )2
iff xx =
= 1, we have ll(w)
> l1(u)
with equality
equalityiff
iffxx=
=11 and u =
= w.
with equality iff
( w )2
( u ) with
w.
Then ifif also
also w
w(
< uu,, by
by symmetry,
symmetry,l(u)
l(u)2>l(w),
l(w),so
souu==ww.
Thatis,
is, (
< isis antiThen
. That
antiFinally,assume
assumeuu(<ww(<v.v.Then
Thenww==xu
and vv =
= yw
yw with 11(w)
_
symmetric. Finally,
xu and
( w )=
= l1(y)
1(w),so
sovv=
=yw
yw =
= yxu
yxu with
lI(x)
( x ) + 1(u)
l(u) and l1(v)
( v )=
( y ) + l(w),
+
+
1(v) =1(y) + 1(w) =1(y) + 1(x) + 1(u).
+
Further ll(yx)
(yx)5< l1(y)
( y ) + l1(x),
( x ) ,while as v =
=yxu,
yxu,
Further
l(yx) > 1(v) -1(u) = 1(y) + 1(w) - (1(w) - l(x)) =1(y) + 1(x),
+
= l1(y)
hence uu (
< v,
v, proving
proving (
< isistransitive.
so l(yx) =
( y ) + ll(x)
( x ) and hence
transitive.
wo. Then by
(2)
respect to
to (
< and
(2) Let w be maximal
maximal with
with respect
and suppose
suppose w # wo.
Exercise 10.3.2,
10.3.2,l1(w)
by 30.12,
30.12, there
there isis aa EE PP with aaw
P. Then
Exercise
( w ) << IP1,
/ P I , so by
w E P.
Then
30.10, l(raw) >
by 30.10,1(rffw)
>1(w),
l(w),so raw
raw >>w,
w ,contradicting
contradictingthe
themaximality
maximality of
of w.
w.
(3) First, (b) and (c) are equivalent by 30.10.
(3)
Next, ifif2(rw)
l(rw) 5
< l(w),
by 30.10,
30.10,ll(w)
= l(rw)
l(rw) + 1,
1(r) =
= 1,
Next,
l(w),then by
(w)=
1, so as l(r)
1, we
have l1(w)
andww=
= r rw,
have
( w ) ==ll(rw)
( r w ) + l1(r)
( r ) and
. r wso
, sorw
rw<(w.w.That
Thatis,
is,(b)
(b) implies
implies (a).
+
+
283
283
Appendix
Appendix
Assume
Thenww==x x rw
rw with
withll(w)=l(rw)+l(x),
Assume rrw
w 5<w.
w . Then
( w )= l ( r w ) + l ( x )so
, soxx==rrand
and
l(rw)=l(w)
1. 1.Thus
l ( r w ) = l ( w ) --1=n
l = n- Thus r w = r z . . . r , with
with rr1i cER,
R , so
=r(rw)=
soww=r(rw)=
rr2
rrz . . .rn.
r,. That
That is,
is, (a)
(a) implies
implies (d).
(d). Finally,
Finally, if (d)
(d) holds then visibly
visibly l(rw)
l ( r w )_=
1(w) --1,1,soso(d)
(d)implies
implies(b).
(b).
1(w)
(4)
As uu<w,
(4) AS
( w , w=xu
w = x u with
with l(w)=l(x)+l(u).
l ( w ) = l ( x ) + l ( u ) .As
As l(wr)<l(w),
l ( w r ) 5 l ( w ) ,l(wr)=
l(wr)=
l ( w )and similarly
similarly l l(ur)
( u r )==l(u)
l(u) Now w
x u r with
with
1(w)
- 11and
- 1.1.Now
wrr =
=xur
1(wr)=1(w) - 1=l(x) + 1(u) - 1=l(x)+1(ur),
so
ur <5wr.
wr.
so ur
Chapter11,
11,Exercise
Exercise5.
5. For
For 11 <
5u <
5v <
5 n,
n , let U,,,
@®
U,+l
€3..®. ®
@ U.
U,.
Chapter
U,,,,,==U,
U®
Un+i
For k =
2 j or
or 2j
2 j--1,1,letletj j(k)
and1(k)
l ( k )==jj + 22ororj j++
respectively. Let
Let
For
= 2j
(k) ==j jand
1,1,respectively.
-Uj,l(k)-1,
u ,,l(k)-l,
.
U' =
U l , j - l ,U2
U 2 =Thus V
V=
= U1
and U3
u3==Ul(k),n.
U1(.),,. Thus
U' ®
€3 U2
U' ®
€3 U3
u3and
and
and
U1
= Ul,j-1,
and
Qk(1+a®y)/,/2sk = I €3 a k €3 6. where 6. = y €3 . . . €3 y and a . = (1 €3 /3 a! €3 y ) / & or
or
(a!+,8)/,/2/3)/& forfork keven
evenororodd,
odd,respectively.
respectively. Let
Let
(a
as
a
a
a!+B ,, bY S c=
b = -YP'
c = - aa! , dd = - Ba
e = - Y a.
aa=- a
e=
+
+
+
1/Z
&'
1/Z'
&'
&'
Check that
y and
Henceya
ya==-ay.
-ay.Similarly,
Similarly,a,8
a/3==
Check
that y/3
y,8==-/3-,8y
and ya!
ya =
= -my.
-ay. Hence
2
pa.Check
Checkalso
also that a,
/3, yy,, and
and aa are
==-- 112
= e2,
ac +
+
--,8a.
a, ,8,
are involutions,
involutions,b2b2
112=e
, ac
c a = a , dde+ed=e,
e + e d = e , cc2 2+ + 1/2
1/2=c,andd
+1/2
1 / 2== d.
d.
ca=a,
= c, and d22 +
( 1 ) As y2= 1, at = ( y @ . . . @ Y ) 2 = Y 2 @ . + .@ Y 2 = 1. Similarly, if
if kis
k is
(1)As
odd then
then Qk =
a2 =
whileififkkisiseven
eventhen
then
odd
= a2
=1,1,while
at
21®,82+a®(.8Y+Y.8)+a2®Y2=1,
I
a;= ( ~1®,8+a®y
@ B ; @ Y ) ~ - ~-@ B ~ + . @ ( B Y + Y B ) + ~ ~ @ Y =
~ 1,
(
)
22
2
!' =
as ,82=
as
= aa2
= y2 =
=11and
and ,8y
/3 y ==-y,8.
-y/3.Therefore
Therefore
Sk=(I ®ak(g k)2=I XQk
Next
let V1=U1,j_1,
V2=Uj,j+i,
V' = U l , j - l , v2
= Uj,j+l,and
and V3=Uj+2,n.
v3= Uj+2,,. IfIf kk==22jj -- 11then
then
Next let
Sk
=I®
sk=I
@(a
( a(9
@ y)
y ) @ 6 L where
9 w h e rke =
6 Ly=®y @
@
yandsk+l=I@ak+l@6k+l.
® .y. .and
Sk+1 = 1 ® ak+1 0 k+1
Thus
Thus SkSk+1
sksk+l ==I 10
@ (a
(a ®y)ak+1
@ y)ak+l 8 6Ltk+1 With
with 6;6k+l=
@ ®...
. . . @®Y2
y 2 ==II as
as
SklSk+1y=2 y2
Y2=
= 1. TThus
to show
show(S.S.+~)~
(SkSk+l)3
= -Iwe
wemust
mustshow
showx x3
~ U Sto
= -I
3 ==- -I,
I , where
where
x=(a®y)ak+l=(a®y)((l®,8+a®y)/,/2-)=a®b+c®1.
+
=x.
x . But
But
Equivalently we must show
show x2
x2 + 11=
x2=a2®b2+(ac+ca)®b+c2®1=-1/2+a®b+c2®1,
as a2 = 1, b2 = -1/2, and ac + ca= a, so as c2 + 1/2 = c, indeed
x2+1=a®b+c2®1+(1®1)/2==a®b+c®1=x,
completing the
the proof
proof that
that (S2j_1S2
(s2j-ls2j)3
completing
j)3==- I-I..
284
Appendix
Similarly
€3 aak
k €30 Ck
and sk+l=
I €3®(1
(1 €3®a)
Similarlylet
letkk==2j.
2j.Then
ThenskSk==I 10
k and
Sk+1 =1
a)€30 Ck+l,
k+1,
sksk+1=1®
0 a)
k+1, and
show y2
y2 + 11=
so sksk+l=
I C ~ ak(1
aJ k ( l €3
a ) 0€3k(k(k+l,
andititremains
remains to show
= y, where
+
y
(1®P + -Y)(1®a)
ark (l0 a)=
=(l
But
y2=1®d2+a®(de+ed)+a2®e2=1®d2-1/2+a®e,
as de + ed = e, a2 = 1, and e2 = - 1/2. Thus as d2 + 1/2 = d,
y2+ 1=a ®e+ 1 ®d2+(1 ®1/2)=a ®e+ 10 d = y,
completing
completing the proof of (1).
(1).
Letkk i
< i --1;1;we
weclaim
claimeither
either
(2) Let
1(k)i<j(i),
i(i), or
(a) l(k>
= 2(j
(b) kk ==22jj and i =
2(j + 1).
1).
For
= j(k)
j (k)then2j
then 2j-- 11i<kk<< ii - 11 <
(i) --11and
Forififjj =
5 22j(i)
j ( i ) - 1,1,so
so jj <5jj(i)
and hence
hence
ll(k)=
(k) =jj ++ ce <i jj(i)unlessc=2(sothatk=2j)and
(i) unless e = 2 (so that k = 2j) andj(i)=
j (i) = jj +
+ 1.
In the latter
1.Inthelatter
case, as
asj(i)=
j (i) =jj + 1 and
and kk==22j
case,
j <
< ii --1,1,we
wehave
haveii== 2(j
2 ( j + 1),
I),establishing
establishing
the claim.
In case
Ui,j-1,
V2V=2 Uj,1(k)-1,
U1(k),j(i)-1,
case(a)
(a)let
letVlVl==
Ul,j-1,
= U;,l(k)-1,V3V3==
Ul(k),j(i)-1,V4
V4 ==Uj(i),i(i)-1,
Uj(i),l(i)-lr
and V5
UI(i),n.
Then
V Vl
= V1
®V
V5
ands,sr==sr,1
Sr,5,
Vg ==Ul(i),n.
Then
V=
€3 . ®
. . €3
5 and
s , l €3®. .0. €3
s,5,where
whereSr
sr
induces
= I for
= ak,
= ai,
induces sr,t
s , , on V.
V, .Further,
Further,Si,,,
s ~=
=Sk,1
,sk,~1 =
for uu <(3,3,sk,2
Sk.2 =
a k ,si,4
s ~=
a
, i ,~and
andsi,5
si,~
and ~
Sk,,,,
y. y.
In In
particular
Sk,wSi,w
for
k ,vv~>>, 2, are of the form y ®
€3... ®.€3
particular
Sk,WSi,w =
=si,wsk,w
Si,WSk,w for
w # 4, whilesk,4si,4
while sk,45i,4
ay
-y,a, and ay=
= -ya.
-yet.
= -=~-si,45k,4,
~ , ~ s ~ , 4 since
, s i n c=
e -ya,
a=y -ya,By,ay
==-yB,anday
==
-I.
Therefore (sksi)'
(SkSi)2
J.
In case (b), let W1=
Wl = Ul,j_1,
Incase
Ul,j-1, W2
W2 =
=Uj,j+2,
U;, j+2,and
andW3
W3 =
=Uj+3,n.
Uj+3,n.Again V
V ==W1®
W1€3
W2 ®W3 ands,
and Sr ==ssr,1
with
of the formyy€3...€3y,and
®... ®y, and
W2@W3
, , 1®Sr,2
€ 3 ~ , ®sr,3
, ~ € 3withs,,~
~ ~ , 3 = I,
I , ssr,3
, , ~oftheform
+
+
+
Sr,1 =
sk,2=(lOp ®Y+a®Y(&Y)1-12, si,2=(l®1®P+1(&a0Y)/J.
~t remains to show
show (siSk)2
(s~s~)'
=
wemust
mustshow
showSk,2Si,2
sk,zsi,z =
This
It
= -I,
-I, soSOwe
= -si,z~k,z.
-si,24,2. This
follows
paPa==-up,
y ==-ya,
and By
followsbecause
because
-up,aay
-ya, and
,ay =
= -yB.
-y,a.
(3)
As (sksi)'
(Sksi)2
fori i >> k + 1,
have -I
-I EE G(1).
LetGG==GI(-I).
G/(-I).
(3) As
==
-I-Ifor
1, we have
G('). Let
< k <<2n),
j I ==mi,j,
mi, j, where
Then G =
=(sk:
(Sk: 1 i
2n),and
andbyby(1)
(1)and
and(2),
(2),Isis
lSiSjI
where mi,i =
=1,
1,
mi,i+1= mi+l,i = 3, and mi, j = 2 for Ji - jjlJ >
Thus by 30.19, G =2Sen.
Let
mi,i+1=mi+l,i=3,andmi9j=2forli> 1.l.Thusby30.19,G
Szn.Let
with
= G(')
n >>2.2.As
As G
G -2Stn
Szn
with2n2n>>4,4,G(1)
G(')=
Ah is a nonabelian simple
(1) 2
= Aen
group by 15.16. Thus
G('), G(1)
G(') is
G(~)
Thus as
as -I
-I EEG(1),
is quasisimple
quasisimpleor
orG(')
G(1)==(-I)
(-I) xxG(2)
= A2n.
on G
G(2)
aninvolution,
involution,
with G(2)
G(') 2
Az,. In
In the
the latter
lattercase
casethe
theprojection
projectionofof5153
$ 1 ~ 3on
( ~isis
) an
as (~1~3)'
(5153)2
-I.
which is impossible, as
==-I.
+
Chapter11,
11,Exercise
Exercise10.
10.We
Wefirst
first observe
observe Z, isissubnormal
subnormalin
in Gy
G, and
and hence
Chapter
also in
for each z EE ,'Iry, since
Let Qy
y, ry .
G,,, foreachz
sinceZX
Z, <iGy,Z
G,,, by
by definition
definitionofofZX
2,.. Let
Q , ==GGy,r,.
also
in Gy,Z
285
285
Appendix
Then Z,
Zx I
<Qy
a Qy
Then
Qy :!S
5 G,G.5<NG(Zx),
NG(Zx),sosoZ,ZxI!
Q, ag Gy,
G,, establishing
establishing the
the obserobservation.
vation.
prime qq such that
that Z,
Zx is
is not
not a q-group,
q-group, and
and for
for H
H5
< G let B(H)
0(H) =
Pick a prime
Oq(H)
=
Oq(H)E(H).
We
next
show
0(Gy)
<
Gx
>
0(Gy,,).
For
as
Zx
8,(H) = Oq(H)E(H). We next show 8(Gy)( G, > 8(GY,,).For as Z, is
is subsubnormal in Gy
we have
have E(G,)E(G,,,)
E(Gy)E(Gy,,)<NGy(Zx)
G, and
and Gy,Z,
G,,,, we
5 NGy(2,) by
by 31.4,
3 1.4, while
while
by hypothesis NG,(Zx)
< Gx.
Zx is not a q-group, 11 # Oq(Zx),
Oq(Zx),
NGy(Zx)(
G,. Similarly, as Z,
so NGy(Oq(Zx))
< G,
Gx and by Exercise 11.9, Oq(Gy)Oq(Gy,z)
< NG,,(Oq(Zx)).
NG,(Oq(Zx))5
Oq(G,)Oq(G,,,) 5
NG,(Oq(Zx)).
This completes
completes the proof of the second claim.
Third we
we show
show8(Gy)
0(Gy)=0(Gx,y)=0(Qy).
<Gx,y,
= Q(G,,,) =8(Qy).By
Byclaim
claim 2,
2, 0(Gy)
8(Gy)I
G,,,, so
SO as
Third
0(Gy) a Gy,
< 0(Gx,y) (cf.
8(Gy)
G,, 0(Gy)
8(Gy)(8(G,,y)
(cf. 31.3).
31.3). By
By symmetry
symmetry between x,
x , y, z and
z,
< Gz,
G, so
z , xx,, y and
and claim
claim 2, 0(Gx,y)
8(GX,,)i
so0(Gx,y)
8(G,,,) <5Qy
Qy<ING(0(Gx,y))
NG(B(G,,,)) and hence
0(G)
<
0
(Q
y
).
Finally,
as
0(Q)
a
G,
0
(Q
y)
<
0
(G
y
), establishing
claim 3.
B(G,,,) I
8(Qy).Finally, as 8(QY)9 Gy,8(Qy)5 8(Gy),
establishingclaim
3.
Now if Zy
Z, is not a q-group then we have symmetry between xx and y, so
so by
0(Gy) =
= 0(Gx,y)
= 8(Gx).
0(Gx). Then if 0(Gx)
1,
claim 3, 8(Gy)
8(G,,,) =
Q(G,) # 1,
Gy < NGy(0(Gy)) = NGy(0(Gx)) < Gx
and by symmetry, Gx
< Gy,
Gy. Thus
G, (
G,, contradicting
contradicting our hypothesis
hypothesis that Gx
G, # G,.
0(Gx) =
= 1, so Oq(Gx)
= E(Gx)
= 1.
8(G,)
Oq(G,) =
E(G,) =
1. But
But Zx
Z, # 1, so F*(Gx)
F*(G,) # 1 and hence
some prime
prime p. However we
we have
have shown
shown Op(Gx)
Oq(Gx)==11 unless
Op(G,) # 1 for some
Op(GX)
Zx
Z, or Zy
Z, is a p-group,
p-group, so
so interchanging
interchanging the roles of x and
and yy ifif necessary,
necessary, we
Z, is a p-group.
may assume Zx
Next Qx 9
a Gx,y,
0(Qx) 5<0(Gx,y)=0(Qy).
Gx,y, so g(Qx)
8(GX,,)= 8(Qy).Hence
Hence
0(Qx) < n 0(Qy) < 0(Zx) =1,
yErs
as Zx
Qx
Z, is
is aap-group.
p-group. That
Thatisis0(Qx)
8(Q,) =1.
= 1.But
ButZyZ,<(
Q,<5Gx,y
G,,, <iNG(Zy),
NG(Z,), so
Op(Zy) # 1,
1,
0(Zy) <
Zy # 1, Op(Zy)
8(Zy)
5 0(Qx)
8(Q,) ==1.1.Therefore
Therefore F*(Zy)
F*(Z,) ==Op(Zy).
Op(Zy).Thus as Z,
p,
and
hence
by
an
earlier
reduction,
so Zy
Z, is not a q-group
q-group for
for any
any prime qq # and hence
0(G,) ==9(Gx,y)
0(G y) =
=11for
y), and
8(G,)
8(GX,,)==8(Gy)
for each
each q ¢#p.
p.Therefore
ThereforeF*(Gx),
F*(G,), F*(G
F*(G,),
F*(Gx,y)
F*(G,,,) are p-groups.
Chapter
Y,F=
.F_ {Gx,
Gy), and
and rr =
Chapter11,
11,Exercise
Exercise11.
11. Let
Let Y
y EE Y,
{G,, G,),
=F(G,
r ( G ,.T)
F )be
be the
the
defined in
in section 3.
3. If
If G,
Gx =
= Gy
= yGx =
= yGy
coset geometry defined
G, then Y
Y=
yG, =
={y},
(y),
contradicting the hypotheses
contradicting
hypotheses that Y
Y is
is nontrivial.
nontrivial. Thus
Thus Gx
G, # G.
G,.
As G
X, G,
Gx is
ismaximal
maximalin
inGGby
by5.19.
5.19.Thus
Thusifif110
G is primitive on X,
# HH ga Gx
G, then
then
of G,,
G, NG(H)
by maximality of
NG(H)==Gx
G, or
orG.
G.But
Butin
in the
the latter case, H <
ikerG.
kerG,(G)
(G)=
=1,
by 5.7, while as G
G is
is faithful
faithful and
and transitive on X, kerGs
kerGx(G)
1, contradicting
contradicting
H # 1.Thus
Thus G,
Gx =
= NG
(H), so
in particular NG,
(H) <5Gx
in the language
NG(H),
soinparticular
NGy(H)
G,.. Thus, inthelanguage
by Exercise
Exercise 11
11.10,
F*(Gx) is a p-group for
Z, # 1, then by
.lo, F*(G,)
of Exercise 11.10,
11.10,if Zx
286
Appendix
Appendix
some prime
1.Thus
Thus Gx
G, is faithful
faithful on
prime p.
p. Therefore we may
may assume
assume 2,
Z, ==1.
0Q == U
U cr,..
yErx
Y € ~ X
Let m
Y I.Then
Then m
m=
= lIr,l=
Fx I =IF,
for each z E r,,
rx , so 101
2 and then,
m ==IIYI.
Ir,/I for
131<m
im2
then, as
as
Gx is faithful
Gx II < m2
Thus we may take
G,
faithfulon
on0,3 1, IG,
m2!.!. Thus
take ff(M)
(m)==M21
m2!..
Chapter
Chapter 12,
12,Exercise
Exercise5.5.(1)
(1)The
Thepossible
possiblecycle
cycle structures
structures for elements
elements of
G are:
1, (1, 2), (1, 2, 3), (1, 2, 3, 4), (1, 2, 3, 4, 5), (1, 2)(3, 4, 5), (1, 2)(3, 4)
giving a set
< ii (
< 7)
71 of
of elements such
I giI =
I =i for
i fori i<
< 7 and
{gi:: 1 I
such that
that Igi
and
set SS=={gi
1971=
by 15.2.4.
15.2.4.Further
Further SS is
isaaset
setof
ofrepresentatives
representativesfor
forthe
theconjugacy
conjugacy classes
classes
lg7
I = 22by
of G by 15.3.2.
15.3.2.
(2) The representation
representation of G on
on X
X is
is of
of degree
degree 55 and
and 2-transitive
Ztransitive by 15.12.2.
15.12.2.
Arguing as in Exercise 5.1, G has aa transitive
of degree 6 on
Arguing
transitive representation
representation of
NA(P)
and
E ESy15(A).
the coset space G/H,
GIH,where
whereHH= =
NA(P)
andP =
P (g5)
= (gg)
Syl,(A). Namely,
Namely,
counting
the number
number of
of 5-cycles, G has 24
counting the
24 elements
elements of order
order 55 and
and hence
hence
6 Sylow
I G::HI
H=
I =6.6.Now
Now the
the only
Sylow 5-subgroups, so by Sylow's
Sylow's Theorem,
Theorem, IG
structure for
for the
the element
elementgg
g5ofoforder
order55on
onthe
the6-set
6-setGG/H
possible cycle structure
I H is one
cycle of
of length
length55and
andone
onefixed
fixedpoint,
point,sosoPPisistransitive
transitiveononGG/H
cycle
I H -- {H}
{H)and
2-transitive on
on G
G/H
hence G is 2-transitive
I H by
by 15.12.1.
15.12.1.
Let 1/r
be the
the permutation
permutation character
character of
of G on X.
@ be
X. By
By Exercise
Exercise 4.5.1,
4.5.1,1/r(g)
@(g)is
points of
of gg Ec G on X, so
the number of fixed points
so
1/r(gi) = 5, 3, 2, 1, 0, 0, 1
for ii =
=1,
1, ..... ., ,7,
respectively.
for
7, respectively.
Similarly let rpipbe
bethe
thepermutation
permutationcharacter
characterof
ofGGon
onGG/
H.Let
Lethh=
= (1,2,4,3).
(1, 2, 4, 3).
I H.
Similarly
By
15.3.1,g5=(2,4,1 1,3,5)=g5,sohEHandH=P(h)
By 15.3.1,g,h=(2,4,
, 3 , 5 ) = g i , s o h ~ Ha n d H = P ( h ) as
as
I P(h)I = I P I I h I =20= IGI/IG : HI =1 H1.
in H are
gl,g4,
g4, g5,
gg, and g7, so
SO for all other
Thus the elements in
are of
of conjugate
conjugate to gi,
= IFixG/ff(gi)l
IFixG/H(gi)I==0.0.By
ByExercise
Exercise2.7,
2.7,ififQQEESyl,(H)
Sylq(H)then
thenIFix(Q)l
IFix(Q)I=
=
gi, ip(gi)
rp(gi) =
II NG(Q)
NG(Q)::NH(Q)l,
NH(Q)I,so
soas
as H
H=
= NG(P),
NG(P), ip(g5)
=
1
while
as
(h)
E
Sy12(H)
with
rp(gs) =
( h )E Sy12(H)
NG((h))
Sy12(G)ofoforder
order8,8,rp(g4)
ip(g4)==Irp(h)l=
I c(h)I =2.2.Thus
Thusg4g4=(a,
= (a,b,b,c,c,d)(e)(f)
d)(e)(f)
N
G((~E
E
) >Sy12(G)
on G/H,
= g:924 =
(a, c)(b,
= 2.
GIH,so
sog7
g7 =
c)(b, d)
d)and
andhence
henceip(g7)
rp(g7)=
= (a,
linear characters.
characters. Let
Let A =
= Alt(X);
(3) By Exercise
Exercise 12.1,
12.1,GGhas
hasI G
I G::G(1)
G(')IIlinear
by 15.16,
A
is
simple,
and
by
15.5,
: 1A=
I 2,
= 2,sosoAA==G(')
GMand
andGG has
has two
15.16,
simple,
15.5, 1lGG: A
linear characters,
characters, the principal
principal character
characterX1 and the sign character
character X2,
~ 2where
,
287
Appendix
by Exercise
ker(X2)=
= A.
A. That
Thatisisx2(a)
X2(a)==1
for aa E
E AAand
andx2(g)
X2(9)=_-11 (the
Exercise 12.1,
12.1, ker(x2)
1 for
(the
primitive 2nd
2ndroot
rootofof1)
1)for
forggEcGG-- A.
primitive
=
((4)
4 ) By Exercise
Exercise 12.1,
12.1, X1
~1 and X2
~2 are irreducible. By Exercise
Exercise 12.6.3,
12.6.3, X3
~3 =
1/r
and x4
X4==rpp - X1
characters. Then
@ -- X1
~1 and
~1 are irreducible characters.
Then by Exercises
Exercises 9.3
9.3 and
and
= x2xi,
X2Xi,i i==33,, 44,, are
are also
also irreducible
irreducible characters.
characters.
9.10, Xi+2
xi+2 =
irreducible characters
characters of G,
This gives 6 irreducible
G , exhibited
exhibitedin the first six rows of the
below. As
As G
G has 7 conjugacy classes, G
G has 7 irreducible characters by
table below.
34.3.1. Thus it remains to determine
Let
ni =
= Xi(1);
35.5.3,
34.3.1.
determine X7.
~ 7Let
. ni
~ ~ ( 1by) 35.5.3,
;
7
120=IGI=
n?
i=1
n7 =
= 66,, giving the first column of the table. Then we use
so n7
use the
the orthogonality
orthogonality
relation 35.5.2
X7(g0 for
for ii >
> 11 to
35.5.2 to calculate x7(gi)
to complete
complete the table.
Character Table
Character
Tableof
ofS5
Ss
91
g2
g3
g4
g5
g6
g7
1
1
X1
1
1
1
1
1
X2
1
-1
1
-1
1
X3
4
2
1
0
-1
X4
5
-1
-1
1
0
-1
-1
-1
X5
4
-2
1
0
-1
1
X6
5
1
-1
-1
0
1
1
X7
6
0
0
0
1
0
-2
1
0
1
0
Chapter 13, Exercise
Exercise 1.
1. ((1)
LetMM =
= CG(g). Then
Then1 H
IHxMI
= (M
Chapter
1 ) Let
xMl =
IM : MH
M H xI,
1,
where M
MHx
the stabilizer
stabilizerofofHHx
in M
M.. Thus
Thus M
MHx
=M
M fl
H x isis the
x in
Hx =
n Hx
H X ==CH=(g),
CHx(g),
so (IHxMI,
CHx(g)
of C
CG(g)
if
( I H x M 1, p) ==1 1ififf
C H I @contains
) containsaaSylow
Sylow p-subgroup
p-subgroup of
G ( g ) iff
p-subgroup of
of cCG(gx-')
iff gX-I
gx-' is extrernal
extremal in
in H
H..
cCH(gx-')
H ( g X - I )contains a Sylow p-subgroup
G ( g X - I )iff
LetAA==HH/K,
a: H
H + A the natural map, and V
I K , a:
V the transfer of G into
((2)
2 )Let
A via a.
a. Recall
A
Recall by
by 37.2
37.2 that
that VV isis aagroup
grouphomomorphism,
homomorphism, so
so as
as A
A is
is abelian
abelian
1.
Gf1>(<ker(V)
ker(V) and hence
hence it suffices to
to show gV
gV # 1.
G(')
Choose
representatives for
for HH in G
G as in
in 37.3.
37.3.As
As IgI
IgI=
= p the
Choose a set X of coset representatives
the
-1
K;;
length
p. If ni ==ppthen
gnh
length ni of
of the
the ith
ith cycle
cycle of g is 1 or p.
then11==
ga, ,so
sognixr
gnlxl ' EE gn'
gni K
that is
is
is
gn'x`
= (gn )a = (ga)n' = 1.
E H.
On the other
other hand
hand ni
ni =
= 11ififfg gfixes
fixesHxi,
H x iin
, inwhich
whichcase
casegxi
gx;'' E
H . Now
Now
g(x`m)-'
=
gx1'
for
CG(g)
permutes
the
fixed
points
of
g
on
G/H
and
M = C G ( g )permutes the fixed points of g on G I H and g("irn)-' = gx;l for
M-1
288
Appendix
each m E M, so if (HyjM: l < j < s} are the orbits of M on FixG/H(g)
(Hx,: l < i < t} then
s
W a = r7 ((gy' )a)k,
=1
j1=11
where kJ = IHy,MJ is the length of the jth orbit. If p divides kj then(gyi')ki =
1 = (ga)ki. On the other hand, if p does not divide ki, then by (1), gyi ' is
extremal in H, so by hypothesis, gy" E g K and hence (gy ' )a = ga, so again
(gyi 'a)ki _ (ga)ki. Thus using 37.3.3,
gV = fl ((gn`)x,')a = fl(ga)n' = (ga)n
i=1
i=1
where n = E; _I n; _ IG : H1. Finally, by hypothesis, (IG : H1, p) = 1 and
g E H - K, so ga 1, and hence IgaI = p. Therefore gV = (ga)n # 1,
completing the proof.
Chapter 14, Exercise 9. (1) As u < w, w = xu for some x e W with
n=1(w)=l(x)+l(u).Thenx=r1 rmandu=rm+1.. rnwith r, ES.By
41.7, (C,: 0 < i < n) is a gallery from C to Cw in 8 where C; = Crn_,+i
rn
By construction, Cl(") = Cn-m = Crm+1 "'rn = Cu.
(2) Let u = sw and n = 1(w). Thus w = su and by hypothesis, 1(u) < 1(w),
so 1(u) = n - 1 and u < w. Hence by (1), Cu is adjacent to Cw, so E _
Cu fl Cw is a wall of the chambers Cu and Cw.
Next BW fixes Cw and hence each subsimplex of Cw. Similarly B" fixes
each subsimplex of Cu, so B" BW fixes E. Thus B fl B"BW CNB(E) = A.
Let X be the set of galleries in 8 from C to E of length n - 1. By 41.7 and
42.3, d(C, Cw) = n and d(C, Cu) = n - 1, so d(C, E) = n - 1. Thus X C E
by 42.3. As A acts on C and E, A acts on X. Further, Cu and Cw are the two
chambers in E through E, so as d(C, Cw) = n, Cu is the terminal member of
each member of X. Thus A < Gcu = B" = BSW
(3) By hypothesis,
l(s SW-1) = l(w-1) = (W)< l(ws) = l(sw-1),
so by (2), B fl B'B' c BW-'. Then conjugating by w, BW n BBS C B.
(4) As u <w, w = xu with n = 1(w) =1(x) +1(u). Induct on l(x). If
l(x) = 0 then x = 1 and the lemma is trivial. Thus x = sy with s e S and
1(y) = l(x) - 1, and u < yu, so by induction B fl By" < B". Also 1(sw) _
l(yu) <1(w),soby(2),Bf1BW <By". Thus BflB' <Bf1By" <B"
Appendix
Appendix
289
(5) Define
Define
nBW.
fl==n
H
BW.
WEW
W
€W
Then H is
is the
the pointwise
pointwise stablilizer in G of E,
C,so
so(a)
(a)and
and(b)
(b)are
areequivalent.
equivalent.
By Exercise
Exercise 10.6.2,
10.6.2,ww 5
< wo
for all
all w
w E W, so by (4), B n BW
°_(<BW.
B'. Thus
By
wo for
BWO
Thus
H
so (b) and (c) are equivalent.
H ==BBni BWO,
l BWO,
equivalent.
(6) LetFT =
= (G, B,
B, i?,
N, S).Recall
S). RecallHH== B
B nn N
N r?a N with
with W
W=
=N
N/H
for
(6)Let
/ H and
andfor
w=nHEW,BW=B".Thus
w
= n H ~ W , B ~ = B ~ . T hH<BWforallwEW,soH<H.
usH_(B~forallw~W,so~_(~.
As
T satisfies BN1, G==(B,
(B,N).ThenasN
N). Then as N
HN=i?
= N _(< G,G
G, G =
= (B,
N).
AsTsatisfiesBN1,G
_( <
HN
(B,i?).
Also
BnN=BnRN =H(BnN)=HH=H.
By construction,
construction,H
R is the
the pointwise
pointwisestabilizer
stabilizerininGGofofC,
E, so
soas
asNN acts
acts on
on C,
E,
By
on H
H and
and hence
henceHH a RN
AlsoH
R inl N
N acts on
AN ==N.i?.Thus
ThusTFsatisfies
satisfies BN1.
BNl. Also
the kernel
kernelof
ofthe
theaction
actionofofNNon
onCE,, while
whileby
by43.5
43.5and
and41.8.2,
41.8.2,WW==NN/H
is
is the
/ H is
regular on
on CE,, so
so H
H inl N ==H.
= N/
H. Then
H.Let
Let4V
w=
#/A.
Then
a
4V=N/H=NH/H=N/HnN=N/H=W,
BN2 isis satisfied
satisfiedwith
with3S==(S:(s":
s ES},
S),where
where
nHfor
forww== nnH
s E
Bw
==n~
H EE S.
S.
so BN2
Notice
< B,
we have wB =
= wB andBB
and BO =
= Bw,
so also BB"'
= Bw.
BI.
~ o t i c ethat
t h a tas
a sHft 5
B,wehaveBB
Bw,soalso
' =
Thus, as T satisfies
T..
satisfies BN3 and BN4, so does F
Therefore
Therefore TF isis aa Tits
Tits system.
system. By
By construction,
construction,
H=nBW=nBO,
WEW
WEW
T is saturated.
so ?
Chapter 14,
Chapter
14,Exercise
Exercise 10.
10. (1)
(1) As
As aw
uw >>0,0,l (sw)
l(sw)>>I (w)
l(w)by
byExercise
Exercise10.6.3,
10.6.3,
l(w-1s) =
l(w-1).
so l(w-'s)
=l(sw)
l(sw) >>l(w)
l(w)==
l(w-').Also
AlsobybyExercise
Exercise10.6.2,
10.6.2, w-1s
w-'s <5wo,
wo,
by Exercise
Exercise 10.3.2,
10.3.2,l(wos)
t(wos)5< l(wo). Therefore
Thereforeby
byExercise
Exercise10.6.4,
10.6.4,w-'
w-1=
=
and by
(w-1s)s <I
wos.
(w-'s)s
wos.Then
Thenby
by Exercise
Exercise 14.9.4,
Ba=BnBWOS<BW'.
(2) As
As aw
aw <<0,0,l(sw)
bybyExercise
l(sw)<<l(w)
l(w)
Exercise10.6.3,
10.6.3,so
soby
byExercise
Exercise 14.9.2,
14.9.2,
(2)
B ni l B''WBW
C
B''W.
Conjugating
this
containment
by
w-1s,
we
conclude
BsWBWg BSW.Conjugating this containment by w-'s, we conclude
BW-i8 n BBS
Bw-Is
BBS C
E B,
B, so
so BW-1S
Bw-Is n BS
B" < B.
Next by hypotheses T is saturated,
saturated, so by Exercise 14.9.5,
14.9.5, B ni l BWO
BWO== H.
Thus
(BanB" )S= (BnBWOSnBW-')S=BSnBW°nBW-'S <BnBWO=H
so BanBW'=HI =H.
290
Appendix
(3) As
Asl(w)
1(w)5<l(ws),
l(ws),l(w-')
l(w-1)5<l(sw-'),
l(sw-1),so
sobyby43.3.1,
43.3.1,SBW-'
sBw-15cBsw-'
Bsw-'B,
(3)
B,
and hence
hence B
B5
c sBsw-1Bw
there exist
existxxEc BS
B' and
s Bsw-' BW ==BSBW.
BSBW.Thus for bb EE B, there
y EE Bw
= xy.
ni? BSB
BWwith bb =
xy. Then
Then y ==x-lb
x-'b E EBw
BW
BSBCcBBbybyExercise
Exercise14.9.3,
14.9.3,
soy
andx=by-1
s o y EBnBW
E B n BWandx
= by-' EEBnBs.
B n BS.
(4) Notice
Noticeaas
= -a
0,0,sosobyby(2),
s=
-a < <
(2),Ba
B, ni? BS
BS==H.
H.Also,
Also,setting
setting w =
=wos,
wos,
1(w)
< l(wo)
= l(ws),
= (B
=) (B
Z(W)5
z(wo) =
~(ws),so
SO by (3), B =
(B nn BS)(B
B ~ ) ( Bn Bw)
B ~=
(B nnBS)Ba.
B~)B,.
(5) If
Then
If Ba
B, =
=HHthen
thenby
by (4),
(4), BB ==H(B
H(Bn i?Bs).
BS).
Thenasass sisisananinvolution,
involution,
BS =
= HS(B
= H(B
HS(Bn BS)S
Bs)S =
H(B nnBS)
BS)==B,
B,contrary
contraryto
toBN4.
BN4.
(6) If aw >>00then
by
then Ba
B, <5Bw-'
BW-'
by(1).
(1).On
Onthe
theother
other hand,
hand, if aw <<0,0,then
then
by (2), H =
=Ba
B, nn Bw-',
BW-',so
SO Ba
B, 6 B'-'
B ~ -by
by
' (5).
(5).
(7)
by p:
gyp:
and w
w EE W; we first
(7)Define
Definegyp:
p: A
A + (D
@ by
B,"Ba
Hi-H
aw.aw.
LetLet
,!? PE En r and
first
show
show
(*)
(* 1
Ba
B," ==Bp
Bp
if
iff
, ! ?==aw.
aw.
=BBgthen
thenasasBBg
=nBBw0'P,
i? BWorB,
B"-'n i?Bw0'pw-' .. Thus
Thus by
First, ifif B,"
Ba =
=B
BaB,= =
Bw-'
(6), aw >>00<<awrewo.
so aw
aw =
= ,!?
awrpwo.Therefore
Therefore awro
awrp <<00by
byExercise
Exercise 10.3, so
by
Similarlyif ifaw
aw== ,P
awrpwo >
> 00 <<caw,
Ba <
by 30.7.
30.7. Similarly
!? then
then awrpwo
aw, so
so by (6), B,
5
Bw-'' nnBw0'fw
B W O ~ ~ W'- ' and
and hence
= =Bp.
by
Bw
henceB,"
Ba'5< BBi? n BWorB
Bw0'P
B.As
As,!?w-'
jw-1 =
= a,a ,by
so=Bp
B,".Thus
Thuswe
wehave
haveestablished
established(*).
(*).
symmetry B;'
B' ' < B,B,,
so B
BC=,W
By
(*), B,"
Ba ==BU = Bp
if awu-'
awu-1 =
= P,!? if
Thus
By (*),
B; ififfBaB,"u-'
Bp iff
iff aw
aw ==3u.
,!?u.
Thus pp
is a well defined injection.
injection.By
By 30.9.3,
30.9.3,pp isis aa surjection.
surjection.As
AsB,"'p
Ba "cp==awu
awu =
=
(Ba
)cpu,
cp
is
W-equivariant.
(B,")pu, p is W-equivariant.
(8) First notice
Bsw = (B n BwOw-'S)S = Bs n BwOw-'.
Next as aw <<0,0,aww0
so
awwo>>0 0bybyExercise
Exercise10.3.
10.3.Hence
Henceby
by(1),
(I),Ba
B, <5Bwow-',
BWow-I,
Ba
B, <BnBw0w-'
5 B n BWOW-I =Bw.
= B,.
Again
as awwo >
> 0,
l(wwo) <5 l(swwo)
by Exercise 10.6.3,sol(wow-'s)
10.6.3, so l(wow-'s) >2
Againasawwo
O,l(wwo)
l(swwo)byExercise
l(w0w-').Then
by Exercise
14.9.3,
<(B,
<
l(wow-'). Thenby
~xercise
14.9.3,BS
BSni? Bw0w-'
BWoW-'
B,so
SOBsw
Biw=
=BS
BSin? Bwow-'
BWoW-'
<
B
Thus BaBsw
Onthe
theother
otherhand
handbyby(4),
(4),BB =
B ni? BwOw
Bwow-' = Bw.
B,. Thus
B,Biw 5S;B,.Bw.On
=
Ba(B
B,(B nnBs),
BS),soSOasasBa
B, <5Bwow_I,
Bwow-I,
Bw = B n Bw0w_' = Ba(B n BwOw-' n Bs) = Ba(B n Bsw) C BaBsw
That
BaBsw== B,.
Bw.Finally
Finallya sas==-a
-a <<0,0,so
= H.
That is, B,Biw
soby
by (2),
(2), Ba
B, n
i? BS
BS =
H. ThereTherefore
= Ba
= H.
fore Ba
B, nn Bsw
BiW=
B, nn BS
BS n Bw°w-'
BWow-'=
H.
(9) We
We apply
apply(8)
(8)with
withww==wi-lwo,
wi_1w0,a a==ai,
al,and
andss=
= riri for
for each
each 115< ii <
< n.
n.
By
10.6.3,aiwi_lwo
= l(wiwo)
By Exercise 10.6.3,
aiwi-lwo <
< 0since
Osincel(riwi_1wo)
l(riwi-lwo) =
l(wiwo) <<l(wi_lwo).
l(wi-lwo).
Thus
Thus by
by (8),
(8),
(*)
Bw;_iwo
Bai Bw
wo
Appendix
291
and
Bai fl B i,,. = H.
(**)
(**>
Claim
k
B"'i_1
(!)
Ba1wl-1
Wi-1W0
$Wk
WkWO
j =i
for l1<(i i<<n nand
k k< <
n. n.
We
andi i<(
Weinduct
inducton
onk.k.By
By (*),
(*),
Bw'-'
w;_Iwo - B 'i_I
a;
w,wO- -Ba;w;_i BW1
Br'wt_l
O
holds when
whenkk =
= i. Assume for kk - 1.1.By
so (!) holds
By (*),
(*),
BWk-1
Wk_IWO
-
B
- akWk-1
B
WkWO -
BWk-1 BrkWk-1
ak
Wk
WkWO'
so by the induction
induction assumption
BWi-1
wi-IwO
-
k-1
k
B
ajwj_I
BWk-1
Wk_1WO -
i=i,
Wk
B ajwj_,
* BWkWO,
1=1
establishing (!). Notice that wn-lwo
wn_lw0 =
= rn,
rn,so
so
BWn-1
W" 1WO
= BW"-'
= BW'
rn
a"
' = Banwn-1
1,we
we get
Substituting this
this equality
equalityinto
into(!)
(!)with
withkk== nn - 1,
n-1
(!!)
BW'
'
Wi_1
WO =
n
H BajWj_1
BW
W"-1' WO =
H B ajwj_;
j=i
j=i
Further
Bwo =
=BB fli?BwOwO
BWow0==BB,
, so
conclude
Further B,,0
soapplying
applying(!!)
(!!)atatii =
= 11 we conclude
n
B = H Baj W j-1
j=1
Next by (**) and (!!),
(!!),
H =
H
= H"
BC:,~
=~Baiwi-1
(B,, n
B ) ~~ n
~ -~ ~ ~
= (Bai
fl B ; ~ ~1 =
~ w i - 1
BwiwO)w'
I
I
~
-
BWiWO
n
Baiwi-1 I
I H Bajwj-i
j=i+1
1Jrr:
i Haiwi-1
aiwi_1isisaabijection
bijectionof
ofII =
= {1,
It remains to show +:
iH
11, ...,
. ..,n}
n} with
with (D+.
@+.
We saw
saw at
at the
thebeginning
beginningofofthe
theproof
proofthat
thataiwi-lwo
aiwi-1w0<<0,
0,so
soaiwi-1
aiwi_1 >
> 0 and
We
c cp+,
= n,
I+ C
@+. By
By Exercise
Exercise 10.3,
10.3, 1cI+J
)@+I =
n,so
soititremains
remainsto
toshow
showiJr is an
hence I1Jr
But ifif aiwi-1
aiwi_1 =
= akwk_1
for some
some kk > i,i, then
injection. But
akwk-1 for
then
+
ai =CYkWk_1Wi 11 =akrk_1...ri,
~
Appendix
292
ri+1, contrary to Exercise 10.6.3, since
so 0 >
> airi
air;==akrk_1
akrk-1 . .ri+l,
l(rk ... ri+i) > l(rk-1 ... ri+i)
Chapter
isisbalanced
Chapter15,
15,Exercise
Exercise4.
4.(1)
(1)To
To show
show H
H ==CG(a)
CG(a)
balanced we
we must
must show
show
Or'
(CH(X))
Or'(H)
Or!(CH
( X ) ) 5< Or/
( H ) for each X of order r in
in X.
X.
But as H isis solvable,
solvable, this
this is
is immediate
immediate from
from 31.15.
3 1.15.
(2)
(2)We must show O0 is balanced. But for a,
a ,bb EE A#
A'
0(b)nCG(a) = Or'(CG(b))nCG(a) = Or'(CCa(a)(X)) < Or'(CG(a)) = 0(a),
isisbalanced
where X =
=(a,
( ab),
,b )as
, asCG(a)
CG(a)
balancedby
bythe
thehypotheses
hypothesesof
of part
part (2).
(2).
(3) Let 80 be the signalizer functor
functor defined in (2);
(2);O0 is
i s aa signalizer
signalizerfunctor
functorby
by
(1) and (2).
(2).By the Solvable
Solvable Signalizer
Signalizer Functor Theorem, O0 is complete;
complete; that
is, there exists
exists an
anrl-subgroup
r'-subgroup YY of
of G
G such
suchthat
thatO(a)
0(a) =
= Cy(a)
Cr(a)for
for each
each a EE A#.
A'.
For B aa noncylic
noncylic subgroup
subgroup of
of A
A define
define
Y(B)
Y ( B )=
=(0(b)
(O(b)::bb EE B#).
B').
Now for gg E G with aa,, ag EE A#,
A',
O(alg
0 , ! ( C ~ ( a )=
Ort(Cc(ag))=
O(ag),
0(a)9 =
= Or,(CG(a))g
=) ~Or'(CG(ag))
= 0(ag),
construction YY(B)
< Y,
so NG(B)
N G ( B )<
5 NG(Y(B)).
NG(Y(B)).Further, by construction
( B )(
Y , while by 44.8.1,
YY <
_( Y(B),
Y ( B ) ,so
SO Y
Y ==Y(B).
Y ( B )Thus
. ThusNG(B)
N G ( B )<5NG(Y),
NG(Y),so
soI'2,A(G)
r z , ~ ( G<5
) NG(Y).
NG(Y).
hypothesis, G
G=
= 1'2,A(G),
so YY <1
< G,
But by hypothesis,
r2,A(G),SO
G ,and
and hence
hence as
as Y
Y is an
an r'-group,
rl-group,
Y
Or'(G).Therefore
ThereforeOr,(CG(a))
Or (CG(a))(
< YY (
< Or'(G).
Y _<
< Orf(G).
Orr(G).
particular if Or!
Or ((G)
(CG
In particular
G )==11then
thenOr
Or/
(CG(a))
(a))==1.1.Hence,
Hence,asasCG
CG(a)
( a )isis solvable,
solvable,
F*(CG(a))
=
F(CG(a))
=
Or(CG(a))
F*(CG(a))= F(CG(a>>= Or(CG(a)).
Chapter 16,
Chapter
16,Exercise
Exercise5.5.(1)
(1)By
Bydefinition
definitionof
of strong
strong embedding,
embedding, H
H isis proper
proper
order, so
so there
thereisis an
an involution
involutioni i EE H.
of even
even order,
H . Indeed,
Indeed, 46.4.3 is one of the
conditions for strong
equivalent conditions
strong embedding,
embedding, so
so
(*)
(HnH91isoddforgEG-H.
for gg E G,
Let jj EE I;I ;ititsuffices
suffices to show j EE iG.
iG.Suppose
Suppose jG _c
E H.
H . Then for
G,
jg EE HHnnHg,
jg
Hg,soSOIH
IHnnHgI
Hglisiseven
evenand
andhence
hence gg EE HHby
by(*).
(*).As
As this
this holds
holds
for each g E
E G,
G ,GG==H,Hcontradicting
, contradictingHHproper.
proper.This
Thiscontradiction
contradictionshows
shows
there is
is aa conjugate
conjugates sofofj j in
in G - H,
H,and
andititsuffices
suffices to
to show
show s EE iG.
iG.But
But if
there
not, then
then by
by 45.2,
45.2, lisl
lisI is even,
even, so by
by 45.2.3
45.2.3 there
there isis an
an involution
involutionzz EE (is)
(is)
not,
centralizing
and s.
s. Now
Now as
as 46.4.1
46.4.1 is
is one
one of the equivalent
equivalent conditions
conditionsdefining
defining
centralizing i and
strong embedding,
embedding,CG(t)
CG(t)5<HHfor
foreach
eacht tEEII nn H
H,, so zz E
< H,
E CG(i)
CG(i)5
H , and
and
strong
then also ss E CG(z)
< H,
CG(z)(
H ,contrary
contraryto
to the
the choice
choice of s.
(2)Leti,t
(2) Let i, t E
E 1InH.By(1),ig
n H . By (I), i g ==tforsomeg
t for some g E
E G,sot
G , so t E
E HnHgand
H fl Hg and
above,gg EE H
H by
by (*). That
That is,
is, H
H is transitive
transitiveon
onII nn H
H..
hence, as above,
293
293
Appendix
(3) As
H,D
D ==HHnnH"
Hu
oddorder
orderby
by(*).
(*).
(3)
As uu 40 H,
isisofofodd
(4) Let
Letjj E J.
J . Then
Then j EE D
D and
and j"
jU==j-1
j-' bybydefinition
definition of
of J.
J.Thus
Thus (uj)2
( ~ j=
=) ~
ujuj =
= j"j
j u j==j-'j-'j
= so
1 , sUj
o uE
j EI.I As
. A sjjEED,
D ,Uj
u j EE uD
u D n II,, sso
o uuJ
J E
ujuj
j = 1,
c
u D nn I.I.On
Onthe
the other
other hand,
~ d=
uD
hand, suppose
supposeddEEDDwith
withud
udEEI.
I. Then
Then 11=
= ((ud)2
=) ~
=dud,
d u d ,so du
d-' and
and hence
D nnI I5c uuJ.
J.
udud =
d" =
= d-1
henceddEEJJ;; that
thatis,is,u uD
(5) Set YY =
= CG(j)(u).
C c ( j ) ( u )As
.ASJJC_CDDand
andI D
ID1I is odd by (3),
(3), Ijl
1 j Iisisodd,
odd, so
so as
j 0# 1,1, 1Ijlj I>> 2.
2. Then
Then as u inverts
inverts jj,, we
we have
have uu 04CG
C c(j( j).) . On the other hand
uE
Y.
E NG((j)),
N c ( ( j ) ) so
,so Y
Y <5 NG((j))
N G ( ( j )and
)and hence
hence CG(j)
C G ( j ) Y.
Suppose
t
E
CG(j)
and
let
K
=
H
n
Y.
By
construction,
Suppose t E C G ( j )and let K = H n Y. By construction, uu EE YY-- H,
H , so
so
K is
is proper
proper in Y
Y and
and tt EE YY so
so K
K has
haseven
even order.
order. Further
K,
K
Further for
for yYEE YY -- K,
y 04 H,
H ,sosoIKIKn nKYI
Kylisisodd
oddby
by(*),
(*),and
andhence
henceKK isisstrongly
stronglyembedded
embedded in
in Y.
Y.
Thus applying
Y, the involutions
involutions uu and
are conjugate
conjugate in Y.
Y. This is
applying (1)
(1) to
to Y,
and t are
impossible, as tt E CG
C G (j)
( j )<1
Y while u E Y
Y --CG
C G(j( j).) .
<Y
Let r,r,ss be
bedistinct
distinctinvolutions
involutions
uD.
Then
uDand
and1 10
rs EE D
D,,
(6) Let
in in
uD.
Then
r DrD
==u D
# rs
so
SO
s=
ErDnI =uDnI =J
by (4). But if rCG(t)
r C ~ ( t==
) SCG(t),
s C G ( ~ also
)also
, rs
rs EECG(t),
CG(t),contrary
contraryto
to (5).
(5).That
That is,
is,
rCG(t)
0
sCG(t).
rCc(t># sCc(t).
(7) and (8): Define nn =
=II G :: H
HII and
C2
{ ( i x,
, x ,y):
y): i cEIIand
and(x,
( x y)
, y)isisaacycleini
0 =={(i,
cycle in i on GIH).
G/H}.
46.4.4 is one
one of
of the
the equivalent
equivalent conditions for strong embedding, i fixes a
As 46.4.4
uniquepointofG/H,soihas(n-1)/2cyclesoflength2.Thus~~~
III(n-1)
unique point of G/H, so i has (n -1)/2 cycles of length 2. Thus 101==III
(n -1)
and
=nm,so
and 1III
11 ==n1I
nlI nnHI
HI =
nm, so I0I
1C21 =n(n
= n(n --1)m.
1)m.
On the other hand, ifif A
A=
=XXxxXXand
and8(x,
6(x,y)y)isisthe
thenumber
numberofofinvolutions
involutions
with cycle (x,
( x ,y), then
8(x, y)
101 =
(x,y) E A
Now,
upto
toconjugation
conjugationininG,
G,we
wehave
have
H,and
andifif6(x,
8(x,y)y)#000then
thenyy=
= Hu
x x==H,
Now, up
for some
someuu Ec I --H.
H .Let
Let86 ==8(x,
6(x,y),
y), u1,
u l , ..... .,,us
us be the involutions with cycle
cycle
(x,
y), and
anddid1==ului
ululfor
for115<i i 5< 6.
8. Then
ThendidlEcDD 5< H, and
and ififii #0 j then
( x , y),
then
by (6),
(61,
dl 1dj = utu1u1u3CG(t) = utu3CG(t) 0 CG(t),
so
diCG(t)
0 dJCG(t).
SO di
CG( t ) #
d j CG(t).Thus
Thus
8<IH:Cx(t)1=ItG1=IInHI=m
by (2).
(2). Hence
n(n-1)m=IQI= i 8(x,y)<IAIm=n(n-1)m,
(x,y) E A
Appendix
Appendix
294
so all inequalities
inequalitiesare
areequalities
equalitiesand
andtherefore
therefore
S(x,
forallallx x#0 y in
S(x,
y) y)
==mmfor
G/H.
GIH.
This
establishes(7).
(7).Further,
Further,asas
diCG(t)
djCG(t)forfori i#0 j, we
This establishes
diCG(t)
# 0djCG(t)
we have
have
ID:
so H ==DCH(t)
andhence (2) implies (8).
ID :CD(t)I
CD(t)l >
2 S6 ==mm ==I H
IH: :CH(t)I,
CH(t)l,so
DCH(t)andhence(2)implies(8).
Chapter
Chapter16,
16, Exercise
Exercise 6.
6. (1)
(1) Let
Let zz be
be an
aninvolution
involution in Z(T).
Z(T). By
By hypothesis,
hypothesis,
CG(Z)
an elementary
elementaryabelian
abelian2-group,
2-group,sosoasasTT <
CG(z) isis an
(CG(Z),
CG(Z),it follows
follows that
CG(Z)
= T.
CG(z) =
T. Thus
Thus (1) holds. Further as T is elementary
elementary abelian, each t EE To
T# is
an involution in
in the
thecenter
centerof
ofT,
T, so
soby
bysymmetry
symmetrybetween
betweent tand
andz,z,TT =
= CG(t).
CG(~).
Therefore
Therefore T is
is aa TI-set
TI-set in G.
G.
(2)
As TT is
is a TI-set
TI-set inin G,
G, NG(S)
NG(S)5<HHfor
foreach
each110
< T, and as T EE
# SS (
(2) As
Sy12(G),this
thisisisone
oneof
ofthe
theequivalent
equivalentconditions
conditionslisted
listedinin46.4
46.4(with
(withkk=
= 1) for
H to
to be
be strongly
strongly embedded
embedded in G.
(3) By (2), we may apply
by part
part (3) of that
apply Exercise 16.5. In particular,
particular, by
exercise,
exercise, D isis of
of odd
oddorder.
order. Thus
Thus ICD(u)I
ICD(u)l is odd,
odd, while
while as
as uu isis an
aninvolution,
involution,
CD(U)
2-group, so
so CD(u)
CD(U)=
= 1. Hence
Hence u inverts D.
CD(u)isis aa2-group,
Adopt the notation of Exercise 16.5.
16.5. Then
Then II n
fl H ==TT'#,
fl H
=
, so
so m =
=IIII n
HII =
q --1,1,as
asTTisiselementary
elementaryabelian
abelianof
of order
orderq.
q.By
By Exercise
Exercise 16.5.2,
16.5.2, H
H is
is transitive
transitive
onIflH,soq-1=IH:CH(t)I=IH:TI,andhence
IHI=ITI(q-1)=
= ITI(q - 1) =
on I n H, so q - 1 = IH: CH(t)l = IH: TI, and hence IHI
q(q --1).
1).As
As uu inverts
inverts D, JJ ==uD
uDis isofoforder
orderIDI,
IDI,while
while JJ isisthe
the set
set of
of
involutions
withcycle
cycle(H,
(H,Hu)
Hu) on
on X,
X, so
so by
by Exercise
Exercise16.5.7,
16.5.7,I JI
IJI =
= m.
m. Thus
involutions with
IDI
= IIJI
J I == m
m =q-l.Therefore,asD
= q - 1. Therefore, as D <5 HHwith
ID1 =
w i t hTTflnD
D ==11as
a sDDis
i s ooff oodd
dd
order, we have
have
ITDI = ITIIDI = q(q - 1) = IHI,
so H ==TD
TDand
andDDisisaacomplement
complementto T in H.
H.
(4) First, as
as T is noncyclic,
noncyclic,qq >
> 2,
D I==qq - 11 >>1.1.Further,
2, so
so II Dl
Further,as
as uu inverts
inverts
D, D
and TT isis abelian,
abelian,while
whileHHIT
- D,
D is
is abelian,
abelian, and
IT S
D,as
asDDisisaacomplement
complement
to TT in
of H
H,, so
in H
H by
by(3).
(3).Thus
Thus HHisissolvable
solvable and
and D
D isis aaHall
Hall 2'-subgroup
2'-subgroup of
so
DG
fl H ==DH
DG n
D~by
byPhillip
PhillipHall's
Hall'sTheorem
Theorem18.5.
18.5.
Let
NG(D)and
anduuEESSEESy12(M).
Sy12(M).As
AsTTisiselementary
elementaryabelian,
abelian,so
soisisS,
S,
Let M
M ==NG(D)
so if S 0# (u)
(u)then
thenSSisisnoncyclic
noncyclicand
and hence D ==(CD(s)
(CD(s): :SsEES#)
s')bybyExercise
Exercise
8.1. This is impossible, as
as D 0#11while
whileCD(S)
CD(s)==1,1,since
sinceCG(S)
CG(S)isis aa 2-group
2-group
odd.
and
D I is
is odd.
and II Dl
Thus
= (u),
39.2, M
M=
= (u)E,
Thus S =
(u),so
so by 39.2,
(u)E, where
where E ==O(M).
O(M). Let
Let A =
=Fix(D)
Fix(D)
be the fixed
fixed points
pointsofofDD on
on X.
X. As
As D~
DGnflHH =
= DH,
D ~M
M, isistransitive
transitive on A
A by
by
5.21. TThus
~ U Sk k==la
1 ==I M
:M n H I .B U ~
5.21.
JAI
IM:MfHI.But
m fl H = NH(D) = NDT(D) = DNT(D) = D,
as H ==TD
T Dand
andND(T)
ND(T)==1,1,asasCD(t)
CD(t)==11for
foreach
each tt EE To.
T'. Thus
Thus MID
M I D is
is
regular
regular on
on A.
A.
295
Appendix
Let F
r be
be the
the set of triples (i, x, y) such that i EE uE
uE and
and(x,
(x,y)
y)isisaacycle
cycle of
of ii
uE=InMisoforderIEI,andasM/D
on
As [El
El isis odd,
= 1,
1,so
on A.. As
odd, CE(U)
CE(u) =
SO U
E = I n M is of order /El, and as M I D
is regular on A, each i EE uE
uEisisregular
regularon
on A.
A. Hence
Hencethere
there are
are k members of F
r
whose first entry is i, so
so
II'I = I EIk = IDIIE : Dik =
IDIk2/2
= (q -
1)k2/2.
On the other hand, there are k(k
k(k - 1)
1)choices
choices for
for (x,
(x, y),
y),and
andby
by Exercise
Exercise16.5.7
16.5.7
there are m
m=
= q -1
involutions
i
with
cycle
(x,
y),
each
in
i
Gx
y
=
i
D
uE,
- 1 involutions i with cycle
in iG,,, = i D c uE,
so IFI
= k(k --1)(q
Irl =
l)(q --1).1).Thus
Thuskk==2,2,sosoDD==E Eand
and(4)
(4)isisestablished.
established.
(5) Let
Let xx =
=H
H EE X,
X, yy ==xu,
xu,and
andzzEEXX--{x}
{x)==X'.
XI.By
ByExercise
Exercise16.5.7,
16.5.7,
there is vv E IIwith
between uu and
and vv,, Fix(Do)
Fix(Do) =
=
with cycle
cycle (x,
(x, z),
z), so
so by symmetry
symmetry between
z) and Do =
=H
(x, z}
H flnH°
H UisisaaHall
Hall 2'-subgroup
2'-subgroup of H.
H . Therefore
Therefore there is hh E H
with Dh
D~ ==Do
DOby
byPhillip
PhillipHall's
Hall's Theorem
Theorem18.5,
18.5,as
asobserved
observed during
during the proof
of (4). Now by (4)
(4)
(x, yh}
Fix(D)h ==Fix(Dh)
F ~ x ( D=~
=Fix(Do)
)Fix(Do)=={x,
{x,z},
z),
yh) =
= Fix(D)h
so yh
yh =
= z.
as HH =
= DT,
= dt
z. Finally,
Finally, as
DT, we
we have hh =
dt for
for some dd E
E D and t E T,
so zz =
= yt,
andand
therefore
T Tis is
transitive
=yh
yh==ydt
ydt
= yt,
therefore
transitiveon
onX.
XI.But
Butby
by46.4,
46.4,
{x} =
= Fix(t)
Fix(t) for
for each
each tt EE TO,
T', so T is
is regular
regular on IX'I.
IX'I.
As T is regular on X',
XI, the
the action
action of D on
on X'
XI is
is equivalent
equivalent to its action on T by
conjugation by
by 15.1
15.11.
Thusas
asDDisisregular
regularon
onT',
T#,DDisisregular
regularon
onXX-- (x, y},
y),
1. Thus
so G is 3-transitive
3-transitive on X
X and
and only
only the
the identity
identity fixes 3 or more points of X.
(6) Let q ==2e,
2",and
andregard
regardTTasasan
ane-dimensional
e-dimensionalvector
vectorspace
spaceover
over GF(2)
GF(2)
subgroup of GL(T)
GL(T) as
(T ). As D is regular on
and D as a subgroup
as in 12.1.
12.1. Let E ==EndD
EndD(T).
T#,
T', D is
is irreducible,
irreducible, so by Schur's Lemma 12.4,
12.4, E isis aa finite
finite division algebra
most e. Therefore by
by 26.1,
26.1, E is aa finite
over GF(2) of dimension at most
finite field and
ff ==I IE:
E :GF(2)I
GF(2)I <5e.e.But
But D#
D' <(E#,
E', so
so
q-1=IDI:5
IEI=2f-1<2e-1=q-1,
group of
of FF,, and D as the
so E ==F.F Thus
. ThusTTcan
canbeberegarded
regardedas
asthe
the additive
additive group
group F'F#,
andthe
theaction
actionofofDDby
byconjugation
conjugationisisa:a:bbH
H aabb for
multiplicative group
, and
Thus
areare
determined
a EE DDand
andbbEET.T.
ThusD Dand
andT T
determinedupuptotoisomorphism
isomorphism and
and
of D on TT isisdetermined
the representation
representation of
determined up
up to
toquasiequivalence,
quasiequivalence, so the
semidirect product
product H is
is determined
determined up to isomorphism
isomorphism by 10.3. Hence there
D7r
= D* =
_ (O(a,
0, 1):
exists an isomorphism
isomorphism7r:
n : H 4 H* with D
n=
(@(a,0,
O,0,
1): aa E F#),
F'},
and the map
,B: HID -+ H*/D*,
Dt H D*(t7r)
296
296
Appendix
Appendix
is aa permutation isomorphism of
of the
the representations
representationsof
ofHH on
on H
HID
I D and
and H*
H* on
on
H *I D*.Finally
Finally the
the map
map
H*/D*.
y: H*/D* - F,
D*0(1, b, 0, 1) H b
is
representations of
of H*
H* on H*/D*
H*/D* and
and F,
F ,so
socomposing
composing
is an
an equivalence of the representations
these two isomorphisms,
isomorphisms, we get the isomorphism
isomorphism aa of
of (6).
(6).
these
(7) As
As D
D is
is transitive
transitiveononXX-- {{H,
thereisisd dEEDDwith
with(la-')ud
(la-1)ud =
=
H , HHu},
u ) , there
la-1.
la-'.Let
Letvv==ud.
ud.Note
Notethat
thatasasuuinverts
invertsDDand
andDDisisabelian,
abelian,vv inverts
inverts D. For
a EE F#,
==O(a,
F', let
let *(a)
@(a)
$(a,0,O 0,
, 0 ,1).
1).Thus
Thus @(a)-' =
='(a-1)
*(a-') and
and *(a)n-1
@(a)n-' EE D,
D,
so
1'(a)n-1
is
inverted
by
v.
Therefore
so @(a)nW'is inverted by v. Therefore
*(a)-1
(aa-1)v = ((lf'(a))a-1)v = (la-1)*(a)n-1v =
_
(la-1)(*(a)-1)n-1
=
(la-1)v(r(a)n-1)v
(lr(a-1))a-1 = a-1a-1
isomorphismof
of sets,
sets, a*: Sym(X)
Sym(Y)isis an
an
(8) As
As aa:: X
X+
- YYisisananisomorphism
Sym(X)+
- Sym(Y)
isomorphism ofofgroups,
- l g a . Thus
L* isisan
anisoisoisomorphism
groups,where
wherea*:
a*:g gHHaa-1ga.
Thusa*:
a*:LL+
- L*
morphism, where
( H ,v),
v ) ,L*
L* ==(H*,
( H * v*),
, v*),and
and v*
v* ==va*.
va*.However,
However, by
by
morphism,
where LL =
= (H,
construction,
v*
fixes
1
and
has
cycles
(0,
oo),
(a,
a-1),
for
a
E
F
to,
11.
construction, v* fixes 1 and has cycles (0, oo), ( a , a-'), for a E F - (0, 1).
1,O)
hn =
( x h ) afor
for all
That is, v*
v*=
= (b(0,
4(0, 1, 1,
0) EE G*. Further
Furthera*
a*extends
extendsnnasasx axahn
= (xh)a
X and h EE H.ThusitremainstoshowthatG
H . Thusitremains to show that G==(H,
( Hv)
, v and
) andG*
G*==(H*,
( H * v*).
, v*).
x EE Xandh
G is
is 3-transitive
3-transitive on X, G
G is
is primitive
primitive on X by 15.14,
15.14, so H isismaximal
maximal
But as G
G by 5.19,
5.19, and hence
( H ,v).
v ) .Similarly
Similarly G* =
=(H*,
(H*,v*).
v*).
in G
hence indeed
indeed GG =
= (H,
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24
G12. Glauberman,
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G13. Glauberman,
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403-20.
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321-40.
Go1 1.
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Gol
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Solvable signalizer functors on finite
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21 (1972),
(1972), 137-48.
13748.
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Alg.11
11(1969),
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1. Gorenstein,
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243-77.
243-77.
Gor
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Gor 2. Gorenstein,
Gorenstein,D.,
D., The
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1983.
Gor 3. Gorenstein,
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Gor 4.
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224-87.
224-87.
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York.
Wi
List of symbols
Symbol
H <I GG
IGI
1GI
H<G
H
gG
xH
Hx, XH
Page
Page
Symbol
Page
2
2
(S)
(S)
G/H
GIH
3
3
2
2
3
3
3
3, 14
4
4
5
5
5
5
6
7
9
11
14
14
19
25
26
28
30
33
38
42
44
44
44
47
54
55
67
76
78
XG
xG
XY
XY
GxH
GxH
nn(G)
(G)
Ep.
EP.
0,
(G), 0(G)
O,(G),
O"(G)
Q
,(G),
S2n(G)
Qn(G),
Qn(G)
aIR
IR
G#
G#
ff
2
2
3
3
33
3
3
4
4
4
5
5
5
5
55
5
5
6
6
6
6
88
9
9
n,
P', n'
p,
65
71
1
I
76
O(V,
O W , fA1,O(V,
O W , Q)
Q>
HG
13
13
19
19
23
23
26,
45
26,45
G(n)
~ ( n )
27
Zn(G)
Z"(G)
28
11
11
33
38
38, 158
Mx
(g)
Mx(g)
42
Tr,
Tr, det
PSL(V),
PSL(V),PGL(V)
PGL(V)
44
PSLn(F), PGLn(F)
PGLn(F)
PSLn(F),
BT
B=
44
Cyca
CYC,
U(G,
U ( G ,V)
V)
nn, lGln
IGI
Rad(V)
Rad(V)
54
1~x7
n,
Zn
z n
mp(G)
mp(G)
Z(G)
Z
(G)
AutG ((X)
Autc
X)
C,
0,R
C,Q,
LQ
Aut(X)
Aut(X)
Sym(X)
Sym(X)
Inn(G)
Inn(G)
Gy,
G Y ,G(Y),
G ( Y )G1'
,GY
Fix(S)
Fix(S)
Sy1p(G)
Sylp(G)
H char
char G
G
Z]
[[X,
X ,Y,
Y , ZI
Ln(G)
Ln(G)
G,, nr))
SS(A,
(A,G
LwrG,
Lwr G ,Lwr,
Lwr,GG
HomR(U,
H
~ ~ R (V)
VU) ,
Fnxn
Fnxn
Pdim
SL(V),
SL(V), PG(V)
PG(V)
SLn(F),
Ln(F)
SLn(F),Ln(F)
vV**
Mov(A)
Mov( A)
Alt(X)
Alt(X)
H'(G,
H ' ( G ,V)
V)
r(G,
T)
U G , F)
GL(V)
G
L(V)
Out(G)
Out(G)
xG
np, IGIp
np,
lGlp
HaaG
[X,
[ X ,Y]
Yl
G*H
G
*H
EndR(V)
E~~R(V)
Soc(V)
Soc(V)
H=G
HLZG
IG:HI
1G:HI
ker(a)
ker(a)
X8
X
NG(X),
NG(X),CG(X)
CG(X)
II; G;
G,
44
47
List
List of
ofsymbols
symbols
300
Symbol
Page
A(V,
A
( V , f,f), A(V, Q)
Q)
AB
A@
sgn(Q),
sgn(Q),sgn(V)
sgn(V)
SO(V),
so(v),O(V),
O(V),Q(V)
Q(V)
0:(9),SOO(R)
so:(9)
OO(R),
PSpn
(q),
(q)
PSpn (9
1, PGUn
PGUn (9
G+
G+
Mode. ,, Dzn
D2
Mod,.
P ' + ~ D8,D:Qr
, Qs
V®U
V
@U
1 3
Pi+2n
Jr
7r
K
Grp(Y :: W)
W)
W(E)
W
(C)
Comp(G)
Comp(G)
F(G)
F(G)
F*(G)
F*(G)
-W(G)
J"
P ((G,
G , V)
V)
char(G)
char(G)
aG
aG
-q(X)
.@(XI
ek (G)o
&,P(G)O
mz.,(G)
m2,p(G)
78
78
78
86, 87
86,87
88, 89
88,89
89
89
89
89
97
97
107
107
110,111
110,111
118
118
119
119
140
140
148
148
157
157
158
158
159
159
162
162
163
163
180
180
188
188
245
245
248
248
261
26 1
Symbol
JJ(X,f)
( X ,f
NF
NF"
Spn(F),
Sp,(F), Sp(V)
SP(V)
SPn(R),
SUn(R)
Spn(q),SUn(9)
QE(9)
cO(R)
POO(R), POO(R)
po:(9),
pQ:(9)
@(HI
(D(H)
SD2,,,
SD211, Q2"
Q2n
L(V1,
L(V1,....
. .,, Vm;
vm; V
V))
UK
u K ,K®FU
,K@~U
Va
x7rua,, vu
l(h),IR(h)
Kh),
l~(h)
C+
E(G)
E(G)
Oc,(G)
O,(G)
On',R(G)
O,~,E(G>
J(G)
J(G)
aaG(B)
c(B)
cl(G)
cl(G)
(Fk(G)
&;(GI
r ~ , k ( G ) , r:,k(G)
rP,k(G),
rp,k(G)
e(G)
e(G)
Page
78
84
84
88
88
89
89
89
89
89
89
105
105
107
107
117
117
119
119
121
121
143
143
149
149
157
157
158
158
159
159
162
162
175
175
179
179
246
247
2611
26
Index
Index
algebraic integer 184
184
algebraic
Alperin's Fusion
Fusion Theorem
Theorem 200
200
Alperin's
alternating group 55
55
alternating
apartment215
215
apartment
automizer55
automizer
automorphism77
automorphism
Baer-Suzuki Theorem
Theorem 204
204
Baer-Suzuki
bar convention
convention 66
bar
basis
basis
dual 47
47
dual
hyperbolic 81
81
hyperbolic
orthogonal 79
79
orthogonal
orthonomal 79
79
orthonormal
BN-pair 218
218
BN-pair
Bore1subgroup
subgroup 219
219
Borel
Br-Property
Bp-Property 263
263
Brauer-Fowler Theorem
Theorem 244
244
Brauer-Fowler
building 215
215
building
Weyl group 219
219
Weyl
Burnside Normal
Normal p-Complement
p-Complement
Burnside
Theorem 202
202
Theorem
Burnside pagb-Theorem
paqb-~heorem187
187
Burnside
Cartan subgroup
Cartan
subgroup 219
219
category 66
category
coproduct 77
coproduct
product 77
product
Cauchy's Theorem
Theorem 20
20
Cauchy's
central product 32
32
central
with identified
identifiedcenters
centers 33
33
with
centralizer 33
centralizer
chamber 209
209
chamber
character 49
49
character
degree 179
179
degree
generalized 180
180
generalized
induced 189
189
induced
179
irreducible 179
character table 183
183
character
characteristic value 127
127
characteristic
characteristic vector 127
127
characteristic
179
class function 179
classical group 88
88
classical
ClassificationTheorem
Theorem 260
260
Classification
Clifford algebra 95
95
Clifford
Clifford group 96
96
Clifford
Clifford's Theorem 41
41
Clifford's
64
cocycle 64
commutator
commutator 26
26
complex
complex 209
209
chamber
chamber209
209
chamber
chambergraph
graph 209
209
connected
connected 209
209
reflection
reflection 212
212
thick
thick 209
209
thin
thin 209
209
ComponentTheorem
Theorem 263
263
Component
composition
composition factors
factors 24
24
composition series
series 23
23
composition
conjugate
conjugate 33
Coprime Action Theorem 73
73
Coprime
covering
covering 168
168
Coxeter
Coxeter complex
complex 211
2 11
diagram
diagram141
141
group
group 142
142
matrix
matrix 141
141
system142,
142,irreducible
irreducible 146
146
system
critical
criticalsubgroup
subgroup 108
108
cycle
cycle 54
54
cycle
cycle structure
structure 54
54
dihedral
dihedral group
group 141
141
direct product 44
direct
distance
distance 88
dual space 47
47
Dynkin diagram 251
25 1
edge
edge 88
exact
sequence 47
exact sequence
47
short
short 47
47
split
split 47
47
Exchange Condition 143
143
Exchange
extension,
extension, central 166
166
perfect 168
168
universal
universal 166
166
extension problem
problem 10
10
FG-homomorphism
FG-homomorphism 36
36
FG-representation 35
35
FG-representation
field,
field, perfect 92
92
Fitting
Fitting subgroup
subgroup 158
158
generalized
generalized 159
159
fixed
fixed point 14
14
flag
flag 88
folding
folding 211
21 1
opposite 212
212
302
form,bilinear
bilinear 75
75
form,
equivalent 78
78
equivalent
hermitian symmetric
symmetric76
76
hermitian
nondegenerate 76
76
nondegenerate
orthogonal 76
76
orthogonal
quadratic 77
77
quadratic
radical 76
76
radical
sesquilinear75
75
sesquilinear
similar78
78
similar
skew hermitian
hermitian 104
104
skew
skew symmetric
symmetric 76
76
skew
symmetric 76
76
symmetric
symplectic 76
76
symplectic
Witt index
index 78
78
Witt
Frattiniargument
argument 20
20
Frattini
Frattinisubgroup
subgroup 105
105
Frattini
Frobeniuscomplement
complement191
191
Frobenius
Frobenius kernel
kernel 191
191
Frobenius
Frobenius Normal
Normal p-Complement
p-Complement
Frobenius
Theorem 203
203
Theorem
Frobeinus Reciprocity
Reciprocity Theorem 190
190
Frobeinus
Frobenius' Theorem
Theorem 191
191
Frobenius'
fusion, control 199
199
fusion,
G-isomorphism 99
G-isomorphism
G-morphism 99
G-morphism
gallery 209
209
gallery
Gaschiitz' Theorem
Theorem 31
31
Gaschatz'
general linear
linear group 99
general
general unitary
unitary group
group 89
89
general
geometry 88
geometry
connected 88
connected
flag 88
flag
flag transitive 88
flag
morphism 88
morphism
oriflamme99
99
oriflamme
polar 99
99
polar
rank 88
rank
residually connected
connected 88
residually
residue 88
residue
type
type 88
Glauberman Z*-Theorem
Z*-Theorem 262
262
Glauberman
graph 88
graph
connected 88
connected
connected component 88
group
group algebra
algebra 36
36
center 55
center
261
characteristic p-type
p-type 261
25
characteristicallysimple 25
characteristically
class 28
28
class
complement 30
30
complement
covering 168
168
covering
exponent 55
exponent
extension 30
30
extension
free 138
138
free
Frobenius 191
191
Frobenius
Lie
Lie type 250
250
k-connected 246
246
k-connected
Index
Index
k-generated
k-generated p-core
p-core 247
247
metacyclic
metacyclic 203
203
nilpotent
nilpotent 28
28
perfect
perfect 27
27
representation by
by conjugation
conjugation11
11
representation
ring
ring 36
36
section
section 261
261
solvable
solvable 27
27
r-subgroup
71
Hall7r
Hall
-subgroup 71
hyperbolic
hyperbolicpair
pair 80
80
hyperbolic
hyperbolic plane
plane 80
80
induction
induction map
map 189
189
inner
innerautomorphism
automorphism group
group11
11
integer
integer
p-part
p-part 19
19
nr-part
r-part 71
71
involution
involution 55
isometry
isometry 78
78
isomorphism
isomorphism 77
Jordan's
Jordan's Theorem
Theorem 58
58
Jordan-Holder Theorem
Theorem 24,
24,37
Jordan-Holder
37
K-group 261
Levi
Levi factor 257
257
Lie
Lie rank
rank 250
250
linear
linearrepresentation
representation 99
irreducible 121
121
absolutely irreducible
dual
dual 47
47
enveloping
enveloping algebra
algebra 42
42
indecomposable
indecomposable 37
37
induced
induced 188
188
irreducible
irreducible 37
37
irreducible constituents
constituents 37
37
irreducible
principal 179
179
182
regular 182
tensor product 119
119
124
written over a field
field 124
linear
lineartransformation
transformation
determinant
determinant 44
44
m-linear
m-linear 118
118
nilpotent 128
128
semisimple
semisimple 128
128
semisimple
semisimple part 131
131
trace
trace 44
44
unipotent
unipotent 128
128
unipotent part 131
131
Maschke's
Maschke's Theorem 40
40
maximal
maximal subgroup
subgroup 55
modular property of groups 66
modular
module
module
absolutely irreducible
irreducible 121
121
absolutely
37
complement 37
37
composition factor 37
Index
condensed 125
125
cyclic 38
38
extension 37
37
homogeneous 40
homogeneous component
component 40
indecomposable 37
37
irreducible 37
permutation 50
38
semisimple 38
simple 37
37
socle 38
38
split extension 37
37
morphism
morphism 6
normal p-complement
p-complement 202
normal
normal series
series 22
normal
A-invariant
A-invariant 22
factors 23
23
length 22
normalizer 33
Odd Order Theorem
Theorem 260
orbit 13
13
orbital 55
55
diagonal 55
55
paired 59
59
paired 59
self paired
ordinary Chevalley
Chevalley group
group 252
orthogonal
orthogonal group
group 88
orthogonality relations
183
relations 183
automorphism group 11
11
outer automorphism
p-group 55
elementary abelian 5
extraspecial 108
108
modular 107
modular
107
special 108
108
symplectic type
109
type 109
p-rank 5
p-rank
2-local 260
2-local
parabolic subgroup 221
221
partition
G-invariant 18
18
nontrivial 18
18
path 88
path
length 88
permutation
permutation
even 55
even
odd 55
55
53
permutation group 53
permutation representation
representation 9
by conjugation 15
15
15
by right multiplication 15
m-transitive
m-transitive 56
56
primitive 18
primitive
18
rank 55
rank
regular 56
regular
semiregular 56
303
14
transitive 14
transitive constituents
16
constituents 16
Phillip Hall's
71
Hall's Theorem 71
1r-group 5
n-group
presentation 140
presentation
140
43
projective geometry 43
dimension
dimension 44
hyperplane
hyperplane 44
lines 44
44
points 44
points
44
projective general
general linear
linear group
group 44
projective special
special linear
lineargroup
group 44
quaternion group 107
107
Rainy Day Lemma
Lemma 216
2 16
reflection 93
93
center 93
93
representation 9
representation
equivalence 99
equivalence
faithful 9
9
faithful
quasiequivalence
quasiequivalence 99
residue 88
group 250
root group
root system 148
148
closed subset
subset 255
ideal 255
255
ordering 149
149
positive system 149
149
simple system 149
149
Weyl group 148
148
root
long
long 251
251
short
251
short 251
scalar matrix
matrix 44
scalar transformation
transformation 44
Conjecture 160
Schreier Conjecture
160
Schur's
38
Schur's Lemma
Lemma 38
Schur multiplier 168
168
Schur-Zassenhaus Theorem 70
70
Schur-Zassenhaus
semidihedral
group 107
semidihedral group
107
30
semidirect product
product 30
229
signalizer functor 229
complete 229
229
solvable 229
solvable
complete 229
solvably complete
similarity 78
simplex 209
209
Sims Conjecture 176
176
Solvable 2-Signalizer Functor
Solvable
Functor Theorem
Theorem 229
space
space
hyperbolic 81
81
orthogonal
orthogonal 77
77
symplectic 77
unitary 77
unitary
special Clifford group
group 97
special linear
linear group
group 44