TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 180, June 1973
ON LAGRANGIANGROUPS
BY
J. F. HUMPHREYS AND D. L. JOHNSON
ABSTRACT.
We study the class
£ of Lagrangian
groups,
that is, of finite
groups
G possessing
a subgroup
of index
tz for each factor
zz of
G . These
groups and their analogues
were considered
by McLain in |_4j and the object
of
the present
work is to extend
the results
in this article.
We study the classes
(G) = [HIC X H £ £ ! and also the closure
of J? under wreath products.
We also
consider
the two classes
Ï and 1) introduced
in L2] and [4j respectively.
0. Introduction.
every
factor
The interesting
purpose
here
a number
Lagrangian
of \G\
groups,
a subgroup
survey
article
to extend
some
of new results
that
of that
[3] contains
such
have
by £
the class
of Lagrangian
then
two basic
results
uble groups
(proved
by Zappa
[6] in answer
to a question
uble
are charactetised
as
factors
by McLain
in [A]).
respectively,
while
Finally,
the class
in §4 we introduce
|G:E|,
G such
prime-power
index.
in [2], where
corollary
that
but we are unable
The authors
helpful
The class
it is proved
comments,
would
is the class
will
of Ore [5]), and (b) sol-
be discussed
the wreath
two classes
subgroup
if was
introduced
of finite
that
to find a counterexample
like to express
their
groups:
and |G:K|
J
is
n
=n,
have
X was introduced
sj( C i) Ç Ï
to the referee
following
of i_.
integer
(so that
to the conjecture
gratitude
for the example
closure
subgroups
in [A], while
here
(proved
in §§ 1 and 2
H of G and every
meet-irreducible
We prove
of i
ptoduct
K of G with H<K<G
A C S.
of sub-
ö of supersol-
of the elements
about
all of whose
and in particular
of Proposition
results
for every
and by s the operation
(a) si.
the following
that
of groups
groups
a result
there is a subgroup
% is the class
s% = sïi),
of these
in §3 we prove
of groups
dividing
while
Extensions
It is our
in [4] and to prove
analogues.
Denoting
the direct
authors.
of references.
group closure,
groups
for
by many
by McLain
and their
are that
G possessing
studied
list
developed
groups
groups
been
a useful
of the ideas
about
is, finite
index,
i) = A.
for many
the second
2.
Received
by the editors August 19, 1971.
AMS(MOS) subject classifications
(1970).
Key words and phrases.
ible, subgroup
closure.
Supersoluble
group,
Primary 20D30.
wreath
product,
/CG-module,
meet-irreduc-
Copyrighl © 1973, American Mathematical Society
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292
J. F. HUMPHREYS AND D. L. JOHNSON
1. p-Lagrangian
Definition.
dividing
A group
\G\,
Lagrangian
G has
groups
Since
each
the elements
(using
Proof.
a subgroup
are all
considered
in this
Lagrangian
of index
article
if, for every
rz; we shall
tower
a Hall p -subgroup
soluble,
property)
both these
Proposition
sequence
G is called
G £ X. contains
in i,
G have
All groups
denote
will be finite.
positive
integer
the class
n
of
by i_.
of i
the Sylow
contained
that
groups.
that
The necessity
being
is obvious,
p dividing
it is easy
\G\,
to show
ö of all supersoluble
groups
is
proper.
n for each
prime
while
elementary
prime
hand
G to be Lagrangian,
of index
of the following
for each
on the other
the class
inclusions
1. For a group
a subgroup
while
it is necessary
power
n dividing
the sufficiency
result:
if E and
and sufficient
\G\.
is an immediate
K are subgroups
con-
of G of
coprime index, then \G : H O K\ = \G : H\ \G : K\.
This
result
leads
Definition.
each
in a natural
A group
nonnegative
integer
the class
of such
denote
of subgroup
closure;
p-Lagrangian
groups
It follows
£
Proposition
soluble
group
G is said
a with
groups
to be p-Lagrangian
p j \G\,
by i.
.
Let
to be si.
and the p-soluble
and
sit
2. A group
possessing
of strongly
i., sJl
G is strongly
a chain
members
soluble
The necessity
groups.
K and an element
So let
G be a p-soluble
By the p-solubility,
x of G such
\GX
is obvious,
(1) is subgroup
of G.
that
group
, : Gx\ = \G ■ , : G .1 = p for
members
\K :L\,
it follows
of (2) leads
that
to a subgroup
—
—
Since
each
1 < i < m.
of type
the class
property
it
of in-
(1) and let
p -subgroup
K
L of
the chain
m
x : K n Gx\,
1 or p,
chain
for the sufficiency
within
possessing
—
1 < z < m.
r. is
while
closed
L < H . Consider
if r. = \K nGx_
lz-lzllz~lz,r
power
over all primes
if and only if G is a p-
we can find a Hall
0 —
of K, where,
respectively.
= H,
K=KnGx>--->KnGx=L
oí subgroups
and strongly
\H\ is prime to p.
of the condition
(2)
We
for the operation
of si
p-Lagrangian
and
to show that the property
be any subgroup
if, for
p .
of subgroups,
\G ._ . : G .| = p, I < i <m,
Proof.
s stand
Lagrangian
are the intetsections
G = Go>...>Gm
is enough
of index
tespectively.
(1)
such that
p is a prime)
a subgroup
the prefix
the classes
1 that
(where
G contains
we define
from Proposition
p of the classes
way to the
I < i < m, we have
r.
;
isa
Thus,
divisor
deleting
(1) for K as required.
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r. <
of the pr
the repeated
ON LANGRANGIAN GROUPS
Corollary
chain
1. G £ si.
of subgroups
Prool.
of these
This
chains
This
already
if and only if, for each
of type
follows
o = si
(see
2. Lagrangian
Consider
a chain
above
group
of siC)
corollary
that
the existence
groups,
since
we
of cube-free
order
are supersoluble.
of subgroups
G, where
s 77!! where
value
and the fact
of supersoluble
G = G0>--->Gm
least
G has a
§0).
groups
(C)
max {s .,•••,
1'
\G\,
of G.
a characterisation
Corollary
of the p-soluble
p dividing
from the proposition
the solubility
provides
know that
prime
(1) above.
at once
entails
corollary
293
as
rp
E is a Hall
p -subgroup
of G, and write
s(C)
=
l = \G
■ ,1 : G Z1
.1,' 1 —
< i —
< ttz. Defining b s P ÍG) to be the
' Z-
(C) ranges
simply
= H
asserts
over all such
that
chains
of subgroups
if s AG) = 1 for all primes
of G, the
p dividing
\G\,
then rAG)
= 1 fot all p,
where r p (G) is the p-rank of G. It is clear that s p ÍG) —
<
p
r
r ÍG) for all
symmetric
p, however
group
the group
G which
symbols
by a cyclic
on three
is the standard
group
wreath
product
of order three
has
of the
rAG)
= 3,
but s (G) = 1, so that r ÍG) / s (G) in general.
We have
the following
Proposition
a
p ", where
then
3. Let
a
result
G be a p-soluble
•••, a
G possesses
for p-soluble
group
are all nonzero
a partial
(3)
Sylow
groups
tower,
with
G satisfying
s ÍG) = 1.
s (G) = 1 and
\G\ = p.
•••
P
and
p,>p~>--->p.=p>...>p
that
is, a chain
, say;
of subgroups
E = G0<Gx<---<Gi_x<G
where
each
Proof.
member
is normal
We proceed
|G . : G ._ . | = p .',
by induction
that i > 1. By hypothesis,
sAH) = 1
in G,
on
\G\ and,
there is a subgroup
by Proposition
2.
By induction,
1 <;' < ¡i — 1, and
to avoid
triviality,
we assume
H of G with |G:E|
H has
a normal
\E\ = 1.
= p, and
subgroup
G.,
of index
a 1
ip,
, which
follows
is normal
by applying
2. Direct
a > 0 and
For,
in G by Sylow's
the inductive
products.
collection
series
for Z
of subgroups
hypothesis
If G is a p-soluble
ip, m) = 1, then
it is obvious
if E is a Hall p -subgroup
position
theorem
a_x,
that
of G and Z
then the set
of all possible
(since
p < p.).
The result
to G/Gy
group
such
the group
that
G xZ
_x = Za_ ,>•••>
\G\ = p m, where
a_
is p-Lagrangian.
Z« = E is a com-
ÍE x Z n, G x Z„ \ 0 < ß < a - jj
p-power
now
indices
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in G x Z
a_ ..
forms
This
a
294
J. F. HUMPHREYS AND D. L. JOHNSON
result
characterises
soluble
leads to the following
Definition.
For a p-soluble
integer
77 for which
integer
exists,
Lemma
group
groups
G x Z
H of index
the direct
factors
G.
n.
group
G, denote
by rz (G) the least
is p-Lagrangian.
and is less
1. Let
as
of Lagrangian
groups
and
definition.
than
and
a, where
G2 be groups
Then
there
exist
(By the above
p
value
remark,
of the
such
an
|| |G|.)
such
that
subgroups
G. x G 2 possesses
H
a sub-
< G ,, H? < G? such
that
H x H 2 has index n in G. x G,.
Proof.
Let
H1 be the projection
of E on G ■ and
E.
be the intersection
of H
with G¿, i = 1, 2. Then E. « H, i = 1, 2, and Hl/Hx » H2/H2 a H/Hx x Hr
Thus
IE1 x H2\ = IE1! \H2\ = |E: E, x E2| |E,| \H2\ = \H\,
as required.
Proposition
4. Let
• • • < n , be the set
G be a p-soluble
of integers
n such
group
that
and let
{«,,•••,
n \, where
G has a subgroup
of index
n.
p".
<
Then
if P\\G\,
(A)
n (G)=
max
1<Z<7-
(» +1 - n ) - 1,
1
while n AG) = 0 otherwise.
P
Proof.
side
We assume
that
of (4) by 77, we first
group
of G of index
or
subgroups
of G x Z
that
p z,> 1 —
< i —
< r,> so that
of indices
P"','-'
respectively.
ß < zz +.,
Now let
since
are finished,
Therefore
o_
p T
and if not we have
ß - n . < n . + , — n . <n
Denoting
is p-Lagrangian.
such
that
\\\G x Z
\.
p
If ß
\\G x Z
n . < ß < n . + . for some
+ 1 and there
It remains
Since
\, so that
is one of the integers
integer
is a subgtoup
ttz is less
r — 1, such that
(6)
E . x K is a subgroup
to prove
than
that
of G x Z
for any integer
the right-hand
side
m < n,
of index
G x Z
of (A), we can find
m + 1 <n+l
-nr
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z with
K of Z
o
whence
E . be a subare
+ 1apn'+n
.
"',
the right-hand
Let
H ,1 x Z ni , • • • , H r xZ p„' ,E r xE
,p"r,p"T
ß be any integer
77„ = 0 and
triviality.
G x Z
r
(5)
pp
p | \G\ to avoid
show
77. <
(5) we
1 < i < r.
of index
P
p , as required.
is not p-Lagrangian.
an i, where
1 < z' <
295
ON LANGRANGIAN GROUPS
Assume,
for a contradiction,
has a subgroup
of index
that
_,
p
.
G xZ
is p-Lagrangian,
f
6
6
pm
By Lemma
1, there
so that
>
are subgroups
G x Z
pm
H < G,
K<Z
such that \G: H\ \Z
: K\ = p"!' + 1_1. Setting |Z
:K| = ps, we have
p "'
p"*
p
0 < s < m and |G :H\ = p 1+ l ' S. But using (6), n . < n ... - 1 -m<
n ,-1-s
< 7z
, , contradicting
The first
to yield
the definition
and second
the two parts
Proposition
halves
of the
of the
of the following
5. Let
For any p-soluble
is p-Lagrangian].
it would
anything
be said
tetise
those
about
elements
class.
\H\;
then
JEJO x E
Also
two propo-
For example,
E of (G) such
that
(a) E is not p-Lagrangian,
is it possible
can
to charac-
and (b)
a = npÍG)?
result
of this
three
type.
properties
of a p-soluble
group
G with
Then
G x E is
a > 0, are equivalent:
(a) some
direct
power
(b) some
direct
power
(c) G has subgroups
Proof, (a) => (b).
(b) => (c).
Lagrangian
has a subgroup
after
process
yields
For
product
p ~ .
power
a subgroup
G or E has
in this way until,
in (G),
p and
E is a direct
in the second,
(c) => (a).
of G lies
Obvious.
Suppose
lemma that either
of G is p-Lagrangian,
of indices
and so contains
done, while
Let
this
of groups
in the above
G £ ÍG)?
6. The following
the direct
with pa\\
is p-Lagrangian,
that
Proposition
A similar
H a group
G xH
of (G) are given
to know more about
one further
G or E.
immediately
G such
We mention
|| |G|,
and
then
by (G) the class
properties
those
generalise
then a >n ÍG).
G, we denote
be interesting
pa\\ \H\, where
p
group
While basic
sitions,
group
H is p-Lagrangian,
(b) if G xH is p-Lagrangian,
proof
result.
G be a p-soluble
(a) if a > « (G) and
tz..
preceding
of G lying
of index
a subgroup
of index
It follows
p.
from the above
In the first
we let E = EQ = H x x G, and deduce
of index
finitely
p.
Since
E
many steps,
a subgroup
D of a - 2 copies
ß < (a - l)2,
LXr x Ky-q x GXs,
where
on the other
a-2<r
p
case
again
we are
that either
we can continue
of index
p in G.
that,
a > 4,
in G.
We prove
fot
of G is p-Langrangian.
of G of indices
we have
p , pa~
theorem
0<zj<a-l.
, p respectively
then
Lx(r"s)
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and let
ß = qia - l) + r, 0 <r <
Ifr
s = (a - 2) - (r + q) is a subgroup
+ z7<2a-3,
s,
a subgroup
p-Lagrangian.
p@\ \D\ so that 0 < ß < a(a - 2). By Euclid's
a - 1, and since
= G for some
we obtain
of index
a < 3, G is already
E, K, L be subgroups
hand,
p.
in (G).
+ t?<a-2,
of D of index
x Kxig~s)
x EXs
then
pß.
has
If,
the
296
J. F. HUMPHREYS AND D. L. JOHNSON
required
index,
Note
rather
where
that
than
s = r + q - (a - 2).
the results
of this
just p-Lagrangian
3. Wreath products.
Proposition
group
of order
W= H
(a)
Let
that
of all possible
\G\ and
p\
for each
integer
submodules
field
bers.
Since
as required.
of Lagrangian,
result.
of exponent
then
e and
the standard
77 between
indices
0 and
whenever
Thus,
\G\,
that the right
H be a cyclic
wreath
with
it will be enough
B the base
product
Since
those
of LG,
from p
to show
that
subgroup
of G over
e divides
W has
different
a G-invariant
over
of VI.
Lagrangian,
q is a prime
representation
modules
to prove
group
of G and as such
it will be sufficient
B contains
regular
dimensions.
correspondence
and
by the elements
VI/B as G, and is therefore
is a p-power).
of all possible
Thus
group
ip — l);
for G, and thus the irreducible
preserving
a single
of p elements
g-power
|ß|
or, equivalently,
we prove
via conjugation
to kG itself.
(since
in terms
group.
zeG-module
subgroups
p"
e divides
k be the field
B is a right
is isomorphic
section,
G be a Lagrangian
\s G is a Lagrangian
Proof,
Then
p such
D is p-Langrangian
may be stated
groups.
In this
7. Let
Thus,
section
(p - l),
where
C denotes
kG-
k is a splitting
kG are in one-to-one
the corresponding
of order
k contains
degree-
the complex
assertion
for each
numsoluble
group G over the field C.
(b) We proceed
for abelian
groups.
G and let
G/G
and
. Denote
G
be the last
the remaining
Since
where
irreducibles
Now write
<ÍG = X®Z,
Now let
Then
X contains
Z = Z
'
CG has
X has
,.,
a submodule,
viz.
by Itô's
Z„,
where
LG-submodules
(see
n between
X involves
|G|,
*
Y say,J
Y © Z
of dimension
, of dimension
of
from
z , ■■• , z „,
[l])
that
only the
of all possible
0 and
clear
series
of degrees
theorem
series
X.
dimensions
for Z,
and
we can find an ttz
where tz - dim Z m —
< dim Z TTz+r
,,/Z
777+ 1'
a submodule,
being
by inflation
> ZQ = (0) be a composition
for any integer
such that dim Z m —
<77<dimZ
—
so that
> ...
obtained
by Z .,-•■,
¿ = |G/G(/)|.
Z ., so that
of G, the result
term of the derived
LG-modules
we have
Zy + , = (ÍG.
that
nontrivial
is abelian,
Z only the
above,
length
G
by induction.
let
Let
on the derived
X. , • ■• , X a be the irreducible
respectively.
z.\t
by induction
tz — dim Z
7Z, and this
< t,' by' the
m —
m
. It follows
completes
the
proof.
Note.
It would
tion of the class
in (b) above.
be interesting
(l of groups
U properly
SL(2, 5) has distinct
the other
hand,
character
of degree
contains
irreducible
U contains
to have
possessing
the class
characters
no nonabelian
2 for such
a purely
a group
group-theoretical
the property
of soluble
of degrees
simple
is ttivial
group,
proved
groups,
characterisa-
for soluble
groups
as the group
1, 2, 2, 3, 3, 4, 4, 5, 6. On
since
the only complex
and so is not in the regular character.
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297
ON LANGRANGIAN GROUPS
4. The classes
meet-irreducible
groups.
every
have
subgroup
that
for every
each
of I
that
the following
Lemma.
prime-power
that
the
X consists
A of groups all of whose
consists
precisely
exist
and
solely
are pairwise
found
of those
groups G having
X.,
X..
groups
of supersoluble
of meet-irreducible
subgroups
E = lln_.
of those
|G :X.|
(tobe
The following
index
H oí G there
to consist
result
the class
G is an intersection
is a prime-power
J
with the restriction
in §0,
of a group
subgroup
|G:X.|
fine the subclass
proves
As noted
of G, it is obvious
the property:
such
J.
subgroups
Since
subgroups
X and
This
G having
coprime.
•••, X
leads
us to de-
the same
Since
of G
property
1( Ç S, one easily
in [4]).
two conditions
on a group
G are equivalent:
(a) G £ y,
(b) for any subgroup
subgroup
H of G and any integer
K of G such that H < K and
It is unknown
to us whether
ot not
n dividing
\G :H\,
there
is a
\G : K\ =77.
X = li; we give
three
partial
to show that
sX C 3,
results
in
this direction.
Proposition
Proof.
ties
8. sX = si).
Since
A cX
of the operation
counterexample.
Let
of subgroups
minimal
and
We assume
with respect
tower
property,
since
sX
I
I
to this
since
a.
coprime
normal
is closed
a 1
px
...p
ar
|G:X|
of |G:E|
= p ,.
that
But NH,
lies
Now let
of |G:E|
which
that
H = Y, and so
M be a maximal
entails
subgroup
the existence
by induction
contains
H = \]s-,
U . where
each
X containing
a.
all nonzero),
\
E, lies in a subgroup
|X : Y| = p
closure
X
of sX,
. If E / Y, we can
X x, ■■• , X r of G with
where
E.
p , is a prime
S containing
power,
less
than
p.
If M / H, the minimality
T of index
p in G containing
E with
\T : S\ = p
M = E and |G :E| = pp ,.
|G : U .| is a prime
follows,
H < NH and
H = ZH CY-\ X ., a contradiction.
of
Thus,
|G|
by the Sylow
and H < Y < Z, by the minimality
= p p.,
a subgroup
that H = S OX, a contradiction.
and the
in subgroups
|G :E|
of a subgroup
dividing
that
of G and the subgroup
|G : X .| = p . l, 1 < i < T, by the above, whence
We conclude
we have
and hence
H < Y and
E,
be
Now if N < H, the tesult
Thus
NH,
|G :Z| = p
and hence
prime
|G :E|
an N exists
■
By the minimality
Y of X such
E is not the intersection
p. are all distinct
that
proper-
G be a minimal
in G and let
p (such
by [2]).
by standard
and let
p be the largest
images.
of |G :E|
Z of G such that
again.
Let
the
that
indices
of G of order
(where
of G with
find a subgroup
property.
homomorphic
by the minimality
is a subgroup
of G such
G is supersoluble
under
to be false
prime-power
subgroup
we deduce
there
the result
E be a subgroup
of pairwise
N a minimal
if |G:E|=p
it is sufficient
s.
and so
Since
H,
~
so
G £ X,
E = E O X -
n^ = 1 ÍU . O X), so for some i, say i = 1, E = U . O X. Now \G : U x\ is a prime
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298
J- F- HUMPHREYS AND D. L. JOHNSON
power
that
and so is either
X has
p orpj.
p cosets
If \G : U x\ = p x, H has
in XU x Ç G.
This
leads
p cosets
|G : U A = p, which
final
contradiction
Proposition
9.
Let
G £ X be a group
of fourth-power-free
Assume
that
the result
and let
Proof.
As in the
preceding
proof,
not the intersection
Let
p.
of |G:E|,
= q
, where
that
H is contained
power
Since
Now let
E = fl"_.
for some
i.
1,0
p.
i,l
These
of each
member
assume
that
intersect
¡G:X.|,
can be assumed
since
otherwise
f|m=i
Y . with
say,
G = X..
to p, the representation
we can assume
chains
of subgroups,
p2, we must have
|G:E|
integet
tradiction.
p-subgroup
an integer
that
again
f l™_.
that
kp
from this
then
|G|
last
Since
chains
being
< p , proving
I 1" = 2 X ■=
a contradiction.
to p,
Other-
argument
with
and if E < X
Thus,
< p2.
D Y.
E = X x C\ Y
is either
p or
But if the p-part
otherwise
that
X.
proper
|G : Y .| is
of |G :E¡
|G :E|
the index
we can
If each
of |G :E|.
the p-part
smaller
E = fl"=1
By the above
X of G, while
= |G :E|
Hence
we can write
is prime
so that
p divides
r
inclusion
Y . yields
say.
the minimality
subgroup
i'
prime-powers.
\X, C\ Y ,:H\
using
theorem,
that
the representation
of |G :E|,
|G : Y A,
to show
. I is a prime
is impossible.
there
|G :E|
is a
= p , a con-
is proved.
that,
if G is a minimal
is divisible
by p , whete
n such
10. Let
G £ X, p be the largest
of G and
that,
g be an arbitrary
counterexample
p is the largest
for each
x £ P,
prime dividing
element
xs = x"
of G.
modiQiP)).
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|
p or p .
We claim
i,l-
|G:X
X of G with
of G to construct
i,l
of G
to the
prime
|G|.
Proposition
Sylow
H =X,d
it suffices
from E up to G with
Since
coprime
p divides
the proposition
at once
'
a chain
in G.
if N £ H, by the
is either
power.
H is
subgroup
and by Hall's
|X . . : X . .
p, which
is irredundant.
k such
XcH,
dividing
than
is the required
Thus
It follows
assertion
each
that
of G with
Thus
|G :E|
the supersolubility
where
|G:X1|=p=|G:Yj|,
is p, X.
positive
that
we see
a contradiction,
of
X
subgroup
p | |G:E|,
a prime
By the minimality
wise,
representation
that
E < f l" = 2 X. < G, the
prime
and this
q /= p.
in some
the p-part
|G : Y .| pairwise
we have
while
is a p-powet.
irredundant,
the
by induction,
for all
G £ A.
indices
follows
to yield
in the next less
prime-power
E lies in a subgroup
|G:X.|
=G
i,a-j
chains
coprime
then
such
normal
we know
and so
counterexample.
index
N a minimal
by hypothesis,
If not, we use
G be a minimal
of minimal
of p -index
X ., with each
order,
and
|| |G : E|,
N ¿ E,
Thus,
X. = X.n<X.,<«..<X.z
q
as a subgroup
a p-power.
p | |E|.
than
|G|,
implying
p <p ,,
the proposition.
of pairwise
dividing
NH, and hence
the prime
we have
|G :X.|
is false
If N < H, a contradiction
minimality
|G:X|
prime
proves
E be a subgroup
of subgroups
p be the largest
of order
let
in (/,,
to the contradiction
Then
\G\,
there
P be a
exists
of
ON LANGRANGIAN GROUPS
Proof.
Since
that
for each
that
xg ^ <x>.
have
that
A. = qX and
x £ P,
<x>
Writing
x
^ <x>.
P < G, we can assume
< G.
Suppose
this
g = g xg2, with
|G :X.|
$(P)
and
= E.
We claim
and choose
gj
g £ G such
a p'-element
of G, we
Now write
7=1
each
that
to be false,
g2 £ P,
(x)= ñ X. nñ
where
299
is a p-power
and
»
Y.
,= 1
'
|G : Y .| is a prime
power
coprime
to p.
Since
P < G, P < Y. for all 7, and we have
1
(x) = (x) n P = Ó X. n H Y n p = [ f| x. ] n p.
7=1
Now for any
there
i, X . contains
is an a . £ P such
a normal
member
abelian
a Hall
that
subgroup
of X. for all
«
7=1
'
V=1
p -subgroup
of G and so, by Hall's
theorem,
g x £ X ..
We let g A = g ■ and obtain, since
P is
a¿
x ' = x
=x81 ^ <x>,
but xgl is a
of G, that
i and also
lies
in P.
Hence,
x
£ (fl'_.
X.) O P = <x>,
a contradiction.
Thus,
to prove
for each
that
* i yS = y7
if not,
let
x £ P, g £ G, we have
i is independent
with
(xy)g
from the linear
of x.
of x and
y that
the composite
À is induced
that
(p - l))
Im Áp lies
and since
the Hall
Corollary
above
notation,
Proof.
intersection
free,
2. Let
since
Let
G/P
p is the usual
of order
homomorphism.
(which
is cyclic
of Aut(P),
dividing
(p — l).
of
it follows
Since
is proved.
is cyclic
dividing
of order
of minimal
of pairwise
Since
\G\ and P a Sylow
canonical
of Aut ÍP/QÍP))
of minimal
of |G:E|
/ = k, as required.
on the p -elements
of Im À is cyclic
E be a subgroup
X = qX.
i = /, while
= x y , and it now follows
prime dividing
G be an element
of subgroups
from the minimality
and
in the centre
the corollary
at once that
xg =
Aut(P/0(F)),
p is one-to-one
p -subgroup
Im À = G/C AP),
i; it remains
homomorphism
by conjugation
By the proposition,
order
integer
g £ G, and let
is cyclic of order dividing p —1.
G -^Aut(P)^
where
x'y'
i = k and
1. Let G £ X, p be the largest
Consider
and
it follows
, 0 < zs < p - 1, so that
p-subgroup of G. Then G/PC AP)
Proof.
x8 = x1 for some
x, y £ P\E
0 < z, 7 < p - 1. If y e <x>,
= (xy)
independence
Corollary
Let
\HOp(G):H\
that
at least
and
in X\o;
then,
with
the
(p — 1)-
index
coprime
order
in G such
indices
|E0
in G.
,(G):E|
that
Then
E is core-
are coprime,
one of O ÍG), O .(G)
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E is not the
lies
it follows
in E,
and
300
J. F. HUMPHREYS AND D. L. JOHNSON
so we deduce that 0p,ÍG)
from Corollary
= E.
But PCQiP) = P xOpAG),
and the result follows
has
of a group
1.
Addendum.
The first
author
constructed
an example
of order
2 • 36 which is in X but not in V.
REFERENCES
1. B. Huppert,
Endliche
Gruppen.
I, Die Grundlehren
134, Springer-Verlag, Berlin and New York, 1967.
2. D. L. Johnson,
A note on supersoluble
der math.
Wissenschaften,
Band
MR 37 #302.
groups, Cañad. J. Math. 23 (1971), 562—564.
MR 43 #7513.
3. D. J. McCarthy,
groups, Trans.
A survey
of partial
4. D. H. McLain,
The existence
Contributions
to the theory
to Langrange's
theorem
on finite
586-594.
of subgroups
Cambridge Philos. Soc. 53 (1957), 278-285.
5. O. Ore,
converses
New York Acad. Sei. 33 (1971),
of given
order
in finite
groups,
Proc.
MR 19, 13.
of groups
of finite
order,
Duke
Math.
J. 5 (1939),
431-460.
6. G. Zappa, Remark on a recent paper of 0. Ore, Duke Math. J. 6 (1940), 511—512.
MR 2, 1.
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF LIVERPOOL, LIVERPOOL, ENGLAND
(Current
address
of J. F. Humphreys)
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILLINOIS, URBANA, ILLINOIS 61801
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF NOTTINGHAM, NOTTINGHAM, ENGLAND
(Current
address
of D. L. Johnson)
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