PERFECT ISOMETRY GROUPS FOR
BLOCKS OF FINITE GROUPS
A thesis submitted to the University of Manchester
for the degree of Doctor of Philosophy
in the Faculty of Engineering and Physical Sciences
2011
Pornrat Ruengrot
School of Mathematics
Contents
Abstract
6
Declaration
7
Copyright Statement
8
Acknowledgements
10
1 Introduction
11
2 Preliminary
16
2.1
p-modular system . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
2.2
Notations and basic definitions . . . . . . . . . . . . . . . . . . . . . .
18
2.3
Block theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
2.3.1
Definition of blocks . . . . . . . . . . . . . . . . . . . . . . . .
19
2.3.2
Defect groups . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
2.3.3
First main theorem . . . . . . . . . . . . . . . . . . . . . . . .
23
2.3.4
Blocks and normal subgroups . . . . . . . . . . . . . . . . . .
23
2.4
Grothendieck groups . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
2.5
Equivalences of blocks . . . . . . . . . . . . . . . . . . . . . . . . . .
27
2.5.1
Morita equivalence and Picard group . . . . . . . . . . . . . .
27
2.5.2
Derived equivalence and derived Picard group . . . . . . . . .
29
3 Perfect Isometries
3.1
32
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
32
3.2
Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
3.3
Examples of perfect isometries . . . . . . . . . . . . . . . . . . . . . .
40
4 Perfect isometry groups
44
4.1
Structure of PI(B) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
4.2
Standard subgroups of PI(B0 ) for a principal block B0
. . . . . . . .
52
4.3
RK (B ⊗O B ◦ ) as a ring . . . . . . . . . . . . . . . . . . . . . . . . . .
54
4.4
Connection to Picard groups and derived Picard groups . . . . . . . .
57
4.5
Factor blocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
4.6
Isometries extended from a normal subgroup . . . . . . . . . . . . . .
63
4.6.1
Direct Product . . . . . . . . . . . . . . . . . . . . . . . . . .
63
4.6.2
Isometries commuting with induction . . . . . . . . . . . . . .
66
5 p-groups
69
5.1
Abelian p-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
5.2
Extra special p-groups . . . . . . . . . . . . . . . . . . . . . . . . . .
73
5.2.1
Irreducible characters of G . . . . . . . . . . . . . . . . . . . .
73
5.2.2
The proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
5.2.3
Picard group and derived Picard group . . . . . . . . . . . . .
79
6 Blocks with cyclic defect groups
81
6.1
The group D o E and its character table . . . . . . . . . . . . . . . .
83
6.2
Perfect isometry groups . . . . . . . . . . . . . . . . . . . . . . . . . .
86
6.3
Special case when defect group is Cp . . . . . . . . . . . . . . . . . .
93
6.4
Comparision with Picard groups and derived Picard groups . . . . . . 101
7 Blocks with TI defect groups
106
7.1
Some Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
7.2
Suzuki group Sz(q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
7.2.1
Principal block of G . . . . . . . . . . . . . . . . . . . . . . . 109
7.2.2
Principal block of NG (D) . . . . . . . . . . . . . . . . . . . . . 115
7.2.3
Non-existence of perfect embedding when q > 8 . . . . . . . . 118
3
7.2.4
7.3
7.4
PIs (A) ∼
= PIs (B) . . . . . . . . . . . . . . . . . . . . . . . . . 119
McLaughlin group M cL . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.3.1
Principal block of G . . . . . . . . . . . . . . . . . . . . . . . 121
7.3.2
Principal block of NG (D) . . . . . . . . . . . . . . . . . . . . . 122
The group P SU3 (q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
7.4.1
P SU3 (3), p = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 123
7.4.2
P SU3 (5), p = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . 125
8 Sporadic groups
128
8.1
Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
8.2
Mathieu group M11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
8.3
Mathieu group M12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
8.4
Mathieu group M22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
8.5
Mathieu group M23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
8.6
Mathieu group M24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
8.7
Janko group J1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
8.8
Janko group J2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
8.9
Janko group J3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
8.10 Higman-Sims group HS . . . . . . . . . . . . . . . . . . . . . . . . . 137
8.11 McLaughlin group M cL . . . . . . . . . . . . . . . . . . . . . . . . . 139
8.12 Held group He . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
8.13 Rudvalis group Ru . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
9 Other blocks
145
9.1
Blocks with defect group C2 × C2 . . . . . . . . . . . . . . . . . . . . 145
9.2
Blocks with defect group C3 × C3 and cyclic inertial quotients of order 4146
10 Isotypies
148
10.1 Cyclic p-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
10.2 Suzuki group Sz(q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
Bibliography
156
4
A Computer-aided results
162
A.1 Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
Word count 43453
5
The University of Manchester
Pornrat Ruengrot
Doctor of Philosophy
Perfect Isometry Groups for Blocks of Finite Groups
December 21, 2011
Our aim is to investigate perfect isometry groups, which are invariants for blocks
of finite groups. There are two subgoals. First is to study some properties of perfect
isometry groups in general. We found that every perfect isometry has essentially a
unique sign. This allowed us to show that, in many cases, a perfect isometry group
contains a direct factor generated by −id.
The second subgoal is to calculate perfect isometry groups for various blocks.
Notable results include the perfect isometry groups for blocks with defect 1, abelian
p-groups, extra special p-groups, and the principal 2-block of the Suzuki group Sz(q).
In the case of blocks with defect 1, we also showed that every perfect isometry can be
induced by a derived equivalence. With the help of a computer, we also calculated
perfect isometry groups for some blocks of sporadic simple groups.
Apart from perfect isometries, we also investigated self-isotypies in the special
case where CG (x) is a p-group whenever x is a p-element. We applied our result to
calculate isotypies in cyclic p-groups and the principal 2-blocks of the Suzuki group
Sz(q).
6
Declaration
No portion of the work referred to in this thesis has been
submitted in support of an application for another degree
or qualification of this or any other university or other
institute of learning.
7
Copyright Statement
i. The author of this thesis (including any appendices and/or schedules to this
thesis) owns certain copyright or related rights in it (the “Copyright”) and s/he
has given The University of Manchester certain rights to use such Copyright,
including for administrative purposes.
ii. Copies of this thesis, either in full or in extracts and whether in hard or electronic copy, may be made only in accordance with the Copyright, Designs and
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8
ac.uk/library/aboutus/regulations) and in The University’s policy on presentation of Theses
9
Acknowledgements
First and foremost, I would like to thank my supervisor, Dr. Charles Eaton, for
his full support, advices, and patience throughout my PhD studies. Without him, I
would be lost.
Thanks to the Royal Thai Government, for providing the generous scholarship
that has supported me during my 10-year studies in England.
Thanks to my office mates: John, Stavros, Stephen for helping me with some
math questions. Special thanks to Stephen for teaching me how to remote access
Linux servers in the department, so I can leave my programs running remotely.
Thanks also to my family and my friends for their support and the fun activities
we did together all these years.
Special thanks to Jom for her support during the difficult times.
10
Chapter 1
Introduction
Let p be a prime number. Let G be a finite group whose order is divisible by p.
Let O be a complete discrete valuation ring whose residue field has characteristic p.
In order to study the modular representation theory of the group algebra OG, it is
convenient to decompose OG into a direct sum of smaller objects called blocks, and
study the representation theory of each block individually.
Many equivalences between blocks have been defined and studied. For example,
let B be a block of OG and A a block of OH, where H is another finite group, a
Morita equivalence between B and A is a relation of the two blocks via an invertible
(B, A)-bimodule (if it exists). A derived equivalence (as triangulated categories) is a
relation via an invertible complex of perfect (B, A)-bimodules, called a tilting complex,
(if it exists). Another concept of equivalence of blocks, which is the main focus of this
thesis, was defined by Broué [8] in 1989, relating two blocks via a perfect isometry at
the level of characters. These equivalences can be related by the following chain of
implications:
Morita equivalence =⇒ Derived equivalence =⇒ Perfect isometry.
When two blocks are equivalent, many invariants associated to blocks are also
carried over. For example, the set of invertible (B, B)-bimodules are shown to form a
group Pic(B) called the Picard group. If two blocks A and B are Morita equivalent,
then Pic(A) ∼
= Pic(B). Similarly, the Derived Picard groups DPic(B) of invertible
11
CHAPTER 1. INTRODUCTION
12
tilting complexes of projective (B, B)-bimodules are invariant under derived equivalences. Since this thesis focuses on perfect isometries, we shall study invariants under
perfect isometries. Broué showed in Theorem 1.5 in [8] that if A and B are two perfectly isometric blocks, then, amongst other things, k(A) = k(B) and l(A) = l(B).
Here k(B) denotes the number of irreducible characters in B, and l(B) the number
of irreducible Brauer characters in B.
Similar to Picard groups and Derived Picard groups, we wish to study structured
invariants under perfect isometries other than the numerical invariants k(B), l(B).
Therefore, we will introduce a perfect isometry group, denoted by PI(B), of all perfect
isometries in the block B. We will prove some structural properties of PI(B) and
calculate and compare it for many blocks. It is not difficult to see that perfectly
isometric blocks have isomorphic perfect isometry groups. The converse, however,
does not hold in general as we found the following counter-example: let B be the
principal 3-block of OG where G is the Ree group G = 2 G2 (3) and let A be the
principal 3-block of ONG (P ) where P is a Sylow 3-subgroup of G, then PI(A) ∼
=
PI(B) ∼
= D12 × C2 but there is no perfect isometry between A and B. In this case,
P ∼
and NG (P ) = P o C2 .
= 31+2
−
Let A be a block of ONG (P ) with a defect group P . Let B be the Brauer
correspondent block of OG with defect group P . Some information about B can
often be derived from that about A. We state two conjectures relating the two
blocks.
Conjecture 1.0.1 (Alperin-McKay). Let A be the unique block of ONG (P ) with
Brauer correspondent block B of OG. Then the number of irreducible characters of
height zero in A and B are equal.
Conjecture 1.0.2 (Broué). Let A be the unique block of ONG (P ) with Brauer correspondent block B of OG. Suppose P is abelian. Then the two blocks A, B are derived
equivalent (as triangulated categories).
Since a derived equivalence implies the existence of a perfect isometry, a weaker
conjecture (which is also still unproved) can also be stated:
CHAPTER 1. INTRODUCTION
13
Conjecture 1.0.3 (Broué). Let A be the unique block of ONG (P ) with Brauer correspondent block B of OG. Suppose P is abelian. Then the two blocks A, B are perfectly
isometric.
Conjecture 1.0.3 has been verified in many cases including when P is cyclic [30,
43, 44], when B is a principal 2-block by Fong and Harris [18], when B is the principal
block of GL2 (pn ) with p 6= 2 [8], when B is a block of a symmetric group by Rouquier
[42], when B is the principal blocks of sporadic simple groups by Rouquier [42] and
when the inertial quotient E of B is small by Puig and Usami [36], [37], [50], [51],
[52].
In order to find the perfect isometry group of a block B, it is sometimes easier
to work with a block A that is perfectly isometric to B but has a simpler structure.
Linckelmann [30] showed that if B is a block of OG whose defect group P is cyclic
and if E is the inertial quotient of B, then B and O(P o E) are derived equivalent,
and consequently perfectly isometric. Our task of finding the perfect isometry group
of B then reduces to finding the perfect isometry group of P o E whose structure is
well understood.
Apart from proving existence of a perfect isometry, showing the non-existence of a
perfect isometry is also interesting. Cliff [10] showed that the centers of the principal
2-block B of the Suzuki group and the principal 2-block A of its Sylow 2-subgroup P
are isomorphic as algebras over k but not isomorphic as algebras over O, where O is
a discrete valuation ring of characteristic 0 whose residue field is k. As we shall see
later that a perfect isometry induces an algebra isomorphism Z(OA) −→ Z(OB),
this implies that there is no perfect isometry between A and B. At the same time,
Robinson [40] showed that, under certain conditions (when NG (P )/Op0 (NG (P )) is a
Frobenius group), the existence of a perfect isometry requires that there are at least
l(B) irreducible characters of B which take constant (non-zero) values on p-singular
elements. This condition immediately rules out the case of the Suzuki group above.
We will show in this thesis that the two blocks A and B have visibly distinct perfect
isometry groups, and therefore, there cannot be a perfect isometry between them.
CHAPTER 1. INTRODUCTION
14
The knowledge of invariants under perfect isometries, especially perfect isometry
groups, may therefore be a useful tool when showing the non-existence of a perfect
isometry.
This thesis can be summarized as follows.
In Chapter 2 we introduce the basic definitions and standard results from modular
representation theory, character theory and block theory as far as required in this
thesis. Some notations introduced in this chapter will be fixed and used throughout
the thesis.
In Chapter 3 we give a definition of a perfect isometry, as defined by Broué [8],
between blocks A, B of finite groups. Some properties and numerical invariants under
perfect isometries will be given, most of which are results due to Broué [8]. We will
also show that every perfect isometry has essentially a unique sign. Examples of
perfect isometries that occur naturally will be given.
In Chapter 4 we restrict our attention to the case where A = B and define the
perfect isometry group PI(B). The structure of PI(B) will be studied. In particular
we will show that, in many cases we have PI(B) = H × h-idi where h-idi is the
subgroup generated by −id and H is isomorphic to a subgroup of the symmetric
group Sk(B) . In the case where B is the principal block, we will show that PI(B)
contains subgroups that can be related to the structure of Aut(G) and Hom(G, O× ).
A relationship between perfect isometry groups and Picard groups and derived Picard
groups will be explained here.
In Chapter 5-6 we calculate the perfect isometry groups for abelian p-groups,
extra special p-groups and blocks with cyclic defect groups. In the case of blocks with
cyclic defect groups, we obtain a complete result for perfect isometry groups when
the defect is 1. This allows us to compare them with the corresponding Picard groups
and derived Picard groups. In particular, we show that the canonical homomorphism
DPic(B) −→ PI(B) is surjective if B is a block with defect 1. When B is a block
with cyclic defect group and the defect is greater than 1, we obtain a partial result
for the perfect isometry group. If G is an extra special group p1+2 , we show that the
CHAPTER 1. INTRODUCTION
15
homomorphism DPic(OG) −→ PI(OG) is not surjective.
In Chapter 7 we study some blocks with TI defect groups. One of the properties
of blocks with TI defect groups is that we have k(B) = k(A) where B is a block of
OG with TI defect group D, and A is a block of ONG (D) such that AG = B. This
guarantees that there is at least an isometry between B and A. We will calculate
both PI(B), PI(A) and compare them. When B is the principal 2-block of Sz(q), we
will show that PI(B) and PI(A) are visibly distinct. This gives yet another proof
that there is no perfect isometry between B and A in this case.
In Chapter 9 we calculate perfect isometry groups for some blocks of sporadic
groups. And in chapter 10 we calculate perfect isometry groups for some blocks
that have been classified or shown to be perfectly isometric to blocks of groups with
simpler structures.
Finally, in Chapter 10 we define isotypies as in [8] and calculate the subgroup of
PI(B) consisting of isotypies I for cyclic p-groups and Suzuki group Sz(q).
Chapter 2
Preliminary
2.1
p-modular system
Let p be a prime number. Let O be a complete discrete valuation ring of characteristic
0. Let p be the unique maximal ideal of O. Let k = O/p be the residue field of
characteristic p and ∗ : O −→ k be the natural projection map. Let K be a field of
fraction of O. The triple (K, O, k) is called a p-modular system. Let G be a finite
group. We say that a p-modular system (K, O, k) is sufficiently large for G if K, O
and k contain a primitive m-th root of unity where m is the exponent of G.
One example of a p-modular system for G to have in mind is as follows. Let
ζ = e2πi/|G| be a primitive |G|-th root of unity. Let Qp be the field of p-adic numbers.
It is the completion of Q with respect to the usual p-adic norm k · kp . Let K := Qp (ζ)
be a finite extension of Qp by adjoining the element ζ. The norm k·kp can be extended
uniquely to a norm on K, still denoted k · kp , by
kxkp = (kNormK/Qp (x)kp )1/[K:Qp ] ,
for x ∈ K
(2.1)
where
NormK/Qp (x) = (−1)[K:Qp ] F (0)[K:Qp ]/d
where F is the minimal polynomial of x over Qp and d = deg(F ). Similarly, the usual
16
CHAPTER 2. PRELIMINARY
17
p-adic valuation vp on Qp can be extended to K via
vp (x) =
1
vp (NormK/Qp (x)),
[K : Qp ]
(2.2)
for x ∈ K × and define vp (0) = +∞. The (extended) p-adic norm and the (extended)
p-adic valuation are related by
kxkp = p−vp (x) ,
for x ∈ K. The reader is referred to [48] or [20, Section 5] for a more detailed
discussion on extensions of p-adic numbers.
With the above setup, the valuation ring is then
O = {x ∈ K : kxkp ≤ 1} = {x ∈ K : vp (x) ≥ 0}.
(2.3)
The maximal ideal is
p = {x ∈ K : kxkp < 1} = {x ∈ K : vp (x) > 0},
and the invertible elements in O is
O× = {x ∈ K : kxkp = 1} = {x ∈ K : vp (x) = 0}.
An important consequence due to the uniqueness of norm extension [20, Corollary
5.3.2] is the following. Suppose we have field extensions Qp ⊂ K ⊂ L and | · |K , | · |L
are the norms on K, L extending the norm k · kp on Qp . If x ∈ K then we must have
|x|K = |x|L . In other words, the extended p-adic norm of x does not depend on the
field where x lives in. The following lemma will be useful when calculating perfect
isometry groups (to be defined later).
Lemma 2.1.1. Let Qp ⊂ K ⊂ L be field extensions of Qp and let OK , OL be the
valuation rings of K and L with respect to the extended norm k · kp . Then
OL ∩ K = OK .
Proof. Let x ∈ OK . Then x ∈ K ⊂ L and kxkp ≤ 1. Since kxkp does not depend on
the context, we still have kxkp ≤ 1 in L and therefore x ∈ OL . On the other hand, if
x ∈ OL ∩ K then x ∈ K and kxkp ≤ 1 in L (and so in K). Hence x ∈ OK .
CHAPTER 2. PRELIMINARY
18
Assumption 2.1.2. Throughout this thesis we will assume the following. When
working with a finite group G, we will work over a p-modular system (K, O, k) where
we assume that p divides |G|, the order of the group G. We will also assume that the
p-modular system is sufficiently large for G. If we are working with two finite groups
G and H at the same time then we will also assume that (K, O, k) is sufficiently large
for both groups.
Remark 2.1.3. The symbols K, O, p, p will be reserved for their respective meanings
as defined above throughout the thesis. The symbol k may clash with some wellknown notations (for example k(B)). In some proofs, we might use k for a dummy
variable (for example, as summation indexes). Since in this thesis we rarely work
over k, this should be clear from the context and no confusion should arise.
2.2
Notations and basic definitions
Let G be a finite group and p a prime number. Denote by
• |G| the order of the group,
• cl(G) the set of conjugacy classes of G,
• cclG (x) the conjugacy class of an element x ∈ G,
• CG (x) the centralizer in G of an element x ∈ G,
• Z(G) the center of G,
• Aut(G) the automorphism group of G.
An element g ∈ G is called a p-singular element if p divides the order of g. If p
does not divide the order of g, it is called a p-regular element. If the order of g is a
power of p, then g is called a p-element. Denote by Gp0 the set of p-regular elements
of G. If g ∈ G, we can write g = gp gp0 where gp is a p-singular element and gp0 is a
p-regular element.
CHAPTER 2. PRELIMINARY
19
Let n be a positive integer. Write n = pa m where p - m. Then pa is called the
p-part of n, denoted by np = pa , and m is the p0 -part of n, denoted by np0 = m.
All modules are assumed to be left modules by default.
An OG-lattice is an OG-module having a finite O-basis. Let V be a finitely
generated KG-module. A full OG-lattice M in V is an OG-lattice M contained in
V , such that V = KM ∼
= K ⊗O M . It follows from [12, (16.15)] that every finitely
generated KG-module contains full OG-lattices.
Let M be an OG-lattice. The K-character afforded by M is the character of the
KG-module K ⊗O M . Denote by Irr(G) the set of irreducible K-characters of G.
The trivial character will be denoted by 1. A generalized character of G is a Z-linear
combination of irreducible characters in Irr(G). The set of generalized characters of
G is denoted by ch(G). Denote by CF(G; K) the space of K-valued class functions
on G. Then Irr(G) is a K-basis of CF(G; K). Denote by CFp0 (G; K) the subspace of
CF(G; K) consisting of class functions vanishing on p-singular elements. The set of
irreducible Brauer characters of G is denoted by IBr(G).
2.3
Block theory
Since in most of the thesis we will work on blocks of a finite group G. We will give
some background materials on blocks theory in this section. The main references for
this section are [34], [12], [13].
2.3.1
Definition of blocks
Let G be a finite group. We can decompose the group algebra OG as
OG = B1 ⊕ . . . ⊕ Bn
where Bi are primitive two-sided ideals. This corresponds to a decomposition of
1 ∈ Z(OG) into primitive idempotents of Z(OG): 1 = e1 + . . . + en where ei ∈ Bi
and OGei = Bi . Each Bi is called a block of OG and ei a block idempotent.
CHAPTER 2. PRELIMINARY
20
Similarly, we can define blocks for KG and kG. By our settings of the p-modular
system, there is a one-to-one correspondence between blocks of OG and blocks of
kG. For a block B of OG we set KB := K ⊗O B and kB := k ⊗O B. In this thesis,
a block will always mean a block over O. The set of all blocks of OG is denoted by
Bl(G).
If M is an indecomposable OG-module, then M = Bi M for some unique block
Bi , and we say that M belongs to Bi . If V is an irreducible KG-module affording a
character χ ∈ Irr(G), then there is an indecomposable full OG-lattice M in V such
that V = KM . We say that χ belongs to a block to which M belongs. This is
independent of the choice of M .
Let S be a simple kG-module affording a Brauer character ϕ. If S belongs to a
block corresponding to a block B of OG, then we also say that ϕ belongs to B.
For any block B ∈ Bl(G) we denote by Irr(B) the set of irreducible characters
χ ∈ Irr(G) belonging to B, and by IBr(B) the set of irreducible Brauer characters
ϕ ∈ IBr(G) belonging to B. The unique block containing 1 is called the principal
block, and will be denoted by B0 . As in standard notations, we use k(B) for | Irr(B)|
and l(B) for | IBr(B)|.
There is an equivalent way of deciding when a character χ ∈ Irr(G) belongs to
which block: we say that χ belongs to Bi if χ(ei ) = χ(1) and χ(ej ) = 0 for all j 6= i.
Another way to determine whether two irreducible characters are in the same
block is via central characters. If χ ∈ Irr(G), then χ uniquely determines an algebra
homomorphism ωχ : Z(KG) −→ K defined by
ˆ = |C|χ(xC ) ,
ωχ (C)
χ(1)
P
where C ∈ cl(G), xC ∈ C and Cˆ = x∈C x.
ˆ is in O, so we can define a homomorphism
It is well known that the value ωχ (C)
λχ : Z(kG) −→ k by setting
ˆ = ωχ (C)
ˆ ∗.
λχ (C)
Definition 2.3.1. [34, Definition 3.1] Let B ∈ Bl(G). Then Irr(B) is an equivalence
CHAPTER 2. PRELIMINARY
21
class in Irr(G) under the relation χ ∼ ϕ if λχ = λϕ for χ, ϕ ∈ Irr(G). That is, if
|C|χ(xC )
χ(1)
∗
=
|C|ϕ(xC )
ϕ(1)
∗
for all C ∈ cl(G). We write λB = λχ for some χ ∈ Irr(B).
Recall that the primitive idempotents of Z(KG) are {eχ | χ ∈ Irr(G)} where
eχ =
χ(1) X
χ(g −1 )g.
|G| g∈G
If B ∈ Bl(G) has a block idempotent eB ∈ Z(OG), then
X
eB =
eχ .
χ∈Irr(B)
If χ, ϕ ∈ Irr(G), then we have ωχ (eϕ ) = δχϕ and so ωχ (eB ) = 1 if and only if
χ ∈ Irr(B), and ωχ (eB ) = 0 if χ 6∈ Irr(B).
P
Let θ = χ∈Irr(G) hθ, χiχ be a class function where h·, ·i is the usual inner product
of class functions. For B ∈ Bl(G), we let
θB =
X
hθ, χiχ.
χ∈Irr(B)
We say that θB is the B-part of θ.
Lemma 2.3.2. [34, Lemma 3.31] Let θ ∈ CF(G; K) be a class function on G. Let
B ∈ Bl(G) with a block idempotent e. Then for all x ∈ KG,
θ(xe) = θB (x).
For a block idempotent e ∈ Z(OG), let
Irr(G, e) = {χ ∈ Irr(G) : χ(g) = χ(ge) ∀g ∈ G}.
Let χ ∈ Irr(G), then by the previous lemma χ ∈ Irr(G, e) if and only if χ(g) =
χB (g) ∀g ∈ G, that is, Irr(G, e) = Irr(B).
CHAPTER 2. PRELIMINARY
2.3.2
22
Defect groups
Let G be a finite group. Let B ∈ Bl(G) with block idempotent eB ∈ Z(OG). Write
(eB )∗ =
X
ˆ C.
ˆ
aB (C)
C∈cl(G)
Then
1 = λB ((eB )∗ ) =
X
ˆ B (C).
ˆ
aB (C)λ
C∈cl(G)
ˆ
This means that there is at least one conjugacy class C ∈ cl(G) such that both aB (C)
ˆ are not 0. We call such a class a defect class for B. A defect group of B
and λB (C)
is then defined to be a Sylow p-subgroup of CG (x) where x is in a defect class of B.
This is well-defined and all defect groups of B are G-conjugate.
If D is a defect group of B, define the defect of B to be the non-negative integer
d(B) such that |D| = pd(B) . In fact, d(B) can also be determined by the irreducible
characters contained in B as follows.
For χ ∈ Irr(G), define the defect of χ to be the integer d(χ) such that
pd(χ) χ(1)p = |G|p .
Then d(B) can be defined as
d(B) = max{d(χ) : χ ∈ Irr(B)}.
The height of χ ∈ Irr(B) is defined to be the integer h(χ) such that
h(χ) = d(B) − d(χ).
An irreducible character χ ∈ Irr(G) is said to have p-defect zero if χ(1)p = |G|p .
A defect group of the principal block of G is a Sylow p-subgroup of G. So the
principal block has the maximal defect whereas a block of defect zero has trivial
defect group and contains only one irreducible character (cf. [34, Theorem 3.18]).
We denote by Bl(G|D) for the set of blocks of OG with defect group D and denote
the set of defect groups of a block B by δ(B).
CHAPTER 2. PRELIMINARY
2.3.3
23
First main theorem
Suppose that D is a p-subgroup of G and let CG (D) ≤ H ≤ NG (D). We can define
a homomorphism BrD : Z(kG) −→ Z(kH) by
X
ˆ =
BrD (C)
x,
x∈C∩CG (D)
ˆ = 0.
for C ∈ cl(G) and extend linearly. If C ∩ CG (D) = ∅ then we set BrD (C)
If H is a subgroup of G and A ∈ Bl(H), we can extend λA : Z(kH) −→ k to a
linear map λG
A : Z(kG) −→ k by
!
ˆ
λG
A (C) = λA
X
x ,
x∈C∩H
G
ˆ
for C ∈ cl(G) and extend linearly. If C ∩ H = ∅ then we set λG
A (C) = 0. If λA is an
algebra homomorphism then there is a unique block AG ∈ Bl(G) such that
λG
A = λAG .
In this case we say that AG is defined and call AG the induced block of A.
Theorem 2.3.3. Suppose that D is a p-subgroup of G and let H be a subgroup of G
satisfying DCG (D) ≤ H ≤ NG (D). If A ∈ Bl(H), then AG is defined. Moreover, if
B ∈ Bl(G), then B = AG for some block A ∈ Bl(H) if and only if D is contained in
some defect group of B.
Proof. See [34, Theorem 4.14].
Theorem 2.3.4 (Brauer’s First Main Theorem). The map A 7→ AG defines a bijection from Bl(NG (D)|D) to Bl(G|D). Furthermore, if eA , eAG are block idempotents
of A, AG respectively, then BrD (eAG ) = eA .
Proof. See [34, Theorem 4.17].
2.3.4
Blocks and normal subgroups
Let G be a finite group. Let σ ∈ Aut(G) and let H ≤ G be a subgroup such that
−1
H σ = H. Then σ acts on CF(H; K) via ϕσ (h) = ϕ(hσ ), where ϕ ∈ CF(H; K) and
CHAPTER 2. PRELIMINARY
24
h ∈ H. The next lemma shows that the action by σ preserves the block structure of
H.
Lemma 2.3.5. Let σ ∈ Aut(G) and H ≤ G be such that H σ = H. If χ ∈ Irr(H)
then χσ ∈ Irr(H). Furthermore, if χ, ϕ ∈ Irr(H) are in the same block then χσ , ϕσ
are also in the same block.
Proof. First, if χ ∈ Irr(H), then
hχσ , χσ iH =
1 X
1 X
−1
χ(hσ )χ(hσ−1 ) =
χ(h)χ(h) = hχ, χiH = 1.
|H| h∈H
|H| h∈H
So χσ ∈ Irr(H). Suppose χ, ϕ ∈ Irr(H) are in the same block. Then we have
χ(hC ) ϕ(hC )
−
≡ 0 mod p,
|C|
χ(1)
ϕ(1)
for all C ∈ cl(H), where hC ∈ C. If C ∈ cl(H), denote by C σ ∈ cl(H) the class of (hC )σ
−1
for some hC ∈ C. Since |C σ | = |C|, we have
|C|
χσ (hC ) ϕσ (hC )
− σ
χσ (1)
ϕ (1)
−1
= |C
σ −1
|
−1
χ((hC )σ ) ϕ((hC )σ )
−
χ(1)
ϕ(1)
!
≡0
mod p,
for all C ∈ cl(G). Hence χσ , ϕσ are in the same block.
Let A be a block of OH. We denote by Aσ another block of H with Irr(Aσ ) =
{χσ : χ ∈ Irr(A)}. If D is a defect group of A then Dσ is a defect group of Aσ ([34,
Problem 4.4]).
Suppose now that N C G. Then G acts on Irr(N ) by conjugation. We say that
A1 , A2 ∈ Bl(N ) are G-conjugate if A2 = Ag1 for some g ∈ G. If A ∈ Bl(N ) we denote
the stabilizer of A in G by IG (A) = {g ∈ G : Ag = A}.
Definition 2.3.6. We say that a block B ∈ Bl(G) covers a block A ∈ Bl(N ) if any
of the following equivalent conditions hold:
(a) If χ ∈ Irr(B), then every irreducible constituent of χN lies in a G-conjugate of
A.
(b) There is a χ ∈ Irr(B) such that χN has an irreducible constituent in A.
CHAPTER 2. PRELIMINARY
25
A proof of the above equivalence can be found in [34, Theorem 9.2].
Theorem 2.3.7 (Extended First Main Theorem). If B ∈ Bl(G) has defect group
D, then there exists a unique NG (D)-orbit of blocks AD ∈ Bl(DCG (D)) such that
(AD )G = B. All these blocks have defect group D and ((AD )NG (D) )G = B.
Proof. See [34, Theorem 9.7].
The block AD ∈ Bl(DCG (D)) from the last theorem is called a root of B.
Theorem 2.3.8 (Brauer’s Third Main Theorem). Let H ≤ G and let A be a block of
OH. Suppose that AG is defined. Then, A is the principal block of OH if and only
if AG is the principal block of OG.
Proof. See [34, Theorem 6.7].
Definition 2.3.9. Let B ∈ Bl(G) be a block with defect group D and AD ∈
Bl(DCG (D)) be a root of B. We define the inertial quotient of B to be the quotient group E = ING (D) (AD )/DCG (D). The order of E is called the inertial index of
B.
Since all the roots of B are conjugate in NG (D) by the Extended First Main
Theorem, the inertial index is well defined. We will use inertial quotient and inertial
index extensively when we study blocks with cyclic defect groups.
Theorem 2.3.10. Let B ∈ Bl(G). Then the inertial index of B is not divisible by p.
Proof. See [13, (61.15)].
2.4
Grothendieck groups
In what follows, all modules will be finitely generated left modules, unless explicitly
stated otherwise.
Definition 2.4.1. Let A be a ring. The Grothendieck group R(A) is the abelian
group generated by the symbols [M ] corresponding to isomorphism classes (M ) of
CHAPTER 2. PRELIMINARY
26
A-modules M with relations
[M ] = [M 0 ] + [M 00 ]
for each short exact sequence 0 → M 0 → M → M 00 → 0 of A-modules.
Let G be a finite group. The structures of R(KG) and R(kG) can be seen by the
following proposition.
Proposition 2.4.2. Let A = KG or kG. Let {S1 , . . . , Sn } be the complete set of
simple A-modules. Then
R(A) =
n
M
Z[Si ].
i=1
Proof. See [12, (16.6)].
If B ∈ Bl(G) we write RK (B) for R(KB), where KB = K ⊗O B.
Now let M, N be two KG-modules. We can define another KG-module by M ⊗K
N . It turns out that we can also define multiplication in R(KG) by [M ][N ] :=
[M ⊗K N ]. If M and N afford characters χ and ϕ respectively, then M ⊗K N affords
the character χϕ defined by (χϕ)(g) = χ(g)ϕ(g) for g ∈ G. The next proposition
shows that we can identify R(KG) with the ring of generalized characters ch(G).
Proposition 2.4.3. Let G be a finite group. There is a ring isomorphism R(KG) ∼
=
ch(KG) defined by [M ] 7→ µ where µ is the character of G afforded by M .
Proof. See [12, (16.10)].
Suppose that V is a KG-module. There is an OG-lattice M such that V = KM ∼
=
K ⊗O M . We then have a kG-module k ⊗O M ∼
= M/pM .
Proposition 2.4.4. The map sending [V ] ∈ R(KG) to [k ⊗O M ] ∈ R(kG) where
V = KM gives a homomorphism of abelian groups d : R(KG) −→ R(kG).
Proof. See [12, (16.17)].
The above homomorphism is called the decomposition map associated with the
p-modular system (K, O, k).
CHAPTER 2. PRELIMINARY
2.5
27
Equivalences of blocks
Let G, H be two finite groups. Let B ∈ Bl(G) and A ∈ Bl(H). In this section we
will discuss briefly some types of equivalences of blocks that can be related to perfect
isometries later on. Good references for this section can be found, for example, in
[9], [23], [28], [5].
To each of these equivalences is associated an invariant group which can be obtained when we let A = B. In some cases where these groups can be determined, we
shall compare them to our invariant group of perfect isometries, which will be defined
and studied in more details later.
Unless stated otherwise, all modules in this section are assumed to be finitely
generated left modules. Recall that an (A, B)-bimodule is an abelian group M which
is simultaneously a left A-module and a right B-module such that (am)b = a(mb)
and rm = mr for all a ∈ A, b ∈ B, m ∈ M and r ∈ O. Denote by A◦ the opposite
algebra of A. An A◦ -module can be regarded as a right A-module and a (B ⊗O A◦ )module can be regarded as a (B, A)-bimodule. Denote by Mod(B) the category of
finitely generated B-modules. So Mod(B ⊗O A◦ ) is the category of finitely generated
(B, A)-bimodules.
2.5.1
Morita equivalence and Picard group
We begin with the Morita equivalence. The following well-known theorem is extracted
from [9].
Theorem 2.5.1 (Morita Theorem). The following are equivalent
1. The categories Mod(A) and Mod(B) are equivalent.
2. There exist
• a (B, A)-bimodule M , which is projective as a left B-module and as a right
A-module,
• a (A, B)-bimodule N , which is projective as a left A-module and as a right
B-module,
CHAPTER 2. PRELIMINARY
28
• an (B, A)-compatible O-duality between M and N such that M ⊗A N ∼
=B
in Mod(B ⊗O B ◦ ) and N ⊗B M ∼
= A in Mod(A ⊗O A◦ ).
Definition 2.5.2. We say that A and B are Morita equivalent if Mod(A) and
Mod(B) are equivalent.
If A and B are Morita equivalent then, with the notation of Theorem 2.5.1, the
functors
M ⊗A − : Mod(A) −→ Mod(B)
N ⊗B − : Mod(B) −→ Mod(A)
provide inverse equivalences of module categories. Moreover, they send simple modules to simple modules ([23, Remark 1.5]).
By tensoring everything with K, Morita equivalence of A and B also implies
Morita equivalence of K ⊗O A and K ⊗O B. Similarly, tensoring with k gives Morita
equivalence of k ⊗O A and k ⊗O B. We summarize some of the well-known invariants
preserved by a Morita equivalence [9].
Proposition 2.5.3. Suppose that A and B are Morita equivalent. Then
1. Z(A) ∼
= Z(B).
2. k(A) = k(B).
3. l(A) = l(B).
4. A and B have the same decomposition matrix.
5. A and B have the same Cartan matrix.
6. A and B have the same defect.
Definition 2.5.4. A (B, A)-bimodule M is called invertible if it gives a Morita
equivalence between A and B.
Now suppose A = B and consider all invertible (B, B)-bimodules.
CHAPTER 2. PRELIMINARY
29
Definition 2.5.5. Let B ∈ Bl(G). Denote by (M ) the bimodule isomorphism class
of a (B, B)-bimodule M . The Picard group of B relative to O, denoted by PicO (B), is
the multiplicative group consisting of all classes (M ) of invertible (B, B)-bimodules.
Multiplication in PicO (B) is defined by
(M )(M 0 ) = (M ⊗B M 0 ).
The identity element is (B), and the inverse of (M ) is (M )−1 = (M −1 ).
In this thesis, we shall write Pic(B) for PicO (B).
Lemma 2.5.6. Let G, H be finite groups. Let B ∈ Bl(G), A ∈ Bl(H). If B and A
are Morita equivalent, then Pic(B) ∼
= Pic(A).
Proof. Let M be a (B, A)-bimodule and N be an (A, B)-bimodule such that M, N
give a Morita equivalent between B and A. It is elementary to check that the map
P 7→ N ⊗B P ⊗B M is a required isomorphism Pic(B) −→ Pic(A).
2.5.2
Derived equivalence and derived Picard group
Morita equivalence is a strong type of equivalences between blocks. It does not happen often, for example, in non-abelian simple groups. A somewhat weaker type of
equivalence, and a generalization of a Morita equivalence, is called a derived equivalence.
Let Db (B) be the derived category of bounded complexes of B-modules. A complex of B-modules is called perfect if it is quasi-isomorphic to a bounded complex of
finitely generated projective B-modules. The following theorem, due to Rickard [38]
[39], is essential.
Theorem 2.5.7. The following are equivalent
1. The categories Db (A) and Db (B) are equivalent as triangulated categories.
2. There are a complex T ∈ Db (B ⊗O A◦ ) whose restriction to B and A◦ are perfect
and a complex S ∈ Db (A ⊗O B ◦ ) whose restriction to B and A◦ are perfect such
CHAPTER 2. PRELIMINARY
30
that
T ⊗A S ∼
= B in Db (B ⊗O B ◦ ) and S ⊗B T ∼
= A in Db (A ⊗O A◦ ).
Definition 2.5.8. We say that A and B are derived equivalent if Db (A) and Db (B)
are equivalent as triangulated categories.
The complexes T and S as in Theorem 2.5.7 are called (two-sided) tilting complexes, inverse to each other. The functors
T ⊗LA − : Db (A) −→ Db (B)
S ⊗LB − : Db (B) −→ Db (A)
then give inverse equivalences between Db (A) and Db (B).
There is a full embedding of Mod(B) (respectively Mod(B ⊗O A◦ )) into Db (B)
(respectively Db (B ⊗O A◦ )) by sending M to the complex with component M concentrated in degree 0. We will identify M with this complex. Using this, it is clear that
if M ∈ Mod(B ⊗O A◦ ) is an invertible bimodule, then it is also a tilting complex
when considered as a complex in Db (B ⊗O A◦ ). So a Morita equivalence induces a
derived equivalence.
Suppose A and B are derived equivalent. Similar to Morita equivalence, tensoring
everything with K and k gives derived equivalences between K ⊗O A and K ⊗O B
and between k ⊗O A and k ⊗O B respectively.
We will consider now the case A = B.
Definition 2.5.9. Let B ∈ Bl(G). Denote by (T ) the isomorphism class of T ∈
Db (B ⊗O B ◦ ). The derived Picard group of B relative to O, denoted by DPicO (B), is
the multiplicative group consisting of all classes (T ) of tilting complexes in Db (B ⊗O
B ◦ ). Multiplication is defined by
(T )(S) = (T ⊗LB S).
The identity element is (B). That this is a group follows from [45, Proposition 2.4].
CHAPTER 2. PRELIMINARY
31
In this thesis, we shall write DPic(B) for DPicO (B). Note that the usual embedding of Mod(B ⊗O B ◦ ) into Db (B ⊗O B ◦ ) gives the canonical injection
Pic(B) −→ DPic(B).
Let Sh(B) be a subgroup of DPic(B) generated by B[1]. Then Sh(B) is an infinite
cyclic group and lies in the center of DPic(B). The direct product Pic(B) × Sh(B)
is a subgroup of DPic(B).
Lemma 2.5.10. Let G, H be finite groups. Let B ∈ Bl(G), A ∈ Bl(H). If B and A
are derived equivalent, then DPic(B) ∼
= DPic(A).
Proof. Let T ∈ Db (B ⊗O A◦ ) and S ∈ Db (A⊗O B ◦ ) be tilting complexes giving inverse
derived equivalences between Db (B) and Db (A). It is elementary to check that the
map X 7→ S ⊗LB X ⊗LB T is a required isomorphism DPic(B) −→ DPic(A).
Chapter 3
Perfect Isometries
Let G be a finite group. Let B be a block of OG. We identify RK (B) with the abelian
group generated by Irr(B), the irreducible characters of simple KB-modules. Let H
be a second finite group and A be a block of OH. We will identify RK (B ⊗O A◦ )
with the abelian group on the irreducible characters of B ⊗O A◦ -modules.
3.1
Definitions
Let µ ∈ RK (B ⊗O A◦ ). We can regard µ as a generalized character of G × H by
µ(g, h) = µ(g ⊗ (h−1 )◦ ).
Definition 3.1.1 ([8]). We say that a generalized character µ of G × H is perfect if
the following two conditions hold:
(i) [Integrality] For all (g, h) ∈ G × H we have
µ(g, h)
∈O
|CG (g)|
and
µ(g, h)
∈ O.
|CH (h)|
(ii) [Separation] For all (g, h) ∈ G × H, if µ(g, h) 6= 0 then, g is p-regular if and
only if h is p-regular.
Perfect characters can arise from bimodules that are projective on both sides, as
is shown by the following proposition.
32
CHAPTER 3. PERFECT ISOMETRIES
33
Proposition 3.1.2. [8, Proposition 1.2] Let G, H be two finite groups. Let M be an
O(G, H)-bimodule which is projective as left OG-module and as right OH-module.
Then the character afforded by K ⊗O M is perfect.
Following Broué [8], each µ ∈ RK (B ⊗O A◦ ) determines two linear maps
Iµ : RK (A) −→ RK (B) ,
Rµ : RK (B) −→ RK (A)
defined by
Iµ (β)(g) =
1 X
µ(g, h−1 )β(h) = hµ(g, ·), βiH
|H| h∈H
(3.1)
Rµ (α)(h) =
1 X
µ(g −1 , h)α(g) = hµ(·, h), αiG
|G| g∈G
(3.2)
and
for α ∈ RK (B) and β ∈ RK (A). Here, h·, ·i is the standard inner product of class
functions.
Lemma 3.1.3. Given µ ∈ RK (B ⊗O A◦ ). The linear maps Iµ , Rµ defined by (3.1)
and (3.2) are adjoint to each other with respect to h·, ·i. That is,
hIµ (β), αiG = hβ, Rµ (α)iH
for all α ∈ RK (B), β ∈ RK (A).
Proof. It is sufficient to prove for the irreducible characters. Let α ∈ Irr(B), β ∈
Irr(A). Then
hIµ (β), αiG = |G|−1
X
= |G|−1
X
Iµ (β)(g)α(g −1 )
g∈G
|H|−1
g∈G
X
µ(g, h−1 )β(h)α(g −1 )
h∈H
!
= |H|−1
X
|G|−1
= |H|−1
µ(g −1 , h−1 )α(g) β(h)
g∈G
h∈H
X
X
Rµ (α)(h−1 )β(h)
h∈H
= hβ, Rµ (α)iH .
CHAPTER 3. PERFECT ISOMETRIES
34
If X is a KB-module and Y is a KA-module. Then they determine a K(B ⊗O A◦ )module X ⊗O Y v where Y v := HomK (Y, K) is a right KA-module. If χ is the image
of X in RK (B) and ϕ is the image of Y in RK (A), then we denote by χ · ϕv the image
of X ⊗O Y v in RK (B ⊗O A◦ ). Now let I : RK (A) −→ RK (B) be a linear map, we
P
can regard I as an element in RK (B ⊗O A◦ ), denoted µI , by µI = ϕ∈Irr(A) I(ϕ) · ϕv .
The following lemma shows that the map induced by µI is I.
Lemma 3.1.4. Let I : RK (A) −→ RK (B) be a linear map. Define
X
µ=
I(ϕ) · ϕv .
ϕ∈Irr(A)
Then µ ∈ RK (B ⊗O A◦ ) and I = Iµ . When viewed as a generalized character of
G × H we have
X
µ(g, h) =
I(ϕ)(g)ϕ(h).
(3.3)
ϕ∈Irr(A)
for g ∈ G and h ∈ H. Furthermore, µ is unique.
Proof. It is clear by definition of I that µ ∈ RK (B ⊗O A◦ ). Since (χ · ϕv )(g ⊗ h◦ ) =
χ(g)ϕv (h) = χ(g)ϕ(h−1 ), we have
X
µ(g, h) = µ(g ⊗ (h−1 )◦ ) =
I(ϕ)(g)ϕv (h−1 ) =
ϕ∈Irr(A)
X
I(ϕ)(g)ϕ(h)
ϕ∈Irr(A)
for g ∈ G and h ∈ H. Let β ∈ RK (A) and g ∈ G, then
1 X
µ(g, h−1 )β(h)
|H| h∈H


X
X
1

=
I(ϕ)(g)ϕ(h−1 ) β(h)
|H| h∈H
ϕ∈Irr(A)
!
X 1
X
−1
=
ϕ(h )β(h) I(ϕ)(g)
|H| h∈H
ϕ∈Irr(A)
X
=
hβ, ϕiI(ϕ)(g)
Iµ (β)(g) =
ϕ∈Irr(A)

= I

X
hβ, ϕiϕ (g)
ϕ∈Irr(A)
= I(β)(g).
CHAPTER 3. PERFECT ISOMETRIES
35
To show uniqueness of µ, suppose there is τ ∈ RK (B ⊗O A◦ ) such that I = Iτ . Let
X
τ=
X
(Cχϕ )χ · ϕv ,
Cχϕ ∈ Z.
χ∈Irr(B) ϕ∈Irr(A)
The same arguments as above applied to β ∈ Irr(A) shows that
X
(Cχβ )χ = I(β).
χ∈Irr(B)
That is,

τ=
X

X

ϕ∈Irr(A)
(Cχϕ )χ · ϕv =
χ∈Irr(B)
X
I(ϕ) · ϕv = µ.
ϕ∈Irr(A)
The following result of Masao Kiyota [47] is useful and convenient when we want
to check an integrality condition of a generalized character µ of G × H once we know
that µ satisfies the separation condition. The author would like to thank Dr. Charles
Eaton for showing him this paper.
Theorem 3.1.5 (Kiyota). Let G, H be finite groups. Let B ∈ Bl(G) and A ∈ Bl(H).
Let µ ∈ RK (B ⊗O A◦ ) be a generalized character of G × H satisfying the separation
condition. Then, in order to check the integrality condition, it is sufficient to check
only for p-singular elements g ∈ G and h ∈ H.
Let CF(G, B; K) be the subspace of CF(G; K) of class functions generated by
Irr(B). Let CF(G, B; O) be the subspace of CF(G, B; K) of O-valued class functions.
Let CFp0 (G, B; K) be the subspace of class functions α ∈ CF(G, B; K) vanishing
outside p-regular elements.
If µ ∈ RK (B ⊗O A◦ ), the linear maps Iµ , Rµ defined in (3.1),(3.2) can be extended
in the usual way to the linear maps
Iµ : CF(H, A; K) −→ CF(G, B; K) ,
Rµ : CF(G, A; K) −→ CF(H, B; K).
Proposition 3.1.6. [8, Proposition 4.1] A character µ ∈ RK (B ⊗O A◦ ) is perfect if
and only if
CHAPTER 3. PERFECT ISOMETRIES
36
(i*) Iµ gives a map from CF(H, A; O) to CF(G, B; O) and Rµ gives a map from
CF(G, B; O) to CF(H, A; O).
(ii*) Iµ gives a map from CFp0 (H, A; K) to CFp0 (G, B; K) and Rµ gives a map from
CFp0 (G, B; K) to CFp0 (H, A; K).
Corollary 3.1.7. Let Iµ : RK (A) −→ RK (B) be a perfect isometry. Let α ∈
CF(H, A; K) be a class function vanishing on p-regular elements. Then Iµ (α) also
vanishes on p-regular elements.
Proof. Let g ∈ G be a p-regular element. Consider
Iµ (α)(g) =
1 X
µ(g, h−1 )α(h).
|H| h∈H
If h is p-regular then α(h) = 0 while if h is p-singular then µ(g, h−1 ) = 0 as µ is
perfect. So Iµ (α)(g) = 0.
We already know that a character µ ∈ RK (B ⊗O A◦ ) defines Iµ and Rµ which are
adjoint to each other. If, in addition, Iµ is an isometry, then we can say more about
Rµ .
Lemma 3.1.8. Let µ ∈ RK (B ⊗O A◦ ). Suppose that Iµ is an isometry. Then Rµ is
also an isometry and (Iµ )−1 = Rµ .
Proof. Let β ∈ Irr(A). Since Iµ (β) is a Z-linear combination of characters in Irr(B)
(by identification of RK (B)), we have Iµ (β) ∈ ± Irr(B). Let α ∈ Irr(B). Since Iµ
and Rµ are adjoint,
hIµ (β), αiG = hβ, Rµ (α)iH .
As the left hand side takes values in {0, ±1}, this implies that Rµ (α) ∈ ± Irr(A). By
adjointness again,
hIµ (Rµ (α)), αiG = hRµ (α), Rµ (α)iH = 1.
Since Iµ (Rµ (α)) ∈ ± Irr(B), this forces Iµ (Rµ (α)) = α. Similarly,
1 = hIµ (β), Iµ (β)iG = hβ, Rµ (Iµ (β))iH
implies that Rµ (Iµ (β)) = β. Hence Rµ = (Iµ )−1 and Rµ is an isometry.
CHAPTER 3. PERFECT ISOMETRIES
37
Definition 3.1.9. We say that I : RK (A) −→ RK (B) is a perfect isometry if I is an
isometry and I = Iµ where µ ∈ RK (B ⊗O A◦ ) is perfect.
Definition 3.1.10. Two blocks A, B are said to be perfectly isometric if there is a
perfect isometry I : RK (A) −→ RK (B).
If I : RK (A) −→ RK (B) is a (perfect) isometry, then I(χ) ∈ ± Irr(B) for all
χ ∈ Irr(B). So I defines a bijection I + : Irr(A) −→ Irr(B) and a sign function
εI : Irr(A) −→ {±1} such that I(χ) = εI (χ)I + (χ) for χ ∈ Irr(A). This gives a
bijection with signs between Irr(A) and Irr(B).
We denote the sign function εI such that εI (χ) = 1, ∀χ ∈ Irr(A) by 1 and the sign
function εI such that εI (χ) = −1, ∀χ ∈ Irr(A) by −1.
3.2
Properties
Let B, A be blocks of OG and OH respectively. It it easy to see that if I : RK (A) −→
RK (B) is a perfect isometry then so is −I. It turns out that a perfect isometry has
essentially a unique sign. This means that, if J is another perfect isometry with
J + = I + , then εJ = εI or εJ = −εI , in other words, J = ±I. We need the following
theorem, due to Osima, which is a converse of block orthogonality.
Theorem 3.2.1. [35, Theorem 3] Let S be a set of ordinary characters of G such
that
X
χ(g)χ(h) = 0
χ∈S
for any p-regular element g and any p-singular element h. Then S is a union of
Irr(B) for some blocks B ∈ Bl(G). In other words, if χ ∈ S and χ ∈ Irr(B) for some
block B, then Irr(B) ⊆ S.
Proposition 3.2.2. Let I : RK (B) −→ RK (B) be a perfect isometry such that I + is
the identity map. Then εI = 1 or −1.
Proof. Suppose I = Iµ . Then, for g, h ∈ G, µ(g, h) =
P
χ∈Irr(B) εI (χ)χ(g)χ(h).
B + = {χ ∈ Irr(B) : εI (χ) = 1}
Set
CHAPTER 3. PERFECT ISOMETRIES
38
and
B − = {χ ∈ Irr(B) : εI (χ) = −1}.
Then µ(g, h) =
P
X
χ∈B +
χ(g)χ(h) −
χ(g)χ(h) −
X
P
χ∈B −
χ(g)χ(h). Hence
χ(g)χ(h) = 0 ∀g ∈ Gp0 , ∀h ∈ G − Gp0 .
χ∈B −
χ∈B +
Since id : RK (B) −→ RK (B) is also a perfect isometry,
X
χ(g)χ(h) +
P
χ∈B +
χ(g)χ(h) = 0 ∀g ∈ Gp0 , ∀h ∈ G − Gp0 .
χ∈B −
χ∈B +
Hence, 2
X
χ(g)χ(h) = 0 and so
P
χ∈B +
χ(g)χ(h) = 0 for any p-regular element
g and p-singular element h. By Theorem 3.2.1, B + = B or φ. If B + = B, then
εI = 1. If B + = φ, then εI = −1.
Lemma 3.2.3 (Uniqueness of signs). Suppose I, J : RK (A) −→ RK (B) are perfect
isometries such that I + = J + . Then J = I or J = −I.
Proof. Let δ = εI and τ = εJ . Then I ◦J −1 : RK (B) −→ RK (B) is a perfect isometry
with (I ◦ J −1 )+ = id and ε = εI◦J −1 is given by ε(χ) = δ((J + )−1 (χ)) · τ ((J + )−1 (χ)).
By the previous proposition, ε = 1 or −1 and hence τ = δ or τ = −δ.
Suppose µ induces a perfect isometry Iµ : RK (A) −→ RK (B). Then Iµ defines a
bijection between primitive idempotents of Z(KA) and Z(KB) by χ 7→ eI(χ) . Note
that by definition we have e−χ = eχ . We next follow the result of Broué [8, Theorem
1.5] and define two linear maps
Iµ◦ : Z(KA) −→ Z(KB) ,
Rµ◦ : Z(KB) −→ Z(KA)
by
Iµ◦ (y) =
X
g∈G
where y =
P
h∈H
!
1 X
µ(g −1 , h)y(h) g
|H| h∈H
(3.4)
y(h)h ∈ Z(KA), and
Rµ◦ (x) =
X
h∈H
!
1 X
µ(g, h−1 )x(g) h
|G| g∈G
(3.5)
CHAPTER 3. PERFECT ISOMETRIES
where x =
P
g∈G
39
x(g)g ∈ Z(KB).
Since µ is perfect, the maps Iµ◦ and Rµ◦ also defined O-linear maps on Z(KA) and
Z(KB) respectively. Let χ ∈ Irr(A). Then
Iµ◦ (eχ ) =
|G|/Iµ (χ)(1)
eIµ (χ) .
|H|/χ(1)
It is easy to check that Iµ◦ (eχ Rµ◦ (e)) = eIµ (χ) , where e is the block idempotent of
B. So we have an algebra isomorphism Z(KA) −→ Z(KB) by defining
y 7→ Iµ◦ (yRµ◦ (e))
(3.6)
with the inverse Z(KB) −→ Z(KA) defined by
x 7→ Rµ◦ (Iµ◦ (f )x)
(3.7)
where f is the block idempotent of A. These also give algebra isomorphisms between
Z(A) and Z(B).
Theorem 3.2.4. [8, Theorem 1.5] Suppose µ induces a perfect isometry Iµ : RK (A) −→
RK (B). Then Iµ induces an algebra isomorphism from Z(A) to Z(B). Furthermore,
Iµ preserves the defect of the blocks, the number of irreducible ordinary and Brauer
characters and the Cartan invariants.
Lemma 3.2.5. [8, Lemma 1.6] Suppose Iµ : RK (A) −→ RK (B) is a perfect isometry.
Let χ ∈ Irr(A). Let
rχ =
|G|/Iµ (χ)(1)
.
|H|/χ(1)
Then rχ is an invertible element in O and (rχ )∗ is constant for all χ ∈ Irr(A).
Proof. Let ρµ : x 7→ Rµ◦ (Iµ◦ (f )x) be the algebra isomorphism from Z(B) onto Z(A)
as above. Let x = ρ−1
Rµ◦ (e) ∈ Z(B). Then
µ
Iµ◦ (f )x = Rµ◦−1 (ρµ (x)) = e.
This shows that Iµ◦ (f ) is invertible in Z(B). But
Iµ◦ (f ) =
X |G|/Iµ (χ)(1)
eIµ (χ) .
|H|/χ(1)
χ∈Irr(A)
CHAPTER 3. PERFECT ISOMETRIES
Thus,
|G|/Iµ (χ)(1)
|H|/χ(1)
40
are invertible elements in O and
|G|/Iµ (χ)(1)
|H|/χ(1)
∗
exist for all χ ∈ Irr(A).
So
∗
|G|/Iµ (χ)(1)
.
=
=
|H|/χ(1)
∗
µ (χ)(1)
But λχ = λφ for all χ, φ ∈ Irr(A). Hence |G|/I
is constant for all χ ∈ Irr(A).
|H|/χ(1)
λχ (Iµ◦ (f ))
Since
|G|/Iµ (χ)(1)
|H|/χ(1)
∗
ωχ (Iµ◦ (f ))
∈ Q, it is constant modulo p.
Since rχ is invertible, the previous lemma also shows that a perfect isometry
preserves defects of irreducible characters.
Other numerical invariants preserved by a perfect isometry is given by the following theorem.
Theorem 3.2.6 (Theorem 4.11). [9] Let A ∈ Bl(H) and B ∈ Bl(G) be perfectly
isometric blocks. Then,
• k(A) = k(B).
• l(A) = l(B).
• d(A) = d(B).
3.3
Examples of perfect isometries
Let G be a finite group. In this section we will describe some typical examples of
perfect isometries that arise “naturally”.
Galois Action
Let ζ = exp(2πi/|G|). Denote by Gal(Q(ζ)/Q) the Galois group of the field extension.
For an integer n coprime to |G|, let σn ∈ Gal(Q(ζ)/Q) where ζ σn = ζ n . Since every
irreducible character χ ∈ Irr(G) has values in Q(ζ), the group Gal(Q(ζ)/Q) acts on
Irr(G) by (χσn )(g) = χ(g)σn , ∀χ ∈ Irr(G), ∀g ∈ G. It is clear that χ(g)σn = χ(g n ).
The next lemma shows that the action of Gal(Q(ζ)/Q) preserves the block structure on Irr(G).
CHAPTER 3. PERFECT ISOMETRIES
41
Lemma 3.3.1. Let σ ∈ Gal(Q(ζ)/Q). Suppose χ, ϕ ∈ Irr(G) are in the same block
then χσ , ϕσ ∈ Irr(G) are also in the same block.
Proof. Since χ, ϕ are in the same block we have
χ(gC ) ϕ(gC )
−
≡0
|C|
χ(1)
ϕ(1)
mod p,
for all C ∈ cl(G), where gC ∈ C. Then
σ
σ
χ (gC ) ϕσ (gC )
χ(gC ) ϕ(gC )
|C|
− σ
= |C|
−
≡0
χσ (1)
ϕ (1)
χ(1)
ϕ(1)
mod p,
for all C ∈ cl(G). Hence χσ , ϕσ are in the same block.
Let σ ∈ Gal(Q(ζ)/Q) and B ∈ Bl(G). Denote by B σ the block of G with
Irr(B σ ) = {χσ : χ ∈ Irr(B)}. We can define an isometry Iσ : RK (B) −→ RK (B σ ) by
Iσ (χ) = χσ for χ ∈ Irr(G).
Lemma 3.3.2. Let σ ∈ Gal(Q(ζ)/Q) and B ∈ Bl(G). The isometry Iσ : RK (B) −→
RK (B σ ) is perfect.
Proof. For short, write µσ = µIσ . Let g, h ∈ G. Suppose that g σ = g n where n is
coprime to |G|. Then
µσ (g, h) =
X
Iσ (χ)(g)χ(h) =
χ∈Irr(B)
=
X
X
χ(g)σ χ(h)
χ∈Irr(B)
χ(g n )χ(h) = µid (g n , h)
χ∈Irr(B)
where id : RK (B) −→ RK (B) is the identity map.
We claim that CG (g) = CG (g n ). To see this, first it is clear that CG (g) ⊆ CG (g n ).
Let x ∈ CG (g n ). There is an integer m such that g = (g n )m . Then xgx−1 =
xg nm x−1 = (xg n x−1 )m = (g n )m = g. Thus x ∈ CG (g). Now, since n is coprime to
|G|, we have that g is p-regular if and only if g n is p-regular. Since the identity map
is a perfect isometry,
µid (g n , h)
µid (g n , h)
µσ (g, h)
=
=
∈O
|CG (g)|
|CG (g)|
|CG (g n )|
µσ (g, h)
µid (g n , h)
=
∈ O.
|CH (h)|
|CH (h)|
CHAPTER 3. PERFECT ISOMETRIES
42
To show the separation condition, notice that if exactly one of g, h is p-singular,
then exactly one of g n , h is p-singular. Thus µσ (g, h) = µid (g n , h) = 0. Hence Iσ is a
perfect isometry as claimed.
In fact, Kessar [26] showed that the map Iσ is an “isotypy” (to be defined later).
Automorphism Action
Let σ ∈ Aut(G) and H ≤ G such that H σ = H. Then σ acts on CF(H; K) via
−1
θσ (h) = θ(hσ ). By Lemma 2.3.5, the action by σ permutes elements in the set
Irr(H) and also preserves the block structure of H. Let A ∈ Bl(H). Define Iσ :
RK (A) −→ RK (Aσ ) by Iσ (χ) = χσ for χ ∈ Irr(A).
Lemma 3.3.3. Let σ ∈ Aut(G) and H ≤ G be invariant under σ. Let A ∈ Bl(H).
Then Iσ : RK (A) −→ RK (Aσ ) is a perfect isometry.
Proof. The isometry Iσ is also defined on CF(H, A; K) in the obvious way. Since σ
is an automorphism, an element h ∈ H is p-singular if and only if hσ
−1
is p-singular.
Thus, if θ ∈ CF(H, A; K) vanishes on p-singular elements, then so does Iσ (θ). Also,
if θ ∈ CF(H, A; K) has values in O for all h ∈ H then so does Iσ (θ). Hence Iσ is a
perfect isometry by Proposition 3.1.6.
As the principal block A0 ∈ Bl(H) contains the trivial character which is invariant
under σ, we have (A0 )σ = A and so Iσ is a perfect isometry in A0 .
An important special case is when H is a normal subgroup of G. Then H is
invariant under any inner automorphism of G. So, conjugation by any element of G
gives a perfect isometry in the principal block of H.
Multiplication by a linear character
Let χ, λ ∈ Irr(G) where λ is a linear character. It is easy to see that λχ ∈ Irr(G)
where λχ(g) = λ(g)χ(g), ∀g ∈ G. The map I : χ 7→ λχ then defines a bijection on
Irr(G).
CHAPTER 3. PERFECT ISOMETRIES
43
Lemma 3.3.4. Let λ ∈ Irr(G) be a linear character. Suppose χ, ϕ ∈ Irr(G) are in
the same block then λχ, λϕ are also in the same block.
Proof. Let C ∈ cl(G) and gC ∈ C. Then
|C|λχ(gC )
λχ(1)
∗
−
|C|λϕ(gC )
λϕ(1)
∗
∗
= λ(gC )
|C|χ(gC )
χ(1)
∗
−
|C|ϕ(gC )
ϕ(1)
∗ = 0.
Hence λχ, λϕ are in the same block.
Let λ ∈ Irr(G) be a linear character. For a block B ∈ Bl(G), denote by λB the
block of λχ for some χ ∈ Irr(B). This is well-defined by the previous lemma.
Lemma 3.3.5. Let B be a block of OG. Let λ ∈ Irr(G) be a linear character. Then
the map I : Irr(B) −→ Irr(λB) defined by I(χ) = λχ is a perfect isometry.
Proof. Let θ ∈ CF(G, B; K) be a class function. Then I(θ) = λθ. Thus, if θ vanishes
on p-singular elements, then so does I(θ). And if θ has values in O, then so does
I(θ). Hence I is a perfect isometry by Proposition 3.1.6.
Chapter 4
Perfect isometry groups
Let G be a finite group. Let B be a block of OG. In this chapter we will consider
the perfect isometries from RK (B) to itself. We denote by Γ(B) the group of all
isometries in RK (B) under compositions. Using Proposition 3.1.6, it is clear that the
set of all perfect isometries in RK (B) forms a subgroup of Γ(B).
Definition 4.0.6. The group of all perfect isometries in RK (B) is denoted by PI(B)
and called the perfect isometry group of B.
One of the reasons to study the perfect isometry groups is because they are invariant for perfectly isometric blocks.
Proposition 4.0.7. Let G, H be finite groups and let B, A be blocks of OG and OH
respectively. Suppose there exists a perfect isometry
I : RK (A) −→ RK (B).
Then PI(A) and PI(B) are isomorphic. An isomorphism
θ : PI(A) −→ PI(B)
is given by θ(α) = IαI −1 for α ∈ PI(A).
Proof. We first need to check that θ is well-defined, that is, it maps a perfect isometry
in PI(A) to a perfect isometry in PI(B). Let α ∈ PI(A). Then θ(α) is an isometry
from RK (B) to RK (B), since it is a composition of isometries
I −1
α
I
RK (B) −→ RK (A) −→ RK (A) −→ RK (B).
44
CHAPTER 4. PERFECT ISOMETRY GROUPS
45
Since I, I −1 , α, α−1 are all perfect isometries, θ(α) gives a map
I −1
α
I
CF(G, B; O) −→ CF(H, A; O) −→ CF(H, A; O) −→ CF(G, B; O)
and
I −1
α
I
CFp0 (G, B; K) −→ CFp0 (H, A; K) −→ CFp0 (H, A; K) −→ CFp0 (G, B; K),
whereas θ(α)−1 = Iα−1 I −1 gives a map
I −1
α−1
I
CF(G, B; O) −→ CF(H, A; O) −→ CF(H, A; O) −→ CF(G, B; O)
and
I −1
α−1
I
CFp0 (G, B; K) −→ CFp0 (H, A; K) −→ CFp0 (H, A; K) −→ CFp0 (G, B; K).
So, by Proposition 3.1.6, θ(α) is a perfect isometry and thus lies in PI(B). The map
θ is clearly an isomorphism.
The converse, however, is not true in general.
Example 4.0.8. Let D be a Sylow 2-subgroup of the group G = L2 (17). Let B0 , A0
be principal 2-blocks of G and NG (D) respectively. Calculations in GAP [19] show
that PI(B0 ) ∼
= PI(A0 ) ∼
= D8 × C2 × C2 , but there does not exist a perfect isometry
between RK (B) and RK (A).
We have seen in the previous chapter that the Galois action, automorphism action
and multiplication by linear characters all give rise to perfect isometries from a block
B to its corresponding block. Since the principal block B0 are invariant under these
actions, we always have elements in PI(B0 ) arising from these actions. We will see
later that, in some cases, the subgroups generated by perfect isometries from some
of these actions contribute a significant part in the structure of PI(B0 ).
4.1
Structure of PI(B)
Let I ∈ Γ(B). Then there is a bijection I + : Irr(B) −→ Irr(B) and a sign function
εI : Irr(B) −→ {±1} such that I(χ) = εI (χ)I + (χ) for all χ ∈ Irr(B). For the
CHAPTER 4. PERFECT ISOMETRY GROUPS
46
purpose of investigating the structure of PI(B), we will label the irreducible characters
in Irr(B) as {χ1 , χ2 , . . . , χn }. Then we can regard I + as a permutation σ ∈ Sn .
We will treat elements in Sn as functions, so in a product of elements in Sn , the
rightmost permutation is applied first. We will also regard εI as an element ε =
(εI (χ1 ), . . . , εI (χn )) ∈ (C2 )n . Denote by ε(i) the i-th component of ε. Then
I(χi ) = ε(i)χσ(i) ,
i ∈ {1, . . . , n}.
Multiplications in (C2 )n are pointwise.
Every isometry I ∈ Γ(B) can then be identified with (σ, ε) where σ ∈ Sn and ε ∈
(C2 )n . For convenience, we will denote (1, 1, . . . , 1) ∈ (C2 )n by 1 and (−1, −1, . . . , −1)
by −1.
Definition 4.1.1. An isometry I ∈ Γ(B) is said to have a homogenous sign if I =
(σ, ε) where ε = ±1. It has the all-positive sign if ε = 1. In other words, an isometry
I : RK (B) −→ RK (B) has all-positive sign if I sends Irr(B) to Irr(B).
The group Sn acts on the group (C2 )n by εσ (i) = ε(σ(i)), where σ ∈ Sn and
ε ∈ (C2 )n . Let φ : Sn −→ Aut((C2 )n ) be the corresponding action. The next lemma
gives a structure of the group Γ(B) via this action.
Lemma 4.1.2. Let I ∈ Γ(B) be an isometry with I + = σ ∈ Sn and εI = ε ∈ (C2 )n .
Then the map I 7→ (σ, ε) is a group isomorphism from Γ(B) onto Sn nφ (C2 )n .
Multiplications in Sn nφ (C2 )n are as follows. If (σ1 , ε1 ), (σ2 , ε2 ) ∈ Sn nφ (C2 )n , then
(σ1 , ε1 )(σ2 , ε2 ) = (σ1 σ2 , (εσ1 2 )(ε2 )).
−1
The inverse of (σ, ε) is (σ −1 , εσ ).
Proof. Define f : Γ(B) −→ Sn nφ (C2 )n by f (I) = (σ, ε), where I(χi ) = ε(i)χσ(i) , ∀χi ∈
Irr(B) for some ε ∈ (C2 )n and σ ∈ Sn . We will show that this is an isomorphism. Let
I1 , I2 ∈ Γ(B). Suppose I1 (χi ) = ε1 (i)χσ1 (i) and I2 (χi ) = ε2 (i)χσ2 (i) for i = 1, . . . , n.
CHAPTER 4. PERFECT ISOMETRY GROUPS
47
Then
I1 (I2 (χi )) = I1 (ε2 (i)χσ2 (i) ) = ε2 (i)I1 (χσ2 (i) )
= ε2 (i)ε1 (σ2 (i))χσ1 (σ2 (i))
= ε2 (i)εσ1 2 (i)χσ1 (σ2 (i))
= (ε2 · εσ1 2 )(i)χσ1 (σ2 (i)) .
Therefore, f (I1 ◦ I2 ) = (σ1 σ2 , (ε1 )σ2 · ε2 ). On the other hand,
f (I1 )f (I2 ) = (σ1 , ε1 )(σ2 , ε2 ) = (σ1 σ2 , (ε1 )σ2 · ε2 ).
So f is a homomorphism. Note that f (I) = (1, 1) if and only if I(χi ) = 1(i)χ1(i) =
χi ∀i = 1, . . . , n if and only if I = id. Hence f is injective. Finally, given σ ∈ Sn
and ε ∈ (C2 )n , the map I defined by I(χi ) = ε(i)χσ(i) , i ∈ {1, . . . , n} is clearly an
isometry.
Since PI(B) is a subgroup of Γ(B), it therefore inherits the structure of Γ(B)
and we shall identify a perfect isometry I ∈ PI(B) with (σ, ε) ∈ Sn n (C2 )n and
write I = (σ, ε) when I(χi ) = ε(i)χσ(i) ∀i ∈ {1, . . . , n}. We will use I and (σ, ε)
interchangeably to denote a perfect isometry. The following facts can be easily proved:
• If I = (σ, ε) then −I = (σ, −ε).
• ε−1 = ε.
• (−ε)σ = −(εσ ).
We will now define an important group PI+ (B) which, in many cases, characterizes
the group PI(B). First, it is clear that id and −id are perfect isometries, where id
is the identity isometry and −id(χ) = −χ, ∀χ ∈ Irr(B). We will denote by h-idi the
subgroup generated by −id.
Lemma 4.1.3. There exists a subgroup H ≤ Sn such that PI(B)/ h-idi ∼
= H.
CHAPTER 4. PERFECT ISOMETRY GROUPS
48
Proof. Embed the sign group (C2 )n into Γ(B) = Sn nφ (C2 )n . By Proposition 3.2.2,
PI(B) ∩ (C2 )n = h-idi. By the Second Isomorphism Theorem,
PI(B)
PI(B)
PI(B)(C2 )n
Sn nφ (C2 )n ∼
∼
=
≤
=
= Sn .
h-idi
PI(B) ∩ (C2 )n
(C2 )n
(C2 )n
Definition 4.1.4. Let B be a block of OG and let Irr(B) = {χ1 , . . . , χn }. Let σ ∈ Sn .
We say that σ is a perfect permutation if there is a sign function ε such that I = (σ, ε)
is a perfect isometry. Denote by PI+ (B) the group of perfect permutations of the
block B. If H ≤ PI(B), we say that H contains a perfect permutation σ ∈ PI+ (B)
if (σ, ε) ∈ H for some sign ε.
The group PI+ (B) therefore represents how PI(B) acts on Irr(B) via permutation
(ignoring the signs). Note that PI+ (B) is the projection of PI(B) ≤ Sn nφ (C2 )n onto
Sn . From the previous lemma it is clear that we have a short exact sequence
1 −→ h-idi −→ PI(B) −→ PI+ (B) −→ 1.
We will now show that, in many cases, the sequence splits. First we need the
following lemma.
Lemma 4.1.5. Suppose there exists a (normal) subgroup H of PI(B) of index 2
such that −id 6∈ H. Then PI(B) = H × h-idi and H ∼
= PI+ (B). Conversely, if
PI(B) = H × h-idi for some subgroup H C PI(B) of index 2, then H contains all
perfect permutations and H ∼
= PI+ (B).
Proof. Suppose there is a subgroup H ≤ PI(B) of index 2 and −id 6∈ H. Let I ∈ H.
Then −I 6∈ H, otherwise we would have (−I) ◦ I −1 = −id ∈ H. Since | PI(B)| =
2| PI+ (B)|, and H has index 2, we have that H contains all perfect permutations
with one copy each. Any perfect isometry in PI(B) can therefore be written as I or
−I where I ∈ H. Since −id commutes with every element in H, we have PI(B) =
H × h-idi and the projection (σ, ε) 7→ σ gives a desired isomorphism H ∼
= PI+ (B).
Assume conversely that PI(B) = H × h-idi. If (σ, ε) ∈ H then (σ, −ε) 6∈ H or we
would have (σ, ε)−1 (σ, −ε) = (1, −1) = −id ∈ H, a contradiction, since −id 6∈ H. So
CHAPTER 4. PERFECT ISOMETRY GROUPS
49
H contains at most one copy of each perfect permutation. Thus |H| ≤ | PI+ (B)| =
| PI(B)|/2. Since H has index 2, we have the equality and so H contains all perfect
permutations with one copy each. An isomorphism H ∼
= PI+ (B) is given by the
projection (σ, ε) 7→ σ.
Lemma 4.1.6. Suppose that every perfect isometry I ∈ PI(B) has a homogenous
sign. Then
PI(B) ∼
= PI(B)+ × h-idi .
Proof. If I = (σ, ε) ∈ PI(B), then, by assumption, we have ε = ±1. Let
H = {(σ, 1) : σ ∈ PI+ (B)}.
Then it is clear that H is a subgroup of PI(B) of index 2, and the result follows from
Lemma 4.1.5.
Proposition 4.1.7. Suppose | Irr(B)| = n is an odd integer. Then
PI(B) ∼
= PI+ (B) × h-idi .
Proof. For ε ∈ (C2 )n , define the weight function
wt(ε) = number of -1 in ε.
It can easily be checked that
• wt(−ε) = n − wt(ε)
• wt(εσ ) = wt(ε) ∀σ ∈ Sn
and if we denote by [ε, τ ] the number of positions where ε and τ both have -1, then
wt(ε · τ ) = wt(ε) + wt(τ ) − 2[ε, τ ].
For each σ ∈ PI+ (B), let εσ be such that (σ, εσ ) ∈ PI(B) and (−1)wt(εσ ) = 1.
Since n is odd, (−1)wt(−εσ ) = (−1)n (−1)wt(εσ ) = −1, so the choice of εσ is unique.
Now define
H = {(σ, εσ ) : σ ∈ PI+ (B)}.
CHAPTER 4. PERFECT ISOMETRY GROUPS
50
We claim that H ≤ PI(B) is a subgroup of index 2 and does not contain −id. The
result will then follow from Lemma 4.1.5.
To see this, first it is clear that id ∈ H and −id 6∈ H since n is odd. If (σ, εσ ) ∈ H,
−1
−1
then (σ, εσ )−1 = (σ −1 , (εσ )σ ). But wt((εσ )σ )) = wt(εσ ). So (σ, εσ )−1 ∈ H. If
(σ, εσ ), (τ, ετ ) ∈ H, then (σ, εσ )(τ, ετ ) = (στ, (εσ )τ · ετ ). But
wt((εσ )τ · ετ ) = wt((εσ )τ ) + wt(ετ ) − 2[(εσ )τ , ετ ]
= wt(εσ ) + wt(ετ ) − 2[(εσ )τ , ετ ]
τ ·ε )
τ
So (−1)wt((εσ )
= (−1)wt(εσ ) (−1)wt(ετ ) = 1 and (σ, εσ )(τ, ετ ) ∈ H.
Finally since H contains only one copy of each perfect permutation, we have
| PI(B)/H| = 2.
Let I ∈ PI(B). Define the ratio rI =
I(χ)(1)
χ(1)
∗
for some χ ∈ Irr(B). By Lemma
∗ ∗
I(χ)(1)
=
,
3.2.5, this is well-defined and a non-zero element in k. Since I(−χ)(1)
−χ(1)
χ(1)
∗
the ratio I(χ)(1)
is also constant for all χ ∈ ± Irr(B). It is easy to see that
χ(1)
r−I = −rI for all I ∈ PI(B).
Lemma 4.1.8. The map I 7→ rI is a group homomorphism from PI(B) to k × , the
group of non-zero elements of k.
Proof. First, it is clear that rid = 1. By Lemma 3.2.5, we have rI 6= 0 for all
I ∈ PI(B). Let I, J ∈ PI(B). For χ ∈ Irr(B) let ϕ = I −1 (χ) ∈ ± Irr(B). Then
rI −1 =
I −1 (χ)(1)
χ(1)
∗
=
ϕ(1)
I(ϕ)(1)
∗
=
1
= (rI )−1 .
rI
Let θ = J(χ) ∈ ± Irr(B), then
∗
I(J((χ)))(1)
=
χ(1)
∗
I(θ)(1)
=
J −1 (θ)(1)
∗ ∗
I(θ)(1)
θ(1)
=
θ(1)
J −1 (θ(1))
rI◦J
= (rI )(1/rJ −1 ) = (rI )(rJ ).
This shows that I 7→ rI is a group homomorphism.
CHAPTER 4. PERFECT ISOMETRY GROUPS
51
We make the following observation based on our computer-generated examples.
Conjecture 4.1.9. Let B be a block of OG. Let I ∈ PI(B). Then rI = ±1.
Remark 4.1.10. If B is the principal block of OG with defect group D, and A is
the principal block of ONG (D), then by Sylow Theorem we have |G : NG (D)| ≡ 1
mod p. Let I : RK (A) −→ RK (B) be a perfect isometry. By Lemma 3.2.5, we can
still define a well-defined ratio
∗ ∗
|NG (D)|/χ(1)
I(χ)(1)
rI =
=
,
|G|/I(χ)(1)
χ(1)
for some χ ∈ Irr(A). Broué [8] noted in his remark after Lemma 1.6 that I(χ)(1) ≡
χ(1) mod p, ∀χ ∈ Irr(A). Since I is not specified, this seems to suggest that rI = 1
for any perfect isometry I in this case. (Although in fact, it should be rI = ±1 as
r−I = −rI .)
Proposition 4.1.11. Let G be a finite group. Let p be an odd prime. Let B be a
block of OG. Suppose that Conjecture 4.1.9 holds for B. Then
PI(B) ∼
= PI+ (B) × h-idi .
Proof. Let
H = {I ∈ PI(B) : rI = 1}.
Since p is odd, r−id = −1 6= 1 and so −id 6∈ H. From the proof of Lemma 4.1.8, it is
clear that H is a subgroup of PI(B). Finally, since rI = ±1 for all I ∈ PI(B) we have
that either I ∈ H or −I ∈ H. So H has index 2 in PI(B). The result now follows
from Lemma 4.1.5.
Apart from Conjecture 4.1.9, we also suspect that we always have PI(B) ∼
=
PI+ (B) × h-idi for any blocks of a finite group G. We are not able to prove this
for all blocks so far but we will list the cases where we have proved it.
Proposition 4.1.12. Let G be a finite group. Let B be a block of OG. Then we
have PI(B) ∼
= PI+ (B) × h-idi in the following cases:
• if every perfect isometry I ∈ PI(B) has a homogenous sign,
CHAPTER 4. PERFECT ISOMETRY GROUPS
52
• if | Irr(B)| is an odd integer,
• if p is odd and rI = ±1 for all I ∈ PI(B).
Conjecture 4.1.13. Let G be a finite group. Let B ∈ Bl(G). Then the short exact
sequence
1 −→ h-idi −→ PI(B) −→ PI+ (B) −→ 1
splits.
4.2
Standard subgroups of PI(B0) for a principal
block B0
Suppose now that B0 is the principal block of OG. We have seen that for each
σ ∈ Aut(G) we have a perfect isometry Iσ : χ 7→ χσ . Let Autc (G) be the group
of class-preserving automorphisms of G, that is, the group of automorphisms σ such
that, for each g ∈ G, σ(g) and g are conjugate in G. We then have a sequence of
normal subgroups (cf. [46]):
Inn(G) ≤ Autc (G) ≤ Aut(G).
Lemma 4.2.1. The map σ 7→ Iσ−1 is a group homomorphism from Aut(G) to PI(B0 )
with Autc (G) in the kernel.
Proof. Let σ, τ ∈ Aut(G). For each χ ∈ Irr(B0 ) we have
I(στ )−1 (χ) = χ(στ )
−1
= χτ
− 1σ − 1
−
= Iσ−1 (χτ 1 ) = Iσ−1 (Iτ −1 (χ)) = (Iσ−1 ◦ Iτ −1 )(χ).
This shows that the map σ 7→ Iσ−1 is a group homomorphism. If σ ∈ Autc (G), then
−1
Iσ−1 (χ)(g) = χσ (g) = χ(g σ ) = χ(g) for all χ ∈ Irr(B0 ), all g ∈ G. Thus, Iσ is the
identity map.
e 0 ) the image of the map Aut(G) −→ PI(B0 ). If M is the kernel
Denote by A(B
then Autc (G) ≤ M ≤ Aut(G) by the previous lemma.
CHAPTER 4. PERFECT ISOMETRY GROUPS
53
Remark 4.2.2. It should be noted that M depends on the block B0 and, in fact,
both bounds can be achieved. If B0 contains only 1 then clearly M = Aut(G) since
we must have 1σ = 1 for all σ ∈ Aut(G). On the other hand, if Irr(B0 ) = Irr(G)
then for any σ ∈ M we have Iσ−1 (f ) = f for any class function f on G (since Irr(B0 )
is a basis for class functions on G). Let x ∈ G and δx be the indicator class function
of x (that is, δx (g) = 1 if g and x are conjugate, and zero otherwise). If σ ∈ M , then
−1
1 = δx (x) = Iσ−1 (δx )(x) = (δx )σ (x) = δx (xσ ).
This shows that x and xσ are conjugate. As x is arbitrary, σ is a class-preserving
automorphism. Hence M = Autc (G) in this case.
Let L(B0 ) be a multiplicative group of linear characters in Irr(B0 ). We have seen
that for each linear character λ ∈ L(B0 ), there is a perfect isometry Iλ ∈ PI(B0 )
defined by Iλ (χ) = λχ for χ ∈ Irr(B0 ).
Lemma 4.2.3. The map λ 7→ Iλ is a group monomorphism from L(B0 ) into PI(B0 ).
Proof. Let λ, λ0 ∈ L(B0 ). For each χ ∈ Irr(B0 ) we have
Iλλ0 (χ) = λλ0 χ = λIλ0 (χ) = Iλ (Iλ0 (χ)) = (Iλ ◦ Iλ0 )(χ).
This shows that the map λ 7→ Iλ is a group homomorphism. Let λ be in the kernel.
Then Iλ (1) = 1 = λ · 1. So λ = 1.
e 0) ∼
e 0 ) and A(B
e 0 ) are
Let L(B
= L(B0 ) be the image of the map λ 7→ Iλ . Then L(B
subgroups of PI(B0 ). The next lemma describes their structure inside PI(B0 ).
e 0 ) o A(B
e 0 ) where all
Lemma 4.2.4. The group PI(B0 ) contains the subgroup L(B
e 0 ) o A(B
e 0 ) have all-positive signs.
perfect isometries in L(B
e 0 ) intersects A(B
e 0 ) trivially. Suppose Iλ = Iσ for
Proof. First we will show that L(B
some λ ∈ L(B0 ) and some σ ∈ Aut(G). Then λχ = χσ for all χ ∈ Irr(B0 ). Taking
e 0 ) ∩ A(B
e 0 ) = {id}. The group A(B
e 0 ) acts on L(B
e 0 ) by
χ = 1 we get λ = 1. So L(B
CHAPTER 4. PERFECT ISOMETRY GROUPS
54
e 0 ) and A(B
e 0 ). If
(Iλ )Iσ = IIσ (λ) . Let S ≤ PI(B0 ) be generated by elements in L(B
e 0 ) and Iσ , Iτ ∈ A(B
e 0 ), then for χ ∈ Irr(B0 ),
Iλ , Iλ0 ∈ L(B
(Iλ , Iσ ) ◦ (Iλ0 , Iτ ))(χ) = (Iλ , Iσ )(λ0 · χτ )
= λ · (λ0 )σ · χτ σ
= λ · Iσ (λ0 ) · Iσ (Iτ (χ))
= (Iλ ◦ IIσ (λ0 ) ◦ Iσ ◦ Iτ )(χ)
= (Iλ ◦ (Iλ0 )Iσ , Iσ ◦ Iτ )(χ).
e 0 ) o A(B
e 0 ). Finally, since λχ and χσ are irreducible characters for any
So S = L(B
linear character λ, any irreducible character χ and any automorphism σ of G. We
see that the perfect isometries Iλ and Iσ have all-positive signs.
4.3
RK (B ⊗O B ◦) as a ring
Let G be a finite group and B be a block of OG. We can regard RK (B ⊗O B ◦ ) as
a ring where multiplication is defined as follows. Define a map − ⊗B − : RK (B ⊗O
B ◦ ) × RK (B ⊗O B ◦ ) −→ RK (B ⊗O B ◦ ) by
(µ ⊗B λ)(g, h) =
1 X
µ(g, x−1 )λ(x, h)
|G| x∈G
(4.1)
for µ, λ ∈ RK (B ⊗O B ◦ ) and g, h ∈ G. It can be easily seen that ⊗B is a bilinear
map. Note also that the formula is induced by the tensor product − ⊗K⊗O B − of
bimodules.
The following lemma shows that perfectness are closed under ⊗B
Lemma 4.3.1. Let µ, λ ∈ RK (B ⊗O B ◦ ). If µ, λ are perfect, then µ ⊗B λ is also
perfect.
Proof. Let g, h ∈ G. Then
(µ ⊗B λ)(g, h)
1 X µ(g, x−1 )
=
λ(x, h)
|CG (g)|
|G| x∈G |CG (g)|
=
X µ(g, x−1 ) λ(xi , h)
i
|C
G (g)| |CG (xi )|
x ∈G
i
CHAPTER 4. PERFECT ISOMETRY GROUPS
55
where xi runs over all conjugacy class representatives of G. Since µ, λ are perfect,
µ(g, x−1
i )/|CG (g)| and λ(xi , h)/|CG (xi )| are in O for all xi . So (µ⊗B λ)(g, h)/|CG (g)| ∈
O. Similarly (µ⊗B λ)(g, h)/|CG (h)| ∈ O. So µ⊗B λ satisfies the integrality condition.
If g is p-singular and h is p-regular, then µ(g, x−1 ) = 0 for all p-regular elements x
while λ(x, h) = 0 for all p-singular element x. So (µ ⊗B λ)(g, h) = 0. Similarly, we
can show (µ ⊗B λ)(g, h) = 0 for p-regular g and p-singular h. Hence µ ⊗B λ is perfect.
The motivation for our definition of the product (4.1) is the following.
Lemma 4.3.2. Let µ, λ ∈ RK (B⊗O B ◦ ) and let Iµ , Iλ be the corresponding isometries.
Then Iµ ◦ Iλ = Iµ⊗B λ .
Proof. Let χ ∈ RK (B) and g ∈ G. Then
(Iµ ◦ Iλ )(χ)(g) = Iµ (Iλ (χ))(g)
1 X
µ(g, x−1 )Iλ (χ)(x)
=
|G| x∈G
1 X
1 X
=
µ(g, x−1 )
λ(x, y −1 )χ(y)
|G| x∈G
|G| y∈G
!
1 X 1 X
=
µ(g, x−1 )λ(x, y −1 ) χ(y)
|G| y∈G |G| x∈G
1 X
=
(µ ⊗B λ)(g, y −1 )χ(y)
|G| y∈G
= Iµ⊗B λ (χ)(g).
Corollary 4.3.3. With the multiplication ⊗B , RK (B ⊗O B ◦ ) is a ring with µid as
the identity element, where
µid (g, h) =
X
χ(g)χ(h),
χ∈Irr(B)
for g, h ∈ G.
Proof. Let µ, λ ∈ RK (B ⊗O B ◦ ). Since Iµ ◦ Iλ : RK (B) −→ RK (B) is a linear map, by
Lemma 3.1.4, Iµ ◦ Iλ = Iτ for some τ ∈ RK (B ⊗O B ◦ ). By Lemma 4.3.2, τ = µ ⊗B λ,
CHAPTER 4. PERFECT ISOMETRY GROUPS
56
so µ ⊗B λ ∈ RK (B ⊗O B ◦ ). Since µid induces the identity map, using Lemma 4.3.2
again, it is clear that µid is the identity of the ring RK (B ⊗O B ◦ ).
If µ ∈ RK (B ⊗O B ◦ ) is a perfect character, then by Theorem 3.2.4 the perfect
isometry Iµ induces linear maps Iµ◦ , Rµ◦ : Z(KB) −→ Z(KB) defined by
Iµ◦ (y)
=
X
g∈G
!
1 X
−1
µ(g , h)y(h) g,
|G| h∈G
and
!
X
1
Rµ◦ (x) =
µ(g, h−1 )x(g) h,
|G|
g∈G
h∈G
P
P
where y = h∈G y(h)h ∈ Z(KB) and x = g∈G x(g)g ∈ Z(KB).
X
If e is the block idempotent of B, then the map ρµ , defined by ρµ (y) = Iµ◦ (yRµ◦ (e))
gives an automorphism on Z(KB) which is also an automorphism on Z(B). Note
that if χ ∈ Irr(B), then ρµ (eχ ) = eIµ (χ) where e−χ = eχ .
Lemma 4.3.4. The map Ψ : Iµ 7→ ρµ is a group homomorphism from PI(B) to
AutO (Z(B)) with kernel h-idi.
Proof. Define Ψ̃ : PI(B) −→ AutK (Z(KB)) by Ψ̃(Iµ ) = Ψ(Iµ ). First we show that
Ψ̃ is a homomorphism. Let χ ∈ Irr(B) and let Iµ , Iλ ∈ PI(B). Then
Ψ̃(Iµ ◦ Iλ )(eχ ) = ρµ⊗B λ (eχ ) = eIµ⊗B λ (χ)
= e(Iµ ◦Iλ )(χ)
= ρµ (eIλ (χ) ) = ρµ (ρλ (eχ ))
= (Ψ̃(Iµ ) ◦ Ψ̃(Iλ ))(eχ ).
By linearity, this gives a group homomorphism. Since the image of Ψ̃ preserves Z(B),
Ψ is also a group homomorphism.
Let Iµ ∈ Ker(Ψ). For any x ∈ Z(KB) there is m ∈ O such that mx ∈ Z(B).
CHAPTER 4. PERFECT ISOMETRY GROUPS
57
Then
Ψ(Iµ )(mx) = mx
Ψ̃(Iµ )(mx) = mx
mΨ̃(Iµ )(x) = mx
Ψ̃(Iµ )(x) = x.
Thus Iµ ∈ Ker(Ψ̃), and so
eχ = Ψ̃(eχ ) = ρµ (eχ ) = eIµ (χ) ,
for all χ ∈ Irr(B). This implies that Iµ (χ) = ±χ for all χ ∈ Irr(B). By Proposition
3.2.2, we have Iµ = ±id.
We shall see later in the case of abelian p-groups that the map Ψ, as defined in
the previous lemma, becomes an isomorphism. Nevertheless, if we know the structure
of AutO (Z(B)), this gives a good upper bound for the size of the perfect isometry
group.
Corollary 4.3.5. Let B be a block of OG. Then
| PI(B)| ≤ 2| AutO (Z(B))|.
4.4
Connection to Picard groups and derived Picard groups
We have seen that the perfect isometry group is an invariant for perfectly isometric
blocks. In the same way, Picard group and Derived Picard group are invariants for
Morita equivalent and derived equivalent blocks respectively. In this section, we will
describe how these equivalences and their associated invariant groups can be related.
Let A, B be blocks of OH, OG for two finite groups H, G respectively. As we saw
in Section 2.5, if A and B are Morita equivalent then they are derived equivalent.
This in turn implies that they are perfectly isometric as well.
CHAPTER 4. PERFECT ISOMETRY GROUPS
58
Theorem 4.4.1. [8, Theorem 3.1] Suppose that A and B are derived equivalent.
Then they are perfectly isometric.
The converse does not hold in general.
Example 4.4.2. [29] Let p = 2. The blocks OD8 and OQ8 are perfectly isometric,
but they are not derived equivalent.
Let U be a bounded complex of A-lattices. Set
[K ⊗O U ] =
X
(−1)i [K ⊗O Ui ] ∈ RK (A),
i∈Z
where Ui is the component of U in degree i. Following the discussion after Lemma
3.4 in [27], if U, U 0 are two quasi-isomorphic complexes as above then [K ⊗O U ] =
[K⊗O U 0 ]. Now suppose that A, B are derived equivalent, induced by a tilting complex
X ∈ Db (B ⊗O A◦ ). Then the functor
X ⊗LA − : Db (A) −→ Db (B),
induces an isometry ΦX : RK (A) −→ RK (B) given by
ΦX ([K ⊗O U ]) = [K ⊗O (X ⊗LA U )].
(4.2)
The map (4.2) can be related to a perfect isometry in the following way. Let X
be a bounded complex of (B, A)-bimodules with components Xi in degree i. Suppose
each Xi is projective as a left B-module and a right A-module. Let µi be the character
of K ⊗O Xi . Define the generalize character µ of the complex K ⊗O X by
µ=
X
(−1)i µi ∈ RK (B ⊗O A◦ )
(4.3)
i∈Z
(where, as usual, we identify RK (B ⊗O A◦ ) with the group of generalized characters
of (B, A)-bimodules).
By Proposition 3.1.2, each µi is perfect. Since any O-linear combination of perfect characters is perfect, the generalized character µ is perfect. The map ΦX is then
equal to the map Iµ in (3.1) on the character level. If, furthermore, X ∈ Db (B ⊗O A◦ )
CHAPTER 4. PERFECT ISOMETRY GROUPS
59
is a tilting complex, then [8, Theorem 3.1] implies that ΦX is a perfect isometry.
Suppose now that A = B. If X ∈ DPic(B), then by the above remarks we have
ΦX ∈ PI(B).
Lemma 4.4.3. The map
DPic(B) −→ PI(B)
(X) 7→ ΦX .
is a group homomorphism. Hence, the restriction to Pic(B), considered as a subgroup of DPic(B) via the canonical injection Pic(B) −→ DPic(B), also gives a group
homomorphism
Pic(B) −→ PI(B)
(M ) 7→ ΦM .
Proof. We only need to prove the first homomorphism. Let X, Y ∈ DPic(B) and let
U be a bounded complex of A-lattices. Then
ΦX ◦ ΦY ([K ⊗O U ]) = ΦX ([K ⊗O (Y ⊗LB U )])
= [K ⊗O (X ⊗LB (Y ⊗LB U ))]
= [K ⊗O ((X ⊗LB Y ) ⊗LB U )]
= ΦX⊗LB Y ([K ⊗O U ]).
So (X) 7→ ΦX is a group homomorphism. Since ΦX only depends on [K ⊗O X], if
X, X 0 ∈ Db (B ⊗O B ◦ ) are isomorphic, then ΦX = ΦX 0 . As each component of X, Y
is projective on the right hand side, ΦX , ΦY preserve exact sequences. So the above
map is well defined.
We will refer to the homomorphisms in Lemma 4.4.3 as the canonical homomorphisms DPic(B), Pic(B) −→ PI(B). Actually, we can define the map Pic(B) −→
PI(B) directly, since all the complexes involved are concentrated in degree 0, by
(M ) 7→ ΦM : [K ⊗O U ] 7→ [K ⊗O (M ⊗B U )]
CHAPTER 4. PERFECT ISOMETRY GROUPS
60
for any B-lattices U . This gives a map which is compatible with the character
product defined in (4.1). More specifically, if M, N ∈ Pic(B) afford the K-characters
µ, λ respectively, then on the character level, ΦM ⊗B N = ΦM ◦ ΦN = Iµ ◦ Iλ = Iµ⊗B λ .
If X ∈ DPic(B), then it can be easily checked that
ΦX =
X
(−1)i ΦXi
i∈Z
where Xi is the component in degree i of X. So if each Xi affords a K-character µi
P
and X affords the generalized K-character µ = i∈Z (−1)i µi then, on the character
level,
Iµ =
X
(−1)i Iµi .
i∈Z
If M ∈ Pic(B), then the map ΦM is a perfect isometry that sends irreducible
characters to irreducible characters. The image of Pic(B) in PI(B) are therefore
perfect isometries with all-positive signs. This raises the following question: Given a
perfect isometry I ∈ PI(B) with all-positive sign, is it always induced by a Morita
equivalence? We will see later, when we study blocks with cyclic defect groups, that
the answer is no.
The images in PI(B) of some elements of Pic(B) can be described as follows. Let
M be any (B, B)-bimodule and let σ, π ∈ AutO (B). We define the bimodule σ Mπ to
have the same elements as M but with actions of G defined by σ and π:
g · m · h := (g σ )m(hπ ),
for all g, h ∈ G, m ∈ M . The following proposition is easily verified.
Proposition 4.4.4. [13, (55.10)] Let B be an O-algebra viewed as (B, B)-bimodule.
For each σ, σ 0 π, π 0 , ρ ∈ AutO B, there are bimodule isomorphisms
σ Bπ
∼
= ρσ Bρπ ∼
= 1 Bσ−1 π ∼
= π−1 σ B1 ,
σ 0 (σ Bπ )π 0
1 Bπ
∼
= σ0 σ Bππ0 ,
⊗B 1 Bπ0 ∼
= 1 Bππ0 .
CHAPTER 4. PERFECT ISOMETRY GROUPS
61
Lemma 4.4.5. The Picard group Pic(B) contains a subgroup isomorphic to OutO (B).
Proof. Define a map AutO (B) −→ Pic(B) by σ 7→ (Bσ ). By [13, (55.11)], this map
is a group homomorphism with kernel InnO (B).
The next lemma further explains Lemma 4.2.4.
^
Lemma 4.4.6. Let B be the principal block of OG. Denote by Out
O (B) the image
of OutO (B) (considered as a subgroup of Pic(B)) in PI(B). Then, with the notations
in Lemma 4.2.4,
^
e
e
L(B)
o A(B)
≤ Out
O (B).
Proof. Let Φ : Pic(B) −→ PI(B) be the canonical homomorphism. Let σ ∈ AutO (B)
and U be a B-lattice. Then
ΦBσ ([K ⊗O U ]) = [K ⊗O (Bσ ⊗B U )] = [K ⊗O (σ−1 B ⊗B U )] = [K ⊗O σ−1 U ].
So if U affords a K-character χ then σ−1 U affords the K-character χσ (where (χσ (g) =
−1
χ(g σ ), g ∈ G). So the the perfect isometry induced by Bσ is χ 7→ χσ .
e
Since B is the principal block, we have Aut(G) ≤ AutO (B). So A(B)
is the
collection of perfect isometries of the form Iσ : χ 7→ χσ where σ ∈ Aut(G) ≤ AutO (B).
^
e
But σ ∈ Aut(G) is in InnO (B0 ) if and only if σ ∈ Inn(G). Thus A(B)
≤ Out
O (B).
Let λ be a linear character in Irr(B). Define an element σ ∈ AutO (B) by g σ =
λ(g −1 )g, ∀g ∈ G and extend linearly. Clearly, σ ∈ InnO (B) if and only if λ is the
−1
trivial character. Then Bσ induces the map χ 7→ χσ where χσ (g) = χ(g σ ) =
χ(λ(g)g) = λ(g)χ(g). This is the same as the map Iλ : χ 7→ λχ (cf. Lemma 4.2.3).
^
e
So L(B)
≤ Out
O (B).
We can also consider the canonical homomorphism DPic(B) −→ PI(B). It is
obvious that the homomorphism is never injective as B and B[2] give the identity
isometry. Whether or not it is surjective is, however, a more interesting question. A
perfect isometry induced by a tilting complex does not have to have a homogenous
sign. Therefore, one could be tempted to try to construct a tilting complex inducing
a given perfect isometry. We will show that it is always possible to do this in the case
CHAPTER 4. PERFECT ISOMETRY GROUPS
62
of blocks with defect group Cp . However, when B = OG where G = p1+2 is an extra
special group of order p3 , we will see (Example 5.2.3) that there is a perfect isometry
I : RK (B) −→ RK (B) that cannot be given by any derived equivalence.
4.5
Factor blocks
This section contains results that will be used later. It allows us, under certain
conditions, to work with the quotient group G/N instead of G. Let N be a normal
subgroup of a finite group G whose order is coprime to p. Let G = G/N . Let π :
G −→ G be the quotient map and let πO : OG −→ OG be the algebra homomorphism
induced by π.
Let B be a block of OG. The kernel of B is defined by
Ker B = {g ∈ G : χ(g) = χ(1), ∀χ ∈ Irr(B)}.
If N ≤ Ker B then by [17, Chapter V, Lemma 4.3], B = πO (B) is also a block of OG.
The map B 7→ B is a bijection between the set of blocks B of OG with N ≤ Ker B
and the set of blocks B of OG. The characters χ in Irr(B) are inflation of the
characters χ in Irr(B) under π, and the map χ 7→ χ is a bijection Irr(B) −→ Irr(B).
Lemma 4.5.1. Let N be a normal subgroup of G whose order is coprime to p. Let
B ∈ Bl(G). Suppose that N ≤ Ker B. Let I : RK (B) −→ RK (B) be defined by χ 7→ χ
for χ ∈ Irr(B) and extend linearly. Then I is a perfect isometry. Consequently,
PI(B) ∼
= PI(B).
Proof. We have µI ∈ RK (B ⊗O (B)◦ ) where
µI (g, h) =
X
I(χ)(g)χ(h) =
X
χ(g)χ(h)
χ∈Irr(B)
=
X
χ(g)χ(h)
= µid (g, h).
Since N is a p0 -group, gN is a p-regular element if and only if g is a p-regular element. Thus µI satisfies the separation condition. Since µid is perfect, we have
CHAPTER 4. PERFECT ISOMETRY GROUPS
63
µid (g, h)/(|CG (g)||CN (g)|) ∈ O as 1/|CN (g)| ∈ O. Since |CG (g)| divides |CG (g)||CN (g)|,
µid (g, h)/|CG (g)| ∈ O. Also µid (g, h)/|CG (h)| ∈ O. So µI satisfies the integrality condition. The last statement follows from Proposition 4.0.7.
Corollary 4.5.2. Suppose G contains a normal p0 -subgroup N C G such that G/N
is a p-group. Let B ∈ Bl(G) and suppose that N ≤ Ker B. Then
PI(B) ∼
= PI(O(G/N )).
Proof. Since G/N is a p-group, it has only one block, the principal block. So the
corresponding block B of G/N is this block. The result now follows from the previous
lemma.
4.6
Isometries extended from a normal subgroup
Let G be a finite group and N a normal subgroup of G. Let B, A be blocks of OG
and ON respectively, where B covers A. In this section we will investigate isometries
RK (B) −→ RK (B) that, in some sense, extend perfect isometries RK (A) −→ RK (A).
Under suitable conditions, this allows us to construct perfect isometries in PI(B) from
known perfect isometries in PI(A).
4.6.1
Direct Product
Let G = N × P where P is a p-group. Let B, A be the principal blocks of OG and
ON respectively. Given perfect isometries I¯ ∈ PI(A) and J ∈ PI(OP ), we will show
how to construct a perfect isometry I ∈ PI(B) from I¯ and J.
Let θ ∈ Irr(A). Then θ extends to an irreducible character E(θ) ∈ Irr(B). Since
irreducible characters of G are of the form ϕχ where ϕ ∈ Irr(N ) and χ ∈ Irr(P ), we
can take E(θ) = θ1. Then by [24, Corollary 6.17],
{χ ∈ Irr(B) : χ ↓N = θ} = {W E(θ) : W ∈ Irr(G/N )},
where W E(θ)(g) = W (gN )E(θ)(g), g ∈ G. So,
Irr(B) = {W E(θ) : W ∈ Irr(G/N ), θ ∈ Irr(A)}.
CHAPTER 4. PERFECT ISOMETRY GROUPS
64
Theorem 4.6.1. Let I¯ ∈ PI(A) and J ∈ PI(O(G/N )). Define an isometry I :
RK (B) −→ RK (B) by
I : W E(θ) 7→ ε(θ)J(W )E(σ(θ)),
W ∈ Irr(G/N ), θ ∈ Irr(A),
(4.4)
¯
where I(θ)
= ε(θ)σ(θ) for a sign function ε and a bijection σ on Irr(A). Then
¯ ↑G for
I ∈ PI(B). Furthermore, if J : Irr(G/N ) −→ Irr(G/N ) then I(θ ↑G ) = I(θ)
all θ ∈ Irr(A).
Proof. Let µ ∈ RK (B ⊗ B ◦ ) be the character induced by I. For g, h ∈ G,
µ(g, h) =
XX
=
I(W E(θ))(g)W E(θ)(h)
W
θ
XX
ε(θ)J(W )(gN )E(σ(θ))(g)W (hN )E(θ)(h)
W
θ
"
X
=
ε(θ)E(σ(θ))(g)E(θ)(h)
#"
X
#
J(W )(gN )W (hN ) .
W
θ
For convenience, we will denote
"
#
X
ε(θ)E(σ(θ))(g)E(θ)(h)
θ
by [
P
θ]
and
#
"
X
J(W )(gN )W (hN )
W
P
P
P
by [ W ]. Suppose µ(g, h) 6= 0 then [ θ ] 6= 0 and [ W ] 6= 0. We will show that g, h
are both p-singular or both p-regular elements and hence satisfying the separation
condition.
Since J is perfect, [
P
W]
6= 0 implies that gN, hN are both p-regular elements or
both p-singular elements of G/N . If gN, hN are both p-singular elements of G/N ,
then g, h 6∈ N as G/N is a p-group. Since G = N × P , this implies that g, h are both
p-singular elements of G.
On the other hand if gN, hN are both p-regular elements of G/N , then g, h ∈ N .
In this case
X
ε(θ)E(σ(θ))(g)E(θ)(h) =
X
=
X
θ
ε(θ)σ(θ)(g)θ(h)
θ
θ
¯
I(θ)(g)θ(h).
CHAPTER 4. PERFECT ISOMETRY GROUPS
65
P
Since I¯ is perfect and [ θ ] 6= 0 this implies that g, h are both p-singular elements or
both p-regular elements of G.
To prove the integrality condition, let g = n1 a, h = n2 b where n1 , n2 ∈ N and
a, b ∈ P . By our choice of E, E(σ(θ))(g) = σ(θ)(n1 ), E(θ)(h) = θ(n2 ). So
"
#"
#
X
X
µ(g, h) =
ε(θ)σ(θ)(n1 )θ(n2 )
J(W )(aN )W (bN )
W
#"
#
X
X
¯
=
I(θ)(n
J(W )(aN )W (bN ) .
1 )θ(n2 )
θ
"
W
θ
For convenience, denote
"
#
X
¯
I(θ)(n
1 )θ(n2 )
θ
by [
P
θ]
and
"
#
X
J(W )(aN )W (bN )
W
P
P
by [ W ]. Since I¯ is perfect, [ θ ] is in |CN (n1 )|O and |CN (n2 )|O. Since J is perfect,
P
[ W ] is in |CG/N (aN )|O and |CG/N (bN )|O. But
|CG (n1 a)| = |CN (n1 )| × |CP (a)|
= |CN (n1 )| × |CG/N (aN )|.
Therefore µ(g, h) ∈ |CG (g)|O. Similarly µ(g, h) ∈ |CG (h)|O. This proves the integrality condition.
Finally, suppose that J : Irr(G/N ) −→ Irr(G/N ). Then, for θ ∈ Irr(A),


X
I((θ) ↑G ) = I 
W E(θ)
W ∈Irr(G/N )
=
X
ε(θ)J(W )E(σ(θ))
W ∈Irr(G/N )
= ε(θ)
X
W ∈Irr(G/N )
= ε(θ)σ(θ) ↑G
¯ ↑G .
= I(θ)
W E(σ(θ))
CHAPTER 4. PERFECT ISOMETRY GROUPS
4.6.2
66
Isometries commuting with induction
Let G be a finite group and N a normal subgroup of G. Let B be a block of
OG covering a block A of ON . In Theorem 4.6.1 we have seen a perfect isometry
e ↑G , ∀θ ∈ Irr(A), where Ie ∈ PI(A). Therefore,
I ∈ PI(B) such that I(θ ↑G ) = I(θ)
given a perfect isometry Ie ∈ PI(A), one could try to construct a perfect isometry
I ∈ PI(B) making the following diagram commutative.
RK (B)
O
I
/
Induction
RK (B)
O
RK (A)
(4.5)
Induction
/
Ie
RK (A)
In this section we will look at isometries RK (B) −→ RK (B) with the above
property and show that, under suitable conditions, such isometries already satisfy
the separation condition.
Remark 4.6.2. If we take Ie = id ∈ PI(A), then the set of perfect isometries I ∈
PI(B) making the diagram (4.5) commutative forms a subgroup of PI(B). These are
perfect isometries preserving inductions from N , that is I(θ ↑G ) = θ ↑G , ∀θ ∈ Irr(A).
Lemma 4.6.3. Let N C G. Let B be a block of OG covering a block A of ON .
If α ∈ CF(N, A; K) vanishes on p-singular elements. Then α ↑G ∈ CF(G, B; K)
vanishes on p-singular elements.
Proof. First note that since B covers A, α ↑G ∈ CF(G, B; K). Let g ∈ G be a
p-singular element. If g 6∈ N then α ↑G (g) = 0 by definition. If g ∈ N then
P
α ↑G (g) = |N1 | x∈G α(x−1 gx). But x−1 gx is a p-singular element for all x ∈ G.
Thus α ↑G (g) = 0, proving the lemma.
For an irreducible Brauer character ϕ of G, let Φϕ denote the projective indecomposable character of G associated to ϕ.
Proposition 4.6.4. Let N C G. Let B be a block of OG covering a block A of ON .
Assume that for all ϕ ∈ IBr(B) there exists ϕ
e ∈ IBr(A) such that Φϕ = (Φϕe) ↑G .
Let Ie ∈ PI(A) and I : RK (B) −→ RK (B) be an isometry making the diagram (4.5)
commutative. Then I satisfies the separation condition.
CHAPTER 4. PERFECT ISOMETRY GROUPS
67
Proof. Since {Φϕ : ϕ ∈ IBr(B)} is a K-basis of CFp0 (G, B; K), it suffices to show
that I(Φϕ ) ∈ CFp0 (G, B; K), ∀ϕ ∈ IBr(B). Let ϕ ∈ IBr(B) then, by the assumption,
∃ϕ
e ∈ IBr(A) such that Φϕ = (Φϕe) ↑G . If {dχeϕe : χ
e ∈ Irr(A), ϕ
e ∈ IBr(A)} are
decomposition numbers, then


X
I(Φϕ ) = I (Φϕe) ↑G = I 
dχeϕeχ
e ↑G 
χ
e∈Irr(A)
=
X
dχeϕeI(e
χ ↑G ) =
χ
e
e χ ) ↑G
dχeϕeI(e
χ
e

= Ie 
X

X
dχeϕeχ
e ↑G
χ
e
= Ie (Φϕe) ↑G .
e ϕe) ∈ CFp0 (N, A; K). Thus
Since Ie is perfect and Φϕe ∈ CFp0 (N, A; K), we have I(Φ
e ϕe) ↑G ∈ CFp0 (G, B; K). Hence I satisfies the separation
by the previous lemma, I(Φ
condition.
e and
Example 4.6.5. Let G = Cp2 and N = Cp C G. In this case, IBr(A) = {1}
IBr(B) = {1} and Φ1 = (Φ1e ) ↑G . Suppose G = hgi and N = hg p i. Let λ ∈ Irr(G)
where λ(g) = e2πi/p . Then λ(h) = 1, ∀h ∈ N . So the isometry I : χ 7→ λχ, ∀χ ∈
Irr(G) satisfies I(θ ↑G ) = θ ↑G , ∀θ ∈ Irr(N ). So, by Proposition 4.6.4, I satisfies the
separation condition. (In fact we already know from Lemma 3.3.5 that I is a perfect
isometry.)
Another situation where we can apply Proposition 4.6.4 is the following [22].
Example 4.6.6. Let N be a normal subgroup of a finite group G such that G/N is
a p-group. Let A be a G-invariant block of ON . Let B be a block of OG covering
A. By [22, Lemma 2.1], B is unique.
• Suppose that the map IBr(B) −→ IBr(A) defined by ϕ 7→ ϕ ↓N is a bijection.
Then
Φϕ = (Φϕe) ↑G ,
where ϕ
e = ϕ ↓N [22, Lemma 2.4].
CHAPTER 4. PERFECT ISOMETRY GROUPS
68
• Suppose that | IBr(A)| < p. Then the map IBr(B) −→ IBr(A) defined by
ϕ 7→ ϕ ↓N is a bijection [22, Lemma 2.2].
Chapter 5
p-groups
In this chapter we will study perfect isometries in blocks of p-groups. An interesting
feature of perfect isometries in these group is that they always have homogenous signs.
It therefore follows from Lemma 4.1.6 that we can decompose the perfect isometry
group as the product of subgroup of perfect isometries with all-positive signs and the
subgroup generated by −id.
Lemma 5.0.7. Let G be a p-group and B = OG. Let I ∈ PI(B) be a perfect isometry.
Then
(i) I has homogenous sign.
(ii) Either I(χ)(1) = χ(1) ∀χ ∈ Irr(G) or I(χ)(1) = −χ(1) ∀χ ∈ Irr(G).
Proof. Suppose |G| = pn . If χ ∈ Irr(B), we know that
I(χ)(1)
χ(1)
is an invertible element
in O, by Lemma 3.2.5. Since both I(χ)(1) and χ(1) are powers of p, we must have
I(χ)(1) = ±χ(1). Let µI ∈ RK (B ⊗O B ◦ ) be the generalized character induced by I.
Consider
µI (1, 1)
=
|G|
P
Since µI (1, 1)/|G| ∈ O and |
χ∈Irr(G)
I(χ)(1)χ(1)
pn
2
χ∈Irr(G) (±χ(1) )|
P
2
χ∈Irr(G) (±χ(1) )
.
pn
P
=
≤ pn this means that either
2
χ∈Irr(G) (±χ(1) )
= pn in which case all the signs are positive, or
2
χ∈Irr(G) (±χ(1) )
= −pn in which case all the signs are negative.
•
P
•
P
This proves (i), and (ii) follows from I(χ)(1) = ±χ(1) by above.
69
CHAPTER 5. P -GROUPS
5.1
70
Abelian p-groups
Since the most basic p-groups are the abelian ones, it is natural to start with these
groups. In this section, we will find perfect isometry groups for abelian p-groups.
Then we will deduce a result for the general abelian groups, using Corollary 4.5.2.
Let G be an abelian p-group. We know from Lemma 4.3.4 that every perfect isometry I ∈ PI(OG) gives rise to an, essentially unique, automorphism in AutO (ZOG).
We shall see that the converse also holds in the abelian case. This enables us to prove
the following main result.
Theorem 5.1.1. Let G be an abelian p-group. Then every perfect isometry has a
homogenous sign and
PI(OG) ∼
= (G o Aut(G)) × h-idi .
Furthermore, if Irr(G) = {χ1 , . . . , χ|G| }, then G ≤ PI(OG) is the subgroup of perfect
isometries χi 7→ λχi , ∀i for some λ ∈ Irr(G) and Aut(G) ≤ PI(OG) is the subgroup
of perfect isometries χi 7→ χσ , ∀i for some σ ∈ Aut(G).
Lemma 5.1.2. Let G be a p-group. Then
OutO OG ∼
= Hom(G, O× ) o Out(G).
Proof. We follow the remarks before Corollary 4 in [41]. There is an injective homoe : g 7→ λ(g)g). An image is
morphism Hom(G, O× ) −→ AutO (OG) given by λ 7→ (λ
inner if and only if it is trivial. Let n OutO (OG) be the image in OutO (OG) of the
normalized (augmentation-preserving) automorphisms of OG. Then we can write
OutO (OG) ∼
= Hom(G, O× ) · n OutO (OG).
Since G is a p-group, no prime divisor of |G| is invertible in O. Since G is a p-group,
it is well-known that G is nilpotent. So, [41, Corollary 3] gives
n OutO (OG) = (Outcent(OG))(Out(G)),
CHAPTER 5. P -GROUPS
71
where Outcent(OG) = OutZ(OG) (OG), regarding OG as an algebra over its center
Z(OG).
To find Outcent OG, we first define Outcent(G) = (Outcent(OG)) ∩ Out(G).
Then [41, Corollary 4] implies that Outcent(OG) = Outcent(G). By definition of
Outcent(G), this means that Outcent(OG) = (Outcent(OG)) ∩ Out(G). This implies
that Outcent(OG) ⊆ Out(G), and so n OutO (OG) = Out(G).
−1
Finally, the group Out(G) acts on Hom(G, O× ) by λσ (g) = λ(g σ ) for λ ∈
gσ . With this action, we have
e σ = (λ)
Hom(G, O× ), σ ∈ Out(G), g ∈ G, and so (λ)
OutO OG ∼
= Hom(G, O× ) o (Out(G)) as claimed.
Corollary 5.1.3. Let G be an abelian p-group. Then
AutO (ZOG) ∼
= G o Aut(G).
Proof. When G is abelian, we have the following:
• AutO (ZOG) = AutO (OG),
• AutO (OG) = OutO (OG),
• Hom(G, O× ) ∼
= G,
• Out(G) = Aut(G).
Applying the previous lemma gives the result.
Proof of Theorem 5.1.1. Because G is a p-group, all perfect isometries have homogenous signs. So we can write
PI(OG) = S × h-idi,
where S is a subgroup consisting of all perfect isometries with all-positive signs. Let
L be the multiplicative group of Irr(G). Let Le and Ae be the images of the maps
L −→ PI(OG) and Aut(G) −→ PI(OG) respectively (see Section 4.2). Then, by
e It is clear that we have Le ∼
Lemma 4.2.4, S contains Le o A.
= G. Since G is an
abelian p-group, by Remark 4.2.2, we have Ae ∼
= Aut(G). Hence, by Lemma 4.2.4,
PI(OG) ≥ (G o Aut(G)) × h-idi .
CHAPTER 5. P -GROUPS
72
On the other hand, Corollary 4.3.5 together with Corollary 5.1.3 imply that
| PI(OG)| ≤ 2|G|| Aut(G)|.
Hence we have PI(OG) ∼
= (GoAut(G))×h-idi. The subgroups of PI(OG) isomorphic
to G and Aut(G) are already described in Section 4.2.
The perfect isometry group for a general abelian group can be easily deduced.
Theorem 5.1.4. Let G be a finite abelian group. Let B ∈ Bl(G). Then
PI(B) ∼
= (Op (G) o Aut(Op (G)) × h-idi .
Proof. It suffices to assume that B is the principal block. Since G is abelian, by the
fundamental theorem of finite abelian groups we can write
G = Op (G) × H,
where H is an abelian group whose order is coprime to p. By Corollary 4.5.2, we
have PI(B) ∼
= PI(O(G/H)) and so, by Theorem 5.1.1,
PI(B) ∼
= (G/H o Aut(G/H)) × h-idi .
We will finish this section with the following useful lemma.
Lemma 5.1.5. Let G be an abelian p-group. Let I : RK (OG) −→ RK (OG) be an
isometry with a homogenous sign. If I = Iµ then µ satisfies the separation condition.
Hence, in order to check if µ is perfect, we only need to check µ(g, h) where g, h are
non-identity.
Proof. Let 1 6= g ∈ G. Since χ(1) = 1 for all χ ∈ Irr(G),
µI (1, g) =
X
I(χ)(1)χ(g) = (±1)
χ∈Irr(G)
µI (g, 1) =
X
X
χ(g) = 0
χ∈Irr(G)
I(χ)(g)χ(1) = (±1)
χ∈Irr(G)
X
χ(g) = 0.
χ∈Irr(G)
Since 1 is the only p-regular element of G, µ satisfies the separation condition. The
last statement follows from Theorem 3.1.5.
CHAPTER 5. P -GROUPS
5.2
73
Extra special p-groups
In this section, we will find perfect isometry groups of extra special p-groups. Recall
that a p-group G is called extra special if the center Z(G) is cyclic of order p and the
quotient G/Z is a non-trivial elementary abelian p-group. The main result for this
section is the following theorem.
Theorem 5.2.1. Let G be an extra special p-group of order p1+2n . Let Z = Z(G) be
the center of G. Then we can write
Irr(G) = {λ1 , . . . , λp2n } ∪ {χ1 , . . . , χp−1 }
where {λ1 , . . . , λp2n } can be identified with Irr(G/Z) and {χ1 , . . . , χp−1 } can be identified with Irr(Z) − {1}. With these identifications, every perfect isometry Irr(G) −→
Irr(G) is of the form I¯ ◦ J¯ where I¯ ∈ PI(O(G/Z)) permutes only the characters
{λ1 , . . . , λp2n } in Irr(G) and J¯ ∈ PI(OZ) fixes 1 ∈ Irr(Z) and permutes only the
characters {χ1 , . . . , χp−1 } in Irr(G). In particular,
PI(OG) ∼
= (G/Z o Aut(G/Z)) × Cp−1 × h-idi .
5.2.1
Irreducible characters of G
First we summarize results about the irreducible characters of G. The details of the
proofs can be found in [21, Lemma 3]. Let Z = Z(G), G = G/Z and g 7→ g = gZ
be the canonical projection map. Let λ̄1 , . . . , λ̄p2n be the irreducible characters of Ḡ.
Then there are irreducible characters λ1 , . . . , λp2n of G such that, for each i,
λi (g) = λi (ḡ),
for all g ∈ G. Let χ̂1 , . . . , χ̂p−1 be the non-trivial irreducible characters of Z. For
each i, we extend χ̂i to G by setting χ̂i (g) = 0 if g 6∈ Z. Then there are irreducible
characters χ1 . . . , χp−1 of G such that, for each i,
χi (g) = pn χ̂i (g),
CHAPTER 5. P -GROUPS
74
for all g ∈ G. We then have
Irr(G) = {λ1 , . . . , λp2n } ∪ {χ1 . . . , χp−1 },
where {λ1 , . . . , λp2n } can be identified with Irr(G/Z) and {χ1 . . . , χp−1 } can be identified with Irr(Z) − {1}.
Table 5.1: Character table of G
G
{1} g ∈ Z − {1} g ∈ G − Z
λ1
1
1
1
..
..
..
.
.
.
5.2.2
λi
..
.
1
..
.
1
..
.
λp2n
χ1
..
.
1
pn
..
.
1
χj
..
.
pn
..
.
pn χ̂j (g)
χp−1
pn
λ̄i (g)
0
..
.
0
..
.
0
The proof
We will now prove Theorem 5.2.1. Since G is a p-group, all perfect isometries have
homogenous signs. It is sufficient to prove that the subgroup of PI(OG) generated
by perfect isometries with all-positive signs is isomorphic to (G/Z o Aut(G/Z)) ×
Cp−1 . We denote generic characters in {λ1 , . . . , λp2n } and {χ1 . . . , χp−1 } by λ and χ
respectively and λ̄, χ̂ the corresponding characters in G/Z and Z.
Let I¯ ∈ PI(O(G/Z)) be a perfect isometry with all-positive sign and Jˆ ∈ PI(OZ)
be a perfect isometry with all-positive sign such that Jˆ fixes the trivial character of
Z. Then we can define an isometry I : Irr(G) −→ Irr(G) as follows.
for all g ∈ G.
¯ λ̄)(ḡ)
I(λ)(g) = I(
(5.1a)
ˆ χ̂)(g)
I(χ)(g) = pn J(
(5.1b)
CHAPTER 5. P -GROUPS
75
On the other hand, since λ(1)p = 1 and χ(1)p = pn , every perfect isometry in
PI(OG) must preserve the sets {λ1 , . . . , λp2n } and {χ1 . . . , χp−1 } up to signs. Let
I ∈ PI(OG) be a perfect isometry with all-positive sign (I : Irr(G) −→ Irr(G)),
then it can be written as in (5.1) for some isometries I¯ : Irr(G/Z) −→ Irr(G/Z) and
Jˆ : Irr(Z) −→ Irr(Z) where Jˆ fixes the trivial character.
Since G/Z is an abelian p-group and Z ∼
= Cp , we know from Theorem 5.1.1 that
the group of all-positive-sign perfect isometries I¯ ∈ PI(O(G/Z)) is G/Z o Aut(G/Z)
and that the subgroup of perfect isometries Jˆ ∈ PI(OZ) fixing the trivial character
is Aut(Z) ∼
= Cp−1 . It remains to show that I is a perfect isometry if and only if I¯
and Jˆ are perfect isometries.
In order to simplify the proof we first make some observations and pre-calculations.
Let g ∈ G. The values of |CG (g)|, |CḠ (ḡ)| and |CZ (g)| can be summarized in the
following table.
g∈G
|CG (g)|
|CḠ (ḡ)|
|CZ (g)|
g=1
p2n+1
p2n
p
g ∈ Z − {1}
p2n+1
p2n
p
p2n
p2n
p
g 6∈ Z
Note that if g 6∈ Z then, by the second orthogonality of characters,
|CG (g)| =
X
|φ(g)|2
φ∈Irr(G)
=
X
|λ(g)|2 +
X
λ
χ
X
|λ(g)|2 =
X
=
λ
|χ(g)|2
|λ̄(ḡ)|2
λ̄
= |CḠ (ḡ)| = p2n .
Let I be an isometry as in (5.1) and µI ∈ RK (OG ⊗O OG◦ ) be the corresponding
CHAPTER 5. P -GROUPS
76
character. If g, h ∈ Z, then
µI (g, h) =
X
=
X
I(λ)(g)λ(h) +
X
I(χ)(g)χ(h)
χ
λ
¯ λ̄)(ḡ)λ̄(h̄) + p2n
I(
X
ˆ χ̂)(g)χ̂(h)
J(
χ̂6=1
λ̄
= µI¯(ḡ, h̄) + p2n (µJˆ(g, h) − 1),
whereas if g 6∈ Z or h 6∈ Z or both then
µI (g, h) = µI¯(ḡ, h̄).
If g = h = 1 then
µJˆ(g, h) =
X
ˆ χ̂)(1)χ̂(1) = p.
J(
χ̂
If g = 1, h ∈ Z − {1} or g ∈ Z − {1}, h = 1 then
µJˆ(g, h) =
X
ˆ χ̂)(g)χ̂(g) = 0.
J(
χ̂
If ḡ = h̄ = 1 in Ḡ then
µI¯(ḡ, h̄) =
X
¯ λ̄)(1)λ̄(1) = p2n .
I(
λ̄
If ḡ = 1, h̄ 6= 1 in Ḡ or ḡ 6= 1, h̄ = 1 in Ḡ, then
µI¯(ḡ, h̄) =
X
ˆ χ̂)(ḡ)χ̂(h̄) = 0.
J(
χ̂
The following table summarizes computable values of µI¯(ḡ, h̄), µJˆ(g, h) and µI (g, h).
¯ Jˆ and I (as long as they take the
These values do not depend on the isometries I,
form as in (5.1)).
CHAPTER 5. P -GROUPS
77
g∈G
h∈G
g=1
h=1
p2n
p
p2n+1
µI¯ + p2n (µJˆ − 1)
g=1
h ∈ Z − {1}
p2n
0
0
µI¯ + p2n (µJˆ − 1)
g=1
h 6∈ Z
0
µI¯
0
µI¯ + p2n (µJˆ − 1)
µI¯(ḡ, h̄) µJˆ(g, h) µI (g, h)
0
g ∈ Z − {1} h = 1
p2n
g ∈ Z − {1} h ∈ Z − {1}
p2n
0
µI =
µI¯ + p2n (µJˆ − 1)
g ∈ Z − {1} h 6∈ Z
0
0
µI¯
g 6∈ Z
h=1
0
0
µI¯
g 6∈ Z
h ∈ Z − {1}
0
0
µI¯
g 6∈ Z
h 6∈ Z
µI¯
Proof that if I is perfect then I¯ is perfect
Since Ḡ is an abelian p-group and I¯ has the all-positive sign, by Lemma 5.1.5, we
only need to consider µI¯ for p-singular elements of Ḡ.
Let ḡ, h̄ be p-singular elements of Ḡ. Then g, h 6∈ Z and so g, h are p-singular
elements in G. Thus µI¯(ḡ, h̄) = µI (g, h) ∈ p2n O. Since |CḠ (ḡ)| = |CḠ (h̄)| = p2n , this
satisfies the integrality condition.
Hence, we conclude that I¯ is a perfect isometry.
Proof that if I is perfect then Jˆ is perfect
Since Z is an abelian p-group and Jˆ has the all-positive sign, by Lemma 5.1.5, we
only need to consider µJˆ for p-singular elements of Z.
Let g, h be p-singular elements of Z. Then g, h 6= 1. But ḡ, h̄ = 1 in Ḡ. So
µI¯(ḡ, h̄) = p2n , and hence µI (g, h) = µI¯(ḡ, h̄) + p2n (µJˆ(g, h) − 1) = p2n µJˆ(g, h). But
µI (g, h) ∈ p2n+1 O, so µJˆ(g, h) ∈ pO. Since |CZ (g)| = |CZ (h)| = p, this satisfies the
integrality condition.
Hence, we conclude that Jˆ is a perfect isometry.
CHAPTER 5. P -GROUPS
78
¯ Jˆ are perfect then I is perfect
Proof that if I,
Let g, h be p-regular elements of G. Then g = h = 1 and ḡ = h̄ = 1 in Ḡ. So
µI¯(ḡ, h̄) = p2n and µJˆ(g, h) = p. Hence µI (g, h) = µI¯(ḡ, h̄) + p2n (µJˆ(g, h) − 1) =
p2n +p2n (p−1) = p2n+1 . Since |CG (g)| = |CG (h)| = p2n+1 , this satisfies the integrality
condition.
Let g be a p-regular element, h be a p-singular element of G. There are two cases
to consider.
1. Suppose g = 1, h ∈ Z − {1}. Then µJˆ(g, h) = 0. Also ḡ = h̄ = 1 in Ḡ. Thus
µI¯(ḡ, h̄) = p2n . Hence µI (g, h) = µI¯(ḡ, h̄)+p2n (µJˆ(g, h)−1) = p2n +p2n (0−1) =
0. This satisfies the separation condition.
2. Suppose g = 1, h 6∈ Z. Then ḡ = 1, h̄ 6= 1 in Ḡ and so µI¯(ḡ, h̄) = 0. Hence
µI (g, h) = µI¯(ḡ, h̄) = 0. This satisfies the separation condition.
Let g be a p-singular element, h be a p-regular element of G. Then the same
argument as in the previous paragraph shows that µI (g, h) = 0, which satisfies the
separation condition.
Let g, h be p-singular elements of G. Then g, h 6= 1. There are four cases to
consider.
1. Suppose g, h ∈ Z − {1} then ḡ = h̄ = 1 in Ḡ. So µI¯(ḡ, h̄) = p2n . Since Jˆ is
perfect, we also have µJˆ(g, h) ∈ pO and so µI (g, h) = µI¯(ḡ, h̄) + p2n (µJˆ(g, h) −
1) = p2n µJˆ(g, h) ∈ p2n+1 O. This satisfies the integrality condition.
2. Suppose g ∈ Z − {1}, h 6∈ Z then ḡ = 1, h̄ 6= 1 in Ḡ. So µI¯(ḡ, h̄) = 0. Hence
µI (g, h) = µI¯(ḡ, h̄) = 0. This satisfies the integrality condition.
3. Suppose g 6∈ Z, h ∈ Z − {1} then the similar argument as in case 2 shows that
µI (g, h) = 0, satisfying the integrality condition.
4. Suppose g, h 6∈ Z. Then ḡ, h̄ 6= 1 in Ḡ. Since I¯ is perfect, µI¯(ḡ, h̄) ∈ p2n O.
Hence µI (g, h) = µI¯(ḡ, h̄) ∈ p2n O. Since |CG (g)| = |CG (h)| = p2n , this satisfies
the integrality condition.
CHAPTER 5. P -GROUPS
79
We have now considered all cases, thus I is a perfect isometry as claimed. This
proves Theorem 5.2.1.
5.2.3
Picard group and derived Picard group
We will now determine Pic(OG) and DPic(OG) and show that the homomorphism
DPic(OG) −→ PI(OG) is not surjective when G = p1+2 .
The following theorem gives us the structures of Pic(B) and DPic(B) in the case
where B is local. Note that it applies over O also by [53, Remark 1.12].
Theorem 5.2.2. Let G be a p-group. Then
Pic(OG) ∼
= OutO (OG)
DPic(OG) ∼
= Pic(OG) × Sh(OG),
where Sh(OG) is the subgroup generated by OG[1].
Proof. This follows from [53, Proposition 3.4] since the block OG is local by [12,
(5.25)].
Recall that, by Lemma 5.1.2, we have OutO (OG) ∼
= Hom(G, O× ) o Out(G).
Let L(G) be the group of linear characters in Irr(G). Then it is easy to see that
Hom(G, O× ) ∼
= L(G) ∼
= G/Z. Thus, by Theorem 5.2.2,
Pic(OG) ∼
= G/Z o Out(G)
DPic(OG) ∼
= (G/Z o Out(G)) × Sh(OG).
(5.2)
(5.3)
Example 5.2.3. Let G = p1+2 be an extra special group of order p3 . Let B = OG.
Since Aut(Cp × Cp ) ∼
= GL2 (p), by Theorem 5.2.1, we get
PI(B) ∼
= ((Cp × Cp ) o GL2 (p)) × Cp−1 × h-idi .
By (5.3) we get
DPic(B) ∼
= ((Cp × Cp ) o Out(G)) × Sh(B).
We know that (Cp ×Cp ) ≤ DPic(B) induces perfect isometries given by multiplication
by linear characters. So (Cp × Cp ) ≤ DPic(B) maps bijectively onto (Cp × Cp ) ≤
CHAPTER 5. P -GROUPS
80
PI(B). Furthermore, Sh(B) ≤ DPic(B) maps onto h-idi ≤ PI(B). So the image of
DPic(B) in PI(B) is
^ × h-idi,
((Cp × Cp ) o Out(G))
^ is the image of Out(G) ≤ DPic(B) in PI(B). But if G has exponent
where Out(G)
p then Out(G) ∼
= GL2 (p)) while if G has exponent p2 then Out(G) ∼
= Cp o Cp−1 [15,
§A 20.8]. So,
^ ≤ | Out(G)| < |GL2 (p) × Cp−1 |.
|Out(G)|
^ × h-idi with | PI(B)| shows that the
Comparing the order of ((Cp × Cp ) o Out(G))
canonical homomorphism DPic(B) −→ PI(B) is not surjective.
Chapter 6
Blocks with cyclic defect groups
One of the most studied families of blocks are blocks with cyclic defect groups. Brauer
[6, 7] first studied blocks of defect 1 and applied his results to study groups whose
order contains a prime number to the first power. Later on, Dade [14] extended
Brauer results and gave a structure of any block with a cyclic defect group. The
equivalences for blocks with cyclic defect groups are also well-understood. Broué’s
Abelian Defect Group Conjecture (Conjecture 1.0.3) is shown to be true in this case
[30, 43, 44]. In particular, Linckelmann [30] showed that the derived category of any
block with cyclic defect group is equivalent to the derived category of the semidirect
product of its defect group and its inertial quotient. Since the latter block is easier
to work with, this gives us a very powerful reduction tool when studying objects
invariant under a derived equivalence, for example, perfect isometry groups.
The aim of this chapter is to give a description of the perfect isometry groups for
blocks with cyclic defect groups. When the defect is 1, the perfect isometry groups
are completely determined. When the defect is greater than 1 we give a partial result
based on conditions of the inertial indexes and a conjecture for the remaining cases.
As a consequence of these results, we will also show that the homomorphism from the
derived Picard group to the perfect isometry group is surjective in the cases where
we can completely determine the perfect isometry groups.
Let G be a finite group and B be a block of OG with a cyclic defect group D.
81
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
82
Let E be the inertial quotient of B and e = |E| be the inertial index. We can form a
group D o E with the evident action of E on D. In order to find PI(B) we use the
following reduction.
Theorem 6.0.4 (Linckelmann [30]). The derived categories of B and O(D o E) are
equivalent.
As a consequence, B and O(D o E) are perfectly isometric and so, by Proposition
4.0.7, PI(B) ∼
= PI(O(D o E)). This allows us to work with the group D o E which
has a simpler structure than G. (Note: that O(D o E) is a block follows from [25,
Remark 49.8].) The main result is the following.
Theorem 6.0.5. Let G be a finite group. Let B be a block of OG with a cyclic defect
group D of order pn . Let e be the inertial index of B and let t = (pn −1)/e. There are
{χ1 , . . . , χe } ⊂ Irr(B) and {θ1 , . . . , θt } ⊂ Irr(B) such that Irr(B) = {χ1 , . . . , χe } ∪
{θ1 , . . . , θt }.
(i) If e = 1 then
PI(B) ∼
= PI(OD) ∼
= (D o Aut(D)) × h-idi .
(ii) If e > 1 and t = 1 then | Irr(B)| = p. Every permutation on Irr(B) gives a
perfect isometry (with a choice of sign), and
PI(B) ∼
= Sp × h-idi .
(iii) If e > 1 and t > 1 then every perfect isometry permutes (with signs) the sets
{χ1 , . . . , χe } and {θ1 , . . . , θt } separately. We have,
PI(B) ∼
= Se × X × h-idi
where Se is the symmetric group on {χ1 , . . . , χe } and X is a permutation group
on {θ1 , . . . , θt }. Furthermore, X contains a subgroup isomorphic to Aut(D)/E.
If D ∼
= Cp , then X ∼
= Aut(D)/E ∼
= C(p−1)/e .
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
83
Unfortunately, in the case e > 1 and t > 1 we were unable to give the complete description of PI(B) when B has defect greater than 1. However, based on
computer-generated results of many examples we can speculate that the subgroup X
is isomorphic to Aut(D)/E ∼
= Cpn−1 (p−1)/e .
Conjecture 6.0.6. Let B be a block of OG with a cyclic defect group D of order pn .
Let e be the inertial index of B and let t = (pn − 1)/e. Suppose e > 1 and t > 1.
Then
PI(B) ∼
= Se × Cpn−1 (p−1)/e × h-idi .
The case D ∼
= Cp (n = 1) in Conjecture 6.0.6 is included in Theorem 6.0.5(iii).
We will prove this in Theorem 6.3.1.
As discussed before, it suffices to work with the group D o E. Note that a
derived equivalence between B and O(D o E) implies that there is a bijection with
signs between Irr(O(D o E)) and Irr(B). This means that characteristics of perfect
isometry groups such as every permutation gives a perfect isometry (with some sign),
every perfect isometry permutes two subsets of irreducible characters in the block
separately, will be preserved. It does not, however, preserve properties related to
signs, such as having homogeneous signs. So if e = 1 then PI(B) may not have
homogenous signs even though PI(D) has (by Theorem 5.1.1).
6.1
The group D o E and its character table
Let H = D o E. Let |D| = pn . We first need some remarks on the group H.
Lemma 6.1.1. The following hold.
(i) e is not divisible by p.
(ii) e divides p − 1
(iii) The group H is uniquely determined by e.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
84
(iv) The group E acts frobeniusly on D, that is, if g h = g for g ∈ D, h ∈ E then
g = 1 or h = 1.
Proof. (i) is Theorem 2.3.10.
Let bD be a root of B. Since D is abelian, DCG (D) = CG (D). So the inertial
quotient E = ING (D) (bD )/CG (D) acts faithfully on D and is therefore a subgroup of
Aut(D). If |D| = pn then we have e|pn−1 (p − 1). Since e does not divide p by (i), e
must divide p − 1, proving (ii).
To prove (iii), notice that if e = 1 then H = D. If e > 1 then by (ii), p must
be odd, and so Aut(D) and E are cyclic. Since two automorphisms of D of the
same order are powers of each other and E ≤ Aut(D), the isomorphism class of H is
uniquely determined by the order of E.
Finally, (iv) follows from [14, Section 1].
Since E acts on D, it also acts on Irr(D). Let λ ∈ Irr(D) be a non-trivial character.
Since the action of E on D is faithful, the stabilizer in E of λ is 1. By the orbitstabilizer theorem, the size of orbit of λ is e. So the pn − 1 non-trivial characters of
Irr(D) break up into orbits of equal size e. Let {φ1 , . . . , φt } ⊆ Irr(D) be non-trivial
E-orbit representatives where t = (pn − 1)/e. By [24, Theorem 6.34], each φi induces
to an irreducible character of H, denoted by Φi . We call Φ1 , . . . , Φt the exceptional
characters of H.
On the other hand, for each irreducible character χi of E there is a corresponding
irreducible character χi of H such that χi (g) = χi (g) where g 7→ g is the canonical
map from H to E = H/D. The irreducible characters χ1 , . . . , χe are called nonexceptional characters of H. We label in such a way that χ1 is the trivial character.
We have now described all irreducible characters of H:
Irr(H) = {χ1 . . . , χe } ∪ {Φ1 . . . , Φt }.
Let a, b be generators of D and E respectively. Since E acts frobeniusly on D,
D − {1} breaks up into orbits of equal size e. Let {a1 , . . . , at } be non-trivial E-orbit
representatives of D. The conjugacy classes and the character table of H are as
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
85
follow:
{1},
(ai )G ,
(0 ≤ i ≤ t − 1),
(by )G = {gby : g ∈ D} (1 ≤ y ≤ e − 1).
H
χ1
...
χj
...
...
Φj
...
{1}
1
1
1
1
e
e
e
Table 6.1: Character table of H
...
{ai }
. . . . . . {by } . . .
1
1
1
1
1
1
1
1
1
1
1
1
χj (by )
1
1
1
0
0
0
φj ↑G (ai )
0
0
0
0
0
0
Lemma 6.1.2. Let g ∈ H. Then g is p-singular if and only if g ∈ D − {1}.
Proof. If g ∈ D − {1}, then it is clear that g is a p-singular element. Suppose that
g 6∈ D. We will show that g is p-regular. First write g = xy where x ∈ D, y ∈ E
and y 6= 1. If x = 1, then g = y is clealy a p-regular element. So suppose x 6= 1.
m
Since D C H, yx = xk y for some k 6≡ 0 mod pn . Then y m x = xk y m and so
k o(y) ≡ 1 mod o(x), where o(x), o(y) are the orders of x, y respectively. Note that
g r = x1+k+k
2 +···+k r−1
y r = x(k
r −1)/(k−1)
y r . Suppose g r = 1, then o(y)|r. But if r = o(y),
then k r ≡ 1 mod o(x) and so g o(y) = x0 y o(y) = 1. Hence, g has order o(y) which is
not divisible by p. So g is p-regular.
The next lemma summarizes the values of |CH (g)| for all g ∈ H.
Lemma 6.1.3.
(i) If g = 1 then |CH (g)| = |H|.
(ii) If g 6= 1 and g is p-regular then |CH (g)| = |E|
(iii) If g 6= 1 and g is p-singular then |CH (g)| = |D|
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
86
Proof. (i) is obvious. For (ii), the previous lemma shows that g 6∈ D so Φj (g) = 0, ∀j.
By column orthogonality of irreducible characters,
|CH (g)| =
X
|θ(g)|2
θ∈Irr(H)
=
=
e
X
i=1
e
X
2
|χi (g)| +
t
X
|Φ(g)|2
j=1
|χi (g)|2
i=1
= |CE (g)| = |E|.
For (iii), by the previous lemma, g ∈ D − {1}. Since D is cyclic, it is clear that
D ≤ CH (g). By Lemma 6.1.1(iv), there is no non-trivial h ∈ E with g h = g. Thus,
|CH (g)| = |D|.
6.2
Perfect isometry groups
We will now prove Theorem 6.0.5 for the group H = D o E. Let A = OH. We divide
the proof into 3 cases based on the values of e and t.
Case: e = 1
In this case H = D, and therefore the result on abelian p-groups (Theorem 5.1.1)
applies.
Lemma 6.2.1. Suppose e = 1. Then
PI(A) ∼
= (D o Aut(D)) × h-idi .
Case: e > 1 and t = 1
Since e | p − 1, we can deduce that p is odd. Since t = 1, we have e = |D| − 1,
implying that D ∼
= Cp and e = p − 1. Therefore, H = Cp o Cp−1 . Let χp = Φ1 so that
Irr(A) = {χ1 , . . . , χp }. There is only one p-singular class: {a}. There are e p-regular
classes: {1}, {by } for y = 1, . . . , e − 1.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
87
Table 6.2: Character Table of Cp o Cp−1
G
1 {a} {by }
χ1
1
1
1
... 1
1
χj
1
1
χj (by )
... 1
1
χp p-1 -1
0
Proposition 6.2.2. Suppose that e = p − 1 and t = 1. Then
PI(A) = Sp × h-idi .
Proof. Since | Irr(A)| = p is odd, by Proposition 4.1.7, PI(A) ∼
= PI+ (A) × h-idi. It
is therefore enough to show that, for any σ ∈ Sp there is a sign function ε such that
I = (σ, ε) is a perfect isometry. Let δi = 1 for i = 1, . . . , p − 1 and δp = −1. Define
an isometry I by
I(χi ) = δi δσ(i) χσ(i) ,
i = 1 . . . , p.
We will show that I is perfect.
Let µI be the character corresponding to I. For g, h ∈ H,
µI (g, h) =
p
X
δi δσ(i) χσ(i) (g)χi (h)
i=1
=
X
δi δσ(i) χσ(i) (g)χi (h) − δσ(p) χσ(p) (g)χp (h).
i6=p
Suppose g = h = 1. Then
µI (1, 1) =
X
δσ(i) χσ(i) (1) − δσ(p) χσ(p) (1)(p − 1).
i6=p
If σ(p) = p, then µI (1, 1) = (p − 1) − (−1)(p − 1)2 = (p − 1)p. Since |CH (1)| =
p(p − 1) and 1/(p − 1) ∈ O, this satisfies the integrality condition. If σ(p) 6= p then
µI (1, 1) = (p − 2) + (−1)(p − 1) − (p − 1) = −p. Since |CH (1)| = p(p − 1), this satisfies
the integrality condition.
Suppose g = 1, h 6= 1, h is p-regular. If σ(p) = p, then µI (g, h) =
P
i6=p
χi (h) = 0.
If σ(p) 6= p, then σ(j) = p for some j ∈ {1, . . . , p − 1}. So µI (g, h) = (−1)(p −
P
1)χj (h) + i6=p,i6=j χi (h) = −pχj (h). In both cases, µI (g, h) ∈ pO. This satisfies the
integrality condition.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
Suppose h = 1, g 6= 1, g is p-regular. If σ(p) = p, then µI (g, h) =
88
P
i6=p
χσ(i) (g) =
0. If σ(p) 6= p, then σ(j) = p for some j ∈ {1, . . . , p − 1}. So µI (g, h) = 0 +
P
i6=p,i6=j χσ(i) (g) − (1)χσ(p) (g)(p − 1) = −pχσ(p) (g). In both cases, µI (g, h) ∈ pO.
This satisfies the integrality condition.
Suppose g = 1, h 6= 1, h is p-singular. If σ(p) = p, then µI (g, h) =
P
i6=p (1)(1)(1)−
(−1)(p − 1)(−1) = 0. If σ(p) 6= p, then σ(j) = p for some j ∈ {1, . . . , p − 1}. So,
P
µI (g, h) = (−1)(p − 1)(1) + i6=p,i6=j (1)(1)(1) − (1)(1)(−1) = 0. Both cases satisfy
the separation condition.
Suppose h = 1, g 6= 1, g is p-singular. If σ(p) = p, then µI (g, h) =
P
i6=p (1)(1)(1)−
(−1)(−1)(p − 1) = 0. If σ(p) 6= p, then σ(j) = p for some j ∈ {1, . . . , p − 1}. So,
P
µI (g, h) = (−1)(−1)(1) + i6=p,i6=j (1)(1)(1) − (1)(1)(p − 1) = 0. Both cases satisfy
the separation condition.
Suppose g 6= 1, h 6= 1 and g, h are both p-regular. Then, |CH (g)| = |CH (g)| = e =
p − 1. But µI (g, h) ∈ (p − 1)O. This satisfies the integrality condition.
Suppose g 6= 1, h 6= 1 and g, h are both p-singular. If σ(p) = p, then µI (g, h) =
P
− (−1)(−1)(−1) = p. If σ(p) =
6 p, then σ(j) = p for some j ∈
P
{1, . . . , p − 1}. So, µI (g, h) = (−1)(−1)(1) + i6=p,i6=j (1)(1)(1) − (1)(1)(−1) = p.
i6=p (1)(1)(1)
Both cases satisfy the integrality condition.
Therefore I is perfect as claimed.
Case: e > 1 and t > 1
In this case we have p > 2 (since e divides p − 1). First we show that every perfect
isometry preserves ±{χ1 , . . . , χe } and ±{Φ1 , . . . , Φt }.
Lemma 6.2.3. Let I : RK (H) −→ RK (H) be a perfect isometry. Then I sends
{χ1 , . . . , χe } to ±{χ1 , . . . , χe } and sends {Φ1 , . . . , Φt } to ±{Φ1 , . . . , Φt }.
Proof. Suppose that I(Φx ) = ±χr for some x and r. Without loss of generality, we
can take I(Φx ) = χr . Since, for any i, j, Φi − Φj = 0 on p-regular elements, so
I(Φi ) − I(Φj ) = 0 on p-regular elements as I is perfect. Since b is a p-regular element
and t > 1, there is y 6= x such that I(Φy )(b) = I(Φx )(b) = χr (b).
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
89
Since Φi (b) = 0 for all i and χr (b) 6= 0, we must have I(Φy ) ∈ ±{χ1 , . . . , χe }, say
I(Φy ) = ±χs where s 6= r. Then ±χs (b) = χr (b). But {χ1 (b), . . . , χe (b)} are distinct
e-th roots of unity, we cannot have χs (b) = χr (b). So we must have I(Φy ) = −χs and
−χs (b) = χr (b).
But we also have I(Φx )(1) = I(Φy )(1), since 1 is a p-regular element. This gives
χr (1) = −χs (1), that is, 1 = −1, a contradiction. Hence I sends {Φ1 , . . . , Φt } to
±{Φ1 , . . . , Φt }, and consequently, I sends {χ1 , . . . , χe } to ±{χ1 , . . . , χe }.
The next lemma shows that every perfect isometry has a homogenous sign.
Lemma 6.2.4. Let I : RK (H) −→ RK (H) be a perfect isometry. Then I has a
homogenous sign.
Proof. By Lemma 6.1.2, the p-singular elements of H are in D − {1}. So χi − χj = 0
on p-singular elements for all i, j. Hence, I(χi ) − I(χj ) = 0 on p-singular elements
for all i, j. In particular, I(χi )(a) is constant for all i. Since χi (a) = 1 for all i, this
shows that I has a homogenous sign on {χ1 , . . . , χe }. We can assume without loss of
generality that I has positive signs on {χ1 , . . . , χe }.
The same reasoning also applies to Φi − Φj , vanishing on p-regular elements for
all i, j. So I(Φi )(1) is constant for all i. This shows that I also has a homogenous
sign on {Φ1 , . . . , Φt }. Let r = |{i ∈ {1, . . . , t} : I(Φi ) ∈ {Φ1 , . . . , Φt }|, that is, r is
the number of Φi ∈ {Φ1 , . . . , Φt } whose image has positive sign. By above, r = 0 or
t. We will show that r = t.
Let µI be the corresponding character of I. Then
µI (1, 1) =
e
X
I(χi )(1)χi (1) +
i=1
t
X
I(Φi )(1)Φi (1)
i=1
= e + re2 − (t − r)e2 = e + 2re2 − te2
n
p −1 2
2
e = e + 2re2 − epn + e
= e + 2re −
e
= 2e(1 + re) − epn .
Since I is perfect, we have µI (1, 1) ≡ 0 mod pn . So 2e(1 + re) ≡ 0 mod pn . Since p
is odd and p 6 |e, this yields 1 + re ≡ 0 mod pn and so r 6= 0. Thus r = t and I has
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
90
a homogenous sign on Irr(H) as claimed.
Proposition 6.2.5. Let σ ∈ Sym{χ1 , . . . , χe }. Define I : RK (H) −→ RK (H) by
I(χ) = σ(χ),
I(Φ) = Φ,
∀χ ∈ {χ1 , . . . , χe }
∀Φ ∈ {Φ1 , . . . , Φt }.
Then I is a perfect isometry.
Proof. Let µ = µI be the character corresponding to I and let µid be the character
corresponding to the identity isometry. Let g, h ∈ H.
µ(g, h) =
X
σ(χ)(g)χ(h) +
X
χ
Φ(g)Φ(h).
Φ
If g = 1, then
µ(1, h) =
X
σ(χ)(1)χ(h) +
X
χ
Φ(1)Φ(h) = µid (1, h).
Φ
So µ(1, h) satisfies the conditions of perfect characters.
If h = 1, then
X
σ(χ)(g)χ(1) =
X
σ(χ)(g) =
X
X
χ(g)χ(1).
χ
χ
χ
χ
χ(g) =
So
µ(g, 1) =
X
σ(χ)(g)χ(1) +
X
χ
Φ(g)Φ(1) = µid (g, 1).
Φ
So µ(g, 1) satisfies the conditions of perfect characters.
Assume from now on that g 6= 1 and h 6= 1. If g is p-regular and h is p-regular,
then by Lemma 6.1.3(ii), |CH (g)| = |CH (h)| = e. Since p - e, µ(g, h)/e ∈ O. So
µ(g, h) satisfies the integrality condition.
If g is p-regular and h is p-singular, then g 6∈ D and g 6= 1, h = 1. So
µ(g, h) =
X
σ(χ)(ḡ)χ(1) + 0 =
X
σ(χ)(ḡ) = 0.
χ
χ
So µ(g, h) satisfies the separation condition.
If g is p-singular and h is p-regular, then h 6∈ D and g = 1, h 6= 1. So
µ(g, h) =
X
χ
σ(χ)(1)χ(h̄) + 0 =
X
χ
χ(h̄) = 0.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
91
So µ(g, h) satisfies the separation condition.
Finally, if g, h are p-singular, then g = h = 1 and, by Lemma 6.1.3(iii), |CH (g)| =
|CH (h)| = pn . So
µ(g, h) =
X
σ(χ)(1)χ(1) +
χ
X
Φ(g)Φ(h) =
X
χ(1)χ(1) +
χ
Φ
X
Φ(g)Φ(h)
Φ
= µid (g, h).
So µ(g, h) satisfies the integrality condition. Hence I is perfect.
Corollary 6.2.6. Suppose e > 1 and t > 1. Then
PI(A) ∼
= Se × X × h-idi
where X ≤ St is a permutation group on the exceptional characters.
Proof. Since all perfect isometries have homogenous signs, by Lemma 4.1.6 we have
PI(A) ∼
= PI(A)+ × h-idi where PI+ (A) is a permutation group on Irr(A). By Lemma
6.2.3, PI+ (A) permutes the exceptional and non-exceptional characters separately.
By Proposition 6.2.5, perfect isometries permuting the non-exceptional characters
only is the full symmetric group Se . We deduce that PI(A) ∼
= Se × X × h-idi where
X is a permutation group on the exceptional characters.
Now it remains to find the group X, that is the subgroup of perfect isometries
permuting only the exceptional characters. We are going to show that X contains
Aut(D)/E, the cyclic group of order pn−1 (p−1)/e. Recall from Section 4.2 that there
e
is a homomorphism Aut(H) −→ PI(A). Let A(A)
be the image of this map. Since
∼
e
Irr(A) = Irr(H), by Remark 4.2.2, A(A)
= Aut(H)/ Autc (H).
∼
e
e
Proposition 6.2.7. The group X contains A(A)
and A(A)
= Aut(D)/E, a cyclic
group of order pn−1 (p − 1)/e.
Proof. First we will show that Aut(H) ∼
= DoAut(D). As before, let a, b be generators
of D and E respectively. The group H = D o E can be written as
n
H = ha, b : ap = be = 1, ab = au i,
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
92
where u has order e in Z×
pn . Let θ ∈ Aut(H). Since a is a p-singular element, θ(a)
is also p-singular and so θ(a) = ar for some positive integer r with (r, pn ) = 1. Let
θ(b) = ax by for some x ∈ {0, . . . , pn − 1}, y ∈ {0, . . . , e}. Also
aru = θ(au ) = θ(ab ) = θ(a)θ(b)
x y
x y r
y r
= (ar )a b = aa b
= ab
y
= au r .
y r−ru
This implies au
= 1, and so
(uy − u)r ≡ 0
mod pn
uy − u ≡ 0
mod pn
since (r, pn ) = 1
uy−1 − 1 ≡ 0
mod pn
since (u, pn ) = 1
∴y≡1
mod e since u has order e
∴ by = b.
So, for each θ ∈ Aut(H) we must have,
θ(a) = ar ,
θ(b) = ax b,
for some r with (r, pn ) = 1
for some x with x ∈ {0, . . . , pn − 1}.
For each σ ∈ Aut(D), g ∈ D, define θσ , θg by
θσ (a) = σ(a),
θσ (b) = b
θg (a) = ag = a,
θg (b) = bg .
Then, we see that Aut(H) is generated by elements of the forms θσ , θg for σ ∈
Aut(D), g ∈ D. Moreover, we have θσ θg (θσ )−1 = θσ(g) . Thus
Aut(H) ∼
= D o Aut(D),
(6.1)
where multiplications are defined by (θg , θσ )(θh , θπ ) = (θg θσ(h) , θσ θπ ) for g, h ∈ D and
σ, π ∈ Aut(D).
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
93
Next we show that Autc (H) ∼
= D o E. For convenience, we write g ∼ h for “g is
conjugate in H to h”. Let g ∈ D be such that θg ∈ Autc (H). Then, θg (b) = bg ∼ b.
But the conjugacy class of b is {xb, x ∈ D}. Thus,
{g ∈ D : θg ∈ Autc (H)} = D.
Let σ ∈ Aut(D) be such that θσ ∈ Autc (H). Then, θσ (a) = σ(a) ∼ a. But
σ(a) ∼ a if and only if σ(a) = ay for some y ∈ E. Thus,
{σ ∈ Aut(D) : θσ ∈ Autc (H)} ∼
= E.
Hence,
Autc (H) ∼
= D o E.
(6.2)
Therefore, from (6.1) and (6.2), the image of Aut(H) in PI(A) is
Aut(H) ∼ D o Aut(D) ∼ Aut(D)
∼
e
.
A(A)
=
=
=
Autc (H)
DoE
E
Since p is odd, Aut(D) is cyclic of order pn−1 (p − 1) and hence Aut(D)/E is cyclic
of order pn−1 (p − 1)/e.
e
Finally, we show that A(A)
≤ X. Let σ ∈ Aut(H). Let g ∈ D, h ∈ E. Then, as
in the beginning of the proof, σ −1 (gh) = g 0 h, where g 0 ∈ D. For χ ∈ {χ1 , . . . , χe },
χσ (gh) = χ(σ −1 (gh)) = χ(g 0 h) = χ(gh),
since gh ∼ g 0 h. So the the perfect isometry Iσ : χ 7→ χσ fixes every character
χ ∈ {χ1 , . . . , χe }. Hence Iσ is a permutation only on the exceptional characters and
e
thus, A(A)
≤ X as claimed.
6.3
Special case when defect group is Cp
In this section, we will prove the last statement in Theorem 6.0.5(iii) in the case e > 1
and t > 1. Although we cannot completely determine the subgroup X ≤ PI(B) when
n > 1. The case when n = 1 (that is, D ∼
= Cp ) can be done. Our proof is direct and
elementary.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
94
Theorem 6.3.1. Let B be a block of OG with a defect group D ∼
= Cp . Let e be the
inertial index of B, and let t = (p − 1)/e. Suppose that e > 1 and t > 1. Then
PI(B) ∼
= Se × Ct × h-idi .
As before we will work with the group H = D o E. Let A = OH. Let a, b
be generators of D, E respectively. In order to calculate the perfect isometry group,
we need to describe the group structure and character values in more details than
in the previous section. We will also slightly change the indexes of our irreducible
characters from the previous section, so that the formulas for character values are
easy to manipulate.
Write
H = ha, b : ap = be = 1, ab = au i
×
2
t−1
where u has order e in Z×
} be coset
p . Let S = hui C Zp and let {1, v, v , . . . , v
F
t−1 i
×
representatives of S in Z×
p , so that Zp =
i=0 v S.
The conjugacy classes of G are:
{1},
i
i
(av )G = {av s : s ∈ S} (0 ≤ i ≤ t − 1),
(by )G = {ax by : 0 ≤ x ≤ p − 1} (1 ≤ y ≤ e − 1).
Let ζ = e2πi/e and ω = e2πi/p be primitive roots of unity. Then the irreducible
characters of G are
{χ0 , χ1 , . . . , χe−1 } ∪ {Φ0 , Φ1 , . . . , Φt−1 },
where
χi (ax by ) = ζ iy
Φj (ax by ) = 0 if 1 ≤ y ≤ e − 1
X j
Φj (ax ) =
ω v sx .
s∈S
Here, the non-exceptional characters are {χ0 , χ1 , . . . , χe−1 } and the exceptional characters are {Φ0 , Φ1 , . . . , Φt−1 }.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
G
χ0
...
χj
...
...
Φj
...
95
Table 6.3: Character table of H
{1} . . .
{ax }
. . . . . . {by } . . .
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
ζ jy
1
1
1
1
e
0
0
0
P
v j sx
e
0
0
0
s∈S ω
e
0
0
0
Recall from the previous section that any perfect isometries I ∈ PI(A) has a
homogenous sign. If I has all-positive sign, then I sends exceptional characters
to exceptional characters and sends non-exceptional characters to non-exceptional
characters. Furthermore, we can write
PI(A) ∼
= Se × X × h-idi
where X is the subgroup of PI(A) permuting only the exceptional characters.
Let T = {0, 1, . . . , t − 1}. For each π ∈ Sym(T ), define Iπ : RK (A) −→ RK (A) by
Iπ (χi ) = χi ,
i = 0, . . . , e − 1
Iπ (Φj ) = Φπ(j) ,
j = 0, . . . , t − 1.
Then X consists of perfect isometries of the form Iπ for some π ∈ Sym(T ). Let
µπ ∈ RK (A ⊗O A◦ ) be the character corresponding to Iπ . Theorem 6.3.1 will be
proved once we show that µπ is perfect if and only if π is of the form π(j) = j + α
mod t, ∀j ∈ T where α ∈ T . This will then imply that X ∼
= Ct .
= Zt ∼
It is easy to see from the character table that µπ satisfies the separation condition
(for any π ∈ Sym(T )). Since |CH (ax )| = p, ∀x ∈ Z×
p , we have that µπ is perfect if
and only if µπ (ax , ay )/p ∈ O, ∀x, y ∈ Z×
p , that is, if and only if
kµπ (ax , ay )kp ≤
1
p
∀x, y ∈ Z×
p.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
96
Let x, y ∈ Z×
p . We have
µπ (ax , ay ) =
X
χi (ax )χi (ay ) +
X
i
Φπ(j) (ax )Φj (ay )
j
!
!
= e+
X X
= e+
XXX
j
ωv
π(j) rx
X
ωv
j sy
s∈S
r∈S
ω
v π(j) rx+v j sy
.
(6.3)
j∈T r∈S s∈S
Recall that K is sufficiently large for H, so K is an extension of Qp (ω). It is evident
from (6.3) that µπ (ax , ay )/p ∈ Qp (ω). If kµπ (ax , ay )/pkp ≤ 1, then µπ (ax , ay )/p ∈
Qp (ω) ∩ O. So, by Lemma 2.1.1, µπ (ax , ay )/p is in the valuation ring of Qp (ω). This
ring can be described by the following lemma.
Lemma 6.3.2. The valuation ring of Qp (ω) with respect to k · kp is Zp (ω).
Proof. See the discussion in Section 5.6 in [20].
Lemma 6.3.3. Let α = C0 + C1 ω + · · · + Cp−1 ω p−1 where Ci ∈ Z for all i. Suppose
that kαkp ≤ 1/p. Then Ci ≡ Cj mod p for all i, j.
Proof. Since α/p ∈ Qp (ω) and kα/pkp = pkαkp ≤ 1, we have that α/p is in the
valuation ring of Qp (ω). By the previous lemma, we can therefore write α = pa0 +
pa1 ω + · · · + pap−2 ω p−2 where ai ∈ Zp ∀i, as {1, ω, . . . , ω p−2 } is a basis of Zp (ω) over
Zp . Write ω p−1 = −1 − ω − · · · − ω p−2 . Then
α = (C0 − Cp−1 ) + (C1 − Cp−1 )ω + · · · + (Cp−2 − Cp−1 )ω p−2
= pa0 + pa1 ω + · · · + pap−2 ω p−2 .
Comparing the coefficients, we have Ci − Cp−1 ∈ pZp for i = 1, . . . , p − 2. But
Ci − Cp−1 are integers, we have Ci − Cp−1 ∈ pN and so Ci ≡ Cj mod p for all i, j as
claimed.
We are going to write µπ (ax , ay ) in a nicer form so we can apply Lemma 6.3.3.
For k ∈ {0, 1, . . . p − 1}, define
Fkπ = {(j, r, s) ∈ T × S × S : v π(j) rx + v j sy ≡ k
mod p}.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
97
Then, from (6.3), we see that
µπ (ax , ay ) = C0 + C1 ω + . . . + Cp−1 ω p−1 ,
where C0 = e + |F0π | and Ci = |Fiπ | for i ∈ {1, . . . , p − 1} and
C0 + . . . + Cp−1 = e + tee = e + (p − 1)e
= pe.
We can derive some properties of the coefficients {C0 , . . . , Cp−1 }. For k ∈ T ,
define
βk = |{j ∈ T : π(j) − j ≡ k
mod t}|.
Lemma 6.3.4. Let k ∈ T be a unique integer such that x−1 (−y) ∈ v k S. Then
|F0π | = βk e and so C0 = (βk + 1)e.
Proof. Consider
v π(j) rx + v j sy ≡ 0
mod p
v π(j) rx ≡ v j s(−y)
v π(j)−j rs−1 ≡ x−1 (−y)
mod p
mod p.
Since x−1 (−y) belongs to a unique S-coset, write x−1 (−y) = v k z for a unique k and
z ∈ S. Then
v π(j)−j rs−1 ≡ v k z
mod p.
But v π(j)−j rs−1 belongs to the coset v π(j)−j S. This is possible if and only if π(j) −
j ≡ k mod t (v has order t) and rs−1 ≡ z mod p. Suppose there exists j with
π(j) − j ≡ k mod t, then the number of pairs (r, s) such that rs−1 ≡ z mod p is
equal to e (= |S|). Hence
|F0π | = {(j, r, s) ∈ T × S × S : v π(j) rx + v j sy ≡ 0
mod p} = βk e.
Lemma 6.3.5. If m, n ∈ Z×
p are in the same S-coset, then Cm = Cn .
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
98
Proof. Let m = v k g, n = v k h where g, h ∈ S. If (j, r, s) ∈ Fvπk g , then
v π(j) x(r) + v j y(s) ≡ v k g
mod p
v π(j) x(rg −1 h) + v j y(sg −1 h) ≡ v k h mod p.
So (j, rg −1 h, sg −1 h) ∈ Fvπk h . Conversely, if (j, r, s) ∈ Fvπk h , it is easy to verify that
(j, rh−1 g, sh−1 g) ∈ Fvπk g . So we have a 1-1 correspondence Fvπk g ↔ Fvπk h , given by
(j, r, s) ↔ (j, rg −1 h, sg −1 h). Hence,
Cm = |Fmπ | = |Fvπk g | = |Fvπk h | = |Fnπ | = Cn .
So far, the results about the coefficients {C0 , . . . , Cp−1 } are valid for any π ∈
Sym(T ). However, if π is of the form π(j) = j + α mod t, ∀j for some α ∈ T , then
the coefficients {C0 , . . . , Cp−1 } have a particularly nice property.
Lemma 6.3.6. Suppose π is of the form π(j) = j + α mod t, ∀j for some α ∈ T .
Then Cm = Cn for all m, n ∈ Z×
p.
k
l
Proof. Let m, n ∈ Z×
p . By Lemma 6.3.5, it suffices to assume m = v , n = v . Suppose
(j, r, s) ∈ Fvπk . Then
v π(j) x(r) + v j y(s) ≡ v k
mod p
v j+α x(r) + v j y(s) ≡ v k
mod p
v j+α+l−k x(r) + v j+l−k y(s) ≡ v l
mod p
v π(j+l−k) x(r) + v j+l−k y(s) ≡ v l
mod p.
So, ((j + l − k) mod t, r, s) ∈ Fvπl . Similarly, if (j, r, s) ∈ Fvπl , then ((j + k − l)
mod t, r, s) ∈ Fvπk . So we have a 1-1 correspondence Fvπk ↔ Fvπl . Hence,
Cm = |Fvπk | = |Fvπl | = Cn .
Lemma 6.3.7. Suppose π is of the form π(j) = j + α mod t, ∀j for some α ∈ T .
Then µπ is perfect.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
99
x y
Proof. Let x, y ∈ Z×
p . It suffices to show that kµπ (a , a )kp ≤ 1/p. Using the notation
discussed above, we can write
µ(ax , ay ) = C0 + C1 ω + . . . + Cp−1 ω p−1 ,
(6.4)
where C0 = e + |F0π | and Cm = |Fmπ | for m ∈ {1, . . . , p − 1} and
C0 + C1 + . . . + Cp−1 = pe.
Let k ∈ T be such that x−1 (−y) ∈ v k S. Then, by Lemma 6.3.4,
C0 = (βk + 1)e.
By Lemma 6.3.4 and Lemma 6.3.6, and using the fact that te = p − 1, we have,
(βk + 1)e + (p − 1)C1 = pe
(βk + 1)e + teC1 = pe
p − 1 − βk
te − βk
=
t
t
βk
=e− .
t
C1 =
So, Cm = e − βk /t for all m ∈ Z×
p . Substituting in (6.4) yields
βk
x y
µ(a , a ) = (βk + 1)e + e −
(ω + ω 2 + . . . + ω p−1 )
t
βk
= (βk + 1)e − e −
t
βk
= βk e + .
t
But,
βk = |{j ∈ T : π(j) − j ≡ k
mod t}|
= |{j ∈ T : j + α − j ≡ k
= |{j ∈ T : α ≡ k
mod t}|
mod t}|.
This implies that βk = tδkα (where δ is the Kronecker delta function). Thus,
µ(ax , ay ) = teδkα +
tδkα
t
= (te + 1)δkα
= pδkα .
Therefore, kµπ (ax , ay )kp ≤ 1/p and µπ is perfect as claimed.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
100
Lemma 6.3.8. Suppose that µπ is perfect. Then π is of the form π(j) = j + α
mod t, ∀j for some α ∈ T .
−1
α
Proof. Let α = π(0). Choose x, y ∈ Z×
p such that x (−y) ∈ v S. This is possible,
for example, choose x−1 = v α and y = p − 1. Then
βα = |{j ∈ T : π(j) − j ≡ α
mod t}|
≥ 1 (∵ the set contains 0).
We claim that βα = t. This will imply that π(j) = j + α mod t for all j ∈ T ,
and the lemma will be proved. To see this, as before we can write
µ(ax , ay ) = C0 + C1 ω + . . . + Cp−1 ω p−1 ,
where C0 = e + |F0π | and Cm = |Fmπ | for m ∈ {1, . . . , p − 1} and
C0 + C1 + . . . + Cp−1 = pe.
(6.5)
By Lemma 6.3.4, we have C0 = (βα + 1)e and by Lemma 6.3.5, we have Cm = Cn
whenever m, n ∈ Z×
p are in the same S-coset. Thus, (6.5) becomes
(βα + 1)e + eC1 + eCv + eCv2 + . . . + eCvt−1 = pe
(βα + 1) + C1 + Cv + Cv2 + . . . + Cvt−1 = p.
(6.6)
Since βα ≥ 1 and Cm ≥ 0 ∀m ∈ Z×
p , this implies 0 ≤ Cv i ≤ p − 2 ∀i, and hence,
by Lemma 6.3.5,
0 ≤ Cm ≤ p − 2,
∀m ∈ Z×
p.
Since µπ is perfect, kµπ (ax , ay )kp ≤ 1/p. So Cm ≡ Cn mod p for all m, n ∈
{0, 1, . . . , p − 1}, by Lemma 6.3.3. This means
Cm = Cn
∀m, n ∈ Z×
p.
Therefore, (6.6) becomes
(βα + 1) + tC1 = p
(te + 1) − βα − 1
p − βα − 1
=
t
t
βα
= e− .
t
C1 =
But C1 is an integer and 1 ≤ βα ≤ t, this forces βα = t as required.
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
101
To summarize, we have shown that µπ is perfect if and only if π is of the form
π(j) = j + α mod t, ∀j for some α ∈ T . So the subgroup X ≤ PI(A) consists of
perfect isometries Iπ where π ∈ Sym(T ) is of the the above form. Hence X ∼
= Ct ,
and so PI(A) ∼
= Se × Ct × h-idi. Since PI(B) ∼
= PI(A), this proves Theorem 6.3.1.
6.4
Comparision with Picard groups and derived
Picard groups
Let B be a block of OG with a cyclic defect group D of order pn and inertial quotient
group E of order e. As we already know, a derived equivalence gives rise to a perfect
isometry. It is interesting to know, in this case, if every perfect isometry in PI(B)
can also be given by a derived equivalence, in other words, if the homomorphism
DPic(B) −→ PI(B) is surjective.
Kunugi [29] showed that if e > 1 then, for any perfect isometry I ∈ PI(O(D oE)),
there exists a tilting complex X ∈ DPic(O(D o E)) such that the induced perfect
isometry agrees with I on the non-exceptional characters. However, in the case
D ∼
= Cp and e = p − 1, we also have perfect isometries sending a non-exceptional
character to the (single) exceptional character (Theorem 6.0.5(ii)). These perfect
isometries are not covered by tilting complexes in Kunugi’s result. In this section we
will show that there are also tilting complexes that induce these perfect isometries.
In fact, we will prove the following theorem.
Theorem 6.4.1. If e = 1 or (e > 1 and t = 1), then the homomorphism DPic(B) −→
PI(B) is surjective. If e > 1 and t > 1 and Conjecture 6.0.6 holds (for example when
D∼
= Cp ), then DPic(B) −→ PI(B) is surjective.
Corollary 6.4.2. If B is a block with defect group Cp , then every perfect isometry
in PI(B) can be given by a tilting complex in DPic(B).
Since B is derived equivalent to O(D oE), they are also perfectly isometric. Then
DPic(B) ∼
= PI(O(D o E)) and we have:
= DPic(O(D o E)), and PI(B) ∼
DPic(B) ∼
= DPic(O(D o E)) −→ PI(O(D o E)) ∼
= PI(B).
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
102
So it suffices to show that DPic(O(D o E)) −→ PI(O(D o E)) is surjective.
Let H = D o E and A = OH. We will study Pic(A) and DPic(A) (in some
cases) and their images in PI(A). It is necessary to study A in three separate cases,
depending on e and t = (pn − 1)/e.
Case: e = 1
In this case we have H = D ∼
= Cpn .
Proposition 6.4.3. The homomorphism DPic(OCpn ) −→ PI(OCpn ) is surjective.
Proof. By Theorem 6.0.5(i), we have
PI(A) ∼
= (D o Aut(D)) × h-idi .
By Theorem 5.2.2 and Corollary 5.1.3, we have
Pic(A) ∼
= (D o Aut(D)).
So, by Lemma 4.4.6, we can see that PI(A) ∼
= Pic(A) × h-idi.
The subgroup Sh(A) ≤ DPic(A) contains A[m] for m ∈ Z. If m is even, A[m]
corresponds to id ∈ PI(A) while if m is odd, A[m] corresponds to −id. Since, by
Theorem 5.2.2,
DPic(A) ∼
= (D o Aut(D)) × Sh(A),
and Sh(A) ≤ DPic(A) is mapped to h-idi ≤ PI(A), the proposition is proved.
Case: e > 1 and t = 1
In this case we must have D ∼
= Cp and so H = Cp o Cp−1 .
The Picard groups for blocks with non-trivial inertial quotient are given by Linckelmann in the following theorem.
Theorem 6.4.4 ([32], Theorem 11.4.11). Let B be a block of a finite group G having
a cyclic defect group D with non-trivial inertial quotient E acting on D. Let F be
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
103
the automorphism group of the Brauer tree of B. Then F is cyclic of order dividing
|E| and we have
Pic(B) ∼
= F × Aut(D)/E.
In this case we have E ∼
= Aut(D). Since a block whose cyclic defect group is
normal has Brauer tree a star with exceptional vertex in the center [2, Theorem
19.1], the automorphism group of the Brauer tree of A is Cp−1 . By Theorem 6.4.4,
therefore,
Pic(A) ∼
= Cp−1 .
The Picard group is generated by Bσ , where σ ∈ AutO (A) is defined by g σ =
λ(g −1 )g, ∀g ∈ H, for some non-trivial linear character λ ∈ Irr(A). Note that the
exceptional character is fixed by the images of Pic(A) in PI(A).
We next need the following theorem, due to Rouquier, which we will state in a
form that is useful to us.
Theorem 6.4.5. [43, Theorem 7] Let B be a block of OG for a finite group G.
L
v
Suppose there exists a direct summand P of
i Pi ⊗O Pi , where Pi runs over all
0
the projective indecomposable B-modules, such that 0 → P → B → 0 induces an
isometry from RK (B) to itself. Then there is a complex C = 0 → P → B → 0
inducing a derived equivalence between B and itself.
Proposition 6.4.6. The homomorphism DPic(O(Cp o Cp−1 )) −→ PI(O(Cp o Cp−1 ))
is surjective.
Proof. Let A = O(Cp o Cp−1 ). By Theorem 6.0.5(ii), we have
PI(A) ∼
= Sp × h-idi .
Suppose now that Irr(A) = {χ1 , . . . , χe−1 , θ}, where θ is the exceptional character.
Since the Brauer tree is a star, we label in such a way that the projective indecomposable Pi has the K-character χi + θ.
Let Ci = 0 → Pi ⊗O Piv → A → 0. The K-character of Pi ⊗O Piv is
(χi + θ)(χi + θ) = (χi )(χi ) + (θ)(θ) + (χi )(θ) + (θ)(χi ).
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
104
The K-character of A as (A, A)-bimodule is
X
(χi )(χi ) + (θ)(θ).
i
Thus the K-character of Ci is
"
− [(χi )(χi ) + (θ)(θ) + (χi )(θ) + (θ)(χi )] +
#
X
(χi )(χi ) + (θ)(θ)
i
X
=
(χj )(χj ) − (θ)(χi ) − (χi )(θ).
j6=i
This generalized character of Ci induces a perfect isometry sending χi 7→ −θ, θ 7→
−χi and fixes everything else. By Theorem 6.4.5, Ci is a tilting complex, and so,
Ci ∈ DPic(A), ∀i. The images of {Ci : i = 1, . . . , e − 1} in PI(A) generates Sp ,
whereas A[−1] gives −id. Thus DPic(A) −→ PI(A) is surjective as claimed.
Case: e > 1 and t > 1
By Theorem 6.0.5(iii) and Proposition 6.2.7, we have,
PI(A) ∼
= Se × X × h-idi,
where X contains Aut(D)/E. If D ∼
= Cp , then X ∼
= Aut(D)/E ∼
= Ct by Theorem
6.3.1.
The Picard group is given by Theorem 6.4.4.
Pic(A) ∼
= Ce × Aut(D)/E.
The (isomorphic) subgroup Ce of Pic(A) comes from the automorphisms σ ∈
AutO (A) of the form g σ = λ(g −1 )g, ∀g ∈ H, for some linear character λ ∈ Irr(A).
This corresponds to the perfect isometry χ 7→ λχ for χ ∈ Irr(A).
Let σ be a generator of the cyclic group Aut(D)/E. Then σ can be regarded as
an element in Aut(D o E) that acts trivially on E. The Aut(D)/E part in Pic(A) is
then generated by Bσ whose induced perfect isometry χ 7→ χσ generates the subgroup
Aut(D)/E ≤ X in PI(A). The map χ 7→ χσ fixes every non-exceptional characters
in Irr(A).
CHAPTER 6. BLOCKS WITH CYCLIC DEFECT GROUPS
105
Lemma 6.4.7. Suppose 1 < e < |D| − 1. If Conjecture 6.0.6 holds (for example
D∼
= Cp ), then DPic(O(D o E)) −→ PI(O(D o E)) is surjective.
Proof. Let A = O(D o E). By Conjecture 6.0.6 and Proposition 6.2.7, we have
PI(A) ∼
= Se × Aut(D)/E × h-idi,
where Se is the symmetric group on the non-exceptional characters, and Aut(D)/E is
a permutation group on the exceptional characters (we know that every perfect isometry permutes the non-exceptional and exceptional characters separately by Lemma
6.2.3).
Kunugi [29, Proposition 3.2] showed that, for every permutation of the nonexceptional characters, there is a tilting complex inducing the same permutation on
the non-exceptional characters. The perfect isometries from Kunugi’s complexes cover
the Se part of PI(A), but may incur non-identity permutations on the exceptional
characters. However, since Pic(A) is mapped surjectively onto Aut(D)/E ≤ PI(A),
we can cancel out these effect by composing with complexes from Pic(A) (embedded
in DPic(A)), which fixes non-exceptional characters. So, there are tilting complexes
which permute only the non-exceptional characters. Hence, DPic(A) is mapped onto
Se × Aut(D)/E. Finally, A[1] gives −id ∈ PI(A). Hence, DPic(A) −→ PI(A) is
surjective.
Chapter 7
Blocks with TI defect groups
Let G be a finite group. Let B be a block of OG with defect group D. Let H = NG (D)
and A be a block of OH with Brauer correspondent B. If D is abelian then we
can expect, by Broué’s conjecture, that A and B are perfectly isometric. If D is
not abelian, however, then one cannot expect such perfect isometries, especially if
k(A) 6= k(B). We want to study a case where we do have k(A) = k(B). This
happens, for example, when D is trivial intersection.
A subgroup D ≤ G is called trivial intersection (TI) if Dg ∩ D = 1 for every
g ∈ G − NG (D). An and Eaton [4] showed that if D is trivial intersection then,
amongst other things, we do have k(A) = k(B), l(A) = l(B). This means that, at
least, there are isometries between RK (A) and RK (B). Such isometries may not
necessarily be perfect. Cliff [10] and Robinson [40], for example, showed that there is
no perfect isometry when G is a Suzuki group Sz(q). On the other hands, Narasaki
[33] defined a generalization of a perfect isometry and showed that such isometries
always exist in TI defect group cases.
In this chapter we will investigate the relationship between perfect isometry groups
PI(A) and PI(B) when a defect group D is TI. Blocks with TI defect groups are
classified by An and Eaton in [3]. We will pick only some examples, most notably
the family of the Suzuki groups Sz(q).
106
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
7.1
107
Some Observations
In this section, we will describe some phenomenons that we observed in TI cases.
We were not able to prove these observations for all TI cases, even computationally,
since many of TI blocks have a large number of irreducible characters. This made
our GAP programs take a very long time to compute perfect isometry groups (see
the Index at the end of this thesis).
Perfect embedding
Let G, H be finite groups. Let B ∈ Bl(G), A ∈ Bl(H). Assume that k(B) = k(A).
We already know that if A and B are perfectly isometric, then PI(A) ∼
= PI(B). If
A and B are not perfectly isometric, it may happen that we can embed one perfect
isometry group inside the other.
Definition 7.1.1. Let I : RK (A) −→ RK (B) be an isometry. We say that I is a
perfect embedding (from A to B) if for every α ∈ PI(A), we have IαI −1 ∈ PI(B).
Suppose I : RK (A) −→ RK (B) is a perfect embedding. It is clear that the map
α 7→ IαI −1 gives a monomorphism PI(A) −→ PI(B), hence an embedding of PI(A)
into PI(B). Every perfect isometry is obviously a perfect embedding. An interesting
case is when a perfect embedding is not a perfect isometry. In TI case, we know that
there is an isometry I : RK (A) −→ RK (B). If A and B are not perfectly isometric, it
may happen that a perfect embedding still exists. However, in the case of the Suzuki
groups Sz(q), we will show that there is neither a perfect isometry nor a perfect
embedding between A and B if q > 8 (there is a perfect embedding when q = 8).
Some subgroups of perfect isometry groups
Recall that CFp0 (G, B; K) is the subspace of CF(G, B; K) consisting of class functions
vanishing on p-singular elements. If I ∈ PI(B), then I preserves CFp0 (G, B; K)
(Proposition 3.1.6). So, we can define the following subgroup.
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
108
Definition 7.1.2. Define PIs (B) by
PIs (B) = {I ∈ PI(B) : I(α) = α, ∀α ∈ CFp0 (G, B; K)}
Lemma 7.1.3. PIs (B) is a normal subgroup of PI(B).
Proof. That PIs (B) is a subgroup is clear from the definition. We will show that
PIs (B) is normal in PI(B).
Let I ∈ PIs (B) and J ∈ PI(B). For any class function α ∈ CFp0 (G, B; K),
JIJ −1 (α) = J(I(J −1 (α)))
= J(J −1 (α)) since J −1 (α) ∈ CFp0 (G, B; K)
= α.
Thus JIJ −1 ∈ PIs (B), and so PIs (B) C PI(B) as desired.
If there exists a perfect isometry I : RK (A) −→ RK (B), then the map J 7→ IJI −1
gives isomorphisms PIs (A) ∼
= PIs (B). In TI cases, we observed that PIs (A) ∼
= PIs (B)
even though A and B are not perfectly isometric.
7.2
Suzuki group Sz(q)
In this section, we will study the principal 2-block of the Suzuki group Sz(q) and
the principal 2-block of its Sylow normalizer. The perfect isometry groups for both
blocks will be calculated and we shall see that they are visibly different. This implies
that there is no perfect isometry between the two blocks.
For the rest of the section, we fix the following notations. Let p = 2, q = 22n+1 , r =
2n+1 where n is a positive integer. Let G = Sz(q) be the Suzuki group of order
q 2 (q − 1)(q 2 + 1). Let B be the principal block of OG with defect group D. Let A
be the principal block of ONG (D).
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
7.2.1
109
Principal block of G
The main result for this section is the following theorem.
Theorem 7.2.1. Let p = 2. Let G = Sz(q) where q > 2. Let B be the principal
√
block of OG. Let r = 2q. Then, every perfect isometry in PI(B) has a homogenous
sign. Furthermore, the irreducible characters Irr(B) can be partitioned into
X = {1} ∪ {Xi : i ∈ I},
|X| = q/2
Y = {Yj : j ∈ J },
|Y| = (q + r)/4
Z = {Zk : k ∈ K},
|Z| = (q − r)/4
W = {Wl : l ∈ L},
|W| = 2,
for some index sets I, J , K, L, such that
PI(B) = Sym(X) × Sym(Y) × Sym(Z) × Sym(W) × h-idi
∼
= Sq/2 × S(q+r)/4 × S(q−r)/4 × S2 × h-idi .
Character table
The information about conjugacy classes and irreducible characters in Irr(B) can be
found in the work of Suzuki [49].
Let I = {1, . . . , q/2−1}, J = {1, . . . , (q+r)/4}, K = {1, . . . , (q−r)/4}, L = {1, 2}.
Following Suzuki’s notations [49, Page 141], the irreducible characters in Irr(B) are
denoted by Xi (i ∈ I), Yj (j ∈ J ), Zk (k ∈ K), Wl (l ∈ L) and the trivial character is
denoted by 1.
Table 7.1: Characters, their degrees and the number of characters in each type.
Characters
Degrees
Number of characters
1
Wl
Yj
Xi
Zk
1
r(q − 1)/2
(q − r + 1)(q − 1)
q2 + 1
(q + r + 1)(q − 1)
1
2
(q + r)/4
q/2 − 1
(q − r)/4
The group G contains three cyclic subgroups A0 , A1 , A2 of order q −1, q +r +1, q −
r + 1 respectively. Let πi denote representatives of conjugacy classes of non-identity
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
110
elements of Ai . There is an involution σ and a unique conjugacy class of involutions.
There are two classes of elements of order 4; represented by ρ and ρ−1 . In total, there
are 7 conjugacy classes of G, represented by 1, σ, ρ, ρ−1 , π0 , π1 , π2 . The sizes of classes
are [10, Page 80] 1, (q 2 + 1)(q − 1), q(q
2 +1)(q−1)
2
, q(q
2 +1)(q−1)
2
, q 2 (q 2 + 1), q 2 (q − 1)(q − r +
1), q 2 (q − 1)(q + r + 1) respectively.
Let ω0 be a primitive (q − 1)-st root of unity. Denote by ξ0 a generator of A0 .
Define a character λi0 of A0 by
λi0 (ξ0j ) = ω0ij + ω0−ij .
Let ω1 and ω2 be primitive (q+r+1)-th and (q−r+1)-th roots of unity respectively.
If ξj is a generator of Aj , j = 1, 2, define a character λij of Ai by
λij (ξjk ) = ωjik + ωjikq + ωj−ik + ωj−ikq .
The character table (for Irr(B)) of Sz(q) is the following ([49, Theorem 13]).
Table 7.2: Table of characters in the
|CG (g)|
|G|
q2
2q
g
1
σ
ρ
I
1
1
1
r
ri
W1
r(q − 1)/2
−2
2
W2
r(q − 1)/2
− 2r
− ri2
Yj
(q − r + 1)(q − 1) r − 1
−1
..
..
..
j∈J
.
.
.
2
Xi
q +1
1
1
..
..
..
i∈I
.
.
.
Zk
k∈K
(q + r + 1)(q − 1)
..
.
−r − 1
..
.
−1
..
.
principal 2-block of Sz(q).
q 2 +1
q 2 +1
2q
q−1
q+r+1
q−r+1
ρ−1
π2
π0
π1
1
1
1
1
ri
−2
−1
0
1
ri
−1
0
1
2
j
−1
0
0
−λ1 (π1 )
..
..
..
..
.
.
.
.
i
1
0
λ0 (π0 )
0
..
..
..
..
.
.
.
.
−1
..
.
−λk2 (π2 )
..
.
0
..
.
0
..
.
Proof of Theorem 7.2.1
Let Iµ : RK (B) −→ RK (B) be an isometry, where µ is a character as in (3.3). We
will identify Iµ = (βµ , εµ ), where βµ is a bijection and εµ is a sign function:
βµ : Irr(B) 7→ Irr(B),
εµ : Irr(B) 7→ {±1}
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
111
associated to Iµ such that Iµ (χ) = εµ (χ)βµ (χ) for all χ ∈ Irr(B).
We will partition the characters in Irr(B) into the following four groups:
X = {1} ∪ {Xi : i ∈ I}
Y = {Yj : j ∈ J}
Z = {Zk : k ∈ K}
W = {Wl : l ∈ L}.
Our first result is that, any perfect isometry in PI(B) has a homogenous sign and
preserves the above partition.
Lemma 7.2.2. Let Iµ = (βµ , εµ ) be a perfect isometry. Then βµ preserves the partition {X, Y, Z, W} and εµ = ±1 (homogenous sign).
Proof. For convenience, let us define X0 = 1 and Ie = I ∪ {0}, so that X = {Xi : i ∈
e To show that the sign is homogenous, we assume that εµ (X0 ) = 1 and show that
I}.
εµ = 1.
For any i, j ∈ I, consider the class function
f = Xi − Xj .
Since the p-singular elements of G are cclG (σ) ∪ cclG (ρ) ∪ cclG (ρ−1 ), we see from
the character table that f vanishes on all p-singular elements of G. Since Iµ is
perfect, Iµ (f ) must also vanish on all p-singular elements. In particular, Iµ (Xi )(σ) =
Iµ (Xj )(σ). Since i, j are arbitrary, we have that
Iµ (Xi )(σ) = εµ (Xi )βµ (Xi )(σ) = constant, = c say,
e
∀i ∈ I.
e
This means that there must be q/2 copies of c in the sequence [Iµ (Xi )(σ), i ∈ I].
The possible values of βµ (χ)(σ) for χ ∈ Irr(B) are 1, −r/2, r − 1, −r − 1 with the
number of copies q/2, 2, (q + r)/4, (q − r)/4 respectively. Since εµ (X0 ) = +1, we must
have Iµ (X0 )(σ) 6= −1. Suppose that Iµ (X0 )(σ) 6= 1. Then c 6= ±1. So, εµ must not
be constant (not enough copies of c), and some numbers in {1, −r/2, r − 1, −r − 1}
must be equal to negatives of other numbers in the same set. But the numbers
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
112
±1, ±(−r/2), ±(r − 1), ±(−r − 1) are all distinct, this is impossible. Thus, we must
have c = 1. Hence,
e
βµ (Xi ) ∈ X, ∀ i ∈ Ie and εµ (Xi ) = +1, ∀ i ∈ I.
Next, consider the class function
f = Yi − Yj .
Since f vanishes on p-singular elements of G, so does Iµ (f ). In particular,
e
Iµ (Yi )(σ) = εµ (Yi )βµ (Yi )(σ) = constant, ∀i ∈ J.
The possible values of βµ (χ)(σ) for χ ∈ Irr(B) − X are −r/2, r − 1, −r − 1 with
the number of copies 2, (q + r)/4, (q − r)/4 respectively. Since (q + r)/4 = max{2, (q +
r)/4, (q − r)/4}, the same reasoning as before shows that
βµ (Yi ) ∈ Y, ∀ i ∈ J
and εµ (Yi ) = constant, ∀ i ∈ J .
Next, consider the class function
f = Zi − Zj .
Since f vanishes on p-singular elements of G, so does Iµ (f ). In particular,
e
Iµ (Zi )(σ) = εµ (Zi )βµ (Yi )(σ) = constant, ∀i ∈ K.
The possible values of βµ (χ)(σ) for χ ∈ Irr(B) − {X ∪ Y} are −r/2, −r − 1 with
the number of copies 2, (q −r)/4 respectively. Since (q −r)/4 > 2, the same reasoning
as before shows that
βµ (Zi ) ∈ Z, ∀ i ∈ K and εµ (Zi ) = constant, ∀ i ∈ K.
Finally we must have
βµ (W ) ∈ W,
∀ W ∈ W.
This proves that Iµ preserves the partition {X, Y, Z, W} (with signs). It now
remains to show that εµ (χ) = 1 for all χ ∈ Y ∪ Z ∪ W.
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
113
Let ξ1k be a non-identity element in A1 . By the column orthogonality relation:
X
χ(ξ1k )χ(σ) = 0
χ∈Irr(G)
1 − r/2 − r/2 + (r − 1)
X
[−λj1 (ξ1k ))] = 0
j∈J
∴
X
[−λj1 (ξ1k )] = 1.
j∈J
Consider,
X
µ(σ, ξ1k ) =
Iµ (χ)(σ)χ(ξ1k )
χ∈Irr B
= 1 + εµ (W1 )(−r/2) + εµ (W2 )(−r/2) + (r − 1)εµ (Y1 )
X
[−λj1 (ξ1k )]
j∈J
= 1 + εµ (W1 )(−r/2) + εµ (W2 )(−r/2) + (r − 1)εµ (Y1 )
Since µ is perfect, µ(σ, ξ1k ) = 0. This implies that
εµ (W ) = +1 ∀ W ∈ W,
and εµ (Y1 ) = 1, that is,
εµ (Y ) = +1 ∀ Y ∈ Y.
Let ξ2j be a non-identity element in A2 . By orthogonality relation applied to the
classes of σ and ξ2j , we see that,
X
(−λk2 (ξ2j ))) = 1.
k∈K
Consider,
µ(σ, ξ2k ) =
X
Iµ (χ)(σ)χ(ξ2k )
χ∈Irr B
= 1 + (−1)(−r/2) + (−1)(−r/2) + (−r − 1)εµ (Z1 )
X
[−λk2 (ξ2j )]
k∈K
= 1 + r + (−r − 1)εµ (Z1 ).
Since µ is perfect, µ(σ, ξ2k ) = 0. Therefore,
εµ (Z) = +1 ∀ Z ∈ Z.
So, εµ (χ) = +1, ∀χ ∈ Irr(B), given εµ (X0 ) = +1. Hence εµ is homogenous.
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
114
Next, we will show the converse of the previous lemma.
Lemma 7.2.3. Any isometry Iµ = (βµ , εµ ) such that βµ preserves the partition
{X, Y, Z, W} and εµ = ±1 is perfect.
Proof. Without loss of generality, assume that εµ = 1. Let g, h ∈ G. Suppose exactly
one of g, h is p-singular. Let U ∈ {X, Y, Z, W}. We see from the character table
(Table 7.2) that, either χ(g) is constant for all χ ∈ U or χ(h) is constant for all
χ ∈ U. In any case, we have
X
εµ (χ)βµ (χ)(g)χ(h) =
χ∈U
X
βµ (χ)(g)χ(h) =
χ∈U
X
χ(g)χ(h).
χ∈U
Consequently,
µ( g, h) =
X
X
εµ (χ)βµ (χ)(g)χ(h) =
χ∈Irr(B)
χ(g)χ(h) = µid (g, h) = 0,
χ∈Irr(B)
since id is a perfect isometry. This proves the separation condition.
For the integrality condition, since p divides |CG (g)| if and only if g is a p-singular
element, it remains to consider µ(g, h) when g or h is p-singular. The case where
exactly one of g, h is p-singular has been covered in the separation condition. So
assume now that g, h are both p-singular. Then, up to conjugation, g, h ∈ {σ, ρ, ρ−1 }.
We see from Table 7.2 that
X
βµ (X)(g)X(h) = q/2
(7.1)
X∈X
X
βµ (Y )(g)Y (h) =
Y ∈Y
X
Z∈Z
βµ (Z)(g)Z(h) =
q+r
4
q−r
4
Y1 (g)Y1 (h)
(7.2)
Z1 (g)Z1 (h).
(7.3)
Suppose that g, h 6= σ. Then |CG (g)| = |CG (h)| = 2q. The contribution from Y is
(q + r)/4, the contribution from Z is (q − r)/4, and the contribution from W is ±q.
So
µ(g, h) = q/2 + (q + r)/4 + (q − r)/4 ± q = 0 or 2q.
Thus µ(g, h) is divisible in O by |CG (g)| and |CG (h)|.
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
115
Suppose g = σ and h ∈ {ρ, ρ−1 }. The contribution from Y is (q + r)(1 − r)/4, the
contribution from Z is (q − r)(1 + r)/4, and the contribution from W is 0. So
µ(g, h) = q/2 +
(q + r)(1 − r) (q − r)(1 + r)
+
+ 0 = 0.
4
4
Suppose g ∈ {ρ, ρ−1 } and h = σ. Then, contributions from Y, Z, W are the same
as in the case g = σ and h ∈ {ρ, ρ−1 }. So µ(g, h) = 0.
Suppose g = h = σ. Then |CG (g)| = |CG (h)| = q 2 . The contribution from Y is
(q + r)(r − 1)2 /4, the contribution from Z is (q − r)(r + 1)2 /4, and the contribution
from W is q. So
µ(g, h) = q/2 +
(q + r)(r − 1)2 (q − r)(r + 1)2
+
+ q = q2.
4
4
Thus µ(g, h) is divisible in O by |CG (g)| and |CG (h)|.
Therefore Iµ is a perfect isometry as claimed.
With Lemma 7.2.2 and Lemma 7.2.3, we have proved Theorem 7.2.1.
7.2.2
Principal block of NG (D)
We will now determine PI(A).
Theorem 7.2.4. Let p = 2. Let G = Sz(q) where q > 2 and D ∈ Syl2 (G). Let
A be the principal block of ONG (D). Then, every perfect isometry in PI(A) has a
homogenous sign. Furthermore, the irreducible characters in Irr(A) can be labeled
{ϕ1 , . . . , ϕq−1 , χ1 , χ2 , χ3 } such that,
PI(A) = Sym{ϕ1 , . . . , ϕq−1 } × Sym{χ2 , χ3 } × h-idi
∼
= Sq−1 × S2 × h-idi .
Character table
References for the groups NG (D) and its irreducible characters can be found in [49],
[10] and [33].
We have |D| = q 2 and H = DC where C = hai is a cyclic group of order q − 1.
Let ω = e2πi/(q−1) . The conjugacy classes of H are given as follows. There is a unique
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
116
class of involutions, represented by σ, |CH (σ)| = q 2 . There are two classes of elements
of order 4, represented by ρ, ρ−1 , |CG (ρ)| = |CH (ρ−1 )| = 2q. There are q − 2 classes
of elements of odd orders, namely cclH (aj ), aj ∈ C − {1}, |CH (aj )| = q − 1.
The irreducible characters in Irr(A) are {ϕ1 , . . . , ϕq−1 , χ1 , χ2 , χ3 } and described
as follows. Each irreducible character in Irr(C) extends to ϕi for i = 1, . . . , q − 1,
and we denote the trivial character by ϕ1 . Each non-trivial irreducible character in
Irr(D) induces to χi for i = 1 . . . , 3.
Table 7.3: Table of characters
|CH (h)|
|H|
h
1
ϕi
1
χ1
q−1
χ2
r(q − 1)/2
χ3
r(q − 1)/2
in the principal 2-block of NG (D).
q2
2q
2q
q−1
σ
ρ
ρ−1
{aj }
1
1
1
ω j(i−1)
q − 1 -1
-1
0
r
ri
ri
−2
−2
0
2
ri
− 2r
− ri2
0
2
Proof of Theorem 7.2.4
Let Φ = {ϕ1 . . . , ϕq−1 } and X = {χ2 , χ3 }.
Lemma 7.2.5. If I ∈ PI(A), then I preserves the sets (up to signs) Φ and X. In
other words, I(Φ) ⊂ ±Φ, I(χ1 ) = ±χ1 and I(X) ⊂ ±X.
Proof. Since |ϕi (1)|2 = |χ1 (1)|2 = 1 for i = 1, . . . , q − 1 and |χ2 (1)|2 = |χ3 (1)|2 > 1,
we must have I(X) ⊂ ±X. For i, j = 1, . . . , q − 1, consider the class function
f = ϕi − ϕj .
From the character table (Table 7.3), we see that f vanishes on p-singular elements
of H. Since I is perfect, I(f ) also vanishes on p-singular elements of H. In particular,
I(ϕi )(σ) = constant , ∀i = 1, . . . , q − 1.
This implies that I(Φ) ⊂ ±Φ, and consequently I(χ1 ) = ±χ1 .
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
Let π ∈ Sym{1, . . . , q − 1} and τ ∈ Sym{2, 3}.
117
Define an isometry Iπ,τ :
RK (A) −→ RK (A) by
i = 1, . . . , q − 1
Iπ,τ (ϕi ) = ϕπ(i) ,
Iπ,τ (χ1 ) = χ1
Iπ,τ (χj ) = χτ (j) ,
j = 2, 3.
Lemma 7.2.6. Given any π ∈ Sym{1, . . . , q − 1} and τ ∈ Sym{2, 3}, the isometry
Iπ,τ is perfect.
Proof. Let µ be the generalized character induced by Iπ,τ . Let µid be the character
induced by the identity map id. We know that µid satisfies perfect conditions. Let
g, h ∈ H. Then
µ(g, h) =
q−1
X
ϕπ(i) (g)ϕ(h) + χ1 (g)χ1 (h) +
3
X
i=1
χτ (j) (g)χ(h).
j=2
We consider the separation condition first. From the character table,
µ(1, σ) = µid (1, σ) = 0
µ(σ, 1) = µid (σ, 1) = 0
µ(1, ρ±1 ) = µid (1, ρ±1 ) = 0
µ(ρ±1 , 1) = µid (ρ±1 , 1) = 0.
Let j be such that aj 6= 1. Since
P
i
ω j(i−1) = 0, we see from the character table
that
µ(σ, aj ) =
X
µ(aj , σ) =
X
µ(ρ±1 , aj ) =
X
µ(aj , ρ±1 ) =
X
ω j(i−1) = 0
i
ω j(i−1) = 0
i
ω j(i−1) = 0
i
i
ω j(i−1) = 0.
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
118
Thus, µ satisfies the separation condition. By Theorem 3.1.5, it suffices to consider
µ(g, h) where g, h ∈ H are p-singular elements. From the character table,
µ(σ, ρ±1 ) = µid (σ, ρ±1 ) ∈ q 2 O
µ(ρ±1 , σ) = µid (ρ±1 , σ) ∈ q 2 O.
It now remains to consider µ(g, h) for g, h ∈ {ρ, ρ−1 }. If g, h ∈ {ρ, ρ−1 }, then
µ(g, h) =
q−1
X
ϕπ(i) (g)ϕ(h) + χ1 (g)χ1 (h) +
i=1
3
X
χτ (j) (g)χ(h)
j=2
= (q − 1) + 1 ± r2 /2 = (q − 1) + 1 ± q
= 0 or 2q
∈ q 2 O.
So µ satisfies the integrality condition. Hence Iπ,τ is perfect as claimed.
We have shown that any perfect isometry must preserve (up to signs) the sets Φ
and X, and we found a choice of sign (homogenous sign) such that any permutation
of Φ and X gives rise to a perfect isometry. It follows by the uniqueness of signs that
every perfect isometry has a homogenous sign and, therefore, the group PI(A) is of
the form as described in Theorem 7.2.4.
7.2.3
Non-existence of perfect embedding when q > 8
Consider first when q = 8. By Theorem 7.2.1,
PI(B) = Sym(X) × Sym(Y) × Sym(Z) × Sym(W) × h-idi
∼
= S4 × S3 × S1 × S2 × h-idi .
By Theorem 7.2.4,
PI(A) = Sym{ϕ1 , . . . , ϕ7 } × Sym{χ2 , χ3 } × h-idi
∼
= S7 × S2 × h-idi .
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
We can see that there is a perfect

X
 0

 X1


 X
 2

 X
 3


 Y1


 Y2


 Y3


 Z
 1


 W1

W2
119
embedding PI(B) −→ PI(A) given by



ϕ1






 ϕ2 








 ϕ 
3






 ϕ 

 4 







 ϕ5 

 7→ 
.



 ϕ6 







 ϕ7 







 χ 

 1 







 χ2 




χ3
Now, if q > 8, then |X|, |Y|, |Z|, |W| > 1. So, there is no irreducible character in
Irr(B) that is fixed by every perfect isometry (with all-positive sign) in PI(B). On the
other hand, χ1 ∈ Irr(A) is always fixed by every perfect isometry (with all-positive
sign) in PI(A). Hence there is no perfect embedding PI(B) −→ PI(A).
7.2.4
PIs (A) ∼
= PIs (B)
Lemma 7.2.7. We have
PIs (A) ∼
= PIs (B)
Proof. We will prove by computing both PIs (A), PIs (B).
Let I ∈ PIs (B). Since χi − χj ∈ CFp0 (G, B; K) for all χi , χj ∈ X, we have
I(χi − χj ) = χi − χj for all χi , χj ∈ X. This implies that I is constant on X.
Similarly, I is constant on Y and Z.
If χ ∈ Irr(B), denote by χ◦ the restriction to p-regular elements. Let {dχϕ : χ ∈
Irr(B), ϕ ∈ IBr(B)} be the decomposition numbers. From the character table of G
we have
W1◦ =
X
dW1 ϕ ϕ =
ϕ∈IBr(B)
X
dW2 ϕ ϕ = W2◦ .
ϕ∈IBr(B)
So
dW1 ϕ = dW2 ϕ
∀ϕ ∈ IBr(B).
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
120
Since CFp0 (G, B; K) is spanned by {Φϕ : ϕ ∈ IBr(B)}, we must have
X
X
dχϕ χ = Φϕ = I(Φϕ ) =
χ∈Irr(B)
dχϕ I(χ).
χ∈Irr(B)
Expanding this, and using the fact that I is constant on Irr(B) − W gives
dW1 ϕ W1 + dW2 ϕ W2 = dW1 ϕ I(W1 ) + dW2 ϕ I(W2 )
dW1 ϕ (W1 + W2 ) = dW1 ϕ (I(W1 ) + I(W2 )).
It follows that
PIs (B) = Sym{W1 , W2 } ∼
= S2 .
Next, let I ∈ PIs (A). Since ϕi − ϕj ∈ CFp0 (H, A; K) for all i, j = 1, . . . , q − 1, I
must fix ϕi for all i.
From the character table of H, we have
X
χ◦2 =
d χ2 ϕ ϕ =
ϕ∈IBr(A)
X
dχ3 ϕ ϕ = χ◦3 .
ϕ∈IBr(A)
So
dχ2 ϕ = dχ3 ϕ
∀ϕ ∈ IBr(A).
Since CFp0 (H, A; K) is spanned by {Φϕ : ϕ ∈ IBr(A)}, we must have
X
dχϕ χ = Φϕ = I(Φϕ ) =
χ∈Irr(A)
X
dχϕ I(χ).
χ∈Irr(A)
Expanding this, and using the fact that I is constant on {ϕi : ∀i} gives
dχ2 ϕ χ2 + dχ3 ϕ χ3 = dχ2 ϕ I(χ2 ) + dχ3 ϕ I(χ3 )
dχ2 ϕ (χ2 + χ3 ) = dχ2 ϕ (I(χ2 ) + I(χ3 )).
It follows that
PIs (A) = Sym{χ2 , χ3 } ∼
= S2 .
Hence PIs (A) ∼
= PIs (B).
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
7.3
121
McLaughlin group M cL
From this section until the rest of this chapter, we will use the convention as in
Section 8.1 to describe perfect isometry groups.
Let G = M cL be the McLaughlin group. Let p = 5. Let D be a Sylow p-subgroup
of G and H = NG (D). Let B be the principal block of OG and A be the principal
block of OH. In this section we will calculate PI(B) and PI(A).
7.3.1
Principal block of G
The irreducible characters in B are as follows:
Characters
Degrees Characters
Degrees
χ1
1 χ13
4752
χ2
22 χ14
5103
χ3
231 χ15
5544
χ4
252 χ16 , χ17
8019
χ5 , χ6
770 χ21 , χ22
9856
χ7 , χ8
896 χ23 , χ24
10395
χ10 , χ11
3520
The perfect isometry group of the block B is as follows.
PI(B) ∼
= D8 × S4 × C2 × h-idi,
where
Subgroups
Generators
D8
¯
(5, 6), (2̄, 14)(4,
13)(5, 23, 6, 24)
S4
¯ 17, 21,
¯ 22)
(16, 17), (16,
C2
(7, 8)
Also,
PIs (B) = h(5, 6)i × h(23, 24)i
∼
= S2 × S2 .
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
7.3.2
122
Principal block of NG (D)
The irreducible characters in A are as follow:
Characters
Degrees
χ1 , χ2 , χ3 , χ4 , χ5 , χ6 , χ7 , χ8
1
χ9 , χ10 , χ11 , χ12
2
χ13 , χ14 , χ15 , χ16 , χ17 , χ18
20
χ19
24
The perfect isometry group of the block A is as follows.
PI(A) ∼
= ((S4 × S4 ) o S2 ) × ((S2 )4 o S2 ) × h-idi,
where
Subgroups
Direct factors
Components
(S4 × S4 ) o C2
S4
(1, 2), (1, 2, 7, 8)
S4
(3, 4), (3, 4, 5, 6)
C2
(1, 3)(2, 4)(5, 7)(6, 8)(13, 14)
(C2 )4 o C2
(C2 )4
C2
Generators
(9, 10), (11, 12), (15, 16), (17, 18)
(9, 11)(10, 12)(15, 17)(16, 18)
Also,
PIs (A) = h(15, 16)i × h(17, 18)i
∼
= S2 × S2 .
Notice that we have PIs (B) ∼
= PIs (A).
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
123
A perfect embedding PI(B) −→ PI(A) is given by







1
5








 13 

 2 
 −9 










14 






 3 
 19 







 15 








 4 
 10 














 16 








 5 
 15 




 7→ 
,
7
→



17







 6 
 16 







 21 








 7 
 3 







 22 








 8 
 4 














23










 10 
 13 




24
11
14
7.4

12 

11 


6 



−1 


−2 


7 


8 


17 


18
The group P SU3(q)
Next we investigate a family of blocks with TI defect group. By [3, Theorem 1.1],
the principal p-blocks of the group PSU3 (pm ) have TI defect groups, where m is any
positive integer.
7.4.1
P SU3 (3), p = 3
Let G = PSU3 (3), p = 3.
Principal block of G
Let B be the principal block of OG. The irreducible characters in Irr(B) and their
degrees are as follows.
Characters
Degrees Characters
Degrees
χ1
1 χ7 , χ8 , χ9
21
χ2
6 χ11 , χ12
28
χ3 , χ4 , χ5
7 χ13 , χ14
32
χ6
14
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
124
We have
PI(B) ∼
= S3 × C2 × h-idi,
where
Subgroups
Generators
S3
(2̄, 8̄)(1, 12)(3, 5),
(2̄, 8̄, 9)(1, 12, 11)(3, 5, 4)
C2
(13, 14)
Also,
PIs (B) = {id}.
Principal block of NG (D)
Let D be a Sylow p-subgroup of G. Let H = NG (D) and let A be the principal block
of H.
The irreducible characters in Irr(A) and their degrees are as follows.
Characters
Degrees
χ1 , . . . , χ 8
1
χ9 , . . . , χ12
6
χ13
8
We have
PI(A) ∼
= (C2 )4 o S4 × h-idi,
where
Subgroups
(C2 )4
S4
Generators
(1, 2), (7, 8), (6, 3), (5, 4)
(9, 10)(1, 7)(2, 8),
(9, 10, 11, 12)(1, 7, 6, 5)(2, 8, 3, 4)
Also,
PIs (A) = {id}.
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
125
Again in this case we have PIs (B) ∼
= PIs (A).
A perfect embedding PI(B) −→ PI(A) is given by the following isometry:




 1 
 1 




 2 
 −9 








 3 
 2 












 4 
 3 








 5 
 8 








 6 
 13 








 7  7→  12  .












 8 
 10 








 9 
 11 








 11 
 6 








 12 
 7 








 13 
 5 








14
4
7.4.2
P SU3 (5), p = 5
Let G = PSU3 (5), p = 5 and let B be the principal block of OG.
The irreducible characters in Irr(B) and their degrees are as follows.
Characters
Degrees Characters
Degrees
χ1
1 χ7
84
χ2
20 χ8
105
χ3
21 χ10 , χ11 , χ12
126
χ4 , χ4 , χ6
28 χ13 , χ14
144
We have
PI(B) = Sym{4, 5, 6} × h(1, 10)i × h(11, 12)i × h(13, 14)i × h-idi
∼
= S3 × C2 × C2 × C2 × h-idi .
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
126
Also,
PIs (B) = Sym{4, 5, 6}.
Let D be a Sylow p-subgroup of G. Let H = NG (D) and let A be the principal
block of OH.
The irreducible characters in Irr(A) and their degrees are as follows.
Characters
Degrees
χ1 , . . . , χ 8
1
χ9 , χ10 , χ11
8
χ12 , χ13
20
We have
PI(A) = (Sym{1, 2, 7, 8} × Sym{3, 4, 5, 6}) o h(1, 3)(2, 4)(7, 5)(8, 6)(12, 13)i
× Sym{9, 10, 11} × h-idi
∼
= ((S4 × S4 ) o S2 ) × S3 × h-idi,
where
Subgroups
Direct factors
Components
(S4 × S4 ) o C2
S4
(1, 2), (1, 2, 7, 8)
S4
(3, 4), (3, 4, 5, 6)
C2
(1, 3)(2, 4)(7, 5)(8, 6)(12, 13)
S3
(9, 10), (9, 10, 11)
S3
Generators
Also
PIs (A) = Sym{9, 10, 11}.
So PIs (B) ∼
= PIs (A).
CHAPTER 7. BLOCKS WITH TI DEFECT GROUPS
127
A perfect embedding PI(B) −→ PI(A) is given by the following isometry:




 1 
 1 




 2 
 12 








 3 
 5 












 4 
 9 








 5 
 10 








 6 
 11 








 7  7→  6  .












 8 
 13 








 10 
 2 








 11 
 7 








 12 
 8 








 13 
 3 








14
4
The following table illustrates how the isometry given above embeds PI(B) into
PI(A). We can see that every perfect isometry in PI(B) “appears” in PI(A).
PI(B)
Sym{4, 5, 6}
h(1, 10)i
h(11, 12)i h(13, 14)i
Irr(B)
4
5
6
1
10
11
12
13
14
3
7
2
8
Irr(A)
9
10
11
1
2
7
8
3
4
5
6
12
13
PI(A)
Sym{9, 10, 11}
Sym{1, 2, 7, 8}
Sym{3, 4, 5, 6}
h(12, 13)..i
Chapter 8
Sporadic groups
In this chapter we compute the perfect isometry groups of blocks of the first 12
Sporadic groups in ascending order1 . We do this using our programs written in GAP
[19]. Since we already obtained the results for blocks with defect 1 in Chapter 6, we
will only consider blocks with defect greater than 1.
Table 8.1: First 12 Sporadic groups in ascending order
Group
Order Factorized order
M11
M12
J1
M22
J2 or HJ
M23
HS
J3 or HJM
M24
M cL
He
Ru
7920
95040
175560
443520
604800
10200960
44352000
50232960
244823040
898128000
4030387200
145926144000
1
24 · 32 · 5 · 11
26 · 33 · 5 · 11
23 · 3 · 5 · 7 · 11 · 19
27 · 32 · 5 · 7 · 11
27 · 33 · 52 · 7
27 · 32 · 5 · 7 · 11 · 23
29 · 32 · 53 · 7 · 11
27 · 35 · 5 · 17 · 19
210 · 33 · 5 · 7 · 11 · 23
27 · 36 · 53 · 7 · 11
210 · 33 · 52 · 73 · 17
214 · 33 · 53 · 7 · 13 · 29
If a block B contains too many irreducible characters, it will take a very long time (15+ days)
to compute its perfect isometry group. We found that the practical limit is k(B) ≈ 12.
128
CHAPTER 8. SPORADIC GROUPS
8.1
129
Convention
For convenience, we will introduce the following notation for (perfect) isometries.
Let B be a block of G and suppose that Irr(B) = {χi : i ∈ I} where I ⊂ N. Let
n = max(I). Let I = RK (B) −→ RK (B) be an isometry. Write I = (σ, ε), where
σ ∈ Sn and ε is a sign function.
Let c be a cycle notation of σ. We will denote an isometry I by c̄, defined as
follows. c̄ contains the same numbers in the same order as c except that:
• if i ∈ c and −χi ∈ I(Irr(B)) then i appears as ī in c̄,
• if i 6∈ c and −χi ∈ I(Irr(B)) then add (ī) to c̄.
For example, let Irr(B) = {χ1 , . . . , χ4 }. An isometry (1, 2̄, 3̄) has the permutation
(1, 2, 3) and {−χ2 , −χ3 } ⊂ I(Irr(B)). So, it is the map


 
−χ2 
χ 1 


 
−χ3 
χ 2 


 
.
  7→ 
χ 
χ 
 1 
 3


 
χ4
χ4
An isometry (1̄, 3)(4̄) is the map




 χ3 
χ 1 


 
 χ2 
χ 2 


 
.
  7→ 
−χ 
χ 
 1
 3


 
χ4
−χ4
For short, we will write Irr(B) = I instead of Irr(B) = {χi : i ∈ I}. We will use
the subscripts of irreducible characters as in the ATLAS of finite groups [11].
Sometimes, the structure of a group S is very complicated and has a long description when using the GAP command:
gap> StructureDescription(S);
CHAPTER 8. SPORADIC GROUPS
130
In these cases, we will list important subgroups whose union of generators generate
the whole group, and, if possible, give the identity of S as in the Small Group Library
(SGL) in GAP. It is obtained by the following command:
gap> IdGroup(S);
If B is a block, we denote by δ(B) a defect group of B.
8.2
Mathieu group M11
M11 , p = 2
Block information:
Block B
Defect
B0
4
δ(B)
Irr(B)
k(B)
QD16 {1, 2, 3, 4, 5, 8, 9, 10}
8
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B0
(C2 )3 × h-idi
16
Generators
C2
(3, 4)
C2
(1, 5)(9, 10)(2̄)(3̄)(4̄)
C2
(1, 9)(5, 10)(8̄)
M11 , p = 3
Block information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
2
(C3 )2
{1, 2, 3, 4, 5, 6, 7, 8, 10}
9
Perfect isometry group:
Block
PI(B)
B0
H × h-idi
| PI(B)| Subgroups
2592
H
Generators
¯
(1, 5̄, 10)(2,
4, 7, 8̄, 3̄, 6),
¯
(1, 4)(2, 7, 8̄, 6̄)(5̄, 10)
CHAPTER 8. SPORADIC GROUPS
131
The group H has the identity [1296, 3490] in SGL. Moreover, H contains the
following subgroups whose union of generators generate the whole group H.
Subgroups
8.3
Generators
S3
(1, 3), (1, 3, 4)
S3
(5̄, 6̄), (5̄, 6̄, 7)
S3
¯
(2̄, 8̄), (2, 8̄, 10)
C2
¯ 6̄, 8̄)
(2, 7)(5̄, 10)(
C6
(1, 2, 4, 8̄, 3̄, 10)
Mathieu group M12
M12 , p = 2
Blocks information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
6
D
{1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13}
11
B1
2
C2 × C2
{4, 5, 14, 15}
4
where D ∈ Syl2 (M12 ) and has the identity [64, 134] in SGL.
Perfect isometry groups:
| PI(B)| Subgroups
Block
PI(B)
Generators
B0
C2 × h-idi
4
C2
(2, 3)(9, 10)
B1
S4 × h-idi
48
S4
¯ 14),
¯
(5, 15,
¯ 5, 15)
¯
(4, 14,
M12 , p = 3
Block information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
3
(C3 × C3 ) o C3
{1, 2, 3, 4, 5, 8, 9, 10, 11, 13, 15}
11
CHAPTER 8. SPORADIC GROUPS
132
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B0
S4 × D8 × h-idi
384
Generators
S4
(2, 3),
(2̄, 3, 4̄, 5)
D8
(1, 8),
(1, 9, 8, 10)(11, 13)
8.4
Mathieu group M22
M22 , p = 2
Block information:
Block B
Defect δ(B)
B0
7
D
Irr(B)
k(B)
{1, . . . , 12}
12
where D ∈ Syl2 (M22 ) and has the identity [128, 931] in SGL.
Perfect isometry group:
Block
PI(B)
| PI(B)|
Subgroups
B0
(C2 )2 × h-idi
8
C2
(3, 4)
C2
(10, 11)
Generators
M22 , p = 3
Block information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
2
C3 × C3
{1, 5, 7, 10, 11, 12}
6
Perfect isometry group:
Block
PI(B)
B0
S5 × h-idi
| PI(B)| Subgroups
240
S5
Generators
(1, 5),
(1, 5, 7, 10, 11)
CHAPTER 8. SPORADIC GROUPS
8.5
133
Mathieu group M23
M23 , p = 2
Block information:
Block B
Defect δ(B)
B0
7
D
Irr(B)
k(B)
{1, . . . , 11, 14, 15, 16, 17}
15
where D ∈ Syl2 (M23 ) and has the identity [128, 931] in SGL.
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B0
(C2 )3 × h-idi
16
Generators
C2
(7, 8)
C2
(10, 11)
C2
(3, 4)(14, 15)
It should be noted that the defect group of B0 is isomorphic to the defect group
of the principal 2-block of M22 . In both blocks, the perfect isometry groups are very
small compared to the number of possible permutations on the irreducible characters
in the blocks. This results in a very long runtime on a computer. In the above case,
it took 30 days to calculate PI(B0 )!
M23 , p = 3
Block information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
2
C3 × C3
{1, 2, 5, 9, 10, 11, 12, 13, 17}
9
Perfect isometry group:
Block
PI(B)
B0
H × h-idi
| PI(B)| Subgroups
2592
H
Generators
¯ 11, 12, 9̄, 10),
¯
(1̄, 5̄, 13)(2̄, 17,
¯ 2̄, 11,
¯ 12, 10)(5, 13)
(1̄, 17)(
CHAPTER 8. SPORADIC GROUPS
134
The group H has the identity [1296, 3490] in SGL. Moreover, H contains the following
subgroups whose union of generators generate the whole group H.
Subgroups
8.6
Generators
S3
¯
(1, 9), (1̄, 9, 17)
S3
¯ (2̄, 12,
¯ 13)
(2̄, 12),
S3
(5, 10), (5, 10, 11)
C2
¯
(2̄, 11)(5,
13)(10, 12)
C6
¯ 12, 9̄, 13)
¯
(1̄, 2, 17,
Mathieu group M24
M24 , p = 2
Block information2 :
Block B
Defect δ(B)
B0
10
D
Irr(B)
k(B)
{1, . . . , 26}
26
where D ∈ Syl2 (M24 ).
M24 , p = 3
Block information:
Block B
B0
Defect δ(B)
3
D
Irr(B)
k(B)
{1, 2, 5, 6, 8, . . . , 11, 17, 18, 19, 22, 23}
13
where D ∼
= (C3 × C3 ) o C3 .
Perfect isometry group:
2
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
| PI(B)| Subgroups
Block
PI(B)
B0
(C2 )4 × h-idi
8.7
135
32
Generators
C2
(5, 6)
C2
(10, 11)
C2
¯ 19,
¯ 22)
¯
(1, 8)(9̄, 23)(
C2
¯ 9̄, 23)
¯
(1, 22)(2, 17)(8̄, 19)(
Janko group J1
J1 , p = 2
Block information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
3
(C2 )3
{1, 6, 7, 8, 12, 13, 14, 15}
8
Perfect isometry group:
Block
PI(B)
B0
((C2 )4 o S4 ) × h-idi
| PI(B)| Subgroups
768
(C2 )4
Generators
(6, 12),
(7, 13),
(8, 14),
(1, 15)
S4
(6, 7)(12, 13),
(6, 7, 8, 1)(12, 13, 14, 15)
8.8
Janko group J2
J2 , p = 2
Blocks information3 :
3
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
136
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
7
D
{1, . . . 11, 13, 14, 15, 18, 20, 21}
17
B1
2
C2 × C2
{12, 16, 17, 19}
4
where D ∈ Syl7 (J2 ) and has the identity [128, 934] in SGL.
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B1
S4 × h-idi
48
Generators
¯ 17),
¯
(16, 19,
S4
¯ 16, 19)
¯
(12, 17,
J2 , p = 3
Block information:
Block B
Defect
Irr(B)
k(B)
B0
3
{1, . . . , 5, 8, 9, 12, 13, 16, 17, 20, 21}
13
where δ(B0 ) ∼
= (C3 × C3 ) o C3 .
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B0
C2 × S3 × h-idi
24
Generators
C2
(2, 3)
S3
(4, 5)(8, 9)(16, 17),
¯ 16)(4, 21, 5)(8, 9, 12)
(1̄, 17,
8.9
Janko group J3
J3 , p = 2
Block information4 :
Block B
B0
Defect δ(B)
7
D
Irr(B)
k(B)
{1, . . . , 13, 17, 18, 20, 21}
17
where D ∈ Syl2 (J3 ) and has the identity [128, 934] in SGL.
4
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
137
J3 , p = 3
Block information:
Block B
Defect δ(B)
B0
5
D
Irr(B)
k(B)
{1, . . . , 5, 7, 8, 9, 10, 13, . . . , 19}
16
where D ∈ Syl3 (J3 ) and has the identity [243, 9] in SGL.
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B0
C2 × C3 × S3 × h-idi
72
Generators
C2
(2, 3)
C3
(14, 15, 16)
S3
(4, 5)(7, 8)(17, 18),
¯ 8̄)(9̄, 18,
¯ 17)
(1̄, 5̄, 4)(7, 19,
8.10
Higman-Sims group HS
HS, p = 2
Blocks information5 :
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
9
D
{1, . . . , 13, 16, 17, 19, . . . , 23}
20
B1
2
C2 × C2
{14, 15, 18, 24}
4
where D ∈ Syl2 (HS).
Perfect isometry group:
Block
PI(B)
B1
S4 × h-idi
| PI(B)| Subgroups
48
S4
Generators
(14, 15),
¯ 15, 18, 24)
¯
(14,
5
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
138
HS, p = 3
Blocks information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
2
C3 × C3
{1, 2, 4, 10, 18, 19, 20, 23, 24}
9
B1
2
C3 × C3
{3, 5, 6, 7, 11, 12, 14, 15, 21}
9
Perfect isometry groups:
Block
PI(B)
B0
H0 × h-idi
| PI(B)| Subgroups
2592
H0
Generators
¯ 24, 23, 18),
¯
(1, 4)(2, 19)(10,
¯ 18,
¯ 20,
¯ 10, 24)
(1, 19, 2)(4̄, 23,
B1
H1 × h-idi
2592
H1
¯ 11, 14, 6̄)(5̄, 12,
¯ 15),
(3̄, 7̄, 21,
¯
¯ 21, 7̄)
(3, 11, 12)(5̄, 15)(6,
14,
The groups H0 , H1 are isomorphic and have the identity [1296, 3490] in SGL. The
group H0 contains the following subgroups whose union of generators generate the
whole group H0 .
Subgroups
Generators
S3
¯
(1, 4), (1̄, 4, 20)
S3
¯
(2, 18), (2̄, 18, 24)
S3
¯ 19),
¯ (10, 19,
¯ 23)
¯
(10,
C2
¯
(2̄, 23)(10,
24)(18, 19)
C6
¯ 18,
¯ 4, 24)
¯
(1̄, 2, 20,
The group H1 contains the following subgroups whose union of generators generate
the whole group H1 .
CHAPTER 8. SPORADIC GROUPS
Subgroups
139
Generators
S3
(3, 11), (3, 11, 12)
S3
¯
(5, 6), (5̄, 6, 21)
S3
¯ (7̄, 14,
¯ 15)
(7̄, 14),
C2
¯ 6̄, 14)(
¯ 7̄, 21)
¯
(5̄, 15)(
C6
¯ 6̄, 11,
¯ 21)
(3, 5̄, 12,
HS, p = 5
Block information:
Block B
Defect δ(B)
B0
3
D
Irr(B)
k(B)
{1, . . . , 6, 8, . . . , 12, 14, . . . , 18, 22}
17
where D ∼
= (C5 × C5 ) o C5 .
Perfect isometry group:
8.11
Block
PI(B)
| PI(B)|
Subgroups
B0
(C2 )3 × h-idi
16
C2
(5, 6)
C2
(11, 12)
C2
(14, 15)
Generators
McLaughlin group M cL
M cL, p = 2
Block information6 :
Block B
B0
Defect δ(B)
7
D
Irr(B)
k(B)
{1, . . . , 6, 9, 12, . . . , 20, 23, 24}
18
where D ∈ Syl2 (M cL) has the identity [128, 931] in SGL.
6
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
140
M cL, p = 3
Block information7 :
Block B
Defect δ(B)
B0
6
D
Irr(B)
k(B)
{1, . . . , 13, 15, 18, . . . , 24}
21
where D ∈ Syl3 (M cL) and has the identity [729, 321] in SGL.
M cL, p = 5
Block information:
Block B
Defect δ(B)
B0
3
D
Irr(B)
k(B)
{1, . . . , 8, 10, 11, 13, . . . , 17, 21, . . . , 24}
19
where D ∼
= (C5 × C5 ) o C5 .
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B0
S4 × S2 × D8 × h-idi
768
S4
Generators
(16, 17),
¯ 17, 21,
¯ 22)
(16,
S2
(7, 8)
D8
(5, 6),
¯
(2̄, 14)(4,
13)(5, 23, 6, 24)
8.12
Held group He
He, p = 2
Blocks information8 :
Block B
7
8
Defect δ(B)
Irr(B)
k(B)
B0
10
D
{1, . . . , 11, 13, 17, . . . , 21, 23, . . . , 29, 32, 33}
26
B1
3
D8
{12, 14, 15, 16, 22}
5
B0 contains too many irreducible characters to compute its perfect isometry group.
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
141
where D ∈ Syl2 (He).
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B1
D8 × h-idi
16
Generators
¯ 22),
¯
(16,
D8
¯ 22, 15,
¯ 16)(14)
¯
(12,
He, p = 3
Blocks information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
3
D
{1, 6, 9, 14, 15, 22, 25, 26, 28, . . . , 31, 33}
13
B1
2
C3 × C3
{2, 3, 7, 8, 10, . . . , 13, 16}
9
where D ∼
= (C3 × C3 ) o C3 .
Perfect isometry groups:
| PI(B)| Subgroups
Block
PI(B)
B0
(C2 )4 × h-idi
B1
H × h-idi
32
2592
Generators
C2
¯ 33)
¯
(26,
C2
(30, 31)
C2
¯
(1̄, 22)(6,
14)(15, 29)
C2
¯
(1̄, 29)(9,
28)(15, 22)
H
¯ 13,
¯ 12, 11),
¯
(2̄, 8̄, 16)(3, 7̄, 10,
¯
¯ 12,
¯ 16)
(2, 3)(7̄, 11)(8,
10,
The group H has the identity [1296, 3490] in SGL and it contains the following
subgroups whose union of generators generate the whole group H.
Subgroups
Generators
S3
¯
(2, 3), (2̄, 3, 13)
S3
(7, 8), (7, 8, 12)
S3
¯ 11, 16)
¯
(10, 11), (10,
C2
¯ 10,
¯ 12)
¯
(7, 16)(8̄, 11)(
C6
¯ 13, 8, 3̄, 7̄)
(2̄, 12,
CHAPTER 8. SPORADIC GROUPS
142
He, p = 5
Block information:
Block B
Defect
δ(B)
Irr(B)
k(B)
B0
2
C5 × C5
{1, . . . , 5, 7, 8, 14, . . . , 18, 28, 30, 31, 33}
16
Perfect isometry group:
Block
PI(B)
| PI(B)|
H × h-idi 1658880
B0
Subgroups
S4
Generators
(1, 2),
¯
(1̄, 2, 3, 33)
S4
(7, 8),
(7, 8, 30, 31)
C2
¯
¯ 28)
¯
(1̄, 7̄)(2̄, 8̄)(3̄, 30)(33,
31)(14,
S6
(4, 5),
¯ 16,
¯ 17,
¯ 18)
(4̄, 5, 15,
where H ∼
= ((S4 × S4 ) o C2 ) × S6 .
He, p = 7
Block information9 :
Block B
B0
Defect δ(B)
3
D
Irr(B)
k(B)
{1, . . . , 6, 9, . . . , 14, 16, 19, . . . , 25, 27, 30, 31}
23
where D ∼
= (C7 × C7 ) o C7 .
8.13
Rudvalis group Ru
Ru, p = 2
Blocks information10 :
9
10
B0 contains too many irreducible characters to compute its perfect isometry group.
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
143
Block B
Defect
Defect group
Irr(B)
k(B)
B0
14
D
{1, . . . , 31, 33}
32
B1
2
C2 × C2
{32, 34, 35, 36}
4
where D ∈ Syl2 (Ru).
Perfect isometry group:
| PI(B)| Subgroups
Block
PI(B)
B1
S4 × h-idi
48
S4
Generators
¯ 34),
¯
(32,
¯ 35, 36)
¯
(32, 34,
Ru, p = 3
Block information:
Block B
B0
Defect δ(B)
3
D
Irr(B)
k(B)
{1, 4, 9, 15, 16, 20, . . . , 23, 25, 26, 32, 33, 36}
14
where D ∼
= (C3 × C3 ) o C3 .
Perfect isometry group:
Block
PI(B)
B0
H × h-idi
| PI(B)| Subgroups
144
S3
Generators
¯ 21,
¯ 23)(22,
¯
(4̄, 33)(
26),
¯ 9̄, 26,
¯ 22)(20, 21,
¯ 23)
¯
(1̄, 4, 33)(
S3
(25, 32),
¯ 32, 36)
¯
(25,
C2
(15, 16)
where H ∼
= S3 × S3 × C2 .
Ru, p = 5
Block information11 :
11
B0 contains too many irreducible characters to compute its perfect isometry group.
CHAPTER 8. SPORADIC GROUPS
Block B
B0
Defect δ(B)
3
D
144
Irr(B)
k(B)
{1, . . . , 7, 10, 14, . . . , 19, 21, 22
25
24, 27, 29, 30, 31, 33, . . . , 36}
where D ∼
= (C5 × C5 ) o C5 .
Chapter 9
Other blocks
In this chapter, we will computer perfect isometry groups for some blocks that are
known to be perfectly isometric to O(D o E) where D is a defect group and E is the
inertial quotient of the block. This allows us to consider only the group D o E.
9.1
Blocks with defect group C2 × C2
Let p = 2. Blocks over k with Klein four defect groups have been classified by
Erdmann [16]. More specifically, let B be a block of OG with the Klein four group
D as a defect group. Let B ∗ be the block of kG corresponding to B. Then, Erdmann
[16, Theorem 4] showed that B ∗ is Morita equivalent to one of {kD, kA4 , C ∗ } where
C ∗ in the principal block of kA5 , corresponding to the principal block C of OA5 .
Linckelmann [31], later extended Erdmann’s results to O, by showing that B is Morita
equivalent to one of {OD, OA4 , C} ([31, Corollary 1.4]) and derived equivalent to
either OD or OA4 ([31, Corollary 1.5]). Since perfect isometry groups are preserved
by derived equivalences, we will show that in fact the isomorphism class of PI(B) is
unique. Much of the results are, of course, due to the harder work of Erdmann and
Linckelmann for their classifications of blocks with Klein four defect groups.
Theorem 9.1.1. Let p = 2. Let B be a block of OG with defect group C2 × C2 . Then
| Irr(B)| = 4, and every permutation of Irr(B) gives a perfect isometry (with a choice
145
CHAPTER 9. OTHER BLOCKS
146
of sign). Furthermore,
PI(B) ∼
= S4 × h-idi .
Proof. By [31, Corollary 1.5], B is derived equivalent to either OD or OA4 . It follows
that PI(B) is isomorphic to either PI(OD) or PI(OA4 ), and | Irr(B)| = | Irr(D)| = 4.
Our calculations in GAP [19] show that
PI(OD) ∼
= PI(OA4 ) ∼
= S4 × h-idi .
Hence PI(B) ∼
= S4 × h-idi as claimed.
Remark 9.1.2. If E is the inertial quotient of the block B, then E is isomorphic to
a subgroup of Aut(C2 × C2 ) ∼
= S3 . Since |E| is coprime to 2 (by Theorem 2.3.10),
|E| = 1 or 3. If |E| = 3, then D o E ∼
= A4 . Hence B is perfectly isometric to D o E,
a phenomenon that occurs quite often in blocks with abelian defect groups, including
the cyclic defect group case (cf. Chapter 6).
9.2
Blocks with defect group C3 × C3 and cyclic
inertial quotients of order 4
Another situation where we have that a block is perfectly isometric to the block of
a semidirect product of its defect group and its inertial quotient group is given by
Puig and Usami. Let p = 3. Let B be a block of OG with defect group D. Let E be
the inertial quotient group of B. Denote by A the block O(D o E). Puig and Usami
[37] showed that if D is abelian and E is cyclic of order 4, then there exists a perfect
isometry from RK (B) to RK (A). As a consequence, we have PI(A) ∼
= PI(B).
Theorem 9.2.1. Let G be a finite group. Let B be a block of OG with defect group
C3 × C3 and inertial quotient C4 . Then
PI(B) ∼
= S4 × C2 × h-idi .
Proof. Let D = C3 × C3 be a defect group and E = C4 be the inertial quotient. Since
E is a subgroup of Aut(D) = GL2 (3), it can be checked (for example using GAP)
CHAPTER 9. OTHER BLOCKS
147
that if E1 , E2 ≤ Aut(D) are cyclic groups of order 4 then D o E1 ∼
= D o E2 . Hence
D o E is unique. Let H = D o E and A = OH. By [37], B and A are perfectly
isometric and so PI(B) ∼
= PI(A). We shall now work with the group H instead.
The irreducible characters of H are (numbers in the brackets are their degrees)
(1)
(1)
(1)
(1)
(4)
(4)
Irr(A) = {χ1 , χ2 , χ3 , χ4 , χ5 , χ6 }.
Calculations in GAP show that all perfect isometries have homogenous signs and
(1)
(1)
(1)
(1)
(4)
(4)
PI(A) = Sym{χ1 , χ2 , χ3 , χ4 } × Sym{χ5 , χ6 } × h-idi
∼
= S4 × C2 × h-idi .
Chapter 10
Isotypies
Let G be a finite group and B be a block of OG. In this chapter, we will study
perfect isometries I ∈ PI(B) with an extra condition: these perfect isometries will be
called isotypies. We will consider the case where CG (x) is a p-group whenever x ∈ G
is a p-element. We will then apply our result to finding isotypies in blocks of cyclic
p-groups and the principal 2-blocks of the Suzuki groups Sz(q).
In order to define an isotypy, we need to use subpairs. We will not go into the
details here but instead refer the reader to [1] for more information.
Let x ∈ G be a p-element and let P be the cyclic group generated by x. Let BP
be a block of OCG (P ) such that (P, BP ) is a B-subpair of G (so (BP )G = B). Let
eP be a block idempotent of BP . We can define a linear map
(x,eP )
dG
: CF(G, B; K) −→ CFp0 (CG (P ), BP ; K),
by
(x,eP )
dG
(α)(y) = α(xyeP ),
for all α ∈ CF(G, B; K) and y ∈ CG (P )p0 . If y ∈ CG (P ) is a p-singular element, then
(x,eP )
we set dG
(α)(y) = 0.
Remark 10.0.2. If B is a block of OG with block idempotent e (so B = OGe), we
can define CF(G, e; K) to be the subspace of CF(G; K) of class functions α satisfying
148
CHAPTER 10. ISOTYPIES
149
α(ge) = α(g) for all g ∈ G. This is the notation used in [8] where we will take the
definition of isotypy from. By Lemma 2.3.2, we have CF(G, e; K) = CF(G, B; K).
So, for consistency with our notations used in this thesis, we will continue to use
CF(G, B; K) instead of CF(G, e; K).
Definition 10.0.3. [8, Cf. Definition 4.6] Let B be a block of OG with block
idempotent e and defect group D. Let (D, BD ) be a maximal B-subpair of G. Each
subgroup P ≤ D determines a unique block BP of OCG (P ) such that (P, BP ) ≤
(D, BD ) ([1, Theorem 3.4]). Let eP be a block idempotent of BP .
Let I ∈ PI(B). We say that I is an isotypy if for each cyclic subgroup P ≤ D,
there is I P ∈ PI(BP ) (where I {1} = I) such that the diagram
CF(G, B; K)

(x,e ) 
dG P y
I=I {1}
−−−−→
CF(G, B; K)

 (x,eP )
ydG
(10.1)
CFp0 (CG (P ), BP ; K) −−−→ CFp0 (CG (P ), BP ; K)
IP
is commutative for every generator x of P .
The family of perfect isometries (I P )P ≤D with I {1} = I as above is called a local
system for I.
Remark 10.0.4.
1. In [8, Definition 4.6], Broué considered isotypies in a more
general setting. He defined isotypies between Brauer correspondent blocks B of
OG and A of OH having a common defect group. Here, we will consider only
in the special case G = H and B = A.
2. Let I ∈ PI(B) be an isotypy. A local system for I may not be unique, since for
each P ≤ D, there may be more than one choice of I P ∈ PI(BP ) making the
diagram (10.1) commutative.
(1,e)
3. If P = {1}, then we always have dG
(1,e)
◦ I = I ◦ dG
for any I ∈ PI(B), as is
easily verified.
Definition 10.0.3 says that an isometry I : RK (B) −→ RK (B) is an isotypy if I is
a perfect isometry and I has a local system (I P )P ≤D with I {1} = I, where P runs over
CHAPTER 10. ISOTYPIES
150
all cyclic subgroups of D. Actually, the fact that I is perfect automatically follows if
I has a local system (I P ){1}6=P ≤D . This is explained in the following lemma.
Lemma 10.0.5. Let B be a block of OG with defect group D. Let I : RK (B) −→
RK (B) be an isometry. Suppose there exists a family of perfect isometries
(I P : RK (BP ) −→ RK (BP )){{1}6=P ≤D}
such that, for all non-trivial cyclic subgroup P ≤ D and a generator x of P , we have
(x,eP )
dG
(x,eP )
◦ I = I P ◦ dG
.
Then I is an isotypy and (I P )P ≤D with I {1} = I is a local system for I. In particular,
I is a perfect isometry.
Proof. This follows from the remark after [8, Definition 4.6].
We will now define an isotypy subgroup of PI(B).
Lemma 10.0.6. The set of isotypies in B is a subgroup of PI(B).
Proof. For each non-trivial cyclic subgroup P ≤ D with a generator x, take I P =
(x,eP )
id ∈ PI(BP ). Then dG
(x,eP )
◦ id = dG
(x,eP )
= I P ◦ dG
. So id ∈ PI(B) is an isotypy
with a local system (I P )P ≤D where I P = id, ∀P ≤ D.
Suppose I ∈ PI(B) is an isotypy with a local system (I P )P ≤D . Let hxi = P ≤ D
(x,eP )
be a cyclic subgroup. Let d = dG
. Let χ ∈ Irr(B) and suppose that I(χ) = ϕ ∈
± Irr(B). Then
(I P )−1 (d(ϕ)) = (I P )−1 (d(I(χ))) = (I P )−1 (I P (d(χ))) = d(χ) = d(I −1 (ϕ)).
Since {I(χ) : χ ∈ Irr(B)} spans CF(G, B; K), it follows that the diagram (10.1) is
commutative. Thus I −1 is an isotypy with a local system ((I P )−1 )P ≤D .
Finally, let I, J ∈ PI(B) be isotypies with local systems (I P )P ≤D and (J P )P ≤D
respectively. With similar arguments as above using associativity, it is easy to see
that I ◦ J is also an isotypy with a local system (I P ◦ J P )P ≤D . This completes the
proof.
CHAPTER 10. ISOTYPIES
151
Definition 10.0.7. The subgroup of PI(B) consisting of isotypies is called the isotypy
subgroup of PI(B), denoted by IPI(B).
The following proposition characterizes a local system for isotypies in the case
where CG (x) is a p-group for every p-element x ∈ G.
Proposition 10.0.8. Let B be the principal block of OG with defect group D. Let
P ≤ D be a cyclic subgroup with P = hxi. Suppose that CG (x) is a p-group. Let
I : RK (B) −→ RK (B) be an isometry. Then, the following are equivalent.
(x,eP )
(i) There exists I P ∈ PI(BP ) such that dG
(x,eP )
◦ I = I P ◦ dG
.
(ii) Either I(χ)(x) = χ(x) ∀χ ∈ Irr(B) or I(χ)(x) = −χ(x) ∀χ ∈ Irr(B).
Proof. Note that since CG (x) = CG (P ) is a p-group, BP = OCG (P ), eP = 1 and
Irr(BP ) = Irr(C). The condition that B is the principal block ensures that (BP )G =
B, since by Brauer’s Third Main Theorem, we must have that (BP )G is the principal
(x,eP )
block. For convenience, we will denote the group CG (P ) by C and the map dG
by
d.
(i) ⇒ (ii). Let χ ∈ Irr(B). Since eP = 1, by the definition, we have d(χ)(1) =
χ(x) and d(χ)(h) = 0 for all p-singular elements h ∈ C. So,
I P (d(χ))(1) =
=
µ P (1, 1)d(χ)(1)
1 X
µI P (1, h−1 )d(χ)(h) = I
|C| h∈C
|C|
µI P (1, 1)
χ(x).
|C|
(10.2)
Let δ(I P ) := µI P (1, 1)/|C|. By Lemma 5.0.7,
P
P
P
2
I
(θ)(1)θ(1)
±
θ∈Irr(C)
θ∈Irr(C) θ(1)
P
δ(I ) =
=
= ±1.
|C|
|C|
Since Irr(B) is a K-basis for CF(G, B; K), we have d(I(χ))(1) = I P (d(χ))(1) for
all χ ∈ Irr(B). But, d(I(χ))(1) = I(χ)(x), and I P (d(χ))(1) = δ(I P )χ(x) from (10.2).
Thus,
I(χ)(x) = δ(I P )χ(x),
for all χ ∈ Irr(B). Substituting δ(I P ) = ±1 yields (ii).
CHAPTER 10. ISOTYPIES
152
(ii) ⇒ (i). Suppose I(χ)(x) = δ(I)χ(x) for all χ ∈ Irr(B), where δ(I) ∈ {±1}.
Take I P ∈ {±id} such that µI P (1, 1)/|C| = δ(I). Let χ ∈ Irr(B). Suppose 1 6= y ∈ C,
then
d(I(χ))(y) = 0 = I P (d(χ))(y),
by the definition of d and the fact that I P is perfect. On the other hands, the only
p-regular element in C is 1, so
d(I(χ))(1) = I(χ)(x) = δ(I)χ(x) =
µI P (1, 1)
χ(x) = I P (d(χ))(1).
|C|
So,
d(I(χ))(1) = I P (d(χ))(1).
Hence, we have d(I(χ)) = I P (d(χ)) for all χ ∈ Irr(B). This proves (i).
Note that Proposition 10.0.8 does not require the isometry I to be perfect.
Corollary 10.0.9. Suppose that CG (x) is a p-group for every p-element 1 6= x ∈ G.
Let B be the principal block of OG. Let I : RK (B) −→ RK (B) be an isometry. The
following are equivalent.
(i) I is an isotypy.
(ii) For each p-element 1 6= x ∈ G, either I(χ)(x) = χ(x) ∀χ ∈ Irr(B) or I(χ)(x) =
−χ(x) ∀χ ∈ Irr(B).
Proof. Let D ∈ Sylp (G) be a defect group of B. Suppose I is an isotypy. Let
1 6= x ∈ G be a p-element. Without loss of generality, assume that x ∈ D. Let
(x,eP )
hxi = P ≤ D. Then there exists I P ∈ PI(BP ) such that dG
(x,eP )
◦ I = I P ◦ dG
.
By Proposition 10.0.8, we have that either I(χ)(x) = χ(x) ∀χ ∈ Irr(B) or I(χ)(x) =
−χ(x) ∀χ ∈ Irr(B). This proves (ii).
Now suppose (ii). Let P ≤ D be a non-trivial cyclic subgroup and let x be a
generator of P . By (ii) and Proposition 10.0.8, there exists I P ∈ PI(BP ) such that
(x,eP )
dG
(x,eP )
◦ I = I P ◦ dG
. Since this is true for any non-trivial cyclic subgroup P of D
and generator x of P , by Lemma 10.0.5, it follows that I is an isotypy.
CHAPTER 10. ISOTYPIES
10.1
153
Cyclic p-groups
Let G be the cyclic group Cq where q is a power of a prime p and let B = OG. Fix a
generator x of G and let ω = e2πi/q . All irreducible characters of G are of the form χk
for k ∈ {0, . . . , q − 1}, where χk (xa ) = ω ka . Suppose I ∈ PI(B) is a perfect isometry.
By Lemma 5.0.7, we know that I has a homogenous sign. So I(χk ) = δ(I)χσI (k) , ∀k
where δ(I) ∈ {±1} and σI is a bijective function from the set {0, . . . , q − 1} to itself.
Theorem 10.1.1. The isotypies in PI(B) are as follows.
• If p is even then the isotypies are {±id, ±J} where J(χk ) = χk+q/2
mod q
, ∀k.
So IPI(B) = hJi × h-idi.
• If p is odd then the isotypies are {±id}. So IPI(B) = h-idi.
Proof. Since I ∈ PI(B) is an isotypy if and only if −I is an isotypy, it suffices to find
isotypies with all-positive signs.
Suppose that I ∈ PI(B) is an isotypy with all-positive sign, say I(χk ) = χσ(k)
for some bijective function σ : {0, . . . , q − 1} → {0, . . . , q − 1}. Since CG (g) is a pgroup for every non-trivial p-element g ∈ G, by Corollary 10.0.9, we have χσ(k) (x) =
I(χk )(x) = δχk (x) for all k ∈ {0, . . . , q − 1}, where δ ∈ {±1}.
Suppose δ = 1. Then χσ(k) (x) = χk (x), ∀k. So ω σ(k) = ω k . Thus σ(k) − k is
divisible by q. But σ(k), k ∈ {0, . . . , q − 1}, this implies that σ(k) = k. That is, I is
the identity map.
Suppose δ = −1. Then χσ(k) (x) = −χk (x), ∀k. So ω σ(k) = −ω k . Thus σ(k) 6= k
and q divides 2(σ(k) − k). Suppose p is odd, then q divides σ(k) − k. This implies
that σ(k)−k = 0, a contradiction. Hence, the only isotypies when p is odd are {±id}.
Suppose that p is even. Then q/2 divides (σ(k) − k). Since σ(k) 6= k, ∀k, we have
0 < (σ(k) − k) < q. This implies that σ(k) = k + q/2, ∀k, that is, I = J. To verify
that J is an isotypy, consider
J(χk )(xm ) = χk+q/2
mod q
(xm ) = ω mk+mq/2 = ω mq/2 ω mk = (−1)m χk (xm ).
CHAPTER 10. ISOTYPIES
154
This shows that, for each g ∈ G, we have J(θ)(g) = δθ(g) for all θ ∈ Irr(G), where
δ ∈ {±1}. By Corollary 10.0.9, this implies that J is an isotypy. Hence the isotypies
when p is even are {±id, ±J}.
10.2
Suzuki group Sz(q)
Let p = 2 and G be the Suzuki group Sz(q). Let B be the principal block of OG with
defect group D. We will use the same notations as in Section 7.2.
Theorem 10.2.1. The isotypy subgroup of PI(B) are
IPI(B) ∼
= Sq/2 × S(q+r)/4 × S(q−r)/4 × h-idi .
In other words, every perfect isometry that sends W1 to {±W1 } and sends W2 to
{±W2 } is an isotypy.
Proof. We have
Irr(B) = {1, Xi } ∪ {Yj } ∪ {Zk } ∪ {W1 , W2 }.
From the results in Section 7.2, every perfect isometry has a homogenous sign and
PI(B) = Sq/2 × S(q+r)/4 × S(q−r)/4 × S2 × h-idi,
where Sq/2 , S(q+r)/4 , S(q−r)/4 , S2 are the permutation groups of {1, Xi }, {Yj }, {Zk } and
{W1 , W2 } respectively.
The p-elements of G are, up to conjugation, 1, σ, ρ, ρ−1 . We also have |CG (σ)| =
q 2 , |CG (ρ)| = |CG (ρ−1 )| = 2q. So, Corollary 10.0.9 applies. Let I ∈ PI(B) be a
perfect isometry with all-positive sign. By Corollary 10.0.9, I is an isotypy if and
only if
• I(χ)(σ) = χ(σ) for all χ ∈ Irr(B),
• I(χ)(ρ) = χ(ρ) for all χ ∈ Irr(B) and
• I(χ)(ρ−1 ) = χ(ρ−1 ) for all χ ∈ Irr(B).
CHAPTER 10. ISOTYPIES
155
Looking at the character table of G (Table 7.2), it is clear that we must have
I(W1 ) = W1 , I(W2 ) = W2 . This implies that I belongs to the subgroup Sq/2 ×
S(q+r)/4 × S(q−r)/4 . But I is an isotypy if and only if −I is an isotypy, we have
IPI(B) ∼
= Sq/2 × S(q+r)/4 × S(q−r)/4 × h-idi .
Remark 10.2.2. Let I : RK (B) −→ RK (B) be an isometry with all-positive sign
such that I(X) = X, I(Y) = Y, I(Z) = Z, I(W1 ) = W1 , I(W2 ) = W2 . Then we can
see that
• I(χ)(σ) = χ(σ) for all χ ∈ Irr(B),
• I(χ)(ρ) = χ(ρ) for all χ ∈ Irr(B) and
• I(χ)(ρ−1 ) = χ(ρ−1 ) for all χ ∈ Irr(B).
By Corollary 10.0.9, I is an isotypy. In particular, I is perfect. This gives an
alternative, quicker, and partial proof of Lemma 7.2.3.
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Appendix A
Computer-aided results
Given a group G and a character table of G. Let B is a block of OG and suppose we
know the irreducible characters in Irr(B). Then, in principal, it is always possible to
calculate PI(B). The basic algorithm is as follows.
1. Pick a new permutation σ : Irr(B) −→ Irr(B). If there is no permutation left,
finish.
2. Pick a new sign function ε : Irr(B) −→ {±1}. If there is no sign function left,
that means σ is not a perfect permutation. We must try another permutation.
Repeat from 1.
3. Construct an isometry I : RK (B) −→ RK (B) such that I(χ) = ε(χ)σ(χ).
4. Construct a character µ ∈ RK (B ⊗O B ◦ ) such that
µ(g, h) =
X
I(χ)(g)χ(h).
χ∈Irr(B)
5. Check for all g, h ∈ G if µ is perfect (Definition 3.1.1). For the integrality
condition, we can use use (2.1) and (2.3) to determine if µ(g, h)/|CG (g)| and
µ(g, h)/|CG (h)| are in O.
6. If µ is perfect, record I and −I. Then repeat from 1. (By the uniqueness of
sign, the only other sign τ such that χ 7→ τ (χ)σ(χ) is perfect is τ = −ε.)
162
APPENDIX A. COMPUTER-AIDED RESULTS
163
7. If µ is not perfect, that means the combination (σ, ε) does not work. We must
try another sign. Repeat from 2.
If | Irr(B)| = n, then we see that we need to search through n!2n combinations of
σ and ε. As n gets larger, the task rapidly becomes impractical to do by hand. So
we write programs in GAP [19] to automate the searching process using the above
algorithm.
There are a few tricks to improve the speed of the programs. For example, we can
use Theorem 3.1.5 to check only the p-singular elements for the integrality condition.
Also, if p is odd, then we can use Lemma 3.2.5 to quickly check the validity of a
sign for a given permutation, before checking the perfect conditions of the resultant
isometry.
Nevertheless, when n is very big (typically n > 13), our programs still take a very
long time to compute the perfect isometry groups, due to a large number of possible
permutations. This makes it impractical to deal with large blocks.
A.1
Codes
Let B be a block of OG. Suppose | Irr(B)| = n. Let chind = [i1 , . . . , in ] be the
subscripts of irreducible characters in Irr(B). Since elements in PI(B) are permutation in Sn with signs, we can embed them into S2n as follows.
Let I =
[i1 , . . . in , −i1 , −i2 , . . . , −in ]. Since any isometry J : RK (B) −→ RK (B) permutes
entries in I, we can identify J with a permutation on positions of entries of I.
For example, if chind = [1, 4, 5], then an isometry [1, 4, 5] 7→ [−1, 5, −4] is a
permutation (1, 4)(2, 3, 5, 6) on positions of entries of I = [1, 4, 5, −1, −4, −5]. In
other words, it sends the entry in position 1 of I (1) to the entry in position 4 of
I (−1), sends the entry in position 2 of I (4) to the entry in position 3 of I (5),
sends the entry in position 3 of I (5) to the entry in position 5 of I (−4) etc. Thus
J can be represented as a list [−1, 5, −4] with respect to chind or as a permutation
(1, 4)(2, 3, 5, 6) ∈ S6 with respect to I. We call [−1, 5, −4] a list form of J (with
APPENDIX A. COMPUTER-AIDED RESULTS
164
respect to chind), and (1, 4)(2, 3, 5, 6) a double permutation form of J (with respect
to I).
Our codes consist of 11 custom functions:
Functions
Description
ListToDoublePerm
Convert an isometry in list form to double
permutation form.
MyValuation
The extended p-valuation of a given number, see (2.2).
getTcenord
Get the sizes of centralizers and orders of
representatives of conjugacy classes of G
mueval
Calculate µJ (g, h) corresponding to an isometry J.
chkperf_integral
Check if µ satisfies the integrality condition.
chkperf_sep
Check if µ satisfies the separation condition.
chkperf
Check if µ is perfect.
getSignPerm
Given a permutation σ, get the set of possible choices
of signs ε. Use Lemma 3.2.5 if applicable (p > 2).
chkPerfPermInternal Check (internally) if a given permutation is perfect.
getPerfPermGroup
Calculate the group PI+ (B) of perfect permutations.
getPerfect
Calculate PI(B) and PI+ (B).
The only functions meant to be called by the user are getPerfPermGroup and
getPerfect. Other functions will be called internally, so we shall not explain their
parameters.
ListToDoublePerm
#-----------------------------ListToPerm-------------------ListToDoublePerm:=function(a,chind,I)
local i, seq, spos, smpos, tpos, tmpos;
seq:=[1..Size(I)];
for i in [1..Size(a)] do
spos := Position(I, chind[i]);
APPENDIX A. COMPUTER-AIDED RESULTS
smpos := Position(I, -chind[i]);
tpos := Position(I, a[i]);
tmpos := Position(I, -a[i]);
seq[spos] := tpos;
seq[smpos] := tmpos;
od;
return PermList(seq);
end;
MyValuation
#----------------------------MyValuation-------------------MyValuation:=function(z,p)
local f,deg,const;
f:=MinimalPolynomial(Rationals, z);
deg:=Degree(f);
const:=Value(f,0);
if const=0 then return 1;
else return Valuation(const,p)/deg;
fi;
end;
getTcenord
#--------------------------getTcenord-------------------------getTcenord:=function(tbl)
local Tcen, Tord;
Tcen:=SizesCentralizers(tbl);
Tord:=OrdersClassRepresentatives(tbl);
return rec(cen:=Tcen, ord:=Tord);
end;
165
APPENDIX A. COMPUTER-AIDED RESULTS
mueval
#----------------------------mueval--------------------------mueval:=function(a,i,j,irr,chind,nchars)
local k, temp,T;
temp:=0;
T:=irr;
for k in [1..nchars] do
if a[k] > 0 then
temp:= temp + T[AbsInt(a[k])][i]*T[chind[k]][j];
else
temp := temp - T[AbsInt(a[k])][i]*T[chind[k]][j];
fi;
od;
return temp;
end;
chkperfintegral
#---------------------------chkperf_integral-------------------chkperf_integral:=function(a,p,irr,Tcen,Tord,chind,nchars)
local i,j,re,psingclass, valcen, uij , uji;
re := true;
psingclass:=ListX([1..Size(Tcen)] , x->Tord[x] mod p = 0 , x->x);
for i in psingclass do
valcen:=Valuation(Tcen[i],p);
if valcen > 0 then
for j in psingclass do
uij:=mueval(a,i,j,irr,chind,nchars);
uji:=mueval(a,j,i,irr,chind,nchars);
if uij <> 0 then
166
APPENDIX A. COMPUTER-AIDED RESULTS
if valcen > MyValuation(uij,p) then
return false;
fi;
fi;
if uji <> 0 then
if valcen > MyValuation(uji,p) then
return false;
fi;
fi;
od;
fi;
od;
return re;
end;
chkperfsep
#---------------------------chkperf_sep---------------------chkperf_sep:=function(a,p,irr,Tord,chind,nchars)
local i , j ,re;
re := true;
for i in [1..Size(Tord)]
do
for j in [1..Size(Tord)] do
if mueval(a,i,j,irr,chind,nchars) <> 0 then
if (Tord[i] mod p = 0) and (Tord[j] mod p <> 0) then
return false;
fi;
if (Tord[i] mod p <> 0) and (Tord[j] mod p = 0) then
return false;
fi;
fi;
167
APPENDIX A. COMPUTER-AIDED RESULTS
od;
od;
return re;
end;
chkperf
#---------------------------chkperf----------------------------chkperf:=function(a,p,irr,Tcen,Tord,chind,nchars)
local re;
re := false;
if chkperf_integral(a,p,irr,Tcen,Tord,chind,nchars) then
if chkperf_sep(a,p,irr,Tord,chind,nchars) then
re := true;
fi;
fi;
return re;
end;
getSignPerm
#--------------------------getSignPerm --------------------# for each permutation to check for perfect isometry,
# find unique signs that are compatible with it
# if p=2 or #chind=2 then return a direct product of C_2
# otherwise if not perfect perm, return {[0,..,0]}
# if perfectperm, return {sgn} where sgn is a determined sign
#------------------------------------------------------------getSignPerm:=function(g,chind,nchars,p,irr,allsigns,zerosign)
local k,sgn,s,a;
a:=OnTuples( chind, g );
for k in [1..nchars] do
168
APPENDIX A. COMPUTER-AIDED RESULTS
169
if Valuation(irr[a[k]][1],p) <>
Valuation(irr[chind[k]][1],p) then
return [zerosign];
# For a perfect isometry we must have I(chi)(1)/chi(1) an invertible
# element in O.
fi;
od;
if (p = 2) or (nchars = 2) then
return allsigns;
fi;
sgn:=[1];
s:= irr[a[1]][1]/irr[chind[1]][1] mod p;
for k in [2..nchars] do
if (irr[a[k]][1]/irr[chind[k]][1] mod p) = s then
Append(sgn,[1]);
elif
(-irr[a[k]][1]/irr[chind[k]][1] mod p) = s then
Append(sgn,[-1]);
else
return [zerosign];
fi;
# For a PI we must have I(chi)(1)/chi(1) constant mod p
# so we add +/-1 to try to make it constant if possible.
# If not then the perm is not perfect
od;
return [sgn];
end;
chkPerfPermInternal
#--------------------------chkPerfPermInternal ------------------chkPerfPermInternal:=function(g,irr,chind,nchars,p,I,
APPENDIX A. COMPUTER-AIDED RESULTS
zerosign,allsigns,Tcen,Tord,chkhomog)
local C, a, y, perm, sgn, j, found, homogsgn,chkhomog_local;
found:=false;
homogsgn:=true;
chkhomog_local:=ShallowCopy(chkhomog);
C:=getSignPerm(g,chind,nchars,p,irr,allsigns,zerosign);
for y in C do;
if Size(C) = 1 then
if y = zerosign then
break;
else
sgn:=y;
fi;
else
sgn:=[];
for j in [1..nchars] do
Append(sgn, [(-1)^((2*j)^y)]);
od;
fi;
perm:=OnTuples( chind, g );
a:=List([1..nchars] , x->perm[x]*sgn[x]); # create a with sign
if chkperf(a,p,irr,Tcen,Tord,chind,nchars) then
if not chkhomog_local then
if not ( ForAll(a,x->x > 0) or ForAll(a,x->x < 0) ) then
homogsgn:=false;
chkhomog_local:=true;
fi;
fi;
found:=true;
break;
170
APPENDIX A. COMPUTER-AIDED RESULTS
fi;
od;
if found then
return rec(seq:=a, found:=true, homogsign:=homogsgn );
else
return rec(seq:=[], found:=false, homogsign:=homogsgn,);
fi;
end;
getPerfPermGroup
The following program computes PI+ (B).
#-----getPerfPermGroup (using SubgroupProperty)----------------# tbl = Character table
# chind = Character indexes (subscripts)
# symgp = Permutation group on chind
# PPlist = list of already known perfect permutations.
#
Set to [] if not known
# p = prime number
# I = sequence of +- entries in chind e.g. if chind = [1,2,3] then
#
I = [1,2,3,-1,-2,-3]
#---------------------------------------------------------------getPerfPermGroup:=function(tbl,chind,symgp,PPlist,p,I)
local j, P,x,irr,zerosign, nchars, allsigns,Tcen,Tord,chkhomog,
C2, C1;
irr:=Irr(tbl);
nchars:=Size(chind);
C2:=SymmetricGroup(2); C1:=Subgroup(C2,[]);
allsigns:=C1;
for j in [2..nchars] do
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APPENDIX A. COMPUTER-AIDED RESULTS
172
allsigns:=DirectProduct(allsigns,C2);
od;
zerosign:=List(chind, x->0);
Tcen:=getTcenord(tbl).cen;
Tord:=getTcenord(tbl).ord;
chkhomog:=false;
P:=SubgroupProperty(symgp,
x -> chkPerfPermInternal(x,irr,chind,nchars,p,I,zerosign,allsigns,
Tcen,Tord,chkhomog).found);
return P;
end;
getPerfect
The following program also computes PI+ (B) like getPerfPermGroup does but may
take longer to run (not using the effective SubgroupProperty built-in function in
GAP). For quick runtime, run
gap>PP:=getPerfPermGroup(tbl,chind,symgp,[],p,I);
first to get the group PI+ (B), and then
gap>PI:=getPerfect(tbl,chind,symgp,PP,p,I);
#-------------------------------getPerfect--------------------# tbl = Character table
# chind = Character indexes (subscripts)
# symgp = Permutation group on chind
# PPlist = list of already known perfect permutations.
#
Set to [] if not known
# p = prime number
# I = sequence of +- entries in chind e.g. if chind = [1,2,3] then
#
I = [1,2,3,-1,-2,-3]
APPENDIX A. COMPUTER-AIDED RESULTS
#---------------------------------------------------------------getPerfect:=function(tbl,chind,symgp,PPlist,p,I)
local x,y,i,j, perm, perminfo, Tord, Tcen, nchars, irr, GG, P, L,
ma, PI, zerosign, chkhomog, homogsgn, D, C, found, sgn,
allsigns, C2,C1,sizeD, sizeP, sizesym;
irr:=Irr(tbl);
nchars:=Size(chind);
Tcen:=getTcenord(tbl).cen;
Tord:=getTcenord(tbl).ord;
GG:=SymmetricGroup(2*nchars);
P:=Subgroup(symgp,PPlist);
sizeP:=Size(P);
sizesym:=Size(symgp);
L:=[]; PI:=[];
chkhomog:=false;
homogsgn:=true;
ma:=List([1..nchars] , x->-chind[x]);
PI:=Subgroup(GG,[ListToDoublePerm(ma,chind,I)]);
C2:=SymmetricGroup(2); C1:=Subgroup(C2,[]);
allsigns:=C1;
for j in [2..nchars] do
allsigns:=DirectProduct(allsigns,C2);
od;
zerosign:=List(chind, x->0);
if Length(PPlist) > 0 then
Print( "Initializing...\n");
j:=1;
for x in PPlist do
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APPENDIX A. COMPUTER-AIDED RESULTS
perminfo:=chkPerfPermInternal(x,irr,chind,nchars,p,I,
zerosign,allsigns, Tcen,Tord,chkhomog);
if perminfo.found
then
PI:=ClosureSubgroup(PI,
ListToDoublePerm(perminfo.seq,chind,I));
if not perminfo.homogsign and not chkhomog then
homogsgn:=false;
chkhomog:=true;
fi;
Print( "No. of perfect isometries found so far =
",2*j,"\n");
j:=j+1;
fi;
if Size(PI) = 2*Size(PPlist) then
Print( "No. of perfect isometries found so far =
",2*Size(PPlist),"\n");
break;
fi;
od;
Print( "End of initialization\n");
fi;
D:=DoubleCosetRepsAndSizes(symgp, P, P);
sizeD:=Size(D);
if D[1][1] = One(symgp) then
i:=2;
else
i:=1;
fi;
while i <= sizeD and sizeP < sizesym do
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APPENDIX A. COMPUTER-AIDED RESULTS
x:=D[i][1];
perminfo:=chkPerfPermInternal(x,irr,chind,nchars,p,I,
zerosign,allsigns,Tcen,Tord,chkhomog);
if perminfo.found
then
P:=ClosureSubgroup(P,x);
sizeP:=Size(P);
PI:=ClosureSubgroup(PI, ListToDoublePerm(perminfo.seq,chind,I));
if not perminfo.homogsign and not chkhomog then
homogsgn:=false;
chkhomog:=true;
fi;
Print( "No. of perfect isometries found so far =
",2*sizeP,"\n");
D:=DoubleCosetRepsAndSizes(symgp, P, P);
sizeD:=Size(D);
if D[1][1] = One(symgp) then
i:=2;
else
i:=1;
fi;
else
Print( "No. of permutations left = ",sizeD-i,"\n");
i:=i+1;
fi;
od;
#-------------------exit x loop ---------------
Print( "========================================\n");
Print( "No. of perfect isometries = ",Size(PI),"\n");
Print( "========================================\n");
Print( "Homogenous signs = ",homogsgn,"\n");
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APPENDIX A. COMPUTER-AIDED RESULTS
Print( "Returning the record with component .full and .perm\n");
# Print( "Full PI Group is ", StructureDescription(PI), "\n");;
# Print( "Perm PI Group is ", StructureDescription(P), "\n");;
return rec(full:=PI, perm:=P);
end;
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