THE JACOBSON RADICAL OF GROUP ALGEBRAS
NORTH-MOLIAND MATHEMATICS STUDIES Notas de Matematica ( 115)
Editor: Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro and University of Rochester
NORTH-HOLLAND -AMSTERDAM
NEW YORK
OXFORD TOKYO
135
THE JACOBSON RADICAL OF GROUP ALGEBRAS G regory KARPlLOVSKY Department of Mathematics Universityof the Witwatersrand Johannesburg, SouthAfrica
1987
NORTH-HOLLAND-AMSTERDAM
0
NEW YORK
OXFORD .TOKYO
@
ElsevierScience Publishers B.V., 1987
All rights reserved. No part of this publication may be reproduced, stored in a retrievals ystem, ortransmitted, in any form orby any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.
ISBN: 0 444 70190 7
Publishers:
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Library of Congress Cataloging-inPubliition Data
Karpilovsky, Gregory, 1940The Jacobson r a d i c a l of group algebras. (North-Holland mathematics s t u d i e s ; 135) (Notas de ; 115) mate&ica Bibliography : p Includes index. 1. Group algebras. 2. Jacobson r a d i c a l . 3. Modules (Algebra) I. T i t l e . 11. S e r i e s . 111. Series: Notas de m a t e d t i c s (Rio de Janelpo, Brazil) ; no. 115. QAl.N86 no. 115 CgA1711 510 s E512'.243 86-32918 ISBN 0-444-70190-7 (U.S. )
.
PRINTED IN THE NETHERLANDS
Ila the memory a€ my father
ILIA KARPICSUtiKq
(1915-1942)
This Page Intentionally Left Blank
vii
Preface
Let
G be a f i n i t e group and l e t F be a f i e l d .
r e p r e s e n t a t i o n s of algebra
FG.
radical
JIFG) of
G
over
F
I t i s w e l l known t h a t l i n e a r
can be i n t e r p r e t e d a s modules over t h e group
Thus t h e i n v e s t i g a t i o n of r i n g - t h e o r e t i c s t r u c t u r e of t h e Jacobson
FG
i s of fundamental importance.
During t h e l a s t two
decades t h e s u b j e c t has been pursued by a number of r e s e a r c h e r s and many i n t e r e s t i n g r e s u l t s have been obtained.
The p r e s e n t book i s intended t o g i v e a
systematic account of t h i s work b u t t h e treatment i s by no means intended t o be exhaustive.
The author has t r i e d t o be f a i r l y complete i n what he c o n s i d e r s a s
t h e main body of t h e theory, and t h e r e a d e r should g e t a considerable amount of knowledge of c e n t r a l i d e a s , t h e b a s i c r e s u l t s , and t h e fundamental methods. have t r i e d t o avoid making t h e d i s c u s s i o n t o o t e c h n i c a l .
We
With t h i s view i n mind,
maximum g e n e r a l i t y has not been achieved in those p l a c e s where t h i s would e n t a i l a l o s s of c l a r i t y o r a l o t of t e c h n i c a l i t i e s . The p r e s e n t monograph i s w r i t t e n on t h e assumption t h a t t h e r e a d e r has had t h e equivalent of a standard f i r s t - y e a r graduate a l g e b r a course.
Thus we assume a
f a m i l i a r i t y with b a s i c r i n g - t h e o r e t i c and group-theoretic concepts and an understanding of elementary p r o p e r t i e s of modules, t e n s o r products, and f i e l d s .
For
t h e convenience of t h e r e a d e r , a chapter on a l g e b r a i c p r e l i m i n a r i e s i s included. This chapter provides a survey of t o p i c s needed l a t e r i n t h e book.
There i s a
f a i r l y l a r g e bibliography of works which a r e e i t h e r d i r e c t l y r e l e v a n t t o t h e t e x t o r o f f e r supplementary m a t e r i a l of i n t e r e s t . A word about n o t a t i o n .
As is customary, Theorem 4.3.2 denotes t h e second
r e s u l t of Section 3 of Chapter 4 ;
however, f o r s i m p l i c i t y , a l l r e f e r e n c e s t o t h i s
viii
PREFACE
r e s u l t within Chapter 4 i t s e l f a r e designated a s Theorem 3 . 2 . A systematic d e s c r i p t i o n of t h e m a t e r i a l is supplied by t h e i n t r o d u c t i o n s t o
i n d i v i d u a l chapters and t h e r e f o r e w i l l not be repeated here. I would l i k e t o express my g r a t i t u d e t o my wife f o r t h e encouragement she has
given me i n t h e p r e p a r a t i o n of t h i s book.
My thanks t o P.Brockhaus,
E.Fomanek,
S.Koshitani, B.Kiilshammer, K.Motose, Y.Ninomiya, Y.Tshushima and D.A.R.Wallace
for
sending me r e p r i n t s and p r e p r i n t s of t h e i r work, and a l s o t o Lucy Rich f o r her e x c e l l e n t typing.
Gregory Karpilovsky
ix
Contents
vii
PREFACE CHAPTER 1. RING-THEORETIC BACKGROUND 1. Notation and terminology 2. Artinian and noetherian modules Completely reducible modules 3. 4. Direct decomposition of rings Matrix rings 5. 6. The radical and socle of modules and rings 7. The Krull-Schmidt theorem 8. Projective, injective and flat modules 9. Projective covers 10. Algebras over fields
CHAPTER 2.
CHAPTER 3 .
1 1 5 12 15
17 22
36 40
46 52
GROUP ALGEBRAS AND THEIR MODULES
67
1. Group algebras 2. Central idempotents 3. The number of irreducible FG-modules 4. The induced modules 5. Relative projective and injective modules 6. Vertices of FG-modules
67 73 79 82 91 100
THE JACOBSON RADICAL OF GROUP ALGEBRAS: FOUNDATIONS OF THE THEORY 1. 2. 3. 4. 5.
6. 7. 8. 9. 10. 11. 12.
Elementary properties Direct products A characterization of elements of J(FG): the general case Conlon's theorem, Fong's dimension formula and related results A characterization of elements of J(FG); G is p-solvable A characterization of elements of J(Z(FG)) Frobenius groups Upper and lower bounds for A characterization of dimFJ(Z( dim# G ) ) and its applications Morita's theorem ~n application: criteria for J V G ) = FG*J(Z(FG)) Group algebras with radicals of square zero
105
105 116 122 140
171 177 183 191 204 228 243 248
CONTENTS
X
13. 14.
15. 16. 17.
CHAPTER 4:
Group algebras with central radicals Commutativity of the radical of the principal block Criteria for the commutativity of J ( F G ) The radical of blocks and normal subgroups Group algebras with radicals expressible as principal ideals
GROUP ALGEBRAS OF p-GROUPS OVER FIELDS OF CHARACTERISTIC
p
1. Dimension subgroups in characteristic p > 0 and related results 2. Computation of t ( P ) for some individual p-groups P characterization of groups P of order pa with 3. t(P) 4.
CHAPTER 5:
CHAPTER 6 :
CHAPTER 7:
a(p-1) + 1 ,
t ( P ) = pa, t ( P )
=
301
301 315
p'-l+p-l,
t ( P ) = pa-' and t ( P ) = (a+l)(p-1) + 1 Characterizations of p-groups P with t ( P )
322 328
<7
THE JACOBSON RADICAL AND INDUCED MODULES
333
1. Annihilators of induced modules 2. Simple induction and restriction pairs 3. Applications 4. p-Radical groups
333 3 36 343 3 54
THE LOEWY LENGTH OF PROJECTIVE MODULES
375
1. Preliminary results 2. The Loewy length of projective covers 3. The Loewy length of induced modules 4. Groups of p-length 2
375 386 396 405
THE NILPOTENCY INDEX
413
1. 2. 3. 4. 5. 6. 7.
CHAPTER a:
=
250 253 259 266 284
Some results on p-solvable groups Upper and lower bounds for t ( G ) Groups G with t ( G ) = a(p-1) + 1 Computation of t ( G ) with M(p) E Syl (G) Characterizations of groups G with !(GI =pa-' Groups G with t ( G ) < (a+2) (p-1) + 1 Characterization of groups G with t ( G ) =pa-',
+ p-1 p odd
RADICALS OF BLOCKS 1. 2. 3. 4.
5. 6. 7. 8.
A lower bound for t ( B ) in terms of the exponent of An upper bound for t ( 2 ( B ) ) Defect groups of covering blocks Regular blocks The Fong correspondence The Khshammer's structure theorem Applications A survey of some further results
413 418 424 430 439 446 448 453
6(B)
453 459 464 466 47 1 478 488 493
BIBLIOGRAPHY
501
NOTATION
5 20
INDEX
527
1
1 Ring4heoretic background This chapter reviews some basic ring-theoretic facts and techniques as well as fixes conventions and notation for the rest of the book.
Because we presuppose a
familiarity with various elementary ring-theoretic terms, only a brief description of them is presented.
Many readers may wish to glance briefly at the contents of
this chapter, referring back to the relevant sections when they are needed later. Occasionally, results are stated somewhat more generally than necessary for later use. 1. NOTATION AND TERMINOLOGY.
All rings in this book are associative with
1 # 0
assumed to have the same identity element as R .
and subrings of a ring R
are
Each ring homomorphism will be
assumed to preserve identity elements. Let R
be a ring.
positive integer n.
R
J"
An ideal J
E
R
in R
is nilpotent if xn
=
is the product of J
with itself n
are orthogonal, if uv = vu = 0.
times.
R.
Z(R1.
such that
J"=0, of
A nonzero idempotent is primitive if it
An idempotent e
is a primitive idempotent of
is nil-
Two idempotents u,V
cannot be written as a sum of two nonzero orthogonal idempotents.
Z ( R ) the centre of
0 for some
is nil if every element of J
is nilpotent if there is a positive integer n
potent, while J where
An element x
We denote by
is called centrally primitive if e
The ring of all nxn-matrices over R
is
denoted by M n ( ( R ) . Let R
be a ring.
opposite ring of R. are just those of R.
x*y = yx , x,y
E
Clearly, Z ( R )
R.
=
From this we construct a new ring Ro,
called the
Both the underlying set and the additive structure of Ro But the multiplication, denoted by It is easy to verify that Ro
Z(Ro) and R
*,
is qiven by
is a ring with theseoperations
is commutative if and only if R = Ro.
CHAPTER I
2
Let
A be a ring, R a commutative ring, and A
morphism.
The resulting system
convenient to suppress the over R.
R
+
2
there is an ideal I of R with R/I Setting ra = A ( r ) a , r E R,a € A ,
Conversely, if A
It will be
as an R-algebra or as an algebra
is an R-algebra (with respect to some 1)
Thus A
(A) a ring homo-
(A,R,A) is called an R-algebra.
and speak of A
A
:
if and only if
isomorphic to a subring of
it follows that A
Z(R).
is an R-module such that
is an R-module for which the above equalities hold, then the
map R - + Z ( A ) , r W r - 1
is a ring homomorphism and thus A
is an R-algebra.
By a
homomorphism of R-algebras we understand a ring homomorphism which is also a homomorphism of R-modules. Let
nRi. %€I
(R.),i E I be a family of rings and let R be the direct product set We can define addition and multiplication on R
(+.)+(y.)
z
z
=
,
(x.+y.) z z
is a ring; we shall refer to R
It is straightforward to verify that R direct product of the f a m i l y
by the rules
(Ri),i E I.
By an R-module we always understand a unital R-module; the symbols
MR will be used to underline the fact that M respectively.
as the
IT
Or
is a left or a right R-module,
We usually consider left R-modules, and, in this case, speak
simply about an R-module. Let
V and W be R-modules.
We denote by Hom(V,W)
the additive abelian
R
V+W.
group of all R-homomorphisms and
W are right R-modules.
The same notation will be used in case
In case V
=
v
W we shall write End(V) instead of R
Hom(V,V). Note that End(V)
R
is a ring under the multiplication
R (fg) ( U ) = f(pW))
( f , g E End(V) $0 E
R A
nonzero R-module
The module Let
V
i s irreducible if
V i s reducible if V
#
0 and
V
v) V
are the only submodules of
0 and V i s not irreducible.
( V ; ) , i E I, be a family of modules over the same ring R ,
and let
V
NOTATION AND TERMINOLOGY
nVi.
be the direct product set
3
Then under the operations
iEI
V
becomes an R-module. (Vi)
family
,i E I .
V
The submodule of
as the d i r e c t product of the
V
We shall refer to
consisting of all
( x . ) with finitely
x . # 0 is called the external d i r e c t sum (or simply d i r e c t sum) of the
many
2
family of modules
(V.)
W.
submodule of an R-module family V i E Vi
(V.)
if each
@ . Vi. Assume that each Vi is a .LEI We say that W is an ( i n t e r n a l ) d i r e c t sum of the
and is denoted by
w
E
W can be written uniquely in the form
for all i E I and with finitely many
internal direct sums of
(Vi),i E I ,
to denote both of them.
@ . Vi
u
i
#
0.
1 Ui
W =
with
iEI Since external and
are isomorphic, we shall use the same symbol
The precise situation will always be clear from
ZEI
the context.
A nonzero R-module
submodules X
and
V
is said to be indecaposable if for any two
Y of V , the equality V
=
X @ Y implies X
= 0
or
Y = 0.
A partially ordered set in which any two elements have a supremum and an infimum is called a l a t t i c e . V
V,
For any R-module
the set of all submodules of
forms a lattice under the partial ordering by inclusion, with intersection as
infimum and sum as supremum.
For any three submodules V 1 , V 2 , V 3
V
of
the
following general relation is called the modular law : if
V1
5V3,
Let Lat(V)
then
V
+
(V
2
n V
) = 3
(V
1
+
V
) 2
n V
3
denote the lattice of all submodules of
is said to be a l a t t i c e homomorphism if f ( V +'J 1
f(Vl n V2) = f(Vl)
n f(V2)
jection,we say that f
for all
) =
2
fCV
V I , V 2 E Lat(V).
is a lattice isomorphism.
bijection with an order-preserving inverse between
A map f
V. )
+ f(V
: Lat(V)+Lat(W)
and
In case f
is also a bi-
Note that an order-preserving Lat(V) and
Lat(W)
is always
a lattice isomorphism. A partially ordered set is said to be inductive if it is nonempty and every chain in it has an upper bound.
According to Zorn's lemma, every partially ordered set
which is inductive has a maximal element.
Let
V
be an R-module.
A submodule
W of V is called maximal if W is a proper submodule (i.e. W + V ) and W is not contained in any proper submodule of
V.
4
CHAPTER I
An 8-module
vi
E V
v
(Vi)$ E I, if there exist elements
is said to have an R-basis
v
such that each
E V
can be written as a finite sum
uniquely determined coefficients r
i
E R.
v
=
with 1r.v z i
V are called free R-
Such modules
moduZes. Unless explicitely stated otherwise, all groups are assumed to be multi-
plicative.
The multiplicative group of a field F G,<X>
is a subset of
will denote the subgroup of
is denoted by
is the centralizer and normaZizer of
As usual, Z ( G )
denotes the centre of
X
X in G, respectively.
G.
H be a subgroup of a group G.
element from each left coset zH
If
G generated by X, while
C(X) and h ' ( X ) G G Let
F*.
A
subset of
G
containing just one
is called a left transversal of H in G,
and
right transversals are defined correspondingly. Let g be an element of a finite group that g
G and let p be a prime.
is a p-element if g has order a power of p; we say that g
element (or is p-regulur! if its order is prime to g = g g
written in a unique way element, and g
P
p
P P"
and g p ,
and p'-parts of g ,
where gp
commute.
respectively.
p.
Each g E G
We say is a p ' -
can be
is a p-element,
The elements gp
gP' is a p'and g p r are called the
The symbols CR(G)
and
Syl(G1
denote
P respectively the sets of conjugacy classes and Sylow p-subgroups of conjugacy class C
of
G.
A
G is called p-regular (respectively,p-singuZar) if the C is not divisible by p
order of the elements of
(respectively, divisible by
P) . Let X of
G
be an arbitrary set.
An action of
into the permutation group of
suppress the symbol 8
stabilizer G ( z )
and the orbit
3:
E
on
X is a homomorphism 0
When the action is understood, we shall
and instead write z i+'z
of z E X
Gz of
X.
G
is defined by
X is defined by
for
8 ( g ) ,g E G,z E X.
The
ARTINIAN AND NOETHERIAN MODULES
One r e a d i l y v e r i f i e s t h a t
Sn
Throughout t h e book, degree
x'
a c t s transitivezy i f
G
We say t h a t
f o r some (and hence a l l )
G
i s a subgroup of
G(s)
An
and
X
=
2 E
X.
IGxl = I G : G ( z )
and t h a t
I.
denote symmetric and a l t e r n a t i n g groups of
zn
n, r e s p e c t i v e l y , while
5
n.
i s a c y c l i c group of o r d e r
2. ARTINIAN AND NOETHERIAN MODULES
V
An R-module
i s s a i d t o be noetherian ( r e s p e c t i v e l y , artinianl i f every
V
ascending ( r e s p e c t i v e l y , descending) chain of submodules of
,#?
We say
i s noetherian ( r e s p e c t i v e l y , artinian) i f t h e r e g u l a r R-module
R
that the ring
stops.
i s noetherian ( r e s p e c t i v e l y , a r t i n i a n l .
V
The module
n
i s called finzyety
.
1
RUi f o r s u i t a b l e 1) ,.. ,vn E V . We say t h a t V i s i=1 f i n i t e l y cogenerated i f f o r every family ( V i ) , i E I of submodules of V with
generated i f
n Vi i€I 2.1.
=
0,
V =
J
there e x i s t s a f i n i t e subset
I
of
such t h a t
n V j
=
PROPOSITION.
(i) V
v
For a module
t h e following statements a r e e q u i v a l e n t
is a r t i n i a n
V
(ii) Every nonempty s e t of submodules of
has a minimal element
(iii) Every factor-module of
V
i s f i n i t e l y cogenerated.
( i )* ( i i ) : Let
X
be a nonempty s u b s e t of submodules of
Proof. assume t h a t
X
does not have a minimal element.
{W' E X l W ' c W}
W
0.
jEJ
W ' with
i s nonempty.
W 3 W'
W
E
X.
. ..
J
of
I.
the s e t
Vi = W,
with
V.
of submodules of
V
and
n V
then W =
(Vi),i E I
f o r some f i n i t e
j€J j
i€I subset
X
w e t h e r e f o r e o b t a i n an
i s a submodule of
W
( i i )=* ( i i i ) : We need only show t h a t i f
V
W E X,
Fixing
W 3 W ' 2 Wrr 2
i s a family of submodules of
WE
and
Thus, by t h e axiom of choice, t h e r e i s a f u n c t i o n
f o r each
i n f i n i t e descending chain
Then f o r each
V
To t h i s end, p u t
Y
=
I n
V
~
5~ I
K i s
kEK Then, by hypothesis, finite.
Clearly,
Y has a minimal element, say W =
n V j€J j
(iii)* ( i ) : Assume t h a t
V
. has a descending chain
V with j€J j
J
5I
and
J
6
CHAPTER I
of submodules and put
.
so there exists n
required. 2.2.
COROLLARY.
W =n&Vn. W
with
V
Let
By hypothesis, V/W Thus V n+i
= Vn.
for i
V,
=
is finitely cogenerated
be a nonzero artinian module.
= 1,2
Then
V
,...,
as
has an irre-
ducible submodule. B
2.3. PROPOSITION. V
(i)
For a module
v
the following statements are equivalent:
is noetherian
(ii) Every nonempty set of submodules of V
(iii) Every submodule of
V
has a maximal element
is finitely generated.
.
The proof of this proposition is dual to that of Proposition 2 . 1 and
Proof.
therefore will be omitted.
.
2.4. COROLLARY.
submodule.
V
Let
Then V
be a nonzero noetherian module.
has a maximal
Note that Corollary 2.4 is also a consequence of the following observation. Let V
2.5. PROPOSITION. submodule of
V
be a finitely generated R-module.
Then every proper
In particular, V
is contained in a maximal submodule.
has a
maximal submodule tf V # 0. Let V
Proof. of
V.
x
be generated by
Consider the set X
If
v = UW
x
W
!J
= V,
wA,
to show.
.
*
-
*
.
lWAr
Thus X
a contradiction.
maximal element.
CW,)
V
which contain W.
is a chain in X, then U W
then each xi E Wx
A'
largest of the modules
and let W be a proper submodule
of proper submodules of
is inductive: it contains W and if submodule.
,...,x
for some
xi.
Let W
1-I
be the
contains x ,...,xp and so 1-I is inductive and, by Zorn's lemma, it has a
This is a maximal submodule of
V
containing W, as we wished
In particular, this result can be applied to the ring R module, R
is a
A A
i Then W
This
is generated by the single element
itself.
As
left R-
1, and the submodules are left
ideals, hence we obtain the 2.6.
COROLLARY.
Let R
be an arbitrary ring.
Then any proper left ideal of R
7
ARTINLAN AND NOETHERIRN MODULES
i s contained i n a maximal l e f t i d e a l of
R.
I n particular,
R has a maximal
l e f t ideal.
2.7.
PROPOSITION.
Let
o-+u-+v-+w-o be a n e x a c t sequence of R-modules.
U
i f both
Proof.
W
U
i s a r t i n i a n ( n o e t h e r i a n ) i f and only
a r e a r t i n i a n (noetherian).
V
Assume t h a t
V,
module of
W
and
V
Then
is artinian.
i s a r t i n i a n by t h e d e f i n i t i o n .
V, W
i s isomorphic t o a factor-module of Conversely, suppose t h a t both
a r t i n i a n , w e may assume t h a t
U C_ V
U
U is
Then s i n c e
W
and
Because every factor-module of
i s a r t i n i a n , again by t h e d e f i n i t i o n .
W
and
isomorphic t o a sub-
are artinian.
=
V/U.
v
To prove t h a t
is
Let
v I - 3- v2 -3 ... 3-- vn -2 ... be a descending chain of submodules of integer
rn
U is
=
vn
= =
2.8.
and only i f each
2.9.
PROPOSITION.
noetherian.
Vn
vn n ( v n + u ) = vn vn+i + (vn+i n u)
)
( i = 1,2, ...
)
such t h a t
n u
>_
( i = 1,2, ...
, we
Vn+i
n
(vnlli + u )
=
vn + i
have f o r each
=
i
=
1,2,
...
vn+i + (vn n u)
The proof of t h e noetherian c a s e i s d u a l .
is a r t i n i a n .
COROLLARY.
> rn
n
a r t i n i a n , t h e r e i s an i n t e g e r
Taking i n t o account modularity and t h a t
V
is artinian, thereisan
vrn+i + u
vn n u = vn + i
Thus
V/U
Because
such t h a t
vm + u Since
V.
... 8 vn. {1,2,. . . , n } , i s
Let
V = V 8
V,,i
E
Then
is a r t i n i a n (noetherian) i f
a r t i n i a n (noetherian)
Assume t h a t t h e nonzero module
Then V i s t h e d i r e c t sum
V
v
.
i s e i t h e r a r t i n i a n or
8
CWTEI?
v
=
v
L
...
@
'n
@
of a finite set of indecomposable submodules. Proof.
For each nonzero module
that does not have a finite indecom-
I/
posable decomposition choose a proper decomposition
v V'
where
= V' 0 X'
Assume that V
has no finite indecomposable decomposition.
nonzero and is not a finite direct sum of indecomposable modules.
is a sequence of proper decompositions.
.
proving that V true.
V 3 V ' 2 V'' 2
and
is neither artinian nor noetherian.
Let V be a nonzero module.
A
Then
Therefore there exist infinite chains
...
X' c X'@ X" c
is
...
So the proposition is
finite chain of n + l
V
submodules of
v = v0 3 v1 3 . . . 3 v n = o is called a composition series of length n is irreducible
(i =
1,2,
in its predecessor.
...,n), i.e.
V
for
provided that each Vi-l/Vi
provided each term in the chain is maximal Vo/V1, V l / V 2 ,
The irreducible modules
..., Vn-l/Vn
are
called the composition factors of the series. 2.10.
V
PROPOSITION.
A
nonzero module
V
has a composition series if and only if
is both artinian and noetherian. Proof.
module of
V
Assume that
V
is both artinian and noetherian.
is artinian, it follows from Corollary 2.2.
Since any factor-
that there is an
ascending chain
oc of submodules of
V
v1
c
v2
c
0 . .
where each term in the chain is maximal in its successor.
ARTINIAN AND NOETHERIAN MODULES
V
Since
v
Thus
is noetherian, the chain terminates at
tion on n , V
where n
has a composition series.
We shall use induc-
is the minimum length of all such series.
is irreducible and there is nothing to prove.
v = v0
3 v 1
.
has a composition series of length n - 1 .
fore a consequence of Proposition 2.7.
If n
=
1,
Otherwise, if
=o
>...>v
is a composition series of minimal length for V, V1
after a finitely many terms.
has a composition series.
Conversely, suppose that V
then
V
9
then
V/Vl
is irreducible and
The desired conclusion is there-
Assume that
v = v0 3 v1 3 . . . 3 v n = o and
v = w0 are two composition series of case n
V.
3 w 1
>...>w m
Then these two series are called eqUiVaZent in
and there is a permutation cr
= rn
2.11. PROPOSITION.
= o
of
(The Jordan-HElder Theorem).
1,2,. .
.,721
such that
composition series, then every pair of composition series for We use induction on c ( V ) ,
Proof.
composition series for c(V) =
V.
where c ( V )
v
has a are equivalent.
is the minimum length of
The case c ( V ) = 1 being trivial, we may assume that
> 1 and that any module with a composition series of smaller length has
YZ
all of its composition series equivalent.
v = vo
U)
v
If a module
2
v
2
Fix a composition series
... 3 vn = 0
of minimal length and choose any other composition series for
v
(2)
If V
1
= =
w0
2
w1
2
... 3 wrn=
0
W , then by the induction hypothesis and in view of c ( V 1
) 4
n-1,
(1)
CHAPTER L
10
and (2) are equivalent. V +W 1
= 1
V
V/Vl
(3)
whence
Vl
since V Wl/(Vl
n
W1)
and
V/W
1
1
V /(V 1
w1
n W1 is maximal in both
W
f
1
V.
is a maximal submodule of
1
2
V
Hence we may assume that
1
in which case
It follows that
I n W 1 )
V
and
1
.
By Proposition 2.10, wl
has a composition series
v 1 n w 1 = x0 3 x 1 3 . . . 3 x k = o and therefore V1 3 Xo 3
... 3 Xk
are composition series for composition series for
V1
V
1
=
0
and
W
and W1.
2
Xo
2
... 2 X k
e(v
Because
)
< n,
=
0
every two
are equivalent, so the two series
v = v0 3 v1 3 v 2 > . . . 3 v n = o and
v = v o 3 v1 are equivalent.
2 x
0
3 . . . > xk = o
It follows that k < n - 1
two composition series for
W1
and e(W ) < n , proving that every 1
are equivalent.
The conclusion is that the two
series
v = w o 2 w1
2...2w
m
= o
and
v = w0 > w 1 are equivalent.
Since, by ( 3 ) ,
3 x
V/Vl
0
3...3xk = o
W1/Xo
2
and
V/Wl
2
Vl/Xo,
the series (1)
and ( 2 ) are also equivalent, a s required. A
module
Zength.
v
that is both artinian and noetherian is said to be of f i n i t e
In view of the Jordan-Hglder Theorem, for such a module
define its (composition) Zength
c(V)
unambiguously by
V
we can
ARTINAIN AND NOETHERIAN MODULES
c(V)
=
I
0
if V = O
n
if
has a composition series of length n.
V
... @ Vn
Let R R = V 8
2.12. PROPOSITION.
11
with V . 2
f
0
for i = 1,2
,..., n,
and write 1
...
e +
=
1
Then
V.
=
+
(ei E V . )
en
2
{e ,...,e is a set of pairwise orthogonal idempotents in R and n Re 1 Q i G n. Conversely, if { e ,. ..,e 1 is a set of pairwise orthogonal i'
idempotents in R, then
n
R(
1 ei) =
i=1
Given r E R, we have r
Proof.
=
n 8 Rei i= 1
r-1 =
1-P =
re +
... +
re,.
Thus in
particular
ei
=
+ e.e 2 1
... + e.e z n
for all i , j . Moreover Re. c V . and 22 z i = Gijei @ Ren which implies that Rei = Vi for all i
proving that e .e
R R = Re 8 1
...
.
Conversely, assume that { e
n potents and put e
n
Re
1 ei.
is a set of pairwise orthogonal idemand eei = eie = ei' for all i. Thus
Then e2 = e
i=1
1 Rei.
=
=
,...,en 1
If
Iriei
= 0
then multiplication on the right by
that r.e = 0 for all j . J j 2.13. COROLLARY. a set {e n
1=
'
,...,e n 1
I ei.
ej implies
n
i=1
Hence Re = 8 Rei, as required. i=1
Assume that R
is artinian or noetherian.
Then there exists
of pairwise orthogonal primitive idempotents of R
with
Furthermore
i=1
RR where each Re Proof.
=
Re 8
... 8 Ren
i is an indecomposable module.
Direct consequence of Propositions 2.12 and 2.9.
We shall refer to the R-modules Re ],...,Ren
in Corollary 2.13 as being the
CHAPTER I
.^
IL
prfncipal MecanposabZe R-modules. 3 . COMPLETELY REDUCIBLE MODULES. An
V
R-module
is said to be completely reducible if every submodule of
direct sunnnand of 3.1. PROPOSITION.
(il
Let V be completely reducible.
every submodule of V
module W '
(i) If W of
Let W
(ii)
verify that U module of thus U
W
module
W 8 K'
=
V
0.
V.
V
V
and
V.
is completely reducible.
contains an irreducible submodule.
Then W
1
V/W'
be a submodule of
V
with
By (i), it suffices to
So assume that
V = V @ V 1
.
2
for some sub-
as required.
and let U = V / K .
is completely reducible. Choose V 2
= W'
V/W
and
then V = K 8 K'
V,
is a submodule of
Then
u
=
U1 =
vl/W
is a sub-
u
(v +
w)/W and
1
8
is completely reducible.
(iii) Let
v=
is isomorphic to a submodule of
is isomorphic to a homomorphic image of
V # 0 then
Proof.
W'
V
Every submodule and every homomorphic image of
(iii) If
is a
V.
Every homomorphic image of
(ii)
V
2,
of
be a nonzero element of V
1
of
v.
for some nonzero submodules
If K '
W ,W 1
it follows that U 4 W 8 Wi maximality of
By Zorn's lemma, there exists a sub-
maximal with respect to the property that
for some submodule kr'
K 8 W2
V.
W.
9
w.
Write
is not irreducible, then
of
2
for either i = 1 or
2,
i
V.
= 2,
Because
contrary to the
Hence W' is irreducible.
The following alternative description of completely reducible modules is often useful. 3 . 2 . PROPOSITION.
For a nonzero module V
the following statements are equiva-
lent: (il
V
is completely reducible
(ii1
V
is the direct sum of irreducible submodules
13
c~PLETELYREDUCIBLE MODULES
(iii) V
is the sum of irreducible submodules. (i) * (ii): Consider the collection of sets of irreducible submodules
Proof. of
V
whose sum is direct.
By Proposition 3.1(ii), it is nonempty and, by
{Vi>, in this collection.
Zorn's lemma, there is a maximal element, say
W
=
W'
8 V
i and let V
W 8 W'.
=
If
W'
0 then by Proposition 3.1 (ii),(iii),
#
Thus W
contains an irreducible submodule V ' .
{Vi}.
to the maximality of
Hence
W'
Let
=
+
V ' = V ' 8 (8 Vi),
V = W,
0 and
contrary
as asserted.
(ii) * (iii): Obvious (iii)
Let W be a submodule of
(i):
a submodule W'
W+W'
of contradiction that W 8 W' # V thesis, V of
=
V
for i =
V
V . n ( W O W'I 3
W 8 W'
We are therefore left to verify that
W8W'.
=
Owing to Zorn's lemma, there exists
maximal with the property that W n W'
V
of
V.
... + Vn 1,2,...,n.
+
+ vi -
V . n ( W 8 W') = 0. 3
where
Since
and choose
vi
E Vi
w'
E V
Vi
with
V
=
0.
Hence
v.
9 W 8
Assume by way
w'.
By hypo-
is an irreducible submodule
for some j and so
is irreducible, we conclude that
Thus
W O W' + V . 3
and so W n (W' 8 V . ) 3 3.3. COROLLARY.
and
Hence V . 4 W @ 3
Vj
2,
=
=
Let
= WOW
' 8 vj
0, contrary to the maximality of W'. V # 0 be completely reducible.
.
The following are equi-
valent (i)
V
is artinian
(ii) V
is noetherian
(iii) V
is the direct sum of a finite number of irreducible submodules.
Proof. Direct consequence of Corollary 2.8 and Proposition 3.2. 3.4. PROPOSITION.
.
If R R is completely reducible, then every R-module is
completely reducible. proof. which sends
Let
V
p
to
Proposition 3.1(ii).
be an R-module and let v E V . M.,
is a homomorphism. Since V
5 1 Ro, 6 V
Thus
Then the map from R R to
RV
Rv is completely reducible by
the result follows by virtue of
CHAPTER I
14
Proposition 3.2. As
a preparation for the proof of the next result, we record the following
observation. Let V
3 . 5 . LEMMA.
=
8 Vi where each Vi
irreducible submodule of
V.
Then W
Proof. Choose nonzero w E W then the map sending rW =
RIJ
where j ranges over those i
Vi = Rvi.
to
into
PV
and write w for P E R
i
Since W,Vi
= CV
V
Vi
E Vi.
If
W
2
Vi
# 0
as required.
is said to be homogeneous if it can be
V
The sum of all irreducible
which are isomorphic to a given irreducible submodule of
called a homogeneous component of
Vi
is a nonzero homomorphism from
written as a sum of irreducible isomorphic modules. submodules of
with
i
are irreducible,
completely reducible R-module
A
58 V j
W zz Vi'
for which
W
is irreducible, and let W be an
V
is
V.
Let V = 0 Vi, where each Vi is irreducible and let J 5 I GI be such that V . j E J are the representatives of all isomorphism classes of Vi, j' i E I. For each j E J , denote by Xj the sum of all Vi with Vz.. E Vj' 3 . 6 . PROPOSITION.
Then The X
(i)
jE J
Proof.
V
are all homogeneous components of
if
(i) Apply Lemma 3.5.
lii) Direct consequence of the definition of X j. Let V
be any R-module.
Then a submodule W of
V is said to be fully
i n v a r k n t in V if it admits all R-endomorphisms of V. 3.7. LEMMA.
Let
v
# 0 be a completely reducible R-module.
is fully invariant if and only if of
submodule W of V
sum of certain homogeneous components
v. Proof.
Vi
W is a
A
We keep the notation of Proposition 3.6.
then either f c V i ) = 0
Vj,
Thus f ( X . 1
J
G Xi r
or f ( V . )
2
proving that each X
j'
If f E End(V) and R since Vi is irreducible
V. 2 V z j and hence the sum of certain X j' I?
is
DZRECT DECOMPOSITION OF RINGS
fully invariant.
w
Conversely, assma that
P + Q
= P,
= P @
P-+Q
:
Q and therefore V
J,
of
v
=
@
x
so
Let
V by
0 be a completely reducible R-module and express
of its homogeneous components:
sum
0,
=
P @ Q 0 U for some submodule 1! of V.
=
be the given isomorphism and define an endomorphism
vf
P is
V, we need on1yshow:if
If Q = P, this is clear; otherwise P n Q
then Q C_ W.
3 . 8 . COROLLARY. Let
To prove that
W and Q is an irreducible submodule of V such
an irreducible submodule of
f
W is fully invariant.
is a sum of certain homogeneous components of
that Q
15
V as the
Then
j € J j'
Proof.
(f.1
Any family
of endomorphisms of
X. defines an endormorphism 3
3
f of
V and it is clear that the correspondence
(f.)wf is an injective 3
homomorphism
Because each X
is fully invariant &emma 3 . 7 1 , any endomorphism f
i
into itself. each X j required.
Setting fj = flXj,j E J ,
it follows that
V
of (fj)
maps
*f
as
4. DIRECT DECOMPOSITION OF RINGS. Let
(Ri),iE I be a family of rings and let R be their direct product.
the natural projections Ti : R
xi
:
Ri
-+
R
+
Ri are ring homomorphisms, while the injections
preserve addition and multiplication, but not the identity element,
and so are not ring homomorphisms.
Of course the latter implies that the images
of the R
i are not subrings, although they are ideals in R. Let R ,R ,...lR
4.1. PROPOSITION. (i)
Then
If R = R
1
1
x
... x
2
n
be arbitrary rings
R n l then there exists a decomposition of 1 1
CHAPTER I
16
1= e
... + en
i-
a s a sum of pairwise orthogonal c e n t r a l idempotents such t h a t
R =Re 8 1
...,en
e,, n
If
(ii)
1 ei
1=
with
i=1
,
then
R
Re 0
=
1e
with
i=1 i
ei
. _ .8 Ren
I
z
R
(1 G
i
E
.
e ,. .,e
li, then
R
and
i G n) R
... x Ren .
Re x Re x 1
R such t h a t R
are i d e a l s i n a r i n g
n 1 =
Re.
and
are pairwise orthogonal c e n t r a l idempotents of a r i n g
Il,...,I
(iii) I f
... 8 Ren
= I
8
... 63 In
and
a r e pairwise orthogonal c e n t r a l idem-
n
R such t h a t I. = Re 1 G i G n. I n p a r t i c u l a r , R I x ... x In. i' 1 Proof. (i) L e t xi : Ri* R be t h e n a t u r a l i n j e c t i o n , and l e t e be t h e i image of t h e i d e n t i t y element of R 1 G i G n. Then e ,...,e a r e pairwise i' n n orthogonal c e n t r a l idempotents of R with 1 = 1 E Hence R = Re 63 . .. 0 Ren i=1 i' Since I d . = Rei and ei i s t h e i d e n t i t y element of t h e r i n g Rei, t h e r e q u i r e d
p o t e n t s of
a s s e r t i o n follows. The f i r s t statement i s obvious.
(ii)
P
*
(re
,...,rn
r
any
,...,re,) in
The map
R
4
Re x 1
... x Ren,
is obviously an i n j e c t i v e homomorphism of r i n g s .
R, r e + 1 1
.. . i- rnen
Since f o r
(riel,...,rnen ) ,
i s mapped t o
t h e map
i s an isomorphism.
n
i
(iii) I f
#
j, then
e.e. E I . n I . = 0 and so e R
Hence, by Proposition 2 . 1 2 ,
Ii = Re. z
=
e.R,1 G i z
z
1-3
n.
=
3
Re 8
...
j 8 Ren = e R 8
I t follows t h a t
1
zei
=
e x
1-e j
=
=
=
1 eiej
... 8 enR z
for all
=
.
e?
3 and t h e r e f o r e
i=l
3c E
Ii,
I.I. c I . n I . = 0 f o r i # j , w e deduce t h a t ei i s a z 3 - z 3 The l a s t statement i s a consequence of (ii) c e n t r a l idempotent, 1 4 i G n. 1G
i (n.
Since
.
A ring
R,R
=
R i s c a l l e d indecomposabZe i f f o r any i d e a l s X and Y of
X @ Y implies X
= 0
or
Y = 0.
The following c o n d i t i o n s a r e equivalent.
4.2.
COROLLARY.
(i)
R is indecomposable
(ii) (8) i s indecomposable (iii)R i s not isomorphic t o a d i r e c t product of
(iv) 0
and
1 a r e t h e only c e n t r a l idempotents of
2
R
rings
m
17
MATRIX RINGS
Direct consequence of Proposition 4.1.
Proof.
Assume that R
4.3. PROPOSITION.
*,...,I n
is either artinian or noetherian. such that R = I
of R
exist ideals 11,1
8
... 0 In
Then there
and each Ii
is
an indecomposable ring. Assume that R
Proof. R = I
1
is noetherian.
We may choose a direct decomposition
@ I ‘ in which 1; is (proper) maximal, in which case Il will be 1
Next write I ’
indecomposable.
1
=
I
2
@ 1’, where 2
I
# 0
and I‘
is maximal.
Continuing in this way, we obtain a sequence of decompositions R = 7
o ... 0 I n o r;
Because the associated sequence of ideals II @
... @ I n
terminate, we deduce that I ’ = 0 at some stage.
n
choose In minimal at each stage, then I;
( n = 1.2,
In case R
...)
must
is artinian, we
is a descending chain which must
So in either case we obtain the desired decomposition.
again stop.
4.4. COROLLARY.
Any artinian or noetherian ring is isomorphic to a direct
product of finitely many indecomposable rings. Apply Propositions 4.1 and 4.3.
Proof. 4.5.
Let R ,R
COROLLARY.
Then R
1
,...,R
2
be arbitrary rings and let R = R
is artinian (noetherian) if and only if each Ri
Proof.
X
... x
R.,
is artinian (noetheriad
Direct consequence of Proposition 4.1 and Corollary 2.8.
5. MATRIX RINGS.
be an arbitrary ring and let S = Mn(R).
Let R
..
consisting of all scalar matrices diag (r,. ,r),P E R .
its image in S
every ordered pair entry
1 and
We shall identify R
( i , j ) ,1 < i , j
< n,
let e . .
The elements e i j ,
0 elsewhere.
called the matrix u n i t s ,
satisfy the following properties (i)
eijeks = 0
(ii) 1
=
e
11
+
if j
f
k
and eijeks
=
eij
... + e nn
(iii) The centralizer of
Ie..}
L I
in S
is R.
For
be the matrix with ( i , j ) - t h
23
if j = k
with
18
CHAPTER I
Observe also that R
1
e
S e
11
.
11
(i), (ii) and (iii) determine S 5.1. PROPOSITION. {V..I
23
i,j
1
set in
s.
s
Let
n)
The following simple observation shows that
up to isomorphism.
be a ring that contains a set of elements
satisfying (i) and (ii) and let R
be the centralizer of this
Then the map
Furthermore, R
is an isomorphism of rings and R-modules.
E
v
11
ull*
We know that M (R) is a free R-module with the matrix units
Proof.
Hence e i j
as a basis.
clearly coincides with
++
vij
$.
e.. '3
extends to a homomorphism of R-modules, which
In view of (i), (ii) and Iiii),
Q is easily seen to
be a ring homomorphism. Assume that
t
E
11,
...,nl
1.aijUij
= 0 and let k , s E {1,2,.
i r3
( - 1f i j V i j )
0 = Vtk
vst
r3
=
.. ,n) .
Then for all
ak s v tt
and so
Finally, fix s E
s
and for each i , j P
Then for all Opt
set - v s v ki jk
ij -
we have UPtYij
= pptVkiS
Vjk =
upis v j t
and P { jU p t =
Thus rij foregoing
commutes with a l l
upt
v k i s 'jk
= 'pis
'jt
and therefore P i j E R.
Furthermore, by the
WTRIX RINGS
proving that $
e
11
Since $ ( e l l Mn(R)e
is surjective.
M (R)ell R ,
19
u SO
) = 11
and since
11
11
the result follows.
n
and a positive integer n, we write
Given a (left) R-module V V.
n-th direct power of
v"
for
The following result shows that matrix rings arise as
endomorphism rings of direct powers of modules.
5.2. PROPOSITION.
End($) R
1 Mn(End(V))
R
Given i , j E {1,2,.
Proof.
,.. . ,On) =
EijW
Then $(Ul
E..E
ks
23
,...,V n )
if $
= 1
$
=
1
1
Eij E End(?)
=
by
R
..
(0,. ,vj,o
t i +
,O)
I . . .
and write
1.
,...,
... = qn.
=
define
Let $ E End($) R Qn(On)). Then $ E =~ E..$ ~ -L3 Hence the centralizer of { E .
and
= ($ (U )
2
End(V).
6. J kEis
.. ,n),
for all i,j if and only
.I
is identifiable with
23
Now apply Proposition 5.1.
R
$ as an R-module.
In the preceding discussion we treated consider
v"
elements of
Ax
as an M (R)-module in the following natural way.
n
fl
as column vectors and, for each A E Mn(R)
as the matrix multiplication.
5.3. PROPOSITION.
(i) If
w
and x E
$, define
V,
is a submodule of
then the map
v
b/w W n
onto the lattice of
vM
submodules of End(?)
End(V) R
M, (R) (iii) The map
VC+
ism classes of R given by
We visualize the
With this convention, we now prove
provides an isomorphism from the lattice of submodules of
(ii)
But we can also
?
induces a bijective correspondence between the isomorph-
and M (R)-mcdules.
The inverse of this correspondence is
I J k ell W.
Proof.
(i) The correspondence
Wk w"
is clearly order-preserving.
Therefore we need only show that it has an inverse which is also order-preserving. Consider the projection on the first factor
7T
:
fl+ v
and, for any submodule
20
CHAPTER I
v",
X of
preserving.
f
fl = x,
) =
is order
From the action of Mn(R)
?
=
on
for a submodule 'L
of
proving the assertion.
and fix f E End(?).
Put S = Mn(R)
Then, by the nature of action of
S
v",f
on
(gf)(W) = W.
can be written in the form
(fg)(X) = f ( V
Hence
(ii) S
It is evident that
we see that X
V.
Then the correspondence X * g ( X )
put g ( X ) = TI1 (XI.
has the same projections, say Xf,
any $ E End(V)
on all factors.
determines an element of End(?)
R
Conversely,
whose projections on all
S
factors are equal to $.
Hence the map
f* A
provides the desired
f
isomorphism. (iii) It suffices to prove that e are R
e
?= 11
11
W
1
?
and Mn(R)-modules, respectively. V x 0 x
+
... x
0
V.
Since eklw = 0
modules.
e
11
... + eni wn
2-isomorphism of
f C-+
V
and
11
Note also that
1
,...,e 11Wn )T
W onto (ellW)n .
11
W
=
e
11
W
W = 0,1
... 0 enl
k
n,W E
.
V
and
' W as R-
W, the map
is well defined and is at least an
We are therefore left to verify that
23
latter being a consequence of the action of Mn(R) on follows.
11
@
f ( e . . e w ) = e . . f ( e k l o ) for all w E W and all i , j , k E {1,2 ZJ k i
where
By the definition of e
if and only if e
(e W
w,
(ell
,...,n].
The
( e l l W ) n , the result
Given an ideal I of R, we write
5.4.
PROPOSITION.
ideals of R
(i) The map
Proof.
Mn(I) is a bijection between the sets of
and Mn(R).
M, ( R )/Mn (I)
(ii)
I+
Mn (R/I)
(i) It is clear that M (I) is an ideal of Mn(R) and that the
given map is injective. entries of elements in J .
Let J
be an ideal of Mn(R) and
Then I is an ideal of R
I s R consist of all
such that J = Mn(I).
The natural map R - PR/I induces a surjective homomorphism
(ii)
Mn(R)-+
Mn(R/I)
whose kernel is Mn(I).
For future use, we next record
kr
MATRIX RINGS
5.5.
21
Let V be an R-module and let e be an idempotent of R.
LEMMA.
Hom(Re,V)
eV
Then
as additive groups
R Similarly, if Proof. f h - 4
V is right R-module, then Hom(eR,V) R
f e Hom(Re,V),
then ef(e) = f(e2) = f(e) E eV. Hence the map R is a homomorphism from Hom(Re,V) to eV. Conversely, if v E eV
f(e)
If
Ve
R then the mapping g
2,
map
2)-
:
xe*
is an R-homomorphism from Re
XG
is an inverse of f++ f(e),
g,
to
V.
Since the
the first isomorphism follows.
The
second isomorphism is proved by a similar argument. 5 . 6 . PROPOSITION.
Let e be an idempotent of a ring R. End(Re) R
as rings
and End(eR) a eRe R
(eRe)'
g
Then
In particular End(#) R
3
Ro
and End(R
) g
R
R R
Applying Lemma 5.5 for V = Re, we see that the map f-
Proof.
f(e)
is
anisomorphism of the additive group of End(Re) onto the additive group of eRe. R Given f,g E End(Re), write f(e) = er e and g(e) = er e for some r1,r2E R. R Then
(fg) (e) = f(er e) proving that fhf(e)
=
er er e = (er el (er e) = g(e)f(e), 2
1
reverses the multiplication.
Since e
is the identity
element of the ring eRe, the above map preserves identity elements. proves the first isomorphism.
This
The second isomorphism is proved by applying
Lemma 5.5 for V = eR. 5.7.
LEMMA.
(Schur's lemma).
Let
V be an irreducible module.
Then End(V)
R
is a division ring. Proof. Let f : V + nonzero submodule of and thus Kerf
=
0.
V be
V, f(V)
=
a nonzero R-homomorphism.
V.
Since f(V)
is a
On the other hand, since f # 0 , Kerf #
Hence f is an isomorphism, proving the assertion.
V
22
We are now ready to prove 5.8. PROPOSITION.
Assume that
&?
is completely reducible.
artinian and there exist primitive idempotents e ,e $ 1
integers nl , T I 2 , .
..,nr
,...,er
of R
is
and positive
such that v
R
(eiRei)
iJMn
2
i
i=l
and each ez3ei
2
Then R
is a division ring.
Proof. Since 1 lies in the sum of finitely many irreducible submodules of R R , the same is true for
#.
. . ,en}
exists a set { e l , .
#
Rel @
=
... @ R e n ,
Hence R
is artinian, so by Corollary 2.13, there
of pairwise orthogonal primitive idempotents of R with
where each Rei
...,Rer
We may assume that Re 1 ,Re 2 ,
is indecomposable and hence irreducible.
are all nonisomorphic among the Rei,l G i G n.
. is a homogeneous component of &? corresponding to Rei,
If X
(Proposition 3.7Cii1 1 and X
i
Applying Propositions 5.6,
and therefore R
2
(Rei]
ni
then
fl = 2:
@
...@ Xr
i f1 G i G r.
for some positive integer n
5.2 and Corollary 3.9, we derive
nMni (eiRei).
The fact that eiRe
i=1
.
i is a division rirg being
a consequence of Lemma 5 . 7 and Proposition 5.6, the result follows. 6. THE RADICAL AND SOCLE OF MODULES AND RINGS.
V be a module over an arbitrary ring R .
Let
Then the radical of
J ( V ) , is defined to be the intersection of all maximal submodules of case
V contains no maximal submodule we set J ( V )
J(R)
of R
We say that R
If V
socle of R ,
In
The Jacobson radical
is the intersection of all maximal left ideals of R .
is semidmple if J(R1 = 0.
The sum of irreducible submodules of
V.
V.
is defined by
Expressed otherwise, S(R1
of
= V.
V , written
8, written SocV, is called the socle
contains no irreducible submodules, then we set SocV = 0.
written S o d ,
is defined by
S o d = Soc(#1.
The
THE RADICAL AND SOCLE
6.1. PROPOSITION. (i) If
(ii) If
f
(v)
0
f
f
V.
f
V.
is a finitely generated R-module, then J ( V )
is an artinian module, then Soc(V)
f
0
V is completely reducible, then J ( V ) = 0
(iii) If (iv) V
0
V
23
is completely reducible if and only if
SocV
=
V
SocV is the unique largest completely reducible submodule of Proof.
V
(i) By Proposition 2.5,
(ii) By Proposition 2.1,
V
V.
has a maximal submodule and so J ( V )
has an irreducible submodule and thus Soc(V)
(iii) We may clearly assume that V
in which case V
0,
f
f
0.
is the direct sum of
Hence, by the definition of J ( V ) ,
irreducible submodules (Proposition 3.2). J ( V ) = 0.
(iv) Again we may assume that (v)
f
0.
.
Now apply Proposition 3.2.
Direct consequence of Proposition 3.2.
6.2. PROPOSITION.
If
V - f .0
is an R-module, then the following conditions are
equivalent: V
(i)
is a direct sum of finitely many irreducible modules
(ii) V
is artinian and completely reducible
(iii) V
is artinian and J ( V )
=
0
Proof. The implication (i) * (ii) is a consequence of Proposition 3.2 and Corollary 2.8, while the implication (ii) * (iii) follows from Proposition 6.1(iii). Assume that (iii) holds. Since J ( V ) = 0 , V
,...,V n
By Proposition 2.1,
this implies that
n n V.
=
0
V.
of
Hence
f (J(W)
V
(V/Vi)
and so
V
Now apply Corollary 3.3.
be a homomorphism of R-modules.
with equality if f is surjective and
Then
Kerf C_ J ( W )
socv
(i), If
5 J(V).
V has no maximal submodules, then J ( V )
Assume that M
is a maximal submodule of
f* given by
nn
is isomorphic to a submodule of
Let f : W +
5J ( V )
(ii) f(S0CW)
f(J(W))
for some maximal submodules i=1
6.3. PROPOSITION.
Proof.
is finitely cogenerated.
i=1
is completely reducible, by Proposition 3.1(ii).
(i)
v
:
W+
V/M
= V
V.
and hence Then the map
CHAPTER I
24
is an R-homomorphism.
Since Kerf*
and therefore f ( J ( W ) C_ M .
Thus f ( J ( W ) 1
J(V).
is surjective and Kerf C_ J(W).
Assume that f
submodules, then so does V ,
If
W has no maximal
in which case
Assume that the set { M . l i E I } of all maximal submodules of z By hypothesis, Kerf
C Mi
for all i E
W containing Kerf
submodules of
U M f ( U ) ,U E X,
W,J(W) CKerf*
is a maximal submodule of
I.
and submodules of
is an isomorphism of X
set of all maximal submodules of
If X and
V
onto
Y.
V,
Y
W is nonempty.
are the lattices of respectively, the map
Thus
{f(Mi)
E
I} is the
and
as required. Iii) Direct consequence of Propositions 3.l(ii) and 6.l(v). 6.4. COROLLARY.
(i)
Let V
be an R-module and let
W be a submodule of
v.
C J(V)
J(W1
2 (J(V)+W)/W W 5 J C V ) , then
(ii) J(V/'/w) (iii) If
(iv) SOCW = Proof. where f
J(V/W) = J ( V ) / W
W n SOCV (i) Direct consequence of Proposition 6.3(i) applied for the case
is the inclusion map
(ii) and (iii) Apply Proposition 6.3(i) for the case where f : V
4 V/W
is the
natural homomorphism (iv) By Proposition 6.3(ii), applied for the case where f we have
SocWC SocV and hence SocW 5
W n SocV.
i s the inclusion map,
a consequence of Proposition 3.2(ii), and 6.l(v), the result follows. 6 . 5 . COROUARY.
Let V be an R-module.
for any submodule IJ of
V
Then J ( V / J ( V ) ) = 0
for which J ( V / W ) = 0.
.
The opposite inclusion being
and J ( V )
5W
25
THE RADICAL AND SOCLE
Proof.
.
The first equality is a consequence of Corollary 6.4(iii), while the
second follows from Corollary 6.4(ii). 6.6. COROLLARY.
W
(i)
=
J(V)
Let
V be an R-module and let W be a submodule of V .
if and only if
(ii) W = SocV if and only if Proof.
W C_J(V) and J ( V / W ) W
2. SocV
and
SocW =
=
0
W
(i) Direct consequence of Corollary 6.5
(ii) Direct consequence of Proposition 6.1 (v).
6.7. PROPOSITION.
Proof.
Let
(Vi),i E I be a family of R-modules.
(i) We may clearly assume that each
Then the product M i x
nVj j+i
Vi
Then
has a maximal submodule M
i'
is a maximal submodule in
V = nVi.
Hence J ( V )
iEI
is contained in each product JCVi) x n V j
and the intersection of these is
j+i nJ(Vi),
whence the assertion.
iEI
(ii) By the argument of (i), Mi 8
so that J ( S )
i
E
I.
C
i
:
W-+
8 V.)
@ J(Vi). iEI
s=
is a maximal submodule in
j+i 3 But, by Proposition 6.4(i),
J(vi)
5 J(s)
@ Vi iEI
for each
Hence the opposite containment is also true and the assertion follows.
(iii) We may clearly assume that 71
(
V = 8 V
has an irreducible submodule W.
i V. is the projection map, then W 5 fB ni(W).
Now each
either 0 or irreducible.
Thus
Ti(W)
5 SocVi
is
Ti(W)
ieI and therefore SocV
5
If
@
SocVi.
iEI
The opposite containment being a consequence of Proposition 6.l(v), the result follows.
' For any R-module V, J ( V ) = 0
6.8. COROUARY.
if and only if
V is isomorphic
to a submodule of the direct product of a family of irreducible R-modules.
tVi) ,i E I
Proof. Let
Proposition 6.7(i),
be a family of irreducible R-modules.
J ( n V i ) = 0.
By
Hence the "if" part is a consequence of
iEI
Corollary 6.4.
Conversely, assume that J ( V ) = 0.
nothing to prove.
Hence we may assume that
V
# 0,
If V = 0
then there is
in which case there is a
26
CHAPTER I
of maximal submodules of V
(Wk),k E K ,
nonempty family
V admits an injective homomorphism into
n (V/Wk).
n Wk = 0. MK
with
Thus
V/Wk is
Since each
KK irreducible, the result follows. submodule W
A
x
of
4
V is called essential in V in case for every submodule
v,
of
W n X = 0 implies
Dually, a submodule W
of
X
=
0
V is called superfluous in V in case for every
X of V
module
W +X 6.9. PROPOSITION.
Let V
=
V implies X
=
V
be an R-module.
v
is
is the unique largest superfluous submodule of
V.
(i) J(V) is the sum of all superfluous submodules in finitely generated, then J(V)
v.
Furthermore, if
(ii) SocV is the intersection of all essential submodules in if
5 J(V),
K + W of
V.
Furthermore,
V is artinian, then SocV is the unique smallest essential submodule of V. Proof. (i) Let W
W
sub-
=
be a superfluous submodule in
we may assume that V
V and hence K
=
module of
V with
module K
of
occw.
RX iW =
V , then either
implies that RX
.
If W E K, then
Thus every superfluous submodule
On the other hand, let x E
V with W S K and x 9 K
Thus x E J ( V )
To prove that
has a maximal submodule K.
V, a contradiction.
V is contained in J ( V 1 .
V.
w=v If x
v.
If
w
is a sub-
or there is a maximal sub-
s J ( V ) , then the latter cannot
is a superfluous submodule of
v,
proving the first statement. Now assume that V verify that J ( V ) submodule of But
is finitely generated.
is a superfluous submodule of
V with
v = w + J(V).
By the foregoing, we need only
V.
So assume that W
Then, by Corollary 6.4(ii),
V/W is finitely generated, so by Proposition 6.1(1) v/W
=
0.
is a
J(v/w) = v/w. Thus
w=v
as required. (ii) Let M
be the intersection of all essential submodules in V .
To prove
THE RADICAL AND SOCLE
that Socv EMI
V has an irreducible submodule W.
we may assume that
V, then L n W # 0 and so W
is an essential submodule of
and therefore we need only verify that M of M.
submodule N
N n N ’ = 0, then
N
5 M 5 N @ N ’,
This N
+
N’ = NO N ’
is a direct summand of
v
Thus SocV
Since V
M and therefore SocV
So fix a
V.
But then
=
M.
We need only verify that Socv is an
W n SocV
Assume that
is artinian, so is W
has an irreducible submodule X.
0 for some nonzero submodule
=
But then X @ SocV is a completely reducible
is essential in V, as required. 6.10. PROPOSITION.
Let
Hence SocV
’
V be an R-module generated by if and only if for all r
Then U E J(V)
W
and hence, by Proposition 2.1,
module strictly containing Socv, contrary to Proposition 6 . l ( v ) .
V.
5M
V maximal with respect to
is an essential submodule of
is artinian.
v.
essential submodule of
U
If L
and by modularity
Now assume that
W of V.
2L.
is completely reducible.
If N’ is a submodule of
N
27
i
E
U
,...,U n ,
R, 1 G i
f
and let
n, the elements
ui + r .u, 1 G i G n, generate V. Proof. Let U E J(V) elements U .
+ riu,
and let
i Q n.
1
W be the submodule of V generated by the W + J(V)
It is clear that
=
V so that W
= V,
by Proposition 6 . 9 ( i ) . Conversely, assume that U
V such that i
U
4 W.
there exists r
i
J(V).
Then there is a maximal submodule W
Hence we must have RU E
thus the elements U . z
R
such that ui E
+ r z.u,
1
Q
i
-
+W
=
+ W,
r.U
V.
of
It follows that for each
whence U i + riU
n , do not generate V.
W and
’
We now turn our attention to the radical and socle of rings. Let
V be an R-module.
It is clear that ann(V)
The annihiZator of
v, written ann(V),
i s an ideal and that
is defined by
V may be viewed as an R/ann(V)-
CBaPTFR I
28
module.
V is faithfd if ann(V)
We say that
An ideal
=
0.
I of R is called primitive if the ring R/I has a faithful Clearly I is primitive if and only if I
irreducible module.
is the annihil-
ator of an irreducible R-module. Let V
6.11. LEMMA.
V
1
R/X
t,
so V
=
Then V
is irreducible if and only if
for some maximal left ideal X of R.
Proof. let
be an R-module.
Clearly
R/X
be a nonzero element of RV.
The map
R
Suppose that V
is irreducible.
--+
V ,P
V. I-+
Then RV 19
is irreducible and
V
is a nonzero submodule of
is a surjective R-homomorphism.
R/XZ V for some left ideal X o$ R.
is irreducible, X
Since V
and
Hence
is
maximal. Every maximal left ideal contains a primitive ideal and every
6.12. LEMMA.
primitive ideal is the intersection of the maximal left ideals containing it. Let X be a maximal left ideal of R.
Proof.
Hence the annihilator of R/X
irreducible R-module.
By Lemma 6.11, R/X
is an
is a primitive ideal
contained in X. Let I be a primitive ideal in R whose annihilator is I . Clearly Mv and so Mv Mv
with
0 # v E V,
J(R)
5I
in V ,
Since V
is a maximal left ideal in R. the result follows.
put Mv = {r E Rlrv = 01.
is irreducible, V = Ru
R/M
.
Let R be a ring. In
is an ideal of R.
is the annihilator of a completely reducible R-module. if I
v
Since I is the intersection of all
is the intersection of the annihilators of irreducible R-modules.
particular, J(R) (ii) J(R1
Given a nonzero v
is a left ideal in R.
6.13. PROPOSITION. (i) J(R)
and let V be an irreducible R-module
Furthermore,
is the annihilator of a completely reducible R-module.
(iii) SocR is an ideal of
R.
Proof. (i) Direct consequence of Lemmas 6.11 and 6.12. (ii) If
v=
@ Vi, iEI
then ann(V) =
n ann(Vi).
Setting
CVi) ,<
E
I to be the
iW
set of representatives of the isomorphism classes of irreducible R-modules, we see
THE W I C A L AND SOCLE
that JtR) = ann(V1.
29
The second statement is a consequence of the fact that any
completely reducible R-module is either
0
or a direct sum of a family of
irreducible submodules. (iiil For a given
RR
21 E
R, the right multiplication by
is an endomorphism of
r
Hence, by Proposition 3.1(ii1, (SocR)r is completely reducible.
(SocRlr 5 S o d ,
by Proposition 6.l(v).
6.14. PROPOSITION. Proof.
Thus
V
If
is an R-module, then J(R)V CJ(V)
V
If
has no maximal submodules, then
to prove.
Let M
R-module.
Since J(R)
V.
be a maximal submodule of
.
annihilates
V and there is nothing
J(v) =
V/M
Then
is an irreducible
we see that J(R1 V
V/M,
5 M.
Thus
J(R)V c JIV~.
Let V
6.15. PROPOSITION. (Nakayama's lemma).
If W
V
is a submodule of
Proof.
with
By Proposition 6.14,
W
+
J(R)V
W + J(V)
= =
be a finitely generated R-module. V,
V.
then W = V . Since J ( V )
is a superfluous
V (Proposition 6.9(i)), the result follows.
submodule of
Let f :
6.16. PROPOSITION.
R1
R
2
be a surjective homomorphism of r i n g s .
J(R21 with equality if Kerf
Then f(J(Rl))
5 Jell.
Let { M i l i E I } be the set of all maximal left ideals of R
Proof.
containing Kerf.
Then
{f(M,) li E 13
is the set of all maximal left ideals of
R2 and
Hence J(R
) =
fl M.
6.17. COROLLARY. (i) J(R/I)
and, by the above, f(J(Rl)) = J(R2).
i q
1
>_
Let I
.
be an ideal of a ring R.
(J(R) + I ) / I
(ii) If 1 5 J(R),
then J(R/I) = J(R)/I.
(iii) If J(R/I) = 0 , (iv) I = J(R)
.
{Mili E I) is the set of all maximal left ideals of R
If Kerf C_ J ( R 1 ) , then
then J(R)
5I
if and only if I C_ J(R1
Proof. (i) and (ii).
In particular, J(R/J(R)) = 0.
and J(R/I)
=
0.
Apply Proposition 6.16 for the case where f : R-
R/I
CHAPTER I
30
is t h e n a t u r a l homomorphism. (iii) D i r e c t consequence of (i) (iv)
Direct consequence of (ii) and C i i i ) .
x
An element
y E R
exists
x
element
x
R
of
yx
such t h a t
R
of
is called a
left
(respectively,
xy
= 1 (respectively,
y E R,
6.18. PROPOSITION.
x
Let
be an element of a r i n g
- rx
r E R, 1
x-l,
denoted by
is a l e f t unit.
right) unit
i f there
unit we mean an
By a
= 1).
which is both a l e f t and a r i g h t u n i t .
is a u n i t i f t h e r e e x i s t s
only i f f o r a l l
. Expressed otherwise,
zy
such t h a t
R.
=
yx
= 1.
x E J(R)
Then
J(R1
In p a r t i c u l a r ,
i f and
contains
no nonzero idempotents. Proof.
1
The element
- PX
i s a l e f t u n i t i f and only i f
So t h e f i r s t statement follows by v i r t u e of Proposition 6.10.
J(R).
an idempotent i n
e
on t h e r i g h t by 6.19.
I of
Hence 6.20.
x E J(R1.
z = 1-y = y
0.
- rx
= 1
f o r some
Assume t h a t
x
E R.
Then
is a u n i t f o r a l l
1
Since
-yx
is in
-5
J(R)
COROLLARY.
Let
R
Multiplying
i s t h e unique l a r g e s t i d e a l
r E R,x E I.
is a l e f t u n i t ,
J(R)
i s a u n i t and t h e r e f o r e
e is
9
By Proposition 6.18, it s u f f i c e s t o v e r i f y t h a t
Proof.
Then
1
=
x(1-el
R be a r i n g .
Let
such t h a t
for all
e
yields
PROPOSITION.
R
By t h e above
R(l-rx) = R.
and hence 1-
x
= y
-1
ytl-5)
1
-x
f o r some
= 1
1 = y'(1-z) = y ' y
is a u n i t
y E R.
f o r some
y'
E
R.
is a l s o a u n i t .
be a r i n g .
(i) J(Rol = J (R 1
(ii) J ( R )
is t h e i n t e r s e c t i o n of a l l maximal r i g h t i d e a l s of
Proof.
COROLLARY.
Proof. f o r any identity
.
( i )D i r e c t consequence of Proposition 6.19.
(ii) Follows from (il. 6.21.
R.
If
I i s a l e f t or r i g h t n i l i d e a l of
Assume t h a t
r E R,rx E I
I
i s a l e f t n i l i d e a l of
and t h e r e f o r e
= 0
R,
R
f o r some
then
and l e t
n a 1.
I C_ JW).
x E I.
Then,
Invoking t h e
31
THE RpIDICTiL AND SOCLE
n-1 i
n-1 (1-y) (
1 - rx
we infer that
c y i=o
c y 1 i=o
= (
1 (1-y)
=
l-yn
Hence, by Proposition 6.18, I s J ( R ) .
is a unit.
is a right nil ideal, then the same argument applied to Ro
.
If I
yields the result.
As a partial converse we have If R
6.22. PROPOSITION.
=
J
k+l
=
...
>
for some k
Setting I
1.
Assume by way of contradiction that I f 0 . with IX # 0, for example X = R. all such X .
Then I ( J M )
=
Let M
( I J ) M = IM # 0
we must have R x =
6.23. PROPOSITION. if I V # V
Let I
Jk
we have I'
and, since JM
5M
=
I.
of R
we have JM
=
M.
is a finitely generated R-
such that T x # 0 and so I ( R x ) # 0.
M, as required.
be a left ideal of R.
Then I E J I R )
if and only
for any nonzero finitely generated module V .
Proof. Let V # 0 be a finitely generated R-module. Hence, by Proposition 6.14, if I
Proposition 6.l(i). IV C - J(R)V
By hypo-
be a minimal element in the set of
By hypothesis, there exists x E M
As Rx E M
=
2 J 2 3 .. . .
There exist left ideals X
By Proposition 6.15, we are left to verify that M module.
is nilpotent.
and consider the chain J
Put J = J ( R )
Proof.
k thesis, J
is artinian, then J(R)
5J ( V )
rated module
V,
# V.
Conversely, if IV # V
6.24. PROPOSITION.
Let
(Ri) ,i E I ,
icI
An
element
n
is1
(pi) E
for all i E I .
6.25. PROPOSITION.
Proof.
For any ring R
Let V
then
V.
be a family of rings.
Thus I C J ( R )
Then
J(Ri)
n R i
.
is a left unit if and only if ri
is1 left unit of Ri
by
for any nonzero finitely gene-
.
J(nRi) =
Proof.
5J(R),
then IV = 0 for any irreducible R-module
by Proposition 6.13 (i).
Then J ( V ) # V
Now apply Proposition 6.18.
and any positive integer n ,
be an R-module and let
v"
be an
Mn (R)-module as in
is a
32
CHAPTER I Then ann(V7 = Mn(ann(V1),
Proposition 5.3.
by the definition of
Vn.
the desired conclusion follows by virtue of Propositions 5.3 and 6.13(i). PROPOSITTON.
6.26.
Let
R be a ring.
Hence H
Then the following conditions are
equivalent: (il
is a direct sum of finitely many irreducible modules
$
(iil
#
(iii)
R
i s completely reducible is semisimple artinian
(ivl Every R-module i s completely reducible (v) R
is a finite direct product of full matrix rings over division rings. The equivalence of (i) - (ivl is a consequence of Propositions 5.8,
Proof.
The implication (iil * (vl follows from Proposition 5.8.
and 3.4.
2.1,6.2
implication (v) * (iii) is a consequence of Proposition 6.25,6.24
and 5.3 together
with Corollary 2.8. A
ring R
is said to be s h p k if 0 and R
PROPOSITION.
6.27.
Let R
be a ring.
are the only ideals of R.
Then the following conditions are
equivalent. (i) R (ii
is simple artinian
&?
(ii
is a direct sum of finitely many isomorphic irreducible modules
R is isomorphic to a full matrix ring over a division ring
)
R is semisimple artinian and a l l irreducible R-modules are isomorphic.
(iv
Proof. Cil *
Wl: Direct consequence of Propositions 6.26 and 5.8
(iil * Liiil: Follows from Propositions 5.2 and 5.6 and the Schur's lemma. (iiil * livl: Follows from Propositions 6.26 and 5.3. (iv)
*
(1):
Follows from Propositions 6.26,5.2
As a partial converse of Proposition 6.14, 6.28.
PROPOSITTON.
Proof.
Let . V
By Proposition 6.14,
J(Rl V
and 5.4.
we prove
be an R-module.
If R/J(R)
5 JLV)
The
is artinian, then
and hence, by Corollary
THE RADICAL AND SOCLE
6.6(i), we need only show that J(V/J(R)V) (Corollary 6.17 (ii)) 6.26). Hence
, every
0.
Since J(R/J(R)1
=
0
R/J(R) -module is completely reducible (Proposition
In particular, the R/JCR)-module V/J(RIV
=
33
V/JCR)V
is completely reducible. = 0,
is a completely reducible R-module, and therefore J(V/J(R1V1
'
by Proposition 6.l(iii)6.29. PROPOSITION.
Let R/J(R)
be an artinian ring.
Then
SocR = { r E RIJ(R)r = 01 Proof. all x
E Sod?.
Since
V
Since SocR
is a completely reducible R-module, J(R1x
Setting V = {r E RIJCRIP = 01,
module (Proposition 6.2.61. by Proposition 6.1(vlI 6.30. PROPOSITION.
V
Proof.
V
Hence
is a completely reducible R/J(R)-
is a completely reducible R-module and so,
Let R be a ring such that every irreducible R-module is Then
The hypothesis on R
for every irreducible R-module
6.31. PROPOSITION.
5V.
5 SocR. '
isomorphic to a submodule of R R
PV = 0
V
can be regarded an R/JCR)-module,
it follows that SocR
0 for
=
Let R
ensures that rSocR = 0 if and only if
v.
.
Now apply Proposition 6.13 (i)
be an artinian ring and let V
8
be an R-module.
Then the following conditions are equivalent: (i)
[I
is of finite length
(ii) V (iii)
is artinian
V
is noetherian
Proof.
By definition, (i) * (ii) and (i) * (iii).
either artinian or noetherian.
By Proposition 6.22, J(RIn = 0
By Proposition 2.1, it suffices to show that the modules
.-.,J(R)n-lV/J(R)nV annihilated by
Assume that
have composition series.
for some
v YZ
is 1.
V/J(R)V, J(R)V/J(R)'V,
However these modules are
J(R) and so can be regarded as R/J(R)-modules.
Furthermore, by
Proposition 6.26, each of these modules is completely reducible and, by hypothesis,
CHAPTER I
34
.
each of them is either artinian or noetherian. fore a consequence of Corollary 3.3. Let R
6.32. COROLLARY.
(i) R
The desired conclusion is there-
be an artinian ring.
is noetherian
(ii) R R has a composition series (iii) Every finitely generated R-module is of finite length. Proof. (i) Apply Proposition 6.31 for V = R R (ii) Apply (i) and Proposition 2.10 (iii) Let V be a finitely generated R-module.
Since
image of a direct sum of finitely many copies of R, V
V
is a homomorphic
is artinian.
Now apply
Proposition 6.31. Let R
6.33. PROPOSITION.
be an arbitrary ring, let V
be an R-module, and let
S be a subring of R. (i) If
V
has a composition series V 2 V 1
2
3
...
In particular, J(Rln = 0 if
then J(RInV = 0.
3
V
Vn+l 3 0
of length n ,
is a faithful R-module.
(ii) If every irreducible R-module is completely reducible as an S-module, then
J(S) 5 J(R). annihilates each factor Vi/Vi-l,
Proof. (i) Because J(R)
J ( R ) n must
annihilate V . (ii) Let
xE
J(S)
and let V
be an irreducible R-module.
.
completely reducible as an S-module, xV = 0. assertion follows.
Hence z E J(R)
Since V
is
and the required
The following consequence is a sharpened version of Proposition 6.22. 6.34. COROLGARY.
length of
RR
Proof.
Let
d be an artinian ring and let n be the composition
Then
By Corollary 6.32,
Proposition 6.33 for
V
=
RR
R R has a composition series.
Now apply
35
THE RADICAL AND SOCLE
Let e
PROPOSITION.
6.35.
0 be an idempotent of a ring R .
f
Then
J(eRe) = eJ(R)e Proof.
Let I be a primitive ideal in R Then eV
irreducible R/I-module.
J(eRe) 5 eRe
5 I.
and so
eV = W .
Let 0 c W = eW
It follows that eV
5 I.
Since I was an arbitrary primitive
Then eue E J ( R )
.
and so, by Proposition 6.18,(1-b)
in which case beae = eue+b.
on both sides yields eue+ebe = ebeue.
Kence
e
Multiplying by
(e-ebel (e-eue) = e
Since eJ(R)e is an ideal of eRe,
is a left unit.
W is an
and so J(eRe) = eJ(eRee1e C_ eJ(R)e
this yield8 J(eRe) C - JCR)
(1-eue) = 1 for some b E R ,
where
is an irreducible eRe-module whence
Thus J(eRe1
Suppose that u E J ( R ) .
5 eV
Hence eV = eRe Kf C_ W
RW since V is irreducible.
J(eRe)V = J(eRe)eV = 0 . ideal in R,
If eV = 0 then
is an eRe-module.
Assume that eV # 0.
Then V =
eRe-module.
and let V be a faithful
so that e-eae
Proposition 6.18 implies
that eJ(R)e 5 J(eRe1. For any idempotent e
6.36. ~RoPosITIoN.
Proof.
Since
R
of R ,
J(Re) = J ( R ) e .
R = Re @ R ( l - e ) , we have J ( R ) = J(Re) @ J ( R ( 1 - e l )
(Proposition 6.7 (ii)) and J ( R 1 = J @ ) e @ J W ) (1-el
JW)e
5 J(Re)
and J ( R ) (1-e)
E R
But, by Proposition 6.14,
Hence J(Re) = J ( R ) e , as required.
Let I be a nilpotent left ideal of a ring R,
6.37. PROPOSITION.
x
5 J(R(1-el).
-
be a nonnilpotent element such that x2-x E I.
contains a nonzero idempotent Proof.
y such that y-x
k = 0, and set
Let
If z1 # 0 ;
and we are done.
z
1
let x
= x2-x E I. 1
commute with each other, and thus if x Hence x
a contradiction.
1
"hen the left ideal Rx
I.
d
= x+z -2xZ
and let
If Z l = 0, 1
E Rx.
choose y =
Then X,X
is nilpotent, then so is X
is a nonnilpotent element of RX,
and 3
X
1
-2
2,
Z
+2XZ
1
and direct
calculation shows that
r 2
1
The element z
2
=
x 12
-x
1
-
2
i
1
423 1
- 322 1
is nilpotent, contains z 2 as factor, and canmutes
1)
CHAPTER I
36
with x
1
.
Continuing in this manner, we may construct a sequence { xz. ) of non-
i
nilpotent elements of R x , choose i
x.
so that 2i 3 k ,
=
.
Moreover, xi # 0
= 0.
x. is the desired idempotent.
If we since
Let I be a nonnilpotent left ideal in a (left) artinian ring
6.38. COROLLARY.
8.
z - x z.
then we have X?
Thus y
is nonnilpotent.
occurs as a factor in x i2- x i .
such that 2'
I contains a nonzero idempotent.
Then Proof.
Let X
be a minimal member of the nonempty set of nonnilpotent left
ideals contained in I.
Then I 2
5I
and so I 2 = I.
Now let Y
in the set of left ideals contained in
x
such that XY # 0 ,
so that Xa # 0.
XU
5Y,
Then X*Xa # 0
and
k
and so a = x a
for some x E X
the nonnilpotent element x.
for all k
Invoking Proposition 6.37, we see that RX
so does I ,
since
0.
Y
=
and choose a E
Xa.
Y
Thus a = m
It follows that Y contains
Setting
Moreover, xa = Y # 0
we have xz-x E L .
whence
be minimal
I 3 X 2 Rx.
so
L cX
and thus L
is nilpotent.
contains a nonzero idempotent, whence
m
7. THE KRULL-SCHMIDT THEOREM
This section deals with the uniqueness of direct sum decompositions.
The main
result (Theorem 7.5) is Azumaya's generalization of a classical theorem of K?zull and Schmidt (Corollary 7.6). A ring R
is said to be a locaz ring if R / J ( R )
what follows, we write 7.1. LEMMA.
(i) R (ii) R
is a division ring.
In
U ( R ) for the group of units of R.
The following conditions are equivalent:
is local has a unique maximal left ideal
(iii) J ( R ) = R-U(R) (iv) The set of nonunits of R
Proof.
(i) * (ii):
Let
is a left ideal
I be a maximal left ideal of R.
Then I / J ( R )
is
37
THE KRULL-SCHMIDT THEOREM
and hence I = J(R).
a proper ideal of R/J(R)
be a unique maximal left ideal of R.
(ii) =) (iii): Let J
Since J(R)
intersection of all maximal left ideals of R, we have J(R) = J , an ideal.
The inclusion J(R)
If RX # R, then RX
so x E J(R).
Z E
(iii) (iv)
=)
=)
of R, fore
and so J
being obvious, assume that x
lies in a maximal left ideal of R, whence Rx then YX
On the other hand, if RX = R
Clearly y Q J(R), some
5 R-U(R)
1 = yx E J W .
otherwise
R and hence z
=
I.
=
E
is
R - U(R).
5J W ) ,
1 for some y E R.
Hence Ry = R,
Thus x E U(R),
is the
so zy = 1 for
a contradiction.
(iv) Obvious (i): Let x Q J(R) then M
x + J(R1
and let I = R - U(R).
I # R, whence M
=
I = J(R).
is a unit of R/J(R).
If M
is a maximal left ideal
Hence x
is a unit and there-
Thus R/J(R)
is a division ring, as
required. 7.2. COROLLARY.
Let R
Then the only idempotents of R
be a local ring.
are
0 and 1. Let e # 1 be an idempotent of R.
Proof. by Lemma 7.1, 7.3. LEMMA. 0
and
e E J(R1. Let V
Then e
is a nonunit and hence,
Now apply Proposition 6.18.
be an R-module.
1 are the only idempotents of
V is indecomposable if and only if
Then End(V).
In particular, by Corollary 7.2,
R if EndCV)
is local, then
V is indecomposable.
R Proof. projection
Assume that
V
=
X @ Y is a proper decomposition.
V - + X is an idempotent of End(V) R
Then the
distinct from 0
Conversely, assume that f # 0,l is an idempotent of End(V).
and
1.
Then V=f(V) $Kerf
R is obviously a proper decomposition. 7.4. LEMMA. (il
If
Let R
be a ring and let V
be an R-module of finite length.
V is indecomposable, then every f E End(V) R
is either a unit or nil-
potent. (iil
V
is indecompsable if and only if EndlV) is a local ring.
R Proof.
(i) We first show that an R-homomorphism f : V-+
and only if it is surjective.
Assume that f
is injective.
V
is injective i f
Then there exists
38
CHAPTER I
n
an i n t e g e r
such t h a t
u E V,
Therefore, f o r each
f
Since
f ( U ) = u E f(V)
is injective,
f
Now assume t h a t
U E V
there e x i s t s
f
and so
is surjective.
f n ( u ) = fn+'(U).
such t h a t
is s u r j e c t i v e .
By hypothesis, t h e r e e x i s t s an i n t e g e r
n
such t h a t
0 c Kerf c
f
Since form
fn)
(and hence
f o r some
U = fn(U)
U E Kerfn+l = Kerfn,
... c K e r y
= KerP+l
u E Kerf
i s s u r j e c t i v e , each U E V.
and thus
f(u) = 0,
Because
u = 0.
may be w r i t t e n i n t h e it follows t h a t
f
This proves t h a t
is injective.
I t w i l l next be shown t h a t
V
Since
i s indecompsable, t h e r e q u i r e d a s s e r t i o n w i l l follow.
n 2 1
integer submodule
fnCV)
injective. u E V
each
V
so
=
=
fn+l(V).
f
Then t h e r e s t r i c t i o n of
t o the
i s s u r j e c t i v e , so by t h e above t h i s r e s t r i c t i o n i s a l s o
This shows t h a t there e x i s t s
f"(V) @ Kerfn,
(ii) Assume t h a t
V
U E V
K e r f n = 0.
Because
fn(u - f n ( U ) )
such t h a t
fn(V)
= 0.
=
f2n(V),
E
i s a local ring.
E.
Then
Let
gf
1-gf i s a u n i t of
by v i r t u e of ( i ) . Thus
I t follows t h a t
w e deduce t h a t
E
E = End(V). R
f
By Lemma 7 . 3 , w e
be a nonunit of
E
and l e t
is a nonunit and hence i s n i l p o t e n t ,
E,
so by Proposition 6.18, f E J ( E )
is local.
W e a r e now ready t o prove
7.5.
THEOREM.
Let
R
for
a s required.
be an a r b i t r a r y element of
Invoking Lemma 7.1,
fn(V) n
is indecomposable and p u t
need only v e r i f y t h a t
g
fnCV)
such t h a t
Choose an
be a r i n g and l e t
V
be an R-module.
Suppose t h a t
39
TBE KRULL-SCHMIDT THEOREM
i,j
where f o r a l l
the rings
End(V.1
and
End(W.)
R Z V.
and a f t e r a s u i t a b l e rearrangement Proof.
rn
W
2
i
t h e p r o j e c t i o n s of
V
n @ W.. j=1 3
=
ei,fj
e f.e
are a l l i n
1 3 1
End(V1).
t h e r e s u l t i s c l e a r s i n c e by
= 1,
Vi,Wj,
onto
e
Because
R
= 1
e f.e
l o c a l , it follows from Lemma 7 . 1 t h a t
respectively.
n
c
j=l
V
morphism of
= e (V).
1
.-I
W e now prove t h a t
x
V.
(X)
= 0
e (y) = 0
as
e
Cf
e
)
(y) = 0 ,
and
f o r some
End(V1)
elfle,
j.
i s an auto-
f
= (f e ) (y)
X
V1 i s
on
y E 8.
f o r some
I 1
elf,el
f (V
proving t h a t
V
i s an automorphism on
.
W.
I t follows t h a t
1
)
n ( Q Vi)
Assume t h a t = 0. i=2 Hence e (U) = (e f e ( W ) f o r )
1
el(v) E e ( V ) = (e f e
Then W E
R
m
I
some
is
End(V1)
i=2
1
then
I 1 V E
and 1
f , (V1) 5 IJ1 and t h e k e r n e l of rn V = f (V @ ( 8 Vi).
Thus (elflel)(y) = el(x) and 2 =
3
Then
R
i=2
Hence
1
Thus
m fl(V1) n 8 vi,
E
e f .e
is a unit i n
1 3 1
Changing n o t a t i o n , i f necessary, it may be assumed t h a t
If
Using induction
i s indecomposable.
Denote by
0.
m = n
Then
...,r n } .
E {1,2,
rn V = @ V.
n
= 1
are local.
3
for a l l
i
or
observe t h a t i f
V
Lemma 7.3
and
z
We may harmlessly assume t h a t
on min{rn,n},
e
R
(V).
1
1
e (v-(f e 1
) 1
(W)l= 0
1
1
U - ( f e 1 (w) E
and
1
1
8 Vi. i=2
Thus m
m proving t h a t Because Lemma 7.3,
V =
f (V 1
Wl
f,(Vll 8 1
5W ,
)
8 Vi). i=2 t h i s implies t h a t (
i s indecomposable.
V
I
and
and hence, by i n d u c t i o n ,
i
E {I,
7.6.
V = W @ 1
rn = n
(
m =
{wl n 8 V i } . i= 2 w e see t h a t W1 =
f, (Vl) 8
f (V ) # 0 ,
Since
1
m i s isomorphic to
w
8 Vi). i=2
1
By
fl(V1)
The conclusion i s t h a t
and a f t e r a s u i t a b l e rearrangement
V.
Wi
...,ml.
COROLLARY.
(Krull-Schmidt theorem).
Let
V
be a nonzero R-module of f i n i t e
m length.
for
Then
V
can be w r i t t e n a s a d i r e c t sum
V = 8 Vi, i=1
where
Vi
are
CHAPTER I
40
n indecomposable submodules.
for each
V
= @ W
is another decomposition of j=1 j and (after possible reordering the W . ) we have Vi Wi 3
this kind, then n = rn
Moreover, if
i. Apply Proposition 2.9, Lemma 7.4Cii) and Theorem 7.5.
Proof.
.
8 . PROJECTIVE, INJECTIVE AND FLAT MODULES.
Let
R be
a ring.
A
sequence of R-modules and R-homomorphisms
v1 is emct at
Vi
-c
2
J1
J 2
v
v2
if Kerf. = Imfi-,
.
c
--t
... 'n-1
n'
The sequence is emct if it is exact at
each place, that is Kerfi = Imfi-, A short exact sequence is an exact sequence of the form
0-x-Y a
(*I Exactness of
(*)
KerB = Ima and
- Bz - 0
is equivalent to the following three conditions: a
8 is surjective.
The short exact sequence
In this case
B
induces an isomorphism
is split if a ( X ) is a direct summand of
(*)
is injective
Y.
This
is equivalent to either of the following: (i) There exists a y E HOm(Y,x) R identity map on X. (ii) There exists a
such that ya = lX, where
6 E Hom(Z,Y)
such that
86
=
lX denotes the
lZ
R The homorphisms y
and
6
are called splitting homomorphisms.
A diagram of R-modules and R-homomorphisms
."x
is conmutative if y = Ba.
Y
The same terminology applies to more complicated
PROJECTIVE,INJECTIVEAND FLAT MODULES
41
diagrams. Let V
be an R-module.
v
Then
is called projective, if for any given
diagram with exact bottom row
V
there exists an R-homomorphism y 8.1. PROPOSITION. (i) ~n R-module
The following conditions are equivalent
V
-
is projective
(ii) Any exact sequence 0 (iii) V
f3 = (Yy.
such that
B Y --+
X
V-0
of R-modules splits.
is a direct sueanand of a free R-module
Proof. ti) * (ii): Because
B
is surjective and
V
is projective, we have
the following commutative diagram:
V
Hence y
is a splitting homomorphism and (ii) follows.
(ii) * Ciiil: Take as
Y a suitable free R-module, and let
splitting homomorphism. (iii) * (i): Let V
Y
E : V-+
F
Then we can let
71
:
the natural injection so that
are R-modules and that
@ : V-
Y
surjective homomorphism, respectively.
-f
Y be a
is a direct summand of Y.
V
be a direct summand of the free R-module F ,
be a free basis for F . and
Then y(Vl
y : V
and
(Y
F+ TE
:
and let
{yi}
V be the natural projection = 1II
.
Now, suppose X and
X-+ Y are
a
homomorphism and a
We then have a natural map
to Y, and we can describe the situation pictorially as follows:
6T
from F
CHAPTER I
42
F
X
Y
0
1
)
t
0-Y
Now, for each generator y i E F , BT(yi) E Y and because a exists xi E X 6 Q .z)
=
xi,
Finally,
BT(yil
with
= cl(xi).
Thus if we define
is surjective there
6
:
F-
X by
then the outer portion of the diagram is commutative, that is, a 6 = 6 T .
E :
V+
F
so that
6E :
V+
X and
Thus the lower portion of the diagram is commutative, and by definition we
conclude that V
is projective.
8 . 2 . COROLLARY.
Let V
be a finitely generated R-module.
projective if and only if
V
Then V
is
is a direct summand of a free R-module of a finite
rank. Proof.
Let v ,v , . . . , I ) 1
module freely generated by surjective homomorphism F-+
.
direct summand of F . result follows.
be a generating set for V
2
x ,x 1
2
V.
,...,xn. Thus if
and let F
Then the map xi* V
vi
i s projective, then
be a free
determines a V
is a
The converse being true by virtue of Proposition 8.1, the
If we reverse the direction of all the arrows in the definition of a projective module, we then obtain the definition of an injective module. module
V
Thus a given R-
is i n j e c t i v e , if for any given diagram with exact bottom row
43
PROJECl'LV, INJECTIVE AND FLAT MODULES
F\ B
\ \y
\ Y-0
a
there exists an R-homomorphism Y 8 . 3 . PROPOSITION.
(i) An R-module
B = ya.
such that
The following conditions are equivalent: V
is injective
(ii) Any exact sequence 0
+
-& Y+
V -%X
0
of R-modules splits.
Because a is injective, we have the following
Proof. (i) ;+ (ii): commutative daigram:
V
I
0-v
(]I
\ *X
Y is a splitting homomorphism and (ii) follows
Hence Cii)
\ A X
*
(i):
Consider a commutative diagram with exact bottom row
v
I* \
\
\r \
0-v
B
01
\
X
We may harmlessly assume that a is the inclusion map, in which case Y direct sumnand of X, by hypothesis. by
y(y+z) = @(y),y E V,z E 2.
Write X = Y e 2
Then y
and define Y : X-
is an R-homomorphism and
B = ya,
required. 8.4. PROPOSITION.
(i)
@
Let
( V i ) , i E I be a family of R-modules.
Vi. is projective if and only if each Vi is projective
&I (ii)
nVi
*I
is injective if and only if each Vi
is a
is injective.
Then
v as
CHAPTER I
44
Proof.
V
(i) If
0 Vi,
=
GI if and only f so is each Vi. and only if each Vi (ii) Let V
V
is a direct summand of a free module
Hence, by Proposition 8.1, V
and let
V
:
7Ti
---t
Vi and
pi
:
Vi
4
projections and injections, respectively. :
V
-+
Bi
:
Vi
X+
Then
Proposition 8 . 3 ,
V is injective.
Conversely, assume that V
:
X
-+
ni =
is injective and let
Then there exists an R-homomorphism
Vi by
:
there
X+
V by
and so, by
= lv
c1 :
Vi-+ X be an injective
B
X
:
Then ycc = ni($a) = nipi
y = "Ti$.
is injective and
Define $
BiC1.
B is an R-homomorphism with @
$(z) = (Bi(z)1 .
R-homomorphism.
with
vi
that each
Then, for each i E I ,
X be an injective R-homomorphism.
exists an R-homomorphism
Define y
Assume
be the natural
V
iEI
let c1
is projective i
is projective.
nVi
=
then
+
V
vi
with
= 1
=
Ba.
and hence, by
'i Proposition 8.3,
V;
is injective.
Given a pair of rings R,S
we say that V
is an
(R,S)-bimoduZe if
is a left R-module and a right S-module, with the actions of
V
R and S on V
commuting
Given a right R-module
V and a left R-module W,
called the tensor product of free &nodule with
V
x W
V
and
the abelian group
W, is defined as follows.
as a basis; then each element of F
V @ W, R
Let F
can be uniquely
written in the form
with finitely many
zij
distinct from 0.
generated by all elements of the form
be a
Let ,T be the subgroup of F
45
PR05T,CTWErINWCTIVE AND FIAT MODULES
U,Ui E V
and #,Wi E
w.
Then
v8W
is defined as the factor group F/T.
The
R (U,W) under the natural homomorphism
image of
Observe that the z-module V 8 W R (Ui E
v,wi
E
F+
F/T
is denoted by
consists of all finite sums
U
8 #.
I U 8~ w,:
W).
Suppose that V
is an (S,R)-bimodule.
V 8 W can be regarded as a R
Then
(left) S-module by putting
S(U
modules.
,...,Wn
V be an (S,R)-bimodule, and let W
Let
V
v
and
(respectively,
-+
S)
R be left R-
W
as S-modules R R are two free modules over a commutative ring R is an R-basis of
(respectively,W),
V
(V €3 W ) is an R-basis of V 8 W. A ! J R Let R be a ring, V,V' two right R-modules, W,W'
v
w,s E
v o w1 8 ... CB V B wn
ib (wl@ ...@wn)
R Furthermore, if
4:
E
as S-modules
R
(UA)
v,w
It is an immediate consequence of the definition that:
(i) V Q R (ii)
(U E
V and W are modules over a commutative ring R, then V 8 W
In particular, if is an R-module.
8 W ) = su 8 w
V',-$
:
W-+
w'
two R-homomorphisms.
and
then
two left R-modules, and
Then the map
defined by
is a homomorphism of additive groups. bimodules and Let R
Furthermore, if
v
and
V'
are (S,R) -
4 is an S-homomorphism, then 4 8 $ is also an S-homomorphism.
be a ring and let V
be a right R-module.
to be f l a t if for every injective homomorphism f
:
the homomorphism lv@f:
V@W'+V@W R
R
W'-+
The module
W
V is said
of left R-modules,
46
CHAPTER I
It is an easy consequence of the definition that, given a family
is injective. (Vi)
,i E I
of right R-modules, the module
,@ Vi
is flat if and only if each Vi
-LEI
is flat.
The following simple observation will be useful for our subsequent
investigations. 8.5. PROPOSITION.
Proof.
Every projective module is flat.
Since projective modules are isomorphic to direct summands of free
modules (Proposition 8-11, we need only show that the regular module Let f
:
PIr---+ W
be an injective homomorphism of left R-modules.
p r : R @ W ' 4 W',r @ o r++ PW',
and ~-r
:
R
R
9 W+
W,r @
PW
W H
RR is flat.
The maps are R-iso-
morphisms which render commutative the following diagram
is injective and RR
Thus l @ f
.
is flat.
9. PROJECTIVE COWFS.
In what follows all modules are assumed to be finitely generated left modules. Let R be a ring.
We know that each R-module is a homomorphic image of a
projective module.
For some modules
of
projective cover of V epimorphism P-+
only if Kerf (ii) If
V ,V 1
(iii) If Pi cover of
V , f(V')
# f(V).
in case P
V1 @
We say that an R-module P
is a
is projective and there is an essential
V.
ti) A homomorphism f
9.1. LEMMA.
an even stronger assertion is possible.
W of R-modules is said to be essential if for every
A homomorphism f : V+
proper submodule V'
v
:
V+
W of R-modules is essential if and
is superfluous. 2
,...,V
are superfluous submodules of 'L
is a projective cover of
... @ vn.
Vi,
then P @
,
then so is V +..+I'
... @ Pn
n
is a projective
PROJECTIVE COVERS
47
(iv) A projective cover of an irreducible module is indecomposable. Proof. (i) Assume that f V
a submodule of
Vf
If Kerf
is a submodule of
V.
Assume that IJ
Then
V1 + (Vp +
2
+
... +
with f(V) = f(V'),
n
(iii) If fi
:
=
then
V f and so f
=
V
f(W) and so W cannot
V = V'
is superfluous.
+
Kerf
.
(V1 +
... +
V
n
)
V;
cible.
qi: @pi-@Vi
=
=
that @Pi
Hence,
is a projective cover of @Vi.
P-+ V is an essential epimorphism, where V is irreduand Pr',
0 since 0 is the only proper submodule of V .
.
+(P') + +(P")= 0, a contradiction.
required.
V.
is
is superfluous, by virtue of (ii)
If P = P' @ PI', for some proper submodules P'
+(PI) = + ( P f ' ) V
:
=
.
@ Kerfi i=1
qi is essential, which means )I
w
is superfluous, we have
is an essential epimorphism, then
an epimorphism whose kernel
(iv) Assume that
Since
+
n by (i),
is
is essential.
such that
and, since V 1
is
The desired assertion now follows by induction on n.
= V.
Pi+
V
is a submodule of
... + Vn + W) = V
+W
V
then f ( V )
Conversely, assume that Kerf
is superfluous, we must have
Cii)
V
V
W + Kerf,
If W
This proves that V = W and hence that Kerf
V.
a superfluous submodule of
W is an essential homomorphism.
V-+
V =
such that
be a proper submodule of
:
Hence, P
then But then
is indecomposable, as
One of the consequences of the following lemma is that if a module does have a projective cover, then it is unique up to isomorphism. 9 . 2 . LEMMA.
be an R-module such that (i) If M (ii)
+
:
P-+ V .
has a projective cover P, and let M
is a homomorphic image of M. is isomorphic to a direct summand of
has a projective cover L,
M
then P is isomorphic to a direct
L
Ciii) If M Proof.
V
is projective, then P
If M
summand of
V
Assume that an R-module
V,
is another projective cover of
(il Fix an epimorphism Since M
4
:
M+
V
then M Z P. and an essential epimorphism
is projective, there is a commutative diagram
CHAPTER I
48
I
0 with exact row and column. 8
is an epimorphism,
Because
$ is an essential epimorphism and
is also an epimorphism.
is a splitting homomorphism
f
:
P+
P
But
M, and hence M
$8
is projective, so there
= Imf @ Ker 8.
Since f is
injective, (i) is established.
(ii) Assume that epimorphisrn.
L is a projective cover of M,
Then
@f
:
and let f
:
L
-+
M be an
V is an epimorphism, hence by (i) P is
L -+
isomorphic to a direct summand of L. (iii) Suppose that M choose
is a projective cover of
4 to be essential.
which will follow that
M and let P'
= BCM').
and hence P' # P .
M
V,
and in the diagram above
It will be shown that
= Imf 2 P .
Assume that
is also essential, from
M'
is a proper submodule of
Then
Thus 9
is essential, as we wished to show.
'
We have accumulated all the information necessary to prove the following fundamental result.
The statement of this result uses the fact that if R
artinian, then V / J ( R ) V is completely reducible (Propositions 6 . 2 and 6.28). From now on we shall write P ( V ) 9.3. THEOREM.
ti)
Let R
for a projective cover of
be an artinian ring and let V
v.
be an R-module.
V has a projective cover
(ii) A projective R-module
W is a projective cover of V if and only if
is
49
PROJECTIVE COVERS
W/J(R)W
V/J(R)V
g
n V / J ( R ) V = @ Vi is a decomposition as a i=1, direct sum of irreducible R-modules, then V = @ P ( V i ) is a decomposition as i=1 a direct sum of indecomposable R-modules. Furthermore, V . 1 V if and only if (iii) If V
is projective and if
z
P(Vi,
(iv) If V V/J(R)V
and
W are projective R-modules, then V
(i) Write
submodule of L.
V
in the form L/S,
Given a submodule W
canonical homomorphism.
Now choose W
where S,
of
We claim that W
W if and only if
W'
is a direct summand of L;
to be a submodule of L
is surjective, and let p
L/W
:
is projective and S
f,
:
to be minimal in
L/W+
L/S
a
be the
such that f,
is
is essential and S is artinian. if sustained, it will follow that
is a projective module and hence that L/W Choose
L
let
essential; such a submodule exists, since f,
W ' L
1
W/J(R)W
Proof.
L/W
j
= P ( V3. 1 .
is a projective cover of
V.
minimal among those whose projection L
+
L/W be the projection.
Since L
is
projective, there is a commutative diagram
L
with exact row and column, and the minimality of Let L'
L'
L/L' and the two factors are essential. W
ensures that q(L) = W'.
The projection f
denote the kernel of q.
mality of
W'
--f
L/W
Because
forces W = L' (i.e. that
3
factors into
L/S
L' is contained in W, the mini-
4 is an isomorphism).
It follows that
CHAPTER I
50
L
=
W '33 W', as required. be a projective module, let f
(ii) Let W and let a af
:
V
5 J(R)W
hand, f ( J ( R ) W ) 5 J ( R ) V and thus V / J ( R ) W
V/J(R)V.
2:
Y
be an essential epimorphism
Because a
5 Keraf.
is essential, so is
V/J(RlV.
Let f
:
V-'
On the other
Hence Keraf
J(R)W
=
W is a projective 8-
Conversely, assume that
V / J ( R ) V be the projection
V / J ( R ) V be the homomorphism induced by the isomorphism Since f is an essential epimorphism and since W
V / J ( R )V .
projective, there exists an epimorphism g'
5 Kerg = J ( R I W
Kerg'
V
(Lemma 9.l(i) and Proposition 6.9(i)).
and therefore J(R)W
module such that W/J(R)W
W/J(R)W
W+
V / J ( R ) V be the projection.
--+
and thus Keraf
and g : W-
:
:
W--t
with fg' = g.
V
W.
is a superfluous submodule of
V,
projective cover of
Therefore,
is
Hence
w
is a
as asserted.
(iii) By Propositions 6.9(i) and 6.28, J ( R ) V is a superfluous submodule of Since
V
is projective, it follows that P(V/J(R)V) = V .
V.
Invoking Lemma 9.l(iii)
and (iv), we therefore infer that V = P(V
where each P(V.)
and, since Vi
P(V.1.
V . 2: V and since V . 2 V z j ~j assertion follows.
and
... @ P ( V n ) ,
Then, obviously, P(Vi)/J(R)P(Vii)~PP(V.)/J(R)P(V.) 3 3 3 is irreducible, V i / J ( R ) V i = Vi. It follows from (ii) that l.
V a
e3
is indecomposable.
Assume that P ( V . 1
(iv) If
)
obviously implies that P ( V i )
W then obviously V / J ( R ) V
W/J(R)W.
W are projective R-modules such that V / J ( R ) V
2
P(V.), 3
the required
Conversely, assume that 2
W / J ( R )W.
By the Krull-
Schmidt theorem, we may write V / J ( R ) V = V1 '33 where each Vi,Wi
... @ Vn
is irreducible and
V.
W/J(R)W 2
=
w1 @
... @ Wn
Wi, 1 < i < n.
Then, by (iii), we
have
W as required.
=
P(W1)@
... @ P ( V n , = P ( V l ) @ ... '33 P ( V n ,
[/
51
PROJECTIVE COVERS
Suppose that R
is an artinian ring and let n
R where the
=
>
...
U @ 1
@
1 be such that
n'
U : are principal indecomposable R-modules.
Schmidt Theorem, the
are uniquely determined up to isomorphism and the order
Ui
By Proposition 2.12, the above decomposition determines a
in which they appear. complete set
Owing to the Krull-
...,e
{el,
Except when J(R) = 0
3
of primitive idempotents in R
such that Ui
=
Rei.
the principal indecomposable R-modules form only a small However, it is the class which plays
subclass of all indecomposable R-modules.
a very important role, as can be seen from the following: Let R
9.4. PROPOSITION.
be an artinian ring
(i) The following conditions are equivalent: V
(a)
is a projective cover of an irreducible R-module
(b) V
is a projective indecomposable R-module
(c) V
is a principal indecomposable R-module
(ii) If Re ,...,Re then Re /J(R)e
are all nonisomorphic principal indecomposable R-modules,
rn
,...,Rern/JCR)ern are all nonisomorphic irreducible R-modules.
Proof. (i) The implications (a) * (b) and (c) * (a) follow from Lemma 9.1 (iv) and Theorem 9.3(iii), respectively.
Assume that (b) holds.
fl
Since V
projective, there exists n 2 1
such that
direct sum of n
Now the indecomposable components of Rn
copies of R.
principal indecomposable R-modules and
V
=
V @ V',
where Rn
is
denotes a are
is indecomposable; hence (c) follows
by virtue of the Krull-Schmidt theorem. (ii) By Ci) , there exist irreducible R-modules V Rei
=
PtVi),
1
i
=Z rn.
By Theorem 9.3(iii),
irreducible 8-modules.
. .,Vrn
such that are all nonisomorphic
By Propositions 6.9Li) and 6.36, J(R)ei
fluous submodule of Re
i and
Because Rei/J(R)ei
V1,.
,.. .,Vm
is a super-
thus
is completely reducible and Rei
is indecomposable, Lemma
52
CHAPTER I
implies that R e i / J W ) e i
9.l(iii) 9.3Ciiil ,
Vi
.
is in fact irreducible.
Rei/J(R)ei and the result follows.
Hence, by Theorem
10. ALGEBRAS OVER FIELDS. Throughout this section, A
denotes a finite-dimensional algebra over a field F.
All modules are assumed to be finitely generated.
v
Let
be a vector space over
F, let L be a field extension of F, and form the tensor product VL L.
which is then a vector space over
If
{V
.,V
Ire'
n
1 form a basis of F, then the elements {la,...,la n
1
L 0 V, F
=
is a basis of
v
VL over L.
With the
over
aid of the injective F-homomorphism.
I v
VL
v -1Qv
we shall often identify V with its image in V L .
With this identification,
n each element of
n
A.
E L
and
C
V,L, has a unique representation of the form
' 'E
X.V.
V
if and only if all
xi
E
F.
C AiVi i=1
with
Therefore, in passing from
V to VL we are extending the field of operators from F to L. Suppose now that A
is an F-algebra and let AL = L @A.
Then AL
is an
F L-algebra and the map
I
-
A -AL a
1Qa
is an injective homomorphism of F-algebras.
image in AL
and denoting by
element of AL moreover,
{a,,
. .. ,arn1
Again, identifying an F-basis of A ,
can be uniquely written in the form
rn C Xiai E A
rn C Xiai
A with its
we see that each with
Xi E L;
i=1
if and only if all
Xi
E F.
i=1 If
V is an A-module, then the vector space VL becomes an AL-module under
a module action
(A By identifying V
and
1
@a) [A 69 V )
=
A l A 2 Q av
(aE A,v E V,A ,A E L) 1
2
A with their images in VL and AL, respectively, the
53
A L G E B W OVER FIELDS
action above can be written as
(A Thus the action of
. , u 3.
2
E L)
AL on VL is just the extension of the action of A on V
by L-linearity. Suppose now that B homomorphism write BL
0
:
L 8 BF
is an F-subspace of A. L @
F
for the image of B
is a subring (ideal) of
10.1 LEMMA.
Let L
Then there is an injective L-
A defined by 8CR 8 bl under
If B
8.
=
R 8 b,!t
E L,b E B.
We
is a subring (ideal), then B
L
AL.
be a field extension of F
and let
A be an algebra over
F. ti)
If
A 1 , A 2 f...fA n are F-algebras, then as L-algebras
(ii) For all n
1,
[MncA)I L (iii) If I is an ideal of A,
= MnUL’
as L-algebras
AL/IL
as L-algebras
then
1
Civl
J(AIL C_ JULI
(v) If
V,W are A-modules with V
3 W,
(V/WI (vi) If
h @a1
n
V is an A-module, then J C V I L
then
2
as A -modules
L
VL/WL
5 JCV,)
is obviously an injective homomorphism of L-algebras.
Since both of
t h e m are of the same L-dimension, the assertion follows.
(iii) The map
h @ a + A @ (a+ll
*
A @ ( a ,.) ( h 8 a . ,) , A E L,uii E A. 23 23 is an L-algebra homomorphism of AL onto
(ii) The required isomorphism is given by
54
CHAPTER I
(A/IIL
whose kernel contains I L .
fact that the algebras
The desired conclusion now follows from the
(A/IIL and AL/IL
(iv) By Proposition 6.22,
J(A)
are of the same L-dimension.
is nilpotent, hence so is J(A)L.
Now apply
Corollary 6.21. (v) The proof is similar to that of (iii) (vi) Applying (iv), we have
as required.
8
10.2. LEMMA. Let V,W be A-modules and let L
be a field extension of F .
Then Hom(V ,W 1
=
AL L L If V
=
W,
A
the two sides of this are isomorphic L-algebras. Given f E Hom(V,W), we have
proof.
as L-spaces
(Hm(V,W)IL
1@
f E Hom(vL,WL).
A
is an injective L-homomorphism. element
in Ho;(VL,WL)
Hence the map
A
It therefore suffices to verify that each
is of the form $ = CX.(l @ f i )
with
i z hi
E
L,fi
E
Hom(V,Wf.
A Let
{xili
I} be an F-basis of L .
E
uniquely written in the form distinct from 0 .
1 hi
8 V i
is1 Define f i : V+ J,(l @
Then obviously each
Then each element with V i E V
3
of
VL can be
and with finitely many
W by V) =
C Xi iEI
Q fi(v)
fi is an A-homomorphism such that J,
(tJ E V )
=
CA.(l@f.). z
8
i z AII
A-module
L of F, VL
V
Vi
is said to be aberoZuteZy irreducibZe if for any field extension
is an irreducible AL-module.
55
ALC3EBWS OVER FEELDS
10.3. PROPOSITION.
V be an irreducible A-module.
Let
Then the following
statements are equivalent:
V is absolutely irreducible
(i)
(ii) If L
is the algebraic closure of F ,
then VL
is an irreducible A
L
-
module. (iii) End(V)
=
F,
i.e.
each A-endomorphism of
V
is a left multiplication by
A an element of F . Proof. (i) * (ii) Obvious (ii) * (iii) We may harmlessly assume that V Then AL
a faithful irreducible A -module.
L
algebraically closed field L, so A Since A
L
2
is faithful, in which case VL is a simple algebra over an
Mn ( L ) for some n > 1 (Proposition6.27). M (D) for some division algebra
is a simple algebra, we also have A
D over F
and some k
>
is
k
Invoking Lemma lO.l(ii), we deduce that
1.
M n (L)
Mk(L 8 D ) . By looking at the centers of both algebras, we have LZ'LQD. F F Hence D = F and, since End(V) 1Do (Propositions 5.6 and 5.8), we conclude that
A End(l') = F .
A (iii) * (i): Again, we may assume that and hence AL
EZ Mn(L),
V is faithful in which case A
for any field extension L of F ,
a minimal left ideal of M n ( F ) ,
we see that VL
Mn ( F )
Identifying V
with
is a left ideal of M ( L ) with
'
n
= n. Hence VL is irreducible and the result follows. LL Let A be an algebra over a field F . Then F is called a splitting field
dimV
for A
if every irreducible A-module is absolutely irreducible.
10.4. COROLLARY.
F
is a splitting field for A
if and only if
for some positive integers 1' n 2 r " ' r n r *
Proof. Let Vl,...,V
D. = End(Vi)
'
A
21
, 1 < i < r.
be all nonisomorphic irreducible A-modules and let By Propositions 5 . 6 and 5.8,
some positive integers n for A
A/J(A)
By Proposition 10.3,
if and only if Di = D?
z
F
for all
F
i E {l,...,~}.
nrMni
(0:) for i=1 is a splitting field
So the corollary is
CHAPTER I
56
true. Let an ideal I
10.5. PROPOSITION.
Then I
potent elements.
of A
possess a basis consisting of nil-
is nilpotent.
Proof. We may harmlessly assume that F is algebraically closed. more, since
( I + J ( A ) ) / J ( A ) is an ideal of A / J ( A )
having a basis consisting of Thus A
nilpotent elements, we may assume that J ( A ) = 0.
Further-
=
nrMni (F)
for some
i=1
positive integers n ,.-.,n r
M
“i iE
(F), 1 g 11,.
i g r.
..,r}.
i the projection of I
71
It clearly suffices to verify that 7T.U) = 0 z
Since T i ( T )
a . U ) = Mn ( F ) .
and we denote by
is an ideal of Mn (F), either
for all
(I)= 0
or
In the latter case it will follow that Mn (F) has a basis
i
i consisting of nilpotent elements.
10.6. PROPOSITION.
Let A
Thus
= 0
as required.
Thus z E I C_ J ( A )
(F)
i
Then J ( Z ( A ) ) = z ( A ) n J ( A ) .
is nilpotent and hence I = Az
is a
.
and therefore J ( Z ( A ) c - Z(A) n J M ) . is nilpotent and hence z E J ( Z ( A ) ) .
Conversely, if z E ZU) n J(AI, then z 10.7. PROPOSITION.
T.(I)
be an F-algebra.
If z E J ( Z U ) ) , then z
nilpotent ideal of A .
.
But then the trace of each matrix in M,
would be zero, which is impossible.
Proof.
Ti
i
into
If F is a splitting field for A ,
then F
is a splitting
field for Z ( A ) . Proof.
By Proposition 10.6, we have ZU)/J(Z(A))
C Z Y l ) + J(A))/J(A) 5 z ( A / J ( A ) )
Invoking Corollary 10.4, we deduce that Z ( A ) / J ( Z ( A ) ) is a subalgebra of a direct product of finitely many copies of F.
But any such subalgebra must also be a
.
direct product of finitely many copies of F.
of Corollary 10.4.
Hence the result follows by virtue
We now turn our attention to direct decomposition of algebras.
First, however,
we must develop our vocabulary. By an F-representation of an F-algebra A we understand any homomorphism A+
EndW
F
(or simply a representation of A )
of F-algebras, where V is a (finite
WEB-
57
OVER FIELDS
If n = dimV, then End(V)
-dimensional) vector space over F.
F with the F-algebra Mn(F). mrphian A+
is identifiable
F
Thus the given representation of A
Mn (F) J we shall refer to any
defines a hano-
such homomorphism as a mutfix
Pepesentation of A . If f : A+
is a representation of A ,
End(V)
define xu
=
f( x ) V for
F x E A,V E
of f.
V.
V becomes an A-module, called the underzying moduZe
In this way
Conversely, if
V is an A-module, define f
..
:A
EndCV) by f( x ) v = xu
F
V.
for x E A,v E
Then
f is a representation of A ;
in case V is the regular
A-module, we shall refer to f as the regular representation of A. Two representations
fd
: A
--C
there exists an F-isomorphism JI :
End(Vi)
,i
F Vl+ V
f b) =
= 1,2, are said to be
such that
w1(3)JI-l
It is easily seen that two representations of A the underlying modules are isomorphic.
f
:
A
+
equivalent if
for all x
E A
are equivalent if and only if
We say that a representation
is CrreQcible (completely reducible, indecomposable) if the
End(V)
F underlying module
V is irreducible Ccompletely reducible, indecomposable); the
same terminology will be applied to matrix representations.
10.8. PROPOGITION.
Let
A be an P-algebra.
(i) There exists a direct decomposition
of A
into indecomposable two-sided ideals Bi # 0 with B . B .
(ii) Write
1 =e
i
... + en
2 3
with
ei E Bi.
Then the ei
=
0 for i # j
are mutually
orthogonal centrally primitive idempotents and Bi = Aei = e$. (iii) Z ( A )
=
ZWe
1
... @ Z(A)en
indecomposable ideals and
is a direct decomposition of Z U )
into
Z(A)ei = Bi n Z U ) .
(iv) J(Z(A)eil = J ( , Z ( A ) ) s i and Z ( A ) s . / J C Z ( A ) e . ) is a finite field extension of 2
2
F. (v) Let Pi
=
Z(Alei/J(Z ( A l e i ) , let Bi
:
Z(A) + Pi be the natural homomorphism
CWTER I
58
and let fi ' i: X
yi
Mn
: Z(A)--t
be the regular representation of Fi.
Mni(F) (F)
by
yi
=
fiei.
Define
Then
i
{yl,.
.. ,y 1
tation of
Z(A).
and
is a complete set of nonequivalent irreducible matrix represenEach yi
satisfies y . ( e . ) z
=
z
y . ( e . ) = 0 for j # i. 2 3 then each ni = 1.
1 and
Furthermore, if F is a splitting field for Z ( A ) Proof.
(i) and (ii).
Apply Propositions I.l(iii) and 4.3
(iii) The first statement and the inclusion Z ( A ) e . z Suppose that Z E B . n Z ( A ) .
Since e
2
have z = ze
i
E Z(A)ei,
n Z(A)
follows from (ii).
i is the identity element of Bi, we
proving that Bi n Z ( A )
(iv) By Proposition 6.35,
5 B z.
5 Z ( A) e i .
J(Z(A)ei) = J ( Z U ) ) e i .
Because each ei
is
primitive, Proposition 9.4(ii) ensures that the Z (/!)-module Z ( A ) e i / J ( Z ( A ) e i ) irreducible.
is
The latter implies that the commutative algebra Z(A)ei/J(Z(A)ei)
is simple and hence is a field.
That this field is a finite extension of
follows from the assumptions that A
F
is finite-dimensional over F .
(v) It is obvious that Keryi = Kere;.
Assume t h a t
Z
E Kerei.
Then Z e i =ze i
for some x E J ( Z ( A ) ) , so
and therefore Kere. c Z ( A ) (l-ei) 2 -
obvious, we conclude that Kere
i
+ =
J(Z(A))ei.
The opposite inclusion being
ZU)(l-eil + J ( Z ( A ) ) e i .
It follows from (iii) and (ivl that
and that each Fi Z(A),
is a field.
then each Fi
Furthermore, if F
and hence ni = 1.
is isomorphic to F
only irreducible representation of P
if
is a splitting field for
the result follows.
We now introduce the important concept of blocks.
n A = .@B 2=1
n
i
=
@Ae
i=1
i
Let
Since fi is the
ALGEBRAS OVER FIELDS
be the decomposition defined in Proposition 10.8.
59
We shall refer to Bi
block and to ei
as a block idempotent of A.
indicate that B
is a block containing the block idempotent e.
as a
B = B(e) to
We shall also write
The represen-
--+
(F) described in Proposition 10.8 is uniquely determined (A) Mni up to equivalence by ei. We call y the irreducible representation of Z U ) tation Y i
:
i
associated with
e
i'
By a central character of A Z(A) 3 F.
y l , ...,yn
If F
we understand any F-algebra homomorphism
is a splitting field for
are all central characters of A;
central character of A
associated with
10.9. PROPOSITION.
Let F
number of blocks of
A.
ei
Z(A), then by Proposition 10.8, we shall refer to yi (or Bi).
be a splitting field for Z(A)
and let n
be the
Then dimJ(Z(A)) = dimZU)
F
F
-
n
As we have seen in the proof of Proposition 10.8(v),
Proof.
as the
copies of F.
is isomorphic to a direct sum of n
Z(A)/J(Z (A))
Hence
and the result follows. 10.10. PROPOSITION.
A
Let A = B(e
1
)
@
... 8 B(en )
be the direct decomposition of
into sum of blocks.
n (i) For every A-module modules.
i E C1,2,.
V, V
In particular, if
8 ei V is a direct decomposition of 'L i=1
=
V
is indecomposable, then V = e . V
into A-
for a unique
.. ,nl
(ii) For each left ideal I of A ,
we have
n
I = i=1 8 (I n B(ei)) In particular, each indecomposable left ideal lies in exactly one of the B ( e i ) . (iii) If
A
and A 2
are ideals of
are direct sums of blocks.
A and A
=
A 1 @ A 2 , then A
and A
In particular, the blocks are the only indecomposable
CHAPTER 1
60
ideals of A . (iv) Let I be an ideal of
of
Z(A)
associated with e
Proof.
Z(A)
Then ei E I
i'
(i) Because ei
and let yi
be the irreducible representation if and only if y i ( l ) # 0.
is central, e . V is an A-module. 2
Since the e i
are orthogonal, we also have
n
n
C e.V = @ e i V i=l
Taking into account that
n
n
the assertion is proved.
n (ii) By (i), I = @ e . A and so i=1
Hence
n
n
Hence for all i
and since B ( e .) 2
is indecomposable we have B ( e i )
5A l
or B ( e i )
5A l .
Thus
as asserted.
(iv) If ei E I then y i ( e i ) = 1 # 0 that yi (1)# 0. have
Because y i ( I )
and hence y . ( 1 ) # 0 . 2
Conversely, assume
is a nonzero ideal of the field y i ( Z ( A ) )
,
we
ALGEBRAS OVER FIELDS
61
It therefore follows,from Proposition 10.8(v), that there
Thus y i ( e i ) E y i ( I ) .
are elements x E I , y E J ( Z ( A ))
x Multiplying both sides by
e
E Z(A)
and
=
+
ei
z(1-e
i yields ei
such that
+y
.)
= xe i - y e i .
Hence, by raising both
sides to the k-th power, we find
ei with u
I.
Since yei
=
u
+
k
k
(-1) ( y e i )
is nilpotent, the result follows.
Let e ,e , . . . , e n be the block idempotents of A 1
i E I1.2,
...,n}
such that
2
B(ei),
and e . V = 0 3
.
Observe that if
V
lies
then
(a) A l l A-modules isomorphic to V t, = 1-t, =
j # i
for all
l i e s i n the block B ( e i )
In this case we say that V
(b)
be an
It follows from Proposition 10.10 that there exists
e.V = V
in the block
and let V
2
indecomposable A-module.
.
lie in B(ei)
for all v E V
e.2,
The above gives a classification of all indecomposable (and in particular, all irreducible) A-modules into blocks. principal indecomposable modules.
Our next aim is to tie together blocks and
The following preliminary observations will
clear our path. Let e
10.11. LEMMA.
be an idempotent of A.
Then
n I = Ie
(i) For any ideal I of A ,
Ae
(ii) Ae/J(A)e
denotes the image of
Proof.
where
-e
(i) Suppose that x E Ae
n I. Then x
Therefore
x = ae = ae2 = xe E Ie
e
in = ae
A'
= A/J(A).
for some a E A .
CHAPTER I
62
proving that Ae n I
5 re.
The opposite containment being trivial, (i) follows
JU), we have
(ii) Applying (i) for I =
.
as required.
Note that if Ae
is a principal indecomposable A-module, then by Proposition
2;
9.4(ii) and Lemma lO.ll(ii),
is an irreducible A-module.
n
10.12. LEMMA.
Let A =
@
composable A-modules. for any A-module
ei
Let
denote the image of
ei
into principal inde-
=
dim HomUe
F
F Z
Then
--
V)
=
rndim End(A e J
F
A
factor of V . By Lemma 5.5, the map Hom(Aei ,v)
A morphism of additive groups;
-
;
A
is the multiplicity of the irreducible A-module
Proof.
1 = A/J(A).
in
V, dime .V
where m
be a decomposition of A
Aei
i=1
jei
as a composition
eiV, f++ f ( e )
is an iso-
Since this map is obviously F-linear, we have
e .V
Hom(Aei,V)
as F-spaces
A The equality dimHomUe
F
V)
= rndim End(2;)
Therefore we may assume that Aei
V is irreducible.
is trivial if
F
i'
A
-
V has a maximal nonzero submodule X. 4 X+
is projective, the exact sequence 0
V - + V/X
0
Because
induces an exact
sequence 0
3
Homue
A
i'
X) --t Hom(Aei,V) 4 Hom(Ae A i' V/X) A
+
0
Thus w e have dim Hom(Ae
F If m, and
and rn
V/X,
A
i'
V) = dim HomUe
F
A
are the multiplicities of
''
X) + dim HomUe V/X) F A
"
zei
as a composition factor of
respectively, then applying induction on dim V ,
we have
X
ALGEBRAS OVER FIELDS
63
dim Hom(f!ei,X) = rn dim EndgZd
F
' F A
A
dim Hom(Aei,V/X) = rn dim End(zii) F A ' F A Adding up these equalities gives dim Hom(Aei,V)
F
as asserted. As
" F A =
.
rndim End (2eil
F
A
an immediate consequence, we derive
10.13. COROLLARY. of e
+m )dim End(Azi)
= (m
A
in
=
Let e be a primitive idempotent of A,
A/J(A)
and let V be an A-module.
e
let
be the image
Then the following conditions
are equivalent: (i)
1;
v
is a composition factor of
(ii) Hom(Ae,V) # 0
A (iiil
eV # 0.
9
U and V be principal indecomposable A-modules.
Let
V are linked, written U
- V,
We say that U
if there exists a sequence U
=
U,.. .,Un
=
and
V of
principal indecomposable modules such that any two neighbouring left ideals in this sequence have a common composition factor.
It is clear that
-
is an
equivalence relation on the set of all principal indecomposable A-modules what follows we denote by X 1
10.14. PROPOSITION.
Let
A
,...,Xrn n = @ Aei
the equivalence classes of be a decomposition of A
i=1 indecomposable modules, and for each jE 11,.
..,rn}
.
In
-. into principal
put
I . = @Ae
,,EXj ti)
...,Irn
11,
are all blocks of
A
(ii) Two principal indecomposable A-modules are linked if and only if they.belong to the same block.
64
CHAPTER I
rn (i) It is obvious that A = 8 I j=l j '
Proof.
suffices to show that each I
j
By Proposition l.lO(iii),
it
is a two-sided ideal contained in a block.
Assume that Ae. c I. and Aek C I k with j # k . Then Aei and Aek have no 2 3 composition factor in common. In particular, Aek has no composition factor Thus, by Corollary 10.13, ez.Aek = 0.
isomorphic to
I.I = 0 for j # k . 3 1 1
This proves that
Hence
rn I A = I.( 8 Ik)= I T c I . 3 3 g=1 3 - 3
and so I
is a two-sided ideal of A.
i
If Aei
--
and Ae
j
have the composition factor Aek in common, then by
Corollary 10.13, ekAei # 0 and exist blocks B,B',B" that 0 # ekAei
E"B
e Ae # 0. By Proposition lO.lO(ii),
k j
such that Aei
5B ,
and hence B
B".
=
Ae. 3 -c B'
and Aek -c E r r .
there
It follows
Similarly, 0 # e Ae . c B"B' k 3 -
and so
Consequently Ae.+Ae. C B and repeated application of this argument -L 3 This establishes the shows that I i s contained in a certain block of A. j required assertion
B' = B".
(ii) Direct consequence of (i). Assume that F is a splitting field for A module.
Then End(V1 = F
and let V
and each element z E ZU)
be an irreducible A-
acts as an element of
A End(V) on V .
Thus there is a function $ : z(A)---+F
such that
A
Clearly J,
is a central character of A.
A afforded by
character of
as the central
V.
Suppose that F
10.15. PROPOSITION.
We shall refer to J,
is a splitting field for
A.
Then two
irreducible A-modules belong to the same block if and only if they afford the same central characters. Proof.
Assume that the irreducible A-modules
characters p
and
JI,
respectively.
If U
and
U and V afford the central
V belong to a block B
=
BCe)
ALGEBRAS OVER FIELDS
then eu = u
and eu
for all u
= U
U,U E V .
E
65
proving that p = $ by virtue of Proposition 10.8(v). p = $,
belong to B ( e ) .
and let U
eu = $(e)u = U
for all U E V .
p(e) = @ ( e )= 1,
Hence
Then P(e)
=
@ ( e )= 1
V
This shows that
Conversely, suppose that and hence
lies in B
and the result
follows. Let e 1
,...,en
be a complete set of primitive idempotents of
A
5
Ae @
Then
... @Aen
A as a direct sum of principal indecomposable modules.
is a decomposition of Let
=
A.
e
denote the image of
i in 2
= A/J(A).
We know, from Lemma lO.ll(ii),
that Aei/J A ei
2
--
A ei
Suppose that the numbering is so chosen that Ae
nonisomorphic irreducible A-modules,
ij
C
C =
(c..) ZJ
j
-_
are all
..,m},
let ‘ij be the as a composition factor of Aei.
are called the Cartan invariants of A,
The nonnegative integers c
m matrix
2;
are all nonisomorphic
2; ,...,Aern
Given i , j E 11,.
multiplicity of the irreducible A-module
x
,...,Aern
Then, by Proposition 9.4,
indecomposable A-modules.
rn
1
is called the Cartan matrix of
A.
and the
The Cartan matrix
is only determined up to a permutation of its rows and columns (depending on
the numbering of Ae 10.16. PROPOSITION.
,. ..,Aem ) . Let F
be a splitting field for A,
A be a decomposition of A
=
Ae @
let
... @Aen
into direct sum of principal indecomposable modules,
and let C = ( C i j ) be the Cartan matrix of (i) cij = dime& F J (ii) The multiplicity of Aei Krull-Schmidt theorem is dim1 F (iii) The multiplicity of
A.
Then
as a direct summand of
A in the sense of the
ei as a composition factor of A
is dimezA F
CHAPTER X
66
proof.
(i) Since F
is a splitting field for A, End(jei) = F .
Lemma 10.12, we deduce that the multiplicity of
Aei is equal to dime l e 3 i'
z ei
Applying
A
as a composition factor of
as required.
$9
(ii) Since F
is a splitting field for A ,
rings over F .
Hence the multiplicity of
Owing to Proposition 9 . 4 ,
Aei
2; i
is a direct sum of full matrix in
is equal to
is determined, up to isomorphism, by Aei
therefore a i d e i also equals the multiplicity of Aei F (iii) Direct consequence of Lemma 10.12. m As
- -F
. and
as a direct sunnnand of
A
an inmediate consequence, we derive
10.17. COROLLARY.
Let F
be a splitting field for A,
nonisomorphic irreducible A-modules and let ni = dimV
if
r
In particular,
r
let V1,...,V 1
i < r.
r be all Then
.
67
2 Group algebras and their modules d i s c u s s a number of t h e i r proper-
I n t h i s chapter w e introduce group a l g e b r a s an t i e s t o be a p p l i e d i n subsequent i n v e s t i g a t i o n s .
These include support of
c e n t r a l idempotents, t h e notion of d e f e c t group of a block, t h e number of irred u c i b l e FG-modules and some formal p r o p e r t i e s of induced modules.
Later c h a p t e r s
w i l l t r e a t v a r i o u s aspects of t h e s e t o p i c s i n g r e a t e r d e t a i l and depth. s e l e c t i n g m a t e r i a l w e have held s t e a d f a s t t o a single-minded purpose:
In t o present
o n l y those r e s u l t s deemed e s s e n t i a l f o r t h e a p p l i c a t i o n t o t h e study of t h e Jacobson r a d i c a l of t h e group algebra
FG.
1. GROUP ALGEBRAS
R
Let
group a l g e b r a
RG
induced by t h a t i n combinations
(ii) Cx * g
9
(iv)
+
be a group, p o s s i b l y i n f i n i t e .
i s t h e f r e e R-module on t h e elements of G.
More e x p l i c i t e l y ,
Cx*g,x
9 ( i ) Cx*g =Cy * g g g
(iii)
G
be a commutative r i n g and l e t
E R,g E G , 9 i f and only i f
RG
9
-
yg
Z
=
for a l l
x # 0 subject t o
9 g E G
Cy * g = CCx +y 1 * g 9 g g
(Ex * g ) (Cyh*h)
=
Czt*t where
9 r(Cx * g ) = C k x ) * g 9 9
for a l l
Cxgyh gh=t
r E R
It i s straightforward t o v e r i f y t h a t these operations define
R-algebra with G,
withmultiplication
c o n s i s t s of a l l formal l i n e a r
with f i n i t e l y many
x -
G,
The
1 = lR*lG where
respectively.
lR and
lG
RG
as anassociative
a r e i d e n t i t y elements of
R
and
With t h e a i d of t h e i n j e c t i v e homomorphisms
we s h a l l i n t h e f u t u r e i d e n t i f y
R
and
G
with t h e i r images i n
RG.
With t h e s e
i d e n t i f i c a t i o n s , the formal sums and products become o r d i n a r y sums and products
CHAPTER I1
68
We shall also adopt the
For this reason,from now on we drop the dot in xg.g. convention that RG Let x = Cz
2
RH means an isomorphism of R-algebras. Then the support of
E RG.
x, Supp, is defined by
It i s clear that Suppx is a finite subset of G 3: =
0.
that i s empty if and only if
We next exhibit some elementary properties of group algebras.
1.1. PROPOSITION.
Let A
be a homomorphism of
f : RG 3 A
G
be an R-algebra and let
into the unit group of A .
Then the mapping
defined by
i s a homomorphism of R-algebras.
In particular, if
R-free with JIG) as a basis, then RG Proof. Because RG of R-modules.
JI is injective and
A
is
A.
is R-free with G as a basis, f i s a homomorphism
Let
x be two elements of RG.
=
Ex
g
g
and y = Cy
s9
Then
as asserted. The augmentation ideal homomorphism from RG
to
I(R,G) consists of all x
I(R,G)
of RG i s defined to be the kernel of the
R induced by collapsing =
ZX
99 E RG
for which
G
to
1.
In other words,
69
QROUP ALGEBRAS
as the augmentation of x.
We shall refer to aug(z)
It follows from the equality CXc$
= CXgCg-l)
+ Cx
is a free module with the elements g-1 (1 # g
that as an R-module, I(R,G)
In the future we shall often suppress reference to R
as a basis.
denote the augmentation ideal of RG by
ideals of RG, respectively, generated by RG*X = C RGx and
and simply
and X*RG for the left and write
X*RG = C xRG ~X
for all x E X, then RG*X = X*RG is a two sided ideal
Of course, if RGx = xRG
For example, if N (n-llg
G)
X, i.e.
6 X
of RG.
E
I(G).
is a subset of RG, we write RG‘X
If X
9
=
g(g
show that RG*IW) = I(N)*RG
G, then the equalities
is a normal subgroup of
-1 ng-1)
and g(n-1) = (gng-l-l)g
is a two-sided ideal of RG.
( n E N,g
E
G)
The signigicance of
this ideal comes from the following fact.
1.2. PROPOSITION.
N
= Kerq.
Let J, : G+
H be a surjective homomorphism of groups and let
Then the mapping f : RG-
RH which is the R-linear extension of J,
is a surjective homomorphism of R-algebras whose kernel is RG*I(N).
In
particular,
Proof.
That f is a surjective homomorphism of R-algebras is a consequence
of Proposition 1.1.
It is plain that RG*I(N) 5 Kerf.
induces a homomorphism
is an isomorphism.
7
:
RG/RG*I(Nl--t RH
.
Thanks to Proposition 1.1,
Consequently, f
The restriction of
I-’
7,
can be extended to a homo-
CHAPTER I1 70
RG/RG*I(N)
morphism RH+
7.
which is the inverse to
Thus Kerf = RG.Z(N),
as desired. Let I be an ideal of RG and let
Then G n (l+I) is the multiplicative kernel of the natural map Ghence a normal subgroup of G.
RG/I,
and
In view of this obsefvation, the next corollary
arises from Proposition 1.2 by taking I = R G * I ( N ) . 1.3. COROLLARY.
Let N
be a normal subgroup of
Let H be a subgroup of G.
G.
Then
Because RH is a subring of RG,
we can view RG
as a left or right RH-module by way of ordinary multiplication. 1.4. PROPOSITION.
Let H be a subgroup of G.
transversal for H
in G, then RG
If T
is a right (left)
is a free left (right) RH-module with T
as
a basis. Proof.
T is a right transversal for H in
Assume that
t E T,IRH)t
G.
Then, for any
is the R-linear span of the coset Ht.
Accordingly, for any n tl,t*,...,tnET,(RH)tl + + (RH)t, is the R-linear span of UHti As is i=1 apparent from the definition of RG, if X and Y are disjoint subsets of G ,
...
their R-linear spans meet at 0.
.
Hence RG = @ (RH)t, proving that RG
is a
FT free left RH-module with where T
T as a basis.
A similar argument proves the case
is a left transversal.
1.5. PROPOSITION.
Let S be a subring of a commutative ring
group G, R B S G
RG
R.
Then for any
as R-algebras.
S
Proof.
Observe that
11 @ g1g E GI
is an R-basis for R 03 SG and that the S
mapping
GROUP ALGEBRAS
For any groups G
1.6. PROPOSITION.
and
R(GxH)
Proof.
Again note that
.
Now apply Proposition 1.1.
is an injective homomorphism.
{X
2
71
H,
RG 8 RH R
8 y l x , y E G}
is an R-basis for RG 8 R H R
and
that the mapping
is an injective homomorphism.
1.7. PROPOSITION. G
Let
Now apply Proposition 1.1.
{cili E 1) be the set of all finite conjugacy classes'of
.
and, for each i E I, let
all
.
Then ZCRG)
is the R-linear span of
.c; Proof.
i E I.
Since for all g E G; g-'C:g
On the other hand, let x
or, equivalently z
i'
EX
99 E
C'
C Z(RG)
for all
z -
Then, for all t E G , t - ' z t = z
Z W ) .
for all t E G,g E Supp.
= z
t-lgt
=
we have
= C+
Hence x
$7'
for which Ci
linear combination of those 6';
S u p p , as required.
.
is an R-
The "if" part of the following result is a classical Maschke's theorem 1.8. PROPOSITION.
istic p
> 0.
order o f
G.
Proof.
be a finite group and let F
Let G
Then FG
is semisimple if and only if p
If p > 0 divides
and x E ZVG).
IGl,
then z
by
W'
and let 8
:
V-'
W
=
Cg
does not divide the
satisfies 'X
SfG
J V G ) # 0.
Hence FGx
let W be a submodule of an FGmodule F-subspace
be a field of character-
Conversely, assume that p V.
Write
V = W 8 W'
be the projection map.
= 0
=
1 IGI,
and
for a suitable
Define $
:
v--+ 'L
CHAPTER I1
72
+ ( y v ) = 1 ~ 1 c- lXex-lya = I G I - ~c (yz)e(yz)-'yu xEG ZEG 1 G I - l C yzez$7EG
=
$
1
u
= y$(~),
i s an FG-homomorphism. Suppose now t h a t
xex-12,
Accordingly,
W"
U E =
2,
W.
Then, f o r any
and
$(Ul = U.
V
is an FG-submcdule of
by t h e above,
- $(u) E
21
W"
Setting
W" n W
such t h a t
V r r , so
x E G,x-'v
U = $(U)
+
= 0.
E W,
-1
= Ker$,
it follows t h a t
u E V.
Finally, l e t
W+W'.
(V-$(U))E
u.
e(3c-l~) = 5
so
Then,
V = d J @ W"
Thus
and t h e r e s u l t follows. 1.9. PROPOSITION.
Let
normal p'-subgroup of
F
be a f i e l d of c h a r a c t e r i s t i c
G
c
e = 1Nl-l
and l e t
x.
Then
p > 0, l e t N be a e
i s a c e n t r a l idem-
xEN
potent of
FG
f a c t o r of
FG.
Proof.
f
:
Kerf = FG-I(N),
Thus
F(G/N).
In p a r t i c u l a r ,
e
it s u f f i c e s t o show t h a t
we a l s o have
FG*I(N) C - FG(1-e)
ne = e
F(G/N)
is a direct
is a c e n t r a l idempotent.
be t h e n a t u r a l homomorphism.
f ( e ) = 1, w e have 1-e E K e r f
n E N,
Given
2
I t i s easy t o v e r i f y t h a t
FG + F(G/N)
Since
FGe
such t h a t
Since
Let
FG = FGe @ FG(1-e)
and
FG*ICN) = FG(1-e). and t h e r e f o r e
FG(1-e)
(n-1)s = 0 and
so t h a t
5 FG*I(N).
FG-I(N)e = 0.
and t h e r e s u l t follows.
We c l o s e t h i s s e c t i o n by providing t h e following p i e c e of information. Let
R
G
be a commutative r i n g and
G
a r e p r e s e n t a t i o n of
V
on
a group.
V,
Given a nonzero R-module
by
we understand a homomorphism
p : G
-t
Aut Cv'l
R W e say t h a t
p
representation
is faithfd i f P
of
G
on
K e r p = 1.
V,
Owing t o Proposition 1.1, t o each
t h e r e corresponds an RG-module s t r u c t u r e on
V
given by
CCx glv = ZX g
Conversely, any nonzero RG-module
V given by
p @ ) O = gv
for a l l
9
p(g)U
V determines a r e p r e s e n t a t i o n
g E G, U E V .
(U E p
of
v, cc9 E G
Hence t h e r e is a b i j e c t i v e
on
R)
CENTRAL IDEMPOTENTS
I3
correspondence between t h e c l a s s of a l l nonzero RG-modules and t h e c l a s s of a l l r e p r e s e n t a t i o n s of
G
on R-modules.
V
Moreover, i f
i s R-free of f i n i t e rank
2 GL(n,R) , t h e l a t t e r being t h e group of a l l n x n nonR Thus t o each r e p r e s e n t a t i o n of G on t h e s i n g u l a r m a t r i c e s with e n t r i e s i n R.
n , then
AutCV)
R-free R-module p* : G -
V
GL(n,R).
n , t h e r e i s a matrix r e p r e s e n t a t i o n
of f i n i t e rank
R = F
In particular, i f
generated F-modules a r e f r e e of f i n i t e rank.
F
I n t h i s case t h e study of FG-
G on F-modules, and matrix r e p r e s e n t a t i o n s of
modules, r e p r e s e n t a t i o n s of over
i s a f i e l d , then a l l f i n i t e l y
G
a r e e s s e n t i a l l y equivalent.
1.10. PROPOSITION.
Let
be a commutative r i n g .
RG.I(H)
H be a subgroup of an a r b i t r a r y group X
If
i s a generating set f o r
C RG(x-1)
=
and
then
I ( H ) R G = C (x-1)RG XGX
XEX
Proof.
H,
G and l e t R
W e s h a l l only e s t a b l i s h t h e f i r s t e q u a l i t y , s i n c e t h e second follows
by a similar argument.
C RG(x-1) 5 R G * I ( H ) . x EX i n El, h-1 E C RG(x-1).
I t i s clear t h a t
h
only show that f o r any n o n i d e n t i t y
Thus we need
h
Now
is a
XEX
group word i n t h e
xi E X and w e proceed by induction on t h e l e n g t h
F i r s t it i s t r u e f o r words of l e n g t h
word.
xYl-1
x -1 E C RG(x-1)
2
XEX Suppose it i s t r u e f o r words t h e form
*1
x; h and both
h-1
=
1 :
-x.-1 [x;-l)
E
C RG(x-1) XEX
h of l e n g t h and
t of t h e
xi1-
t.
Any word of l e n g t h
1 are in
C RG(x-1). XEX
t+l i s of
Hence
a s desired. 2. CENTRAL IDEMPOTENTS.
Throughout t h i s s e c t i o n , istic
p > 0.
G denotes a f i n i t e group and F a f i e l d of c h a r a c t e r -
Our aim i s t o provide information on support of c e n t r a l idem-
p o t e n t s of t h e group a l g e b r a Let
A
FG.
be an F-algebra and d e f i n e
[A,A],
t h e canmutator subspace of
A , to
74
CHAPTER I1
be the F-linear span of all Lie products
[Z,yI =
x p y x with Z,y
E
A.
The
following observation illustrates how the commutator subspace can be brought into argument.
2.1. LEMMA.
Let A
a 1 ,a 2 ,...,a rn
E A
Proof.
be an algebra over a field F of characteristic p > 0.
and if n
5
0 is a given integer, then
By induction argument, we need only prove that
n (a+blP for a l l
If
a,b E A .
n
2
t
n
Suppose first that
n bp (mod I A , A I )
= 1.
Expanding by the distributive
law,
+ bp +
( a t b ) P = ap
la a 1 2
...
where the sum is over a l l products a equal to a
or b.
a of p terms, ai E { a , b } , not all 1 P With each word a a a associate its cyclic permutations 1
a a 1
2
...ap
,a a 2 3
1
we have I-y
=
...ai
ai ai
,y
=
P
2
[y,bl E [ A , A I
... P p l
1
[A,AI.
This is so since for
...aipai ...a'.j-1
a 2. J.a2. j+l
1
where
...a .
y=a.a 1
2
...a a ,...,aPa ...aP-1
All these products are congruent modulo x =
...aP
and
6 = a.
'j-1
2
a.
' j 'j+l
Therefore the sum of these cyclic permutations is px hence it belongs to (2)
[A,A].
1
...ai
...x P
p
modulo
[A,AI,
and
Invoking (1)I it follows that
(xty)P =
B
t yp t 2
(X E
and hence that
@pySIp
(wIp-(&dP(mod = Is,(y3c)p-1E/1 oaoa
[A,All IA,AI I
[AIAI)
75
CENTRAL IDEMPOTENTS
Accordingly
s E IA,AI
(3)
implies
2’ E
[A,AI
n.
The r e s u l t now follows from (2) and (31 by induction on
The n e x t lemma w i l l enable us t o e x p l o i t t h e previous r e s u l t .
follows
denotes t h e set of all conjugacy c l a s s e s of
CR(G1
Proof.
C = {g
Given
n A . E F and 2
c
h
i=1
. = 0. ’
...,g,,
,
E CRCG) n-1
x
E [FG,FGl.
Since
Cgt-3 t
=
IFG,FGI
Let
H be a subgroup of
G
x
=
- t( g t - l )
( t , g E G)
i s spanned by a l l
zy-p with r , y
E G,
-1
q-yx
t h e r e s u l t follows by v i r t u e of the e q u a l i t y
G.
C Aigi, where i=1 and so i n view of t h e i d e n t i t y
assume t h a t
x = c Ai (gi-g,) i= 1
Then
g-tgt-1 we have
,
I n what
= q - 3 ~ (zycy)~.
Then t h e n a t u r a l p r o j e c t i o n
defined by
“8’‘,EHX$
n( c g EG i s obviously an F-linear map.
define
+ X
I n what f o l l o w s , f o r each s u b s e t
X.
t o be t h e sum of a l l elements i n
E FG
X
G,
of
we
The next r e s u l t , due t o
Brauer , w i l l be of fundamental importance. 2.3. THEOREM. subgroup of IT : F G d
P be a p-subgroup of
Let
G
for which
5 H 5 N(P1. G
Let
and assume t h a t
C ,C 1
2
C’
,...,Cp
Z(FG)
Z(FH)
into
n S
=
0,
be a l l conjugacy classes of
G,
with
Xk # 0 f o r k E
By P r o p o s i t i o n 1 . 7 , t h e elements Note t h a t
let
C E CI(G)
(1
H
S = C(P)
and l e t G Then t h e n a t u r a l p r o j e c t i o n
FS induces a r i n g homomorphism of
t h e F - l i n e a r span of a l l Proof.
S
G,
,...,t}
+ + .: Cl,Cp ,..C
C
and
and
Xk
=
0
for
be a
whose k e r n e l i s
let
Xi
= Ci n S
k E {ttl, ...,r ) .
c o n s t i t u t e a n F-basis for
Z @‘GI
.
CHAPTER I1
I6
,...,Xt
X
and t h a t
a r e mutually d i s j o i n t .
H.
union of conjugacy c l a s s e s of
ZWG)
to
Z(FH)
Hence
1T
8 E
i n which case
Ts
Ts with s
++ n(C.C.)
=
2 3
E S.
i,j
0 =
3
=
=
(z,y)
where
x,y
with
E
and
0,
E
I l l . . .,PI,
xy
s)
=
T
and denote by
X . = 0 or
then e i t h e r
2’ #
Now suppose t h a t
0.
xj
=
0,
Then
P a c t s as a
it follows t h a t
(hsh-l,hyh-l).
Because
ranges a l l o r b i t s of
On t h e o t h e r hand, an o r b i t of
(x,y) 1
a s required.
is a
C xy = C C xy (z,yFT SES ( z , y F T s
Ts v i a h ( z , y )
sum of a l l elements zy
of t h e form
=
E Cj
and s E S, hsh-l = 8 ,
hE P
permutation group on
0.
T
If
+ + n(CilTr(.C . I .
++
equal t o
Iz E Ci,y
= {(x,y)
T(C.C.)
Since f o r a l l
i < t,
whose k e r n e l i s of t h e r e q u i r e d form.
S, we p u t
t h e union of a l l
1 <
induces an F-homomorphism from
We a r e t h e r e f o r e l e f t t o v e r i f y t h a t f o r a l l
Given
Xi,
Moreover, each
T
8
Ts
charF = p, of s i z e
the
# 1, i s
has s i z e 1 i f and only i f it i s
The conclusion i s t h a t
S.
.
I n f u t u r e we s h a l l r e f e r t o t h e homomorphism Z(FG1 Theorem 2 . 3 , as t h e Brauer homomorphism.
-f
Z(FH) constructed i n
W e come now t o t h e main r e s u l t of t h i s
section. 2 . 4 . THEOREM.
(Osima (1955)).
Let
e be a nonzero c e n t r a l idempotent of
FG.
Then Suppe c o n s i s t s of p’-elements. Proof. with
Z
=
(Passman(l969a)).
xy
= yz
be t h e subgroup of
where
z
suppn(e].
# 1 i s a P-element and Y i s a P‘-element.
G generated by
c e n t r a l idempotent i n have
X
Fs
z E Suppe
Assume by way of c o n t r a d i c t i o n t h a t
where
s
2.
Then, by Theorem 2.3,
= CAP].
Furthermore, since
Thus we may harmlessly assume t h a t
x
E
n(e1 z
Z(G).
Let
p
is a
5,
we also
Invoking
CE-L
IDEMPOTENTS
77
Lemma 2 . 2 , w e i n f e r t h a t
x
(*I Choose an i n t e g e r
t
If
t gE
= Ctsg,
n with
pn 2 / G I
E suppv
f o r a l l v E IFG,FGI
V
=
pn
and with
lcmodq) and p u t
n n It: gp
[mod IFG,FGI)
n i s a p'-element,
gBn
e
hand, s i n c e
.
Let
x 9 SupptP
it follows from ( * ) t h a t
pn
i s a c e n t r a l idempotent and s i n c e
t, x E Suppt,
But, by t h e d e f i n i t i o n of true.
C be a conjugacy c l a s s of
G
l(modq),
a contradiction.
g
and l e t
E C.
is c a l l e d t h e defect of
HI
and
H
C.
a r e subgroups of
2
indicate t h a t
H
proper subgroup of
6(C)
W e denote by
G,
H
we w r i t e
i s conjugate i n
G
H 2 ) , while H
=
G
' G
.
On t h e o t h e r
w e have
So t h e theorem i s
A Sylow p-subgroup of
C ( g ) i s c a l l e d a defect group of c (with r e s p e c t t o p ) . G groups of C a r e conjugate and so have a common o r d e r , say
d
.
F, then by Lemma 2 . 1 , n tP
Since
t = y-le
Thus a l l d e f e c t
d p
.
The i n t e g e r
C. I f
any d e f e c t group of
H2
(respectively,
H2
t o a subgroup o f
H
w i l l mean t h a t
PG
and l e t
H < H2) t o 'G
(respectively, t o a
H
and
H
are
' G conjugate i n Let
G.
e be a block idempotent of
B
=
B(e).
Since
e E Z (PG) ,
Proposition 1 . 7 implies t h a t suppe =
C
U C
1
f o r some
Ci E Ck(G).
denoted by
8 ( e ) (or
U 2
...
U Ct
6 ( B ) ) , i s c a l l e d a d e f e c t group of
w i l l be shown ( P r o p o s i t i o n 2.6) t h a t a l l d e f e c t groups of
d have a common o r d e r , say p B)
Ci, 1 G i
The l a r g e s t of t h e d e f e c t groups of
.
The i n t e g e r
d
e
( o r of
B)
r,
.
It
e a r e conjugate and so
i s c a l l e d t h e defect of
e
(or of
.
2.5.
LEMMA.
Let
D be
a p-subgroup of
G
and l e t
C ,C2,...,Ct 1
be a l l elements
78
in
CHAPTER I1
C!L(G)
6(Ci)
with
i s an i d e a l of
2 (FG)
Proof.
i
Fix
D.
.
ID(G)
Then t h e F-linear span
+ + cI,c 2,...,ci
of
G
...,t}
E {l,
++ C.C.
I t clearly suffices t o verify t h a t
++#
0
C.C.
We may assume t h a t
2 3
C
and denote by
G.
ID(G).
E
2 3
any conjugacy c l a s s of
.i
i n which case w e may choose
++
g E SuppCiCj.
Because
g = uV
w e have
u E Ci
f o r some
g
conjugacy class containing
P
Then
a c t s on
Ck':
2 3
D
If
ID(G).
V E
C
P be a d e f e c t group of t h e
Let
.i-
and l e t
X by conjugation, so by t h e argument employed i n t h e proof of
.
G E I D ( G ) , a s required.
I[C]
P 4 6(Ci)
Hence
and t h e r e f o r e
G
C E CR(G),
i s a d e f e c t group of The f a c t t h a t
.
C(P) # 0
Ci
Theorem 2.3, w e have Thus
and
G
TIC1
we s h a l l write
i s an i d e a l of
D.
P
ZVG)
i n s t e a d of
w i l l be e x p l o i t e d i n t h e
proof of t h e following r e s u l t . 2.6.
PROPOSITION.
Let
e
be a block idempotent of
i r r e d u c i b l e r e p r e s e n t a t i o n of (i) suppe = C
where
(a)
C
1
U
C
1
U 2
a s s o c i a t e d with
... U Ct U Ct+l
,...,C,arep-regular
For a l l
ZVGI
FG,
U
i E 11 ,...,tj, e E I I C i l
y be t h e
e
... U C, G
c l a s s e s of
and l e t
and
such t h a t
6CC.I = 6(C.) ' G 3
for a l l
i , j E 11,...~t} (b)
Cii)
6(Ck) < 6CCi) If
G C E CRCG)
for all
...,r }
k E {t+l,
i s such t h a t
and
6(C) < 6 ( C i )
i E 11 f o r some
G
,...,t} i E 11,. ..
ycc+, = 0 (iii) There e x i s t s
i
(ivl
i s such t h a t
If
C E CR(G1
E 11,. ..,t}
such t h a t yCC+) # 0,
+ #0 ytCi) then
6 (el
< 6 (C) . G
then
THE NUMBER OF IRREDUCIBLE FGMODLTLES
79
F
Proof.
(i) wing to Theorem 2.4,
...,CF
and some p-regular classes C l , some k E {l, ...,PI y ( I I C k ] ) # 0.
with
F
..,C P1
E hiCif
=
for some nonzero Ai
in F
i=1
of G.
Since Y ( 1 )
But C i E I I C k l ,
=
Y(e) # 0 ,
there is
so we must have
Invoking Proposition l.lO.lO(iv), we infer that e E I I C k ] .
renumbering C1 ,...,C ICl,.
# 0.
y(Cl)
e
in such a way that {Cl
consisting of all Ci
6(Ci)
with
,...,C t } =
By
is the subset of
6 ( C k ) , we deduce that (a) and
G
(b) hold. (ii) By (i) , 6 ( C ) < 6 ( c k ) .
Suppose that y(c+) # 0.
G
have y ( I [ c ] ) # 0
6 (Ckl
G
and so by Proposition l.lO.lO(iv),
Since
c'
e E I[c].
E IIC] we
Hence
6 (C) , a contradiction.
(iii) This was established in (i) (iv) By the argument of (ii), e E I[c1.
The desired assertion is now a
consequence of the definition of I [ c ] . For future use, we shall need a deep result due to Brauer (see Curtis and Reiner(1962)).
The proof will be omitted since it relies on modular character
theory, a topic we do not touch upon in this book. Let F be an algebraically closed field of characteristic p > 0,
2.7. THEOREM. let G
n
be a group of order p m
with defect d.
Then d
(p,rn) = 1 and let B
with
be a block of
is the smallest integer such that pn-d
FG
divides the
dimensions of all the irreducible FG-modules in B. 3 . THE NUMBER OF IRREDUCIBLE FG-MODULES.
Throughout this section G istic p > 0.
denotes a finite group and F a field of character-
Our goal is to determine the number of irreducible FG-modules in
the case where F is a splitting field for FG. 3.1. LEMMA.
Let A
field for A
and let S = l A , A J .
be a finite-dimensional F-algebra, let F
rn
T = {a:€ is an F-subspace of A ,
AIP
be a splitting
Then
s
for some integer
m
11
and the number of nonisomorphic irreducible A-modules
CHAPTER I1
80
.
dim ( A / T ) F As we have seen i n t h e proof of Lemma 2 . 1 , Proof.
equals
2E
[A,A]
S C_
and so
a
T
thus
+
b E T.
T
implies
then f o r a
h E F
and a E
T, Xa
E
T and
J ( A ) i s n i l p o t e n t , it follows from the
Since
T.
?
and
i r r e d u c i b l e A-modules.
=
Because
T/J(AI.
Denote by
r
t h e number of
is a splitting field for
F
A,
Corollary
t e l l s us t h a t
r...,n
f o r some p o s i t i v e i n t e g e r s clearly
5
=
[1,11
Si,Ti
i ?
-
S = (S+J(AI)/J(A).
Next p u t
r'
Then
and
? Let
A.
J(A)
that
1 = A/J(A)
Now put
arb E T
if
It is c l e a r t h a t f o r a l l
i s an F-subspace of
d e f i n i t i o n of
1.10.4
Again, by Lemma 2.1,
E [A&!]
m,
sufficiently large
proving t h a t
T.
U
rn
= {I
E
112'
E
B
f o r some i n t e g e r
m 2
11
be t h e analogous F-subspaces defined f o r t h e F-algebra
A . = Mn (F) 2
i
Then
and so w e a r e l e f t t o v e r i f y t h a t
Let
est
entries
denote t h e
0.
ni x ni
Then f o r any
matrix with ( 8 , t ) - t h e n t r y equal t o 1 and a l l o t h e r
8
# t
est
=
esjejt
-
ejtest
E Si
and
e
88
-
ett = e 8 t e t s
- etaeat
si
THE NUMBER OF IRREDUCIBLE FG-MODULES
contains the nt-1
Hence Si S
# t) and e
- e
ss
linearly independent elements est(l g s,t
( s = Z,..., n i ) .
11
dim(Ai/Ti) = dimAi F F
Because
-
dimT. G n?
Ti # Ai
4
si.
But ep
-
this implies that
(n?-l) = 1
are all n. x ni 2
m 11
si 5 T i ,
ni'
FZ
Si
On the other hand, the elements of In particular, e
81
e
=
11
for all m
matrices with trace 0. 1, so e
11
and therefore dim(A ./T .I = 1. F z 2
11
9 Ti.
Hence
Thus the lemma is proved.
We come now to the demonstration for which the present section has been developed. Let G be a finite group and let F be a field
3.2. TtIEOREM (Brauer (1935)).
0 such that F is a splitting field for FG.
of characteristic p
Then the
number of irreducible FG-modules is equal to the number of p-regular classes of G.
rn and let a;
E
Choose an integer m
T.
be an F-basis of
,I) ,...,Ut
2)
1
m
let T = {x E FGlZ E S for some integer m
Let S = [FG,FG],
Proof.
1)
so large that
2
S for each i E 11,. ..,t) and pm
Sylow p-subgroup of
for all X i E F .
G.
x
=
Ex#
xE T
where, by our choice o f regular classes of
is at least as large as the order of a
Owing to Lemma 2.1, we have
Hence
(1)
T if and only if m m if and only if gp
g mE
E
rn, each
&$
S
or equivalently
E S
is p-regular.
Let C1,.
.. ,Cs
be all p-
GI and put
Applying (1) and Lemma 2.2, we conclude that x = 1 x 8 E T if and only if
for all i E 11, (2)
>
...,
8).
Thus
1 x 3 E T if and only if
C z
9'3 $7
=
0 for a l l
E {lI...,s}
82
CHAPTER I1
JI
Finally, define
...xF ( S copies)
: FG-'
FX
Czgg
6. =
where
,
C z
i
1Q
< s.
-
Then
a s t h e F-linear mapping given by
(B1 I . . .,BS) J,
T
i s c l e a r l y s u r j e c t i v e and by (21
is
SEqg
w.
t h e k e r n e l of
Hence, by Lemma 3.1,
and t h e r e s u l t follows.
4. INDUCED MODULES.
Throughout t h i s s e c t i o n ,
H
G.
a subgroup of
F
If
V
V,
b u t only a c t i o n of
i s defined on
FH
thus a s an F-module,
vH.
Let FG,
We say t h a t an FG-module
g E G,V.+ G
V
Since w e may consider
... @ Vn
a c t s on t h e s e t
i E {l, ...,n ) .
V
{Vl,.
.. V I
}
V.
with
by t h e r u l e
we s h a l l r e f e r t o t h e set
imprimitivity f o r
V.
V
is imprimitive i f
of F-subspaces
@J
W.
This
can be w r i t t e n a s a d i r e c t
n > 1 such t h a t f o r each
{Vl,
...,Vn}.
'Vi
=
cVl
FG
FH
8.
i s a permutation of t h e s e t
gVi
t o a uniquely
a s a subalgebra of
FH
w e can define an FG-module s t r u c t u r e on t h e tensor product
V = V @
vH
vH*
W be any FH-module.
i s t h e induced module and we denote it by
sum
the
This process w i l l be
c a l l e d r e s t r i c t i o n and it permits us t o go from any FG-module determined FH-module
VH
i s an FG-module, then we s h a l l denote by
FH-module obtained by t h e r e s t r i c t i o n of algebra; equals
a f i n i t e group and
G
denotes an a r b i t r a r y f i e l d ,
gVi
Expressed otherwise,
for a l l
,... lVnl
g E G,
as a system of
The following observation r e l a t e s t h e concept of
i m p r i m i t i v i t y t o t h a t of induction.
4.1. LEMMA. and l e t
ti)
g l , ...,gn
Let
H
be a proper subgroup of
be a l e f t t r a n s v e r s a l f o r
i s a system of i m p r i m i t i v i t y f o r
H
G, in
let
G,
W G and t h e a c t i o n of
F/
with
G
be an FH-module,
g E H.
is transitive.
Then
INDUCED MODULES
(ii) Let
V
be an imprimitive FG-module with
imprimitivity and assume that G the stabilizer of
83
. ,vn}
{Vl,.
{ V l,...,Vn}.
acts transitively on
V1 under the action of G.
v = v 1G.
Then
as a system of
y1
Let H
be
is an FH-module and
n
Proof. (i) By Proposition 1.4, FG
@ g .FH i=1
=
and so
'
i= 1
g E G induces a permutation of the set
Since the left multiplication by { g , 8 W ,... ,gn 8 W } G
(ii) Since
and g ( g 8 W1
g 8 Y, the assertion follows.
acts transitively on
V1, it follows that n in G, say g ,
=
,... ,gn
{Vl,
...,V n }
and H
is the stabilizer of
Thus we can choose a left transversal for H
= (G:H).
such that g.V
=
' 1
Vi, 1
i < n.
Since
and
V as F-spaces, the map
8 : V:+
=
V
g V Q 1
v E V
,
v,
n
However, for each g E G
there exist uniquely determined h E H Thus, for all
... @ g n
given by
n
is a vector space isomorphism.
1
and j E {l,
...,n}
and each
i
E 11,.
such that ggi
=
..,n}, gjh.
we have
as required. Let H let j
:
U+
be a subgroup of G ,
let
V be the inclusion map.
U be a submodule of an FH-module V and Since FG
is a free FH-module
CHAPTER I1
84
(Proposition 1.4), it follows from proposition 1.8.5 that FG module.
is a flat
FH-
Hence the canonical homomorphism
For this reason, from now on we shall identify U
is injective.
.
G
image in V
G
with its
The following lemma collects together some elementary properties
of induced modules. 4.2. LEMMA.
V
Let
V
and
(i) VGc VZG if and only if 1 -
G
(ii) V = 1
(iv)
2
VP
(vl n v ~ = v,") ~n
(vl If V Proof.
V
c
1 -
V G if and only if V
I V ~ + V =~+:v) ~
(iii)
=
V
@
1
V
V.
be submodules of an FH-module
1
V
2
= V
2
VP
then 2'
VG= V G @ Vp
We first observe that (ii) is a consequence of (i), that (v) follows
from (iii) and (iv), and that (iv) is a consequence of the fact that FG free FH-module. To prove (i), observe that Vl VGC VG. 1 -
Let T be a transversal for H
2
5 V2
in G
is a
obviously implies that
containing 1.
Because
vc= E T
@ t 8 V
each element in Therefore, if C ' V 1 -
v2
c t
V G can be uniquely written in the form
V1 E
implies
Vl
18
is such that
V.
E V2.
Thus
.
v1 5 v ~ , proving
To prove (iii), observe that V
V2, then V
Vl E
with vt E
@ Vt
Z%T
G
(i) G 1
+ V G c (V + V 2 ) 2 -
1
.
by virtue of (i)
The
opposite containment being a consequence of the equality
t 8 C u + v l = t @ v +t@V2 1
2
( V E V 1
lPV2
EV,tET)
the result follows. We next provide some important formal properties of induced modules.
4.3. PROPOSITION.
(il
@/A)'
2
Let A
FG*A/FG*B
CB
be left ideals of
FH.
Then
2
85
INDUCED MODULES
(ii) Let I
be a right ideal of FG
such that I s FG*A.
If
V
is an FH-
module, then there is a canonical injective map
Proof.
All tensor products are to be taken over FH.
(i) Consider the diagram
-
o
-FG’-B-FG.B/FG.A
Then the diagram is commutative, and since FG
where all maps are canonical.
a flat right FH-module, both rows are exact, and f , g
are isomorphisms.
is
This
proves the required assertion. (ii) Let
0 is exact.
V-
denote the canonical map
-
FG 63 AV+
FG 63 V
Let x E I , y E FG, and
written in the form q = y a 1
and therefore I ( F G 63 V )
4.4. PROPOSITION.
1
+
V E
The sequence
V.
0
Then xy E I s FG*A, with y
i E FG,ui E A .
which finishes the proof.
Let H be a subgroup of G
sequence of homomorphisms of FH-modules. (il
Then the sequence
a FG 63 (V/AV)&
... + ynan
5.Ker(1 63 U) ,
V/AV.
and let U-
h
V
so q
can be
Thus
. A W be a
Then the following properties hold:
CHAPTER I1
86
0
u
3
-LV J L w-
0
is exact if and only if the corresponding sequence of FG-modules o
-
u
G
~
~
F
~
-
-
-
+
o
is exact (ii) Assume that the sequence
(*)
is exact.
Then
splits if and only if
(*)
(**)
I
splits. (iii) If U
is a submodule of
then $/VG
V,
1
(V/U)G
~
I
Proof. (i) That exactness of (*I fact that FG is a flat FH-module.
implies that of
(**)
is a consequence of the
Conversely, suppose that
(**)
is exact.
Then
Since Ker(1 8 1-I)
=
Im(1 8 A ) ,
it follows from Lemma 4.2(ii) that Kerp
A similar argument shows that Ker(1 8 A ) = 0
Im(l 8 p )
=
8
implies TW
(ii) Suppose that 0y : W-
=
W.
implies KerX = 0
Thus the sequence (*I
A vA
=
TmX.
and that
is also exact.
W-+ 0 is a split exact sequence and let
V be a splitting homomorphism.
18 y
Then
:
8- fi
is obviously
a splitting homomorphism for the induced sequence Conversely, assume that the exact sequence
of FG-modules splits.
Let T
denote a transversal for H in G containing 1.
Consider the mappings
.'cv
V
C t 8 vtl---t ET Then
0
and
T
are well defined FH-linear maps.
splitting homomorphism and let
W
E
W.
Let J,
Then J l C l 8 w ) =
:
V]
84VG
be a
C t 8 vt,vt E
ET
V, and
INDUCED MODULES
c
(1 8 ll)qJ(l8 W ) =
07
t 8 l l ( U t ) = 1 8 2J
ET Thus 2J = ! J ( u )
and therefore
It follows that
T$O
:
v is a splitting homomorphism, as required.
W-
.
(iii) Direct consequence of (i) applied to the natural exact sequence
o~u-v-v/u-o. As
4.5.
an immediate consequence, we derive
COROLLARY.
so is
Let
V
be an FH-module.
fi
If
is completely reducible, then
v.
Another useful property of induced modules is given by If L,H
4.6. PROPOSITION [Transitivity of the induction). with H
5 L,
Proof.
are subgroups of
G
then for any FH-module V,
The required assertion follows from
(+)G
= FG
8 IFL 8 I 4
g
(FG 0 FL) 8
FH
FL
FL
V
FH
gFG8V=VG
FH Let S be a subring of a ring R
r E R,@ E
Hom(R,M), S
define r@
:
and let M
M by
R-
be a left S-module.
(r@)(x) = $[zr).
Given
It is evident
The following that the additive group Hom(R,M) becomes a (left) R-module. S result illustrates how a module of this type can be brought into argument.
4.7. PROPOSITION.
Let
V be an FH-module. Hom(FG,V)
FH
Then
v"
as FGmodules
CHAPTER I1
88
Let [ g
Proof.
,
$ E Hom(FG,VI
define
,,..&n 1
e($)
Be a left transversal for H in G I and for any by
E
=
FH
Then 8
is
.i=1
C gi 8 $(gil) = 0 . Then @ ( g i l )= 0 i=1 Since {gl , . . . , g i l) is a right transversal for H
obviously an F-homomorphism. for all i in G,
n C gi 8 @ ( g i l l .
{1,2,
E
...,n ) .
given g E G ,
and thus 0 elements g
@ E Hom(FG,V)
,...,gn
-1
-1
-1 we have g = hgi
is injective.
-1
Assume that
Because FG
for some h
H.
Hence $ ( g ) = h $ ( g i l ) = O
is a free left FH-module with the
as a basis, for every set
such that @ ( g i l l = U i , l
E
i G n.
{Ui E
V(1
Thus 8
i G n } there is a is an F-isomorphism.
FH Given g E G, there is an hi E H and a permutation i H T ( i ) 11
,...,nl
-1
and hence that g g T ( i ) = 9.h..
such that g . g = hign(i)
as asserted. Let
-1
of Thus
2 2
.
V be an FH-module.
Then the mapping
is an injective FH-homomorphism.
In what follows we shall refer to f
as the
canonical i n j e c t i o n . 4.8.
PROPOSITION (Universal characterization of induced modules).
subgroup of
G, let V be an FH-module, and let f
injection.
Then, for any FG-module
W
and any
:
V-+
8
@ E Hom(V,WH)
Let H
be a
be the canonical there exists one
FH and only one
9 E Homt#,W) FG
which renders commutative the following diagram:
INDUCED MODULES
Since g 8 D = g ( l 8 V ) ,g E G,v E V,
Proof.
is uniquely determined by its restriction to one such $.
v
,...,V n
Let T
form an F-basis of
$ c t 8 vi)
=
J,
fl.
J , (8 t
for some t
{ t 8 Vile E T , 1
fi-
~l
E V.
= t @ ( v ) for any
is an FG-homomorphism.
H.
Since
W
Hence there exists at most
be the F-homomorphism
E T,h E
VG+
in G containing 1 and let
Then the elements
Let $
t @ ( v i ) . Then
write g t = t h
whence
V.
any FG-homomorphism
1 8 V.
be a transversal for H
be an F-basis of
89
W
i
n)
determined by Let g E G
and
Then we must have
$of
= @,
the result follows.
We are now ready to prove the following important result. 4.9.
THEOREM.
Let H be a subgroup of a group G, let V
and let W be an FGmdule. (i) Hom(fl,W)
FG (ii) Hom(W,?)
be an FH-module,
Then
Hom(V,wH)
as F-spaces
FH as F-spaces
Hom(WH,V)
FG
FH Proof. ti) Let f
:
V-
fi
be the canonical injection.
Given
CHAPTER I1
90
Hom(fl,w) FG
Ilr
is an injective F-homomorphism.
-
9
@
If
E
Hom(v,WH)
FH
Pf
Hom(v,WH),
then by Proposition 4.8
FH
there exists a unique 9 E Hom(fi,w) such that
0=
$f.
Thus
8
is an iso-
FG morphism. (ii) Owing to Proposition 4.7, we need only verify that Hom(w, Hom(FG,V) FG FH Given @
E
Hom(W,Hom(FG,V)), FG FH
Then the map @++ @
the map
:
@*
2
as F-spaces
Hom(W ,v) H FH
@* Hom(WH,V) by
define
FH
If I) E Hom(WHH,V), then FH
is an injective F-homomorphism.
W - + Horn(FG,V) defined by
FH
is an FG-homomorphism. Let
Since $*
=
I)I
the assertion follows.
U and V be any FG-modules.
Then the intertmining number for U
and
V is defined to be i(U,V) = dim Hom(U,V)
F Let U
be a completely reducible FG-module, and let V be an irreducible FGWe say that a nonnegative integer m
module.
irreducible constituent of modules contains exactly m Let V/J(V)
FG
U
if a decomposition of submodules isomorphic to
V be an FG-module.
as the head of
4.10. THEOREM.
U
V as an
into irreducible sub-
V.
In what follows we shall refer to the factor module
V.
(Nakayama Reciprocity).
be a subgroup of G.
is the muZtipZicity of
Let F be an arbitrary field and let H
Suppose further that V
W is an irreducible FG-module.
Let
is an irreducible FH-module and
m be the multiplicity of W as an
RELATIVE PROJECTIVE AND INJECTIVE MODULES
p
irreducible constituent in the head of
and let n
as an irreducible constituent in the socle of
WH.
91
V
be the multiplicity of
Then*
In particular, (i) If F (ii) V
is a splitting field for FG
is isomorphic to a submodule of
and FH, then m = n
wH
if and only if W
is isomorphic to
n
a factor module of
v".
We first note that Hom(U,W) = 0 FG U P W. Since for any FG-homomorphism
Proof. with
Hom(fi,W) FG
2
for all irreducible FG-modules U
f
:
... @ Hom(W,W)
. .. @ Hom(V,V)
FH =
V*
WH is
WH, we have
HomIv,V) @ Thus i(V,W 1 H
(rn times)
FG
Since the image of any FH-homomorphism
contained in the socle S of
we have
Hom(fi/J(fi) ,w) FG
z Hom(W,W) @ FG Thus i(fi,W) = m i ( W , W ) .
fi-+W, JCr/;) 5 Kerf
(n times)
FH
n i(V,VI and the result follows by virtue of Theorem 4.9(i).
.
5. RELATIVE PROJECTIVE AND INJECTIVE MODULES. Throughout this section, F denotes an arbitrary field, G a subgroup of
a finite group and H
G.
An FG-module
V
is said to be H-projective if every exact sequence of F G
modules
0-u-
w-
v-0
for which the associated sequence of FH-modules
0
-UH
wH-
vH-
0
92
CHAPTER I1
s p l i t s i s a l s o a s p l i t e x a c t sequence of FG-modules.
The FG-modules
V
is
c a l l e d H-injectiue i f every e x a c t sequence of FG-modules
v-
0-
w-
u-0
f o r which
i s a s p l i t e x a c t sequence of FH-modules, i s a l s o a s p l i t e x a c t sequence of FGThus an FG-module i s p r o j e c t i v e ( r e s p e c t i v e l y , i n j e c t i v e ) i f and only
modules.
i f it i s 1-projective ( r e s p e c t i v e l y , 1 - i n j e c t i v e ) .
5.1. LEMMA.
T
Let
modules, and l e t
H
be a t r a n s v e r s a l f o r
f E Hom(8 ,W 1.
G,
in
Then t h e map
f*
:
W
let
V
V
W defined by
+
and
be FG-
FH
f*(u)
C (tft-llu
=
iET i s an FG-homomorphism which i s independent of t h e choice of
Proof.
Assume t h a t
tut
=
St
t E T with ut E H.
f o r each
C (S@;’)U
Then
C t u t f u t-1 t -1u
=
ET
T.
ET
c tUtUt-1ft-lu
=
E T =
proving that for a l l
f*
f*Wl
i s independent of t h e choice of
T.
I t t h e r e f o r e follows t h a t
g E G. f*(9U)
=
c
c g ( g - l t )f (t-lg)u
c t f t - 3 (gu) =
ET
IET
E g C g - h f ( g - l t ) -$ l%T
as required.
Let
5.2. LEMMA. and l e t
f
:
(VH)l)(i
T be a t r a n s v e r s a l f o r +
V
H
in
be given by
f C C t B u t ) = C tvt %T 7ET
G,
let
V
be a n FG-module,
RELATIVE PROJECTIVE AND INJECTIVE MODULES
Then f is a surjective FG-homomorphism and Kerf
is a direct summand of
It is clear that f is a surjective F-homomorphism.
Proof.
f is in fact an FG-homomorphism, fix g t' E T
93
such that g t = t ' u t .
ut E H
and
G.
E
c
g(
t 8 Ut)
e
=
Then for each
c
g t 8 ut =
c
=
To prove that
E T
there exists
It follows that
ET
ET
t
(6jH.
t ' u t 8 ut
E T
t ' 8 UtUt
ET and so
c
f(g(
t 8Ut))
c
=
ET
t'utut
ET =
=
c
gtut
ET
c
qfC
t
Q Ut),
ET proving that f is a surjective FG-homomorphism. canonical injection.
H n Kerf
Obviously
E(V
= 0.
Let
E :
VH + (V,)
G be the
Because
we conclude that
as asserted.
9
5.3. LEMMA.
Let: T be a transversal for H
and let f
:
V*
in G, let V
be an FG-module,
( V H I G be defined by
c
f(U) =
t 8 t-lu
7ZT
Then f is an injective FG-homomorphism which is independent of the choice of the transversal T and Proof.
Assume
(f(V)),
that T'
is a direct summand of
is another transversal for
each t E 1, there exists ut E H
c t 8 t-lu E T
t'
and =
c ET
E
t'ut
((V,) H
)H.
in G.
T' such that t 8 u,l(t')-$
G
= t'ut.
Then, f o r Hence
CHAPTER I1
94
proving that f
is independent of the choice of the transversa
It follows that for all g E G, V
so that f
is an FGhomomorphism.
Since
WHIG = @ t 8
vfl
*T f
is an injection.
G
(CV,) I H ,
Now
t 8 VH
W' = @
is obviously an FH-submodule of
tfT,#H and we show that
If f t u ) = C t 8 t-lu E f ( V ) n W',
then u = 0.
On the other hand, if
ET
t ' E T n H, we have C t But
ET
-
t 8 t-'t'utl
C
E W'
*T
So the lemma is verified.
The next result establishes several important characterizations of H-projective and H-injective modules. 5.4. THEOREM. migman(1954)). transversal for H
in
Let V
be an FG-module and let T be a
G containing 1.
Then the following statements are
equivalent: (i) V (ii) V (iii) V
is H-projective is isomorphic to a direct summand of is isomorphic to a direct summand of
(iv) There exists
JI
E End(VH)
(V,)
G
8,where
W is an FH-module
such that
FH for all u E V
RELATIVE PROJECTIVE W D INJECTIVE MODULES
(v) V
is H-injective
-
Proof. (i) * (ii): modules 0
U
summand of
.
V'
I
(v,) G.
By Lemma 5.2,
CVH)
-+
submodules splits
where
95
f. V
there exists an exact sequence of FG-
0 such that the associated sequence of FH-
-+
v
By hypothesis,
is H-projective, so
Kerf
is a direct
Thus
V.
(ii) * (iii): Obvious (iv): We first prove that any FG-module V
(iii)
W is an FH-module, satisfies (iv).
It is easily checked that $
Define $
:
of the form V
V+
=
#,
where
V by
is an FH-homomorphism.
Moreover] for all w E FI,
we have
Thus J,
satisfies (iv).
Turning to the general case, we may harmlessly assume
that
#=V@V'
[direct sum of FG-modules)
n
Let
71
:
p-+ V
8. Then
since n
ToJ,
be the projection map and let $ induces an FH-homomorphism V - +
is an FG-homomorphism.
(iv) * (v): Suppose that
U = V 0 W for H
VH
H
some
V, and for all U
E V,
we have
The desired implication follows.
V is an FG-submodule of the FC-module U and that
FH-submodule
W.
and let I$E Hom(U,v) be defined by
F
satisfy (ivl with respect to
Denote by
71
the projection of
UH onto
CHAPTER I1
96
where J1
.
End(VH;i) satisfies (ivl
FH U
=
V @ Kerf$.
But, by Lemma 5.1,
FG-submodule.
If v E V,
f$
(v) * (i): Assume that
V
is an FG-homomorphism, so KerG
Then, by Lemma 5.3,
is H-injective.
isomorphic to a direct summand of
(VHl
infer that there exists $ E End(VH) FH
G
.
v+
v
and define
0
:
U
4
u
=
G/
(direct sum of FH-modules) Set
4
= f-l
U by
6
B(U) @Kerf
5 . 5 . COROLLARY.
= d
flW is an FH-isomorphism of W onto VH.
proving that U = e ( U )
Thus
is
is a surjective FG-homomorphism for which
6(u)
Then, by Lemma 5.1,
V
such that
U=Kerf @ The restriction f =
is an
Invoking implication (iii) * (iv), we
(1 t$t-l)v E T :
and thus
V is H-injective.
Hence
Assume that f
then $(U) = u
I: t0Jlft-lu ET
is an FG-homomorphism and
+
Kerf.
Finally, if
B C U ) E 8 ( U ) n Kerf,
then
and the result follows.
(Higman, 1954).
divide the index of H
=
in G.
Assume that the characteristic of
F does not
Then any FG-module is H-projective and H
-
97
RELRTIV’E PROSECTIVE AND INJECTIVE MODULES
injective. proof.
Then J,
V
Let
be an FG-module and let $
4 V
V
:
be defined by
is an FH-homomorphism and for all U E V
c c t$t-l)u
= u
*T where
is a transversal for H
T
Let N
5.6. COROLLARY.
fi
and let V
G
be a normal subgroup of
be an irreducible
N
in G ,
is completely reducible.
VG and let T be a transversal for N
Let W be a submodule of
Proof. in G.
Now apply Theorem 5 . 4 .
If the characteristic of F does not divide the index of
FN-module. then
in G.
Since
Cs;,,= (#)N
it follows that
reducible, and thus WN
Hence WN
is completely reducible.
(#IN.
is a direct summand of
Consequenctly, W
N-injective.
@ t @ V ET is completely
By Corollary 5.5,
is a direct summand of
r/;
W is
and the result
follows. Let
U
be FG-modules with F-bases u
Then U 8 V
respectively.
{ui 8 u .I 1
V
and
1
,...,u
is a vector space over F
and U , . . . , U r n , 1
with an F-basis
F Q
i
G n, 1 & j G m}.
We can define an action of
G
on
g(ui8 u .I 3
for all i , j
=
g u . 8 gv
and g E G, and then extend to
z
j
U 8 V and FG by linearity.
F FG-module U 8 V
is called the (inner) tensor product of
F 5.1. PROPOSITION.
Let
s
U 8 V by
F
3
5H
be subgroups of
U
and
V
The
.
G and let V,W be FG-modules.
WH is 5’-projective, then V 8 W is S-projective F (iil If W is projective, then V 8 W is also projective. F Proof. (il Let X be a transverse1 for H in G and let Y be a (i) If
V
is H-projective and
CHAPTER I1
98
s
transversal for
in H.
Then XY
in G.
is a transversal for S
By
Theorem 5.4, there exist y E End(vH)
and
6
and
C (y6y-l)w L8y
Endtkls) FS
FH such that
(
c
%?yola:
-1 )v =
2,
3SX for all
Then y 8 6 E EndC(V 8 W ) s ) FS F
W.
E v,W
2,
=
w
and a straight-forward
calculation shows that
(
c
q c y 8 6)y-lx-l)
It,8 W )
= t,
8 w
sx,yEy W.
for all U E V,w
The desired conclusion is now a consequence of Theorem
5.4. (ii) Apply ti) for the case H = G
and
S = 1.
9
The following lemma collects together some simple properties of H-projective modules. 5.8. LEMMA.
Let
S 5 H be subgroups of G and let V be an FG-module.
n
8 Vi, where Vi is an FG-module, 1 i=1 projective if and only if each V. is H-projective cil Let V
(iil
If
=
is S-projective, then V
V V
(iii) If
is H-projective and
(ivl For any FH-module W,
i
n.
Then
V is H-
is also H-projective
VH is S-projective, then V is $-projective
the FG-module
8
is H-projective
(vl If kr
is an S-projective FH-module, then any FG-module which is a direct
summand of
d'
"
Proof.
is S-projective.
(i) Assume that
V is &projective.
which satisfies Theorem 5.4Iiv). onto Vi
defined by the decomposition V =
implication (ii) onto Vi,
Let nil
* (iiil of Theorem
5.4,
1
@Vi.
Then there exists $ E End(VH)
FH 6 n, be the projection of
V
Then, as in the proof of
n.03, induces an FH-homomorphism of
which also satisfies Theorem 8.4(.iii).
Thus each
vt Vi is H-projective.
Conversely, if each Jli E EndlViIH satisfies Theorem 5.4.Uv1, then $ E End(VH1
FH
FH
RELATIVE PROJECTIVE AND INJECTIVE MODULES
99
$ l V . = $i, obviously satisfies the same condition.
defined by
-
V
Hence
is
H-projective. (ii) Let 0
U
W-
--L
V -
* UH
the associated sequence 0 sequence 0
us
-+
sequence 0
U
+.
ws
-+
-+
1
VH.
Thus
< <. < P
direct summand of
-+
G
(V,)
+
VH
-+
* 0 splits. *
=
@
WH
-+
Vs
V
(ValG
(iii) By Theorem 5.4,
8''
-+
W
0 be an exact sequence of FG-modules such that
fi
0
splits.
(VSlH
and
V' @
x @ YG,
V
Since
Then the
is S-projective, the
Thus V
must also split.
v' @ X =
0
V " 8 Y,
=
so
is H-projective.
where V' z V
and
is isomorphic to a
Now apply Theorem 5.4.
(iv) This was established in the course of the proof of Theorem 8.4. (v) By Theorem 5.4, (iv),
8S
THEOREM.
(i) If
@
Let H
V
<
W, so that
fi
z
kF 8 f .
is S-projective.
.
is S-projective, again by appealing to (i)
By
.
Hence
G.
be a subgroup of
is a projective FG-module, then VH is a projective FH-module.
V
converse is true, provided (ii) If
W' @ X where W'
is S-projective and therefore, by (i),
any direct summand of 5.9.
=
V
charF
The
) (G:H).
is an FH-module, the induced module
8
is projective if and only if
is projective.
(iii) Let charF = p an FG-module.
7 0,
Then V
Proof. (i) Let
V
let P
be a Sylow p-subgroup of
is projective if and only if be a projective FG-module.
V H is projective and let charF
0-
be an exact sequence of FG-modules.
o + uH splits, since VH is projective. hence the sequence
w-
u-
-
Then
and let V
V-
k
LG:H).
be
is projective.
VH is a projective
Conversely, assume that V
FH-module, by virtue of Lemma 5.8Cii). module such that
Vp
G,
is an FG-
Let
0
Then the associated sequence
w*-
VH-
0
Owing to Corollary 5.5,
V
is ?+projective;
CHAPTER I1
u+w-v-0
0Consequently, V
also splits. (ii) Let V
is a projective FG-module.
be an FH-module and let T
be a transversal for
H in G.
Then
(direct sum of FH-modules)
where, of course, 1 8 projective.
fi
is identifiable with
(#IH
Then, by (i),
Lemma 5.8(i), If
v
V.
Assume now that
V is projective.
Conversely, suppose that
V is a free FH-module, then the FG-isomorphism FG 8 FH FH is a free FG-module.
fi
Hence
is projective.
FG implies that
In the general case, we
have W zz V
U for some free FH-module W and some FH-module U.
8 fl @ 8
and hence
2
fl
Then
.
is projective by Lemma 5.8(1).
(iii) Direct consequence of ti).
6 . VERTICES OF
is
Hence, by (14) and
is also projective.
V is projective.
fi
(14)
FG-MODULES.
Throughout this section, G denotes a finite group and F a field of characteristic p > 0 dividing the order of Let
V and W be FG-modules.
is isomorphic to a direct summand of Assume
that
V
(i) V
(V,) G
is a component of
v
V is
v
restriction that p
component of W
if
V
W.
(vH)'
Then a subgroup Q
for any proper subgroup H of
if and only if
1
Observe that if p
FG-modules have vertex 1.
a
of
G
if the following two properties hold:
is not a component of
Hence 1 is the vertex of module.
We say that
V is an indecomposable FG-module.
is said to be a vertex of
(ii)
G.
v
Q.
is a principal indecomposable
[GI, then by Maschke's theorem all indecomposable
In order to avoid trivial statements, we made the
divides
IGI .
Owing to Theorem 5.4, (i) and (ii) are equivalent to the condition that &-projective, but not H-projective for any proper subgroup
H of Q.
V is
Observe
VERTICES OF FG-MODULES
also that by Theorem 5.4, (i) is equivalent to (i') V
UG for a suitable FQ-module
is a component of
Assume that &?
v
is a vertex of
and write
vQ
=
U 0 i
... d Up
Then
as a sum of indecomposable modules.
and hence, by the Krull-Schmidt theorem, V indecomposable FQ-module U.
U
8
is a component of
U
Any such module
for some
v.
is called a source of
Our aim in this section is to prove that the set of vertices of
v
is a class of
conjugate p-subgroups of G. Let
W be any FH-module, where H
define the F@Hg-')-module
is a subgroup of G.
as follows: gW has the same underlying space as
'W
W and the elements r,x E 9Hg-I
act according to the rule z
*
w = (g-lrg). and L
Let H
6.1. THEOREM (Mackey decomposition).
T be a set of double coset representatives for an FH-module.
F(tHt-' n L )
.
For any g E G,
For each t E T, let Vt
be subgroups of G, let
(L,H)
in G
and let V tV
denote the restriction of
be to
Then (as FL-modules)
Proof.
Choose a left transversal gl = l , g 2,...,gn
P=
for H
in G.
Then
n (as F-spaces)
0 (gi 8 V ) i=l
.
.
Since g (1 8 V) = g 8 V z 2
and since V
18 V,
g . 8 V E giv z
Put S = {g, 8 V,gp 8 V,.
s.
Furthermore, g i @ V
..,g,
8 V}.
and g . 8 3
Then
we have as
F (giHgil)-modules
G, and in particular L ,
acts on
V lie in the same L-orbit if and only if
(*I
LO2
gi
CHAPTER I1
and
g
belong t o t h e same double U,H)-coset.
i
t E T,
Given
Wt
Then each
Wt
denote by
t h e sum
of those
@ V
f o r which
gi
g E LtH.
i
i s an FL-module and G (V l L =
@
*T
wt
.
L
(Vt) L e t X 5 {g,, . . . , g } be such t n t h a t L t H = U XH so t h a t Wt = @ kc 8 Y ) . W e may harmlessly assume t h a t 6 X &X t E X. Now L a c t s t r a n s i t i v e l y on t h e set {;c @ b'[113: E X} and under t h i s W
It therefore suffices t o v er i f y t h a t
t 8 V is
a c t i o n t h e s t a b i l i z e r of
Wt
Thus, by (*) and Lemma 4.1, Let
6 . 2 . LEMMA.
W
that
H
L
and
i s a component of
(V6)L
2
and t h e r e s u l t follows.
be subgroups of
8
w
@
where
G
of
Vi
U. i s Proof. and f o r each
Denote by t E T,
T let
Write
Then t h e r e e x i s t elements
i E 11,. .. ,n), Wi
such t h a t f o r each
g iV
i s t h e r e s t r i c t i o n of
V.
,.. @
a s a d i r e c t sum of indecomposable FL-modules.
,...,g,
V be an FG-module such
and l e t
f o r some FH-module
VL =
g,
G
.
to
FQiHgil n L)
i s a component of
.
In particular,
-1 (giHgi n L)-projective
a set of double c o s e t r e p r e s e n t a t i v e s f o r
Vt
<,
be t h e r e s t r i c t i o n of
'V
U,H)
in
to
F(tHt-l n L ) .
Wi
i s a component of
G
Then,
by Theorem 6.1,
Because
w
is a component of
(8$,. Hence t h e
8,
it follows t h a t each
required a s s e r t i o n i s a consequence of t h e indecomposability of
Wi and t h e Krull-Schmidt theorem. 6.3. LEMMA.
Let
W be an indecomposable FG-module and l e t
Q
be a v e r t e x of
W.
VERTICES OF FG-MODULES
(i) For any g E G, gQil (ii) If W
L
2. gQil
W
is also a vertex of
(WL)'
is a component of
103
for some subgroup L
of
G, then
for some g E G.
Proof.
(i) Let H be a subgroup of G.
Then it is easy to verify that
for all g E G Hence, if W
is H-projective, then for all g
particular, for all g E G,
Q1
=
Q and so gQl;l
W is gQg -projective.
WL
a component of
(g,Qg:'
2 . 2
P k2
=
W
fi @ ... @
for some
@
... @ W,,
If
W is gQ ;'-projective, W, we have
is a vertex of
i.
4
W.
is a vertex of
where each
W. is an indecomposIn particular,
and so, by the Krull-Schmidt theorem, W
Invoking Lemma 5.8(v), we infer that W -1
n Ll-projective.
-1 -1 giQgi = giQgi
Now any subgroup of gQg
-1 (giQgi n L)-projective for some gi E G.
able FL-module which is
W is a component of
Q.
In -1
This proves that gQ;'
gQil.
(ii) Owing to Lemma 6.2,
of
Since Q
is Q -projective. =
W is gH; 1-projective.
-1
is of the form gQ i1 for some subgroup Q then by the above, W
G,
But, by (il, giQgi
is a vertex of
W,
is
is
hence
n L and the result follows.
We are now ready to prove Let W
6.4. THEOREM.
be an indecomposable FG-module with vertex
G is a vertex of W if and only if
and a subgroup H of
is a p-subgroup of G
Then Q
Q.
H is conjugate to Q in G. Proof.
Let P be a Sylow p-subgroup of
projective, so W that P
1 gQil
is a component of
for some g E G.
(W,)
Hence
6.3Ci),
it suffices to show that if H
to Q.
So suppose that
H 2. gQgl for some g
E
G
.
projective and thus H = gQ5'.
By Corollary 5.5,
is a vertex of
Because gQG1
W is P-
Applying Lemma 6.3(ii), we deduce
Q is a p-subgroup of G.
H is a vertex of W.
G.
G.
W,
then H
By Lemma is conjugate
Then, by Lemma 6.3 (ii),
is a vertex of
W,W
is gQ9-l-
This Page Intentionally Left Blank
105
3 The Jacobson radical of group algebras: Foundations of the theory The aim of this chapter is to present a number of general results concerning the Jacobson radical of group algebras.
The material discussed constitutes the
foundations of the theory and will be frequently applied in our subsequent investigations.
After presenting some properties of elementary nature, we examine the We then provide necessary and sufficient
Jacobson radical o f direct products. conditions for an element z = Ex#
E FG
to be in J ( F G ) .
G is p-solvable.
subsequently simplified in the special case where
application, we characterize the elements of J(Z(FG)). Frobenius groups and the dimensions of J V G )
These conditions are
and
As an
Further topics include
J(Z(FG)).
Next we establish
an important decomposition theorem due to Morita and provide a number of applications. J(FG1
Among other topics, we examine criteria for the commutativity of
and J ( B o l
,
where Bo
is the principal block of FG.
The chapter ends
with a beautiful result due to Morita, which provides circumstances under which J(FG)
is expressible as a principal left and right ideal.
1. ELEMENTARY PROPERTIES.
The aim of this section is to present a number of results of elementary nature
Throughout G
which will be frequently applied in subsequent investigations.
denotes a finite group and F spaces over F
a fieZd of characteristic p > 0.
All vector
are assumed to be finite-dimensional.
1.1. PROPOSITION.
Assume that P
is a normal p-subgroup of
G.
Then
(il F G * I ( P ) f J V G ) (ii) J ( P ( G / P ) ) 1 J(FG)/FG.IIP) (iii) If P is a Sylow p-subgroup of Proof.
(i) Given
x
E P, we have
G, then J ( F G )
P
= FG*I(P).
n = 1
for some n 5 1.
Then
CHAPTER' 111
106
n (x-11P
n
=
P -
1
=
which proves, by Proposition 1.10.5, that I ( P )
x-l,x # 1,s E P ,
0
is nilpotent, since the elements Because P 4 G, we have
constitute an F-basis for I ( P ) .
and hence
Thus F G * I ( P )
and so F G * I ( P ) s J ( F G ) .
is a nilpotent ideal of FG
(ii) Direct consequence of (i), Proposition 2.1.2 and Corollary 1.6.17(ii) (iii) If P
is a Sylow p-subgroup of
G, then by Maschke's theorem J ( F ( G / P ) ) = o .
Now apply (ii).
1.2. COROLLARY. ring.
If G is a p-group, then J(FG)
In particular, 0
Proof.
I(G)
and FG
is a local
1 are the only idempotents of FG.
By Proposition l.l(iii),
conclude that FG 1.3. LEMMA.
and
=
JWG)
=
I(G).
Since FG/I(G) 2
F , we
is a local ring.
Let H be a subgroup of G
and let 2'
be a right tranversal for
H in G. li) If I is a left ideal of FH with F-basis an F-basis of FG.1.
X, then { t x l t E T,x E X) is
In particular, dimFG.1 = IG:Hldid
F Cii)
The elements
{ t ( h - 1 ) It E T , h
Proof. ( i ) For any
F E
H,h # 11 constitute an F-basis of FG*I(H)
t E 2',{txlx E XI
is an F-basis of
t.1.
On the other
F G - I = 8 t.1, proving the assertion. *T (ii) Direct consequence of (i) and the fact that {h-llh E H,h # 11
hand, by Proposition 2.1.4,
basis of I(H). 1.4. COROLLARY.
.
Assume that a Sylow p-subgroup P
any transversal T
of
P in G, the set
of
is an
G is normal.
F-
Then, f o r
107
ELEMENTARY PROPERTIES
J(FG).
c o n s t i t u t e a n F-basis f o r
I n particular,
dimJ(FG) = IGI F Proof. 1.3(ii).
.
-
IG:PI
w i n g to P r o p o s i t i o n 1.1(iii), J(FG) = FG.I(P)
1.5. COROLLARY.
Assume t h a t
Invoking Lemma 1.3Cii1,
P
i s a normal p-subgroup of
we also have
dimFG*I(P) = [ G I
-
Now apply Lemma
Then
G
IG:PI,
proving t h e
F assertion.
where
G i s s a i d t o be a pcomplement i f
H of
A subgroup
P E Syl (GI.
Since
P
IGI
=
IHI lPl,
G = H*P with
t h e pcomplement
H
H n P= 1,
i s always a p'-
group.
1.6. PROPOSITION.
(1) J(FG1 n FH
Let
H
be a subgroup of
G.
5 JVH)
IGl
(ii) dimJ(FG) dimJ(FH) + - [HI F F (iii) I f G has a p-complement, then
Proof. i d e a l of
(i)
dimJ(FG) IGI - ,IG:PI F Direct consequence of t h e f a c t t h a t J(FG) n FH
FH.
(ii) By ( i ) ,w e have
as required.
is a nilpotent
108
CHAPTER 111
(iii) L e t
H
be a p-complement.
0 (G)
I n what follows, w e denote by
G n (1
1.7. PROPOSITION.
0 (G)
That
P
n
Hence, t h e r e e x i s t s
C_
0 (G)
1.8. PROPOSITION.
Let
P
1
gp = 1 which proves t h a t J(FG))
+ J(FG)L= o
5G n
n
+
(1
+
by
+
i s a consequence of Proposition l . l ( i ) .
J(FG)).
Then
N
is nilpotent.
n
n = 0.
(g-1)'
G n (1 + J(FG))
.
G
=
(g-1)'
But
i s a p-group.
be a normal subgroup of
gp
-
1,
so
Thus
n.
of index
5 FG*J(FN) C_ J(FG)
If p
(ii) (Villamayor(1959))
g-1 E J(FG)
n
and t h e r e s u l t follows.
P
G.
(G)
J(FG))
such t h a t
(Passman (1970a) 1. J ( F G I n
(i)
S(FH) = 0
is a p'-group,
t h e maximal normal p-subgroup of
P
g E G n (1
Conversely, assume t h a t
G n (1
H
Now apply C i i ) .
Maschke's theorem.
Proof.
Since
) n,
then
and, i n p a r t i c u l a r ,
Ciii)
willems(1976).
Proof.
I t follows t h a t
N
in
V
G.
where each
J ( F G ) = FGqJ(FN1,
(i) Since conjugation by
g-lJ(FN)g = J(FN).
Let
If
Thus
FG*JVN)
FG*J(FN) =
g E G
then
p ) n.
i s an automorphism of
FN,
w e have
J(FN)*FG and t h e r e f o r e
i s a n i l p o t e n t i d e a l and so
be an i r r e d u c i b l e FN-module and l e t
FG*J(FN)
{gl,...,g,
C_ J F G ) .
be a t r a n s v e r s a l f o r
Then
g 8V
i
series of l e n g t h
n.
i s an i r r e d u c i b l e FN-module.
Hence
Since any chain of FG-submodules of
chain of FN-submodules of
(81,.
we i n f e r t h a t
(@),
8
has a composition restricts to a
VG has a composition s e r i e s of
109
ELET@NTARY PROPERTIES
length m <
Hence
I?.
Now FG
by Proposition 1.6.33(1).
g, ,...,gn
as a basis.
y,, ...,y,
in FN
is a free FN-module with the elements
Therefore, given x E J ( F G ) " ,
such that
=glyl Since for any u E V,x (1 8 V )
=
3: E
(*),
FG*J(FN).
yiU = 0,l Q
+
-.*
+
gnyn
0 we have
g, 8 ylu + and so, by
there exist unique
i < n.
... + gn 8 ynu
=
0
It follows that yi E J ( F N )
and hence that
This proves that J ( F G ) " c - FG*J(FN) and since J ( F G I n
c J(FG)", -
we finally deduce that
as required. (ii) Assume that p Then
fl
,/'
n.
Let V
is completely reducible (Corollary 2.5.6),
argument of (i), x E FG*J(FN). J(FG1 = FG*J(FN).
hence xfi = 0 and, by the
Thus J ( F G ) c - FG.J(FN)
and therefore, by (i),
The final assertion follows by virtue of Lemma 1.3(i).
(iii) We may view F(G/N)
as an FG-module by means of x(gN)
x,g E G.
Fix x E J P N ) .
x E F.
Hence, for all g 6 G,
a
be an irreducible FN-module and let x E J ( F G ) .
Since S ( F N )
5 1") ,
x =
c
3c
aEN a
=
xgN, for all
(a-1) for some
Thus by our assumption
Hence F ( G / N )
is a completely reducible FG-module.
a completely reducible F(G/N)-module.
Therefore F(G/N)
The desired conclusion is now a
is also
110
CHAPTER I11
consequence of proposition 2.1.8.
8
is said to be p-soZvable if the composition
A group G
Let p be a prime.
factors of G are either p-groups or ?'-groups.
It is an immediate consequence of
the definition that: Ci) Any solvable group is p-solvable Ciil
Any extension of a p-solvable group by a p-solvable group is p-solvable.
(iiil Subgroups and homomorphic images of p-solvable groups are p-solvable. 1.9. PROPOSITION (Passman(l97oa)). G/N
is p-solvable.
In particular, if G
If
Let N
IG/Nl = np'
be a normal subgroup of
G such that
( n , p ) = 1, then
where
is p-solvable of order npa
where
(n,p)
=
1, then
a
J(FG)P = o Proof.
Owing to Proposition l.ELi), we need only verify that
we proceed by induction on
lG/Nl.
from proposition 1.8(k) and if G/N Proposition 1.8(ii).
Write
is a p'-group, the result follows from
(G/MI
b
= rp
,IM/NI
By induction
b J CFG)
and
is a p-group, the result follows
If neither of these cases occurs, then G / N
proper normal subgroup M/N.
( r , p ) = ( s , p ) = 1.
Tf G/N
5 FG-J
(FM)
= Spc
with
must have a
ELEMENTARY PROPERTIES
b e
Since p p
pa
=
the result follows.
.
Let P be a normal p-subgroup of
1.10. LEMMA.
111
G.
Then the kernel of the
Brauer homomorphism
is a nilpotent ideal. be such that C n C ( P ) =
Let C E CR(G)
Proof.
la.
Invoking Theorem 2.2.3,
G Since FG*I(P) is a nilpotent
it suffices to verify that C+ is nilpotent.
ideal of FG (Proposition l.l(i)), the latter will follow provided we show that C' E FG-I(P).
The group P acts by conjugation as a permutation group of the set Because C n C ( P ) = 0,
c.
the size of each orbit under this action is divisible by
G p.
But for all z E P
Hence
C+ E
and g E C ,
FG*I(P) and the result follows.
1.11. LEMMA.
9
Let P be a normal p-subgroup of
potent of FG.
Then e
€
G and let e
be a central idem-
FCW.
G proof.
e
1
,...,e
We may harmlessly assume that e be a l l block idempotents of FG
be the Brauer homomorphism. 1 = rile )
+
Hence nCel)
in particular, e E
Im.
Let
and let
Then
... + rr(etl)
Furthermore, by Lemma 1.10, Kern zero idempotent.
is a block idempotent of FG.
and
for i # j
n ( e i ) n ( e . )= 0 3
is a nilpotent ideal, so each IT(ei)
,...,IT(en)
is a non-
are all block idempotents of FG
Since for any z E Imn, supp
C
and,
C ( P ) , the result
- G follows. We next present a criterion for FG
to be indecomposable.
First, however,
112
we must develop our vocabulary. In what follows we write 0
P
,CG)
for the maximal normal p‘-subgroup of
The inverse image of 0 (G/U ,(G)) in G
P
is denoted by
P
(GI.
P’tP
(C).
A
group G
is
P’t P
called p-constrained if the inverse image of contained in 0
0
G.
Thus, if G
CGIop
(opr ,p(G)/OP (G))
in G
is
is p-constrained and 0 ,(G) = 1, then
P
It is well-known (see Huppert and Blackburn (1982a)) that any p-solvable group is p-constrained. An ,obviousnecessary condition for FG to be indecomposable is that 0 ,(G) = 1.
P
Indeed, if m = 10 ,(G)I
then
1
7
P
-1
Ex
e = m
6 0
P
is a nontrivial central idempotent.
,IG)
The following result provides a class of
groups for which the converse is also true.
1.12. PROPOSITION.
Then FG
Let G be p-constrained.
is indecomposable if
and only if 0 ,(GI = 1.
P
Proof.
By the foregoing remark, we need only prove that if 0 ,(G) = 1,
P
then FG
is indecomposable.
Since G
is p-constrained, we must have
a central idempotent e
So assume that 0 ,(GI = 1
P
C,(P)
5 P.
of FG, we have e E F C ( P )
and set P = 0 (G). P Hence, by Lemma 1.11, given
5 FP.
Invoking Corollary
G 1.2, we deduce that e = 0
or
e = 1, as desired.
Having discovered a class of groups G
for which FG
is indecomposable, we
now prove 1.13. PROPOSITION.
Let G # 1 be a group such that FG
is indecomposable (e.g.
0 , ( G ) = 1 as G # 1). If C ,C ,..., C P 1 2 r conjugacy classes of G distinct from 1, then the elements
G is p-constrained,
c; constitute an F-basis for J(Z(FG)).
-
ICil*l
are all
(1
i G r)
113
ELEMENTARY PROPERTEIS
proof.
owing to proposition 2.10.8(iv),
be the restriction to
ZWG)
Z(FG)
is a local ring.
homomorphism and hence J (Z(FG)) = Kera.
The elements C!
are linearly independent and have augmentation zero. has dimension r
and is contained in J(Z(FG)).
therefore a consequence of the fact dimZ(FG)
=
Let G = S4
and let p = 2.
- I Ci I -1, 1
is p-constrained and
=
c,
Then C , C ,cz,Cu
(12)(34) and denote by X
G
r,
i
.
r+l.
Put x = (12341, y = (1231, 2 = (12) and u
classes of
is a
TI
The desired conclusion is
Then G
the corresponding conjugacy classes.
Then
Hence their linear span
F Example.
F.
FG+
of the augmentation map
Let
0 ,(G)=l.
P
and x' Y are all conjugacy C
C C ,,
Y
distinct from 1 and
Hence the elements
c; , c+ , c,' , c; + Y
constitute an P-basis for J ( Z ( F G 1 )
.
We next examine the behaviour of J ( F G ) Let E
be a field and K
considered as an extension of
sepambZe over K
1
under extension of ground field.
a subfield; we write
K.
A polynomial f
for the field E
E/K
over a field K
if it is a product of irreducible polynomials in K [ X l ,
of which have only simple roots (in a splitting field of f). is separable over K
E/K
nomial over K
A
field K
if it is algebraic over K
is separable.
every element of E
is called
An extension E/K
all
An element of
and if its minimal poly-
is a s e p a m b l e extension if
is separable over K .
is said to be p e r f e c t if either charK
=
0
or
charK = p > 0
and
K p = K. Let A
denote a finite-dimensional algebra over a field K .
is said to be separable if
VE
An A-module
is a completely reducible A -module for every E
V
114
CHAPTER 111
field extension E
of K .
Thus a separable module is completely reducible; the
converse need not be true as we shall see below. For convenience of reference, we now quote the following two standard facts: 1.14. PROPOSITION. extension of K Proof.
field K
A
is perfect if and only if every algebraic
is separable. See Jacobson(l964)
1.15. PROPOSITION.
v
Let
.
.
be a completely reducible A-module. V
separable if and only if for every irreducible submodule k! of the division algebra End(W)
separable. of K ,
A
is
is separable if for every field extension E
is a semisimple E-algebra.
1.16. PROPOSITION.
(i) A/J(AI
The following conditions are equivalent:
is separable
(iil For every field extension E / K ,
J(AE)
(iii) For every field extension E / K ,
(A/J(A))E s i AE/sr(AE)
In particular, if A/JU)
=
J(A)E
is separable, then so is AE/J(AE)
extension E / K . Proof. Direct consequence of Lemma 1.10.1 (iii),(iv). Let A say that
the centre of
is said to be separabZe if the left regular module AA
Expressed otherwise,
AE
is
.
.
See Bourbaki (1958)
The algebra A
v
is a finite separable field extension of F.
A Proof.
Then
A
be an algebra over a field K is definabZe over L
for all x ,x E X, 2 x 1
2
1
Expressed otherwise, A
for any field
.
and let L be a subfield of K .
if there exists a K-basis X of A
We
such that
is an L-linear combination of the elements of X. 2
is definable over L
if A
K@Al L
for some L-algebra
Al.
1.17. PROPOSITION.
Let
A be an algebra over a field K and assume that A is
definable over a perfect subfield of K . proof. subfield L
By hypothesis, A of K.
2
K @A L 1
Then A/J(A)
is separable
for some L-algebra A l and some perfect
BY Proposition 1.16, we need only prove that
A1/Jul)is a
115
ELEMENTARY PROPERTLES
The latter being a consequence of Propositions 1.14 and
separable L-algebra.
1.15, the result follows.
Let K
1.18. COROLLRRY.
and
be an arbitrary field.
Z(KG)/J(Z(KG)) are separable.
Then the K-algebras KG/J(KG)
In particular, for every field extension E
of K ,
be the prime subfield of K .
Let L
Proof.
KG
1
K 8 LG
and
L
Now apply Propositions 1.17 and 1.16.
Then for any field extension L
A/J(A) is separable. cible A-module
module.
=
=
of K
and any irredu-
and, in particular, dimP(V) K
=
dimP(VL)
L
is a projective A-module, P(V)L is a projective A L-
By Lemma l.lO.l(v1,
0.
is separable, VL
is a completely reducible A L-module and so
Applying Lemma l.lO.l(vi) and Corollary 1.6.5, we conclude that
.
J(P(V)lL = J(P(VIL).
VL.
P('vlL
Since P ( V 1
Since A/JL4)
JCV,)
and assume that
V,
P(VL) Proof.
is perfect and
Z (KG) _1 K 8 Z(LG) L
be an algebra over a field K
Let A
1.19. PROPOSITION.
.
Then L
1.20. PROPOSITION.
Hence, by Theorem 1.9.3,
P(WL
Let G be p-constrained, let N
e
c
= 1~l-l
is a projective cover of
= 0
P
,(G) and let
z
&N Then FGe
2
F(G/N) is the principal block of FG.
Proof.
FGe
2
F(G/N).
By Proposition 2.1.9, e Since G
is a central idempotent of FG
is p-constrained, so is
5=
G/N.
But 0
P'
such that
(z) = 1 and
116
C W T E R III
hence, by Proposition 1.12,
.
Finally, since e
block idempotent of FG.
module, the result follows.
This proves that e
FE is indecompsable.
is a
acts as identity on the trivial FG-
2. DIRECT PRODUCTS.
denotes a finite group and F a field of character-
Throughout this section, G istic p > 0.
Our
aim is to investigate JCFG)
for the case G = G
1
x G
for
2
some G . 4 G, i = 1,2. and B
Let A
be (finite-dimensional) algebras over a field K . and B
customary, we shall identify the algebras A in A 8 B.
18 B
and
is
As
with their images A 8 1
With these identifications, A
and B
become commuting
K subalgebras which generate A 8 B .
K
V and W be two vector spaces over K.
Let
Given f E End(V), g E End(W),
K one can define an element f 8 g
of End(V 8 W 1
K
by setting
K
If f
:
A
-f
End(V1
and g : B
3.
K
End(Wl K
A @ B+
are algebra homomorphisms, then
EndCV 8 W)
K
K
a 8b is also an algebra homomorphism.
I+
f(a) 8 g(b)
This means that if
modules, respectively, then 8 8 W
and
A
and B-
is an A @ B-module by means of
K
Let G
W are
V and
K
H be groups and let V and W be KG and KH-modules, respectively.
Then the outer tensor product underlying vector space is
V# W
of
V and W is a K ( G
X
H)-module whose
V @ W with the module operation given by K
On the other hand, as we have seen above, the vector space
V @ W is an KG K
@
K
KH-
117
DIRECT PRODUCTS
module by means of
Hence if we identify KG 8 KH
with
K(G
X
H)
(see Proposition 2.1.6),
K V
nothing else but the outer tensor product of
2.1. THEOREM.
Let
irreducible
A
is
W.
A and B be algebras over an algebraically closed field K ,
V1,..., Vr and W l , W2,...,Wt are all nonisomorphic and B-modules, respectively, then Vi 8 w 1 G i G P , l j t,
and let C = A 8 B .
K
and
L'8 W K
If
i'
are all nonisomorphic irreducible C-modules. Let fi
A
End(Vi) and g j : B + End(W.) be the homomorphisms K K 3 associated with the modules Vi and W respectively. It is a consequence of Proof.
:
-+
j'
Corollary 1.10.4 that each fi and g
is surjective; hence so is fi 8 g j
j
Vi 8 W
is an irreducible C-module. K j Assume now that U is an irreducible C-module and that $
therefore
and
the homomorphism associated with
U.
Because the C-module U
sfonal, we may choose an irreducible A-submodule A-homomorphisms from
V
to
U.
Then L # 0
V.
Let L
: C+
End(U) is K is finite-dimen-
be the space of a l l
and it is easy to verify that L
is a B-module under the action
Let W be an irreducible B-submodule of L $ : V 8
and let
w 4u
K be the F-linear map defined by
$ ( v 8 W ) = w ( v ) ; the map
$
is in fact a C-homo-
morphism since for all a E A,b E B,v E V,W
E
Now U
V 8 W is irreducible by the foregoing. K
is irreducible, by hypothesis, and
W,
CHAPTER I11
118
Since 0 # 0, we infer that U
V 8 W. K We are therefore left to verify that the isomorphism class of
Vi 8
K mines the isomorphism classes of both Vi Vi 8 W
A-submodule of
and obviously, M.
2
j. bMi.
and
For any b E B,bMi
=
K it follows that as an A-module, V; 8 W
K
W
Let Mi be an irreducible j. is also an irreducible A-submodule
C bMi e B
,
is a direct sum of copies of
j
Thus Vi 8 pi'j K A similar argument shows that Vi 8
from the mull-Schmidt theorem that
Vi.
V. 2
Y
Mi.
K isomorphism class of 2.2. COROLLARY.
W j'
Let K
so the theorem is true.
Wl,Wz,...,Wt
Vi 8 K K j ducible KG-modules.
2.1.61
GI and G2 and let V 1 ,V 2 ,...,Vr
(1
that the map g g
I-+
Let
r, 1
j
tl
and KG -modules, respectively
are all nonisomorphfc irre-
B
I be a nilpotent ideal of a ring R.
By the nizpotency index of I we
understand the least positive integer n such that write
t(G)
and
gl 8 g2,g1 E G1,g2 E G2, induces an isomorphism
KG1 8 KG2 of K-algebras. K
-+
i
be the
Direct consequence of Theorem 2.1 and the fact (see Proposition 1 2
KG
determines the
j
9
be all nonisomorphic irreducible KG1
Furthermore, the
Proof.
W
be an algebraically closed field, let G
(internal) direct product of
Mi.
V. 8 W we deduce z K j' determines the
is identifiable with an irreducible A-submodule of
isomorphism class of
deter-
j
Because
Vi 8 W.
Since Vi
W
for the nilpotency index of J ( F G ) .
f = 0.
In what follows, we
For the sake of convenience, we
Put = FG
We are now ready to provide information on J ( F ( G l x G 2 ) ) .
Since the
DIRECT PRODUCTS
119
corresponding r e s u l t e n t a i l s a f a i r amount of n o t a t i o n , it w i l l be e s p e c i a l l y u s e f u l t o pause t o assemble most of it i n one p l a c e .
G
i s t h e ( i n t e r n a l ) d i r e c t product of
s
i s a f i x e d i n t e g e r such t h a t
T . = {a. .I 1 J(FG )’
z
modulo
J(FGI)
i+l i+l
1G s
t(G)
i s a s u b s e t of
,0
i
G
C
s
G2.
and
J[FG1Ii
which forms an F-basis of
J(FG1)’ = 0 ,
CIf
then we s e t
T. =
0
ri = 0).
and
Ts
1
j G r .-r
23
2
Gl
{askll Q k < r s }
=
set
Ti
T s = k?~ =
{aiRI1 G ?! G
FG
for
r
and
J(FG1ls ( I f J(FGIIS
i s an F-basis of
0, then we
=
= 01
ri-r{+l
Ti = {ai,ll G
and
.f
a r e defined s i m i l a r l y
Pi}
.
2 . 3 . THEOREM (Loncour (19711, Motose(l974b)).
With t h e n o t a t i o n above, t h e
following p r o p e r t i e s hold: S
-
C
(i) dimJ(FG)s =
i=o
F (ii) J(FG)‘
=
J(FGl)iJ(FGz)s-i
+ JCFG
-
tG) = t(C1) + t(G21
(iv) X =
C r.rfz s i+l i=1 and, i n p a r t i c u l a r ,
i=o
JCFG) = J(FG )FG2 (iii)
S
U 2‘27‘’ i+k>s
2
)FG
1
1
form an F-basis f o r
S[FGlS,
TiTi = {tt I t Proof.
C i ) and ( i i ) .
a l g e b r a i c c l o s u r e of
of a l l i r r e d u c i b l e
{Ui 8 V.11 3
< i < a,
EG
and
16
E Ti,t
{Uil 1 6 i G a } and
j < bl
1
Let
bl
1 1
0
-
i=1
By Corollary 2 . 2 ,
b
+ ro(rA- Z
(dimVi)’) i=1 L
L
a
b
(ro- C
(rL- C
(dimUi) i=1 L
(dirnViI2)
i=l
E be an
be t h e s e t
i s t h e s e t of a l l i r r e d u c i b l e LG-modules. U
0
s = 1.
{V.ll 6 j 3
EG -modules, r e s p e c t i v e l y .
r r r + r21’-r r r = rr(ro-C ( d i d i ) ’ ) 1 0
E Ti}
W e f i r s t consider t h e case where and l e t
F
where
L
Thus
CHAPTER I11
120
a
b
L
(by Corollary 1.17)
= dimJ(FG)
F which proves (i) for the case s = 1. To prove ( i i ) ,observe that
It therefore follows that dim(J(FG )FG
F
l
2
+ J(FG 2 )FG1 )
= did(FG1)FG2 + did(FG2)FGl
F
-
F
dim(J(FGl)FG2 n J(FG2)FG1)
F
= r r f+ r r ' - r r ' 1 0
0
1 1
1
= dimJ(FG)
F Since J(FG1)FG2 + J(FG2)FGI
J(FG)
JVG) which proves (ii) for the case
=
,
we conclude that
J(FG lFG2 + J(FG2)FG
8 =
1
1.
Turning to the general case, we have
proving t i i ) .
The property (i) now follows from (ii) upon noting that
.
(iii) Direct consequence of (ii) (iv) Since
1x1
= di~d(FG)~,we need only verify that
F
X
i s linearly independent
DTRECT PRODUCTS
over F.
121
To this end, assume that
where the Xijka
are elements of F.
Setting
it follows that
and that
Now 1~~
=
-B
n J(FG2)FG1 = S(FG11"J(FG2)
E J(FG1)'FGz
and the map
is an isomorphism.
Hence all the coefficients of p
are zero and thus p l = - B 2
is an element of
Invoking the natural isomorphism
-
we conclude that all coefficients of p
.
procedure, we obtain eventually X i j k a proof of the theorem. 2.4.
COROLLARY.
are zero. =
Repeating the above
0 for all i , j , k , ! L .
For any groups G1 tG2,** * tGk
r
This completes the
122
CHAPTER I11
t(G
Proof.
i
G
... x
x
2
Let G
k.
=
Then
z
Gk)
k C t(Gi) - (k-1)
=
i=1
k and Theorem 2.3(iii).
Apply induction on
2.5. COROLLARY.
ni, 1
1
x G
x
z
... x z
x
k ni t(G) =
for some positive integers
nk
P
P 2
P l
9
-
C p
Ck-1)
i=1 Proof.
Let G = < g > be a cyclic group of order p
n
.
Owing to Corollary
2.4, we need only verify that t(G) =
By Proposition 1.1, J ( F G ) = IIG)
i elements g -1, 1
i c p".
=
FG(g-11.
and so J ( P G )
is the F-linear span of all
Since
gi-1 JPGI
pn
=
(g-1) u+g+. * . +gi-1 )
The desired conclusion is therefore a consequence of the
equality
applied for k G n .
.
k g p -1
k =
Ig-lP
3 . A CHARACTERIZATION OF ELEMENTS OF
J ( F G ) : THE GENERAL CASE
Let G be a finite group and let F
be an arbitrary field of characteristic
p > 0.
The problem that motivates this section is to discover necessary and
sufficient conditions for an element z
=
zz
certain additional hypotheses le.g. when G
99 E FG
to be in J ( F G ) .
Under
is p-solvable), these conditions will
be simplified in Sec. 5. To accomplish our aim, we shall have to set up some machinery pertaining to the theory of Frobenius and symmetric algebras.
Many of the results presented
below will only be needed in subsequent investigations.
We present them here in
order not to interrupt future discussions at an awkward stage.
Our method
n CHRRACTERIZATION OF ELEMENTS OF J ( F G ~
123
borrows heavily from an important paper due to ~ulshammer(1981a). Throughout this section, unless explicitely stated otherwise, A finite-dimensional algebra over an arbitrary field F
denotes a
and A* = H o m ( A , F ) .
The
F left (right) A-module A Given a subset X
means the regular left (right) A-module.
of A ,
define A X
and XA
A X = {Caizilai E A , Z ~E
by
XI
XA = {Cr.a.Iai E A,zi E XI 2 2
If
X = {a)
tively. all a E A
(A,$)
that the pair pair
(iii) a
(ii) J,(Aa) = 0
0
=
We shall refer to A
A
and XA,
respec-
the following conditions are equivalent:
$(ad)
(1)
instead of A X
is said to be a Frobenius d g e b r a if for
where $ E A * ,
(A,$),
pair
A
and aA
then we shall write A a
(A,$),
=
0
itself as a Frobenius algebra if there exists J, E A*
such
is a Frobenius algebra. where
is said to be a symetric aZgebra if
$ E A*,
(A,$)
is a Frobenius algebra such that for all r , y E A
J,(zy) = $ ( p )
The algebra A (A,$)
is called a symmetric dgebra if t ?re exists
E A*
such that
is a symmetric algebra.
Let f
: A
x A+
F
be a bilinear form.
Then f
is called nonsingukm if
the following conditions are equivalent. (i) f ( a , A )
=
(ii) f ( A , a ) = 0
0
We shall refer to f
(iii) a = 0
as being associative if for all z,y,Z E A
and symmetric if
f(2,Y)
3.1. LEMMA.
(i) Assume that
(A,$)
=
f(y,z)
i s a Frobenius algebra.
for all z,y E A
Then the map
CHAPTER I11
124
f
:
A x
form.
defined by Q(x,y) = $(.y) is a nonsingular associative F-bilinear
F
A -+
Furthermore, if
(ii) Assume that f : A x and let $ : A+
F
is a symmetric algebra, then f
(A,$)
A-
F is a nonsingular associative F-bilinear form $(x)
be defined by
Furthermore, if f
algebra.
= f(2,l).
is symmetric then
(i) For any x , y , z E A ,
Proof.
=
AfCcc,y).
Suppose that f @ , y ) = 0 I=
0.
A
Then (A,$)
is associative.
is a symmetric algebra.
Given h E F,
A.
we also have
Then $(d) = 0
similar argument shows that if f b , y ) = 0
proving that f
is a Frobenius
Hence f is an F-bilinear form on A.
for all y E
proving that f is nonsingular.
(A,$)
we have
and similarly f(rc,y+z) = f h , y ) + f & , z ) .
and similarly f ( z , h y )
is also symmetric.
and therefore
for all x E A,
then y = O ,
For all x,y,Z E A, we have
Because $key) = $(yr)
implies f ( z , y ) = f ( y , x ) ,
the assertion follows. Lii) It is clear that $ E A * .
f(ab,l)= 0 of f a = 0,
for all b E
A.
implies that a = 0 . proving that
@,$I
Let a E A
Then
However, f(ab,l) = f ( a , b ) , so the nonsingularity A similar argument shows that if $(/la) = 0 is a Frobenius algebra.
for all x,y E A
as asserted. Note that A*
be such that $(d) = 0.
is an (A,Al-bimodule via
If
then
f is symmetric, then
A CHARACTERIZ?iTION OF ELEMENTS OF
for a l l
x,a E A
3 . 2 . LEMMA.
f E A*.
and
(i) Assume t h a t
i s a Frobenius algebra.
(A,$)
f defined by modules.
f ( a ) (3)
125
J(&'G)
$(xu) f o r a l l
=
Furthermore, i f
:
Then t h e map
A-A* i s an isomorphism of l e f t A-
Z,a E A ,
is a symmetric a l g e b r a , then
(A,$)
f
i s an iso-
morphism of ( A , A ) -bimodules.
(iil
f
Assume t h a t
$(a1 = f t l ) ( a ) .
:
A+ (A,$)
Then
i s an isomorphism of l e f t A-modules and s e t
A*
is a Frobenius algebra.
(A,$)
isomorphism of (A,A)-bimodules, then
(i) I t is obvious t h a t f o r a l l
Proof.
for a l l
x,y E A .
proving t h a t
f
f
If
r,x E A ,
a E A , f ( a ) E A*
dh4 = did*
,f
By t h e d e f i n i t i o n of
$,
is a b i j e c t i o n .
F
i s a symmetric a l g e b r a .
(A,$)
and t h a t
then
F Assume that
f i s an
i s a symmetric a l g e b r a .
i s a homomorphism of l e f t A-modules.
i s i n j e c t i v e and s i n c e
Furthermore, i f
Then, f o r a l l
r,x E A,
w e have
a s desired.
$ E A*.
(ii) I t is c l e a r t h a t
f o r all
x E
f o r all
x
X and hence a
E A,
Assume t h a t
w e have
then
f
a
= 0.
=
Suppose t h a t
0.
$(ax)
= 0
for a l l
z E A.
A similar argument proves t h a t i f
Hence
(A,$)
Then
$(xu) = 0
i s a Frobenius algebra.
is an isomorphim of (A,A)-bimodules.
Then, f o r a l l
x,y E A ,
126
CHRPTER 1 x 1
.
as required.
3 . 3 . COROLLARY.
The following conditions are equivalent:
A is a Frobenius (respectively, symmetric) algebra
(i)
(ii) There exists a nonsingular associative F-bilinear form f :
A
x
A+
F
(respectively, there exists a nonsingular, associative, symmetric F-bilinear form
f
:
A x A+
F)
(iifl A
A*
A
as left modules (respectively,
Y
A*
.
as (A,A)-bimodules)
Direct consequence of Lemmas 3.1 and 3.2.
Proof. 3.4. LEMMA.
Let
A ,Az,...,An be F-algebras.
n nAi
Then
is a Frobenius
i=1
(respectively, symmetric) algebra if and only if each
Ai
is a Frobenius
(respectively, symmetric1 algebra. Suppose that $ . f5 A? is such that ( A i , q i ) w z z and define $ : AF by put 4 =
Proof.
is a Frobenius algebra
nAi
i 6 i G n.
i=1
Then
is obviously a Frobenius algebra.
(A,$)
Frobenius algebra, then so is of $
onto Ai.
(Ai,$il,
1
Conversely, if
i G n, with qi
Finally, for a = Ca ,...,an) and b = (b
if and only if $i(uib2) = $icbz&i)
for all
i E 11,.
..,n}.
(A,$)
is a
being the projection
,...,bn),$(ab)=$(ba) Thus the lemma is
proved, We have now accumulated all the information necessary to prove the following important result. 3.5. THEOREM. (Eilenberg and Nakayma (1955)).
If
(i1
(iil
If
A
1
and
A
'F
A is a semisimple F-algebra, then A is symmetric
Proof.
(i) Suppose that
A
1
and
A
are Frobenius F-algebras.
corollary 3 . 3 , there exists an isomorphism fi :
i
= 1,2.
@A
are Frobenius (or symmetric) F-algebras, then so is A=A
Consider the map
f
:
Ai-
A d A* defined by
Then, by
Atz of left Ai-modules,
f
A routine calculation shows that
over, if each
is an isomorphism of left A-modules.
fi is an (Ai,Ail-isomorphism, then f
More-
is an (A,A)-isomorphism.
The desired assertion is now a consequence of Corollary 3.3. (ii) We first assume that A
=
trace map, we conclude that A
Mn (F) for some n 2 1. If D
is symmetric.
Then, by looking at the is a division algebra over
F, then
Invoking Ctl and Lemma 1.4, we may therefore assume that A
J,
Then, for any nonzero
U,$I
in A * , ( A , $ )
is a Frobenius algebra.
J,
in A*.
To prove that
J,(IA,AI) = 0
is a symmetric algebra, it suffices to verify that
nonzero
is a division algebra.
The latter will follow provided we show that
for some
[A,Al # A.
The last statement being independent of the qroundfield, we may assume that F the centre o f Let A
E
E
we have
A. F which is a splitting field for AE.
be a field extension of
Mn ( E l for
is
n 2 1.
some
IAE,AE]
# AE.
For a given subset
Since each matrix in
[AE,AEI = [ A , A I E ,
But
X of A ,
we denote by
so
Then
[Mn&) ,MnlE)l has trace 0,
IA,AI #
A,
as required.
RCY) and r(X) the left and
right annihilators of X defined by
rCYl
It is obvious that
E(X)
ly.
X is a left (respectively, right) ideal of A,
Moreover, if
and
are left and right ideals of
(A,$)
Define the subsets
'X
is a Frobenius algebra and that X and 'X
'X
X It is plain that if
1
of
A
E
then
A($(Xd
X is a subspace of A,
&(XI
is a subset of
by
= {a E A I J 1 W ) = =
respective-
A.
(respectively, r ( X ) ) is a two-sided ideal of Suppose that
A,
=
0) 01
then so are
'X
and
X
L
.
A.
CHRPTER I11
128
3 . 6 . THEOREM.
of
(A,$)
Let
be a Frobenius a l g e b r a and l e t
X
be an F-subspace
A.
(i) (IX)'
(X',
c
=
X
1 dimXl = dim X = d i d - d i d
and
F
X
(ii) If
F
F
i s a l e f t ( r e s p e c t i v e l y , r i g h t ) i d e a l of
R(X) =
(respectively,
I
,
= (1-e)A
(Ae)'
F
x).
A,
then
r(X) = X
I n p a r t i c u l a r , f o r every idempotent
A/(Ae)'
3
eA
X
dimX F If
X
(iv)
A,
i s a l e f t i d e a l of
+
F
X
The mapping
then
dlmr(X) = dim.4
F
i s a r i g h t i d e a l of
A,
A.
of
a s r i g h t A-modules and dimeA = d i d e .
F (iii) I f
e
1
&@(XI) = X
and
F then
k rCY) i s a d u a l i t y of t h e l a t t i c e of l e f t i d e a l s of
A
onto t h e l a t t i c e of r i g h t i d e a l s , t h a t i s
X1,Xp
for a l l l e f t ideals Similarly,
X
n R(X)
of
is a d u a l i t y of t h e l a t t i c e of r i g h t i d e a l s of
t h e l a t t i c e of l e f t i d e a l s of (v)
If
(A,$)
If
(A,$)
Cartan matrix of Proof.
X
of
onto
A.
A.
F
i s symmetric and
A
X.
is a s p l i t t i n g f i e l d for
A,
then t h e
i s symmetric.
(i) Fix an F-basis
is an F-basis of
A
is symmetric, then
f o r any two-sided i d e a l (vil
A.
The map
...,an 1
{al ,
of
A
such t h a t
~U1,...,U
m
},m
n
J(FG)
A CHARACTERIZATION OF ELEMENTS OF
XI consists
i s an F-isomorphism such t h a t t h e image of
A
1
=
1 = 2
... = 1rn = 0.
1
Hence
d i d F I t follows t h a t
I
dim X = d i d - d i d . F F F
did-dimX F F
=
129
of a l l
(Al,.
../In)
with
and, by a similar argument,
dim?'X)l = d i d F F
1 1 t X) 3 X, we i n f e r t h a t
Because
tX)'
=
X,
as required.
The second e q u a l i t y
i s proved s i m i l a r l y .
X is
(ii) Suppose t h a t
$(ax1 = 0 or i f and o n l y if $CaXAl = 0. 0,
R(X1 = 'X.
to
aX
=
of
A
follows by the same argument.
we have
x
E (&)'
Then
A.
a r i g h t i d e a l of
i f and only i f
Since the l a t t e r e q u a l i t y i s equivalent
The e q u a l i t y
= 0
P(X) =
for all
NOW
* Aex
a E 'X
* ex
= 0
1
X
for a l e f t ideal
X
x e A.
*x
E (l-e)A,
so
Ue)' = Il-e)A dfmeA = d i d e ,
Finally
by applying (i)
F
F
and
bii) Direct consequence of (i) and
A/&?)'
eA
.
ui).
It is clear t h a t
(iv)
Invoking C i i i ) , w e conclude t h a t
The correspondkng a s s e r t i o n f o r l e f t i d e a l s follows s i m i l a r l y .
(v)
u,$) i s a
A s s u m e that
X of
A.
,...,en
Let
e
let
C = Cc..)
1
Then
'X
=
XI f o r any s u b s e t
Now apply (ii).
Assume t h a t
(vil
symmetric algebra.
ZJ
(A,$)
is symmetric and that F i s a s p l i t t i n g f i e l d f o r
be a complete s e t of orthogonal p r i m i t i v e idempotents of be t h e Cartan matrix of
A.
Then
A
A. and
CHRPTER 111
130
n
n @ezJl
A = @Ae.= i=1 z
and therefore
n
A = Fix j,k E 11
,...,n}
with j # k .
@
(direct sum of F-spaces)
eAe
i,j=l 3 ; Then
and so, applying (i)and Proposition 1.10.16(1),
c.. = dime .A@ = d i d 3% p 2 j F
we have
- dim(e A e .) I F
z
3
A similar argument shows that c . . G c . . and the result follows. $3 3% We next concentrate on factor algebras and prove the following result. Let X be an ideal of A.
3.7. THEOREM (Nakayama(1939) I .
(il
Assume that
(A,$)
i s a Frobenius algebra.
algebra if and only if there exists a E (iil
Assume that
(A,$)
Then A/X
A such that rlX)
i s a synnnetric algebra.
is a Frobenius =
aA = Aa.
Then A/X
i s a synnnetric
algebra i f and only if there exists a E z(A) such that P(X) = ud. Proof. II : A
3
Ci)
A/X
Suppose that
(A/X,l.l) i s a Frobenius algebra, and let
be the natural map.
exists a E A
such that PIT = $,
show that r(X) = aA = Aa.
Then, by Lemma 3.2Ci) and Corollary 3.3, there where $a(b) = $(ba)
b
E
RCAa).
Thus
b
E
.
Since Jla(X) = $(Xu) = 0 ,
fore ad C_ X1 and, by Theorem 3.6Cii),
aA
5 r(X).
for all b E A . 1 we have a E X
Suppose that
b
We shall There-
€ &(&)
or
Then ba = 0 and hence
X and so both C(u.4)
we infer that
aA
=
and
~ ( A u ) are in X.
r(R(aA)l 2 rU1, proving that rlX)
Applying Theorem 3.6(iii)r =
ad.
Consequently,
A CHARACPERIZATION OF ELEMENTS OF J ( F G )
Aa
5 r(X)
and therefore LUa)
3.X.
It follows that
X
=
131
RUd
and thus
r ( X ) = rR(Aa) = Aa. Conversely, assume that r ( X ) = a4 = A a
Because a E r ( X ) , p
But r(X) =
b E k(u.4).
Suppose that
Then $(Aha) = 0, so ba = 0 and thus
aA implies that X
hence b E X.
= k ( r ( X )) = k ( d ) ;
A similar argument shows that p (Cb+X)( A / X ) 1 = 0
A/X
Define
is well defined and obviously an F-linear map.
is such that V((A/X) (b+X)) = 0.
bE A
for some a E A .
implies b E X, proving that
is a Frobenius algebra.
(ii) Suppose that A / X
By ti), r C X ) = aA
is symmetric.
for some a E A .
On
the other hand,
for all r , y E A .
for all y E A ,
Hence ay = y a
.
Conversely, suppose that a E Z U ) metric.
and so
Then, by Cil
,
a E Z(A).
A/X
is obviously sym-
So the theorem is proved.
We next exhibit some important properties of socles and commutator subspaces of Frobenius and symmetric algebras.
First, however, it will be convenient to
recall the following piece of information, Then the socle of
V,
defined as the sum of all irreducible submodules of
V.
Let
V
be a left A-module.
module is defined similarly. socles of AA
and A A ,
3 . 8 . THEOREM.
SocA
Assume that A
are defined as the
In case the left and right socles of A
to denote both of them. i s a Frobenius algebra.
ti) The left and right socles of A
SocV, is
The socle of a right A-
The left and right socles of A
respectively.
coincides, we shall write
written
coincide.
Furthermore,
132
CHAPTER 111
dim s o d = dim(A/JCA)) F F (ii) For each primitive idempotent e E A,(socA)e suhnodule of Ae. (iii) If A
In particular, soc(Ae) = (SocA)e
is symmetric, then such that SocA = Az
(a) There exists z E Z U )
(c) For any irreducible A-module
socle of A.
Let e
1
Ae/J(A)e
V, V is isomorphic to a submodule of P ( W .
Proof. (i) and (ii): Setting J = J ( A ) , is the left socle of
and J(A) = L(Z)
A, soc(le)
(b) For any primitive idempotent e E
that r ( J 1
is the unique irreducible
A.
it follows from Proposition 1.6.29
By a similar argument, !.(J) Then Je
be a primitive idempotent of A .
By Theorem 3 . 6 , the map X C +
A containing
and the set of right tdeals of
inclusion-reversing bijection
X W X'/(Ae)l
and the set of submodules of A/(Ae) inclusion-reversing bijection
eA.
which implies that e(Je)'
.
XI-+ ex'
and the set of submodules of eA. ducible submodule of
.t
er(J).
is an inclu-
1 CAe)
.
we thus obtain an
between the set of submodules of Ae
Again, applying Theorem 3 . 6 , we obtain an between the set of submodules X of Ae
In particular, e(Je)'
Further, for y E A ,
=
'X
X of A contained in
sion-reversing bijection between the set of left ideals Ae,
is the unique
i s an irreducible module
maximal submodule of Ae, and the quotient Ae/Je (Propositions 1.9.4 and 1 . 6 . 3 6 1 .
is the right
Thus er(JI
is the unique irre-
we have
is the unique irreducible sub-
module of eA. Because J rCJ)J
annihilates all irreducible A-modules, we have e r ( J ) J = 0 .
is annihilated by all primitive idempotents e E
that r ( J ) J = 0 , symmetry.
and therefore r ( J )
5R(J).
A.
Thus
The latter implies
The reverse inclusion holds by
The furthermore assertion being a consequence of Theorem 3 . 6 ( i ) ,
property ti) is established.
The same argument (reversing left and right) also
proves (ii).
(iii) Note that Ca) is a consequence of Theorems 3.5(ii)
, 3.6(iii)
and 3.7111).
A CHARACTERIZATION OF ELEMENTS OF J ( F G )
To prove (b), let $ E A*
be such that
133
$s a syrmnetric algebra.
[A,$)
foregoing, r ( J ) e is the unique irreducible submodule of Ae
By the
and J r ( J ) e = 0.
Hence every f E HomCAe,r(J)e) induces a map A E Hom(Ae/Je,r ( J ) e ) . Since A A Hom(Ae/Je , r ( J ) e ) 2 e r ( J ) e as additive groups (Lemma 1.5.5), we are left to A
verify that e r ( J ) e # 0 . Assume by way of contradiction that e r ( J ) e = 0.
and hence r ( J ) e = 0 ,
which is impossible.
To prove (c), observe that V idempotent e
This proves (b). and P ( V )
Ae/JCA)e
of A(propos$tion 1.9.4)
.
Then
By (b), V
Ae Socue)
for some primitive
5 Ae
and the
result follows. As a preliminary to the next result, we prove 3.9.
LEMMA.
(i) Assume that
[A,$)
is a Frobenius algebra.
is the largest left ideal of A
set X of A, E ( X )
contained in ' X
is a symmetric algebra. Then Z ( A ) 1 1 Proof. (i) By the definition of X, we have $(XI 5 x .
(ill Assume that
left ideal of A
so
IX
=
(A,$)
contained in
0 and thus
I5
'x.
Then, for any sub-
=
IA,Al
1
Let 1 be any
Then
?,(A'), as required.
(ii) For any z E A, z E [A,A]'
if and only if for all z,y E A
or if and only if
Since the latter is equivalent to yz-zy = 0 for all y E A, established.
Fran now on, we assume that c h F
=
p
3
0
and s e t
the result is
CHARACTER I11
134
TU) is
Then
A
an F-subspace of
containing
we a r e now ready t o prove
[A,AI.
t h e following important r e s u l t .
3.10. THEOREM (Kbhammer (1981all
. With
the n o t a t i o n above, the following
p r o p e r t i e s hold: (i) T ( A ) = [A,AJ
+ SU)
and
JU) is
t h e l a r g e s t l e f t i d e a l of
A
contained i n
TU).
(iil
A
If
A
(iii) If
T(A)
(a)
1
(bl S o d
= ZU)
('2 (A)
=
A
is a symmetric a l g e b r a , then
I
(TU)1'
(c)
1
S o d = T(A)
is a Frobenius algebra, then
n SocA and, i n p a r t i c u l a r ,
n
TCA)'
i s a n i d e a l of
ZU)
socA)A
= ZcB1l @
.. . d Z ( B P ) ,
B 1 ,. ..,By a r e a l l blocks of
where
A
which a r e sfmple F-algebras. (d)
F is a s p l i t t i n g f i e l d f o r A ,
If
a r e simple a l g e b r a s is equal t o
then t h e number of b l a c k s of
A
that
dim(T(Al')2
P (i1
Proof. any
ZU)
z E
both
[A,A]
W e f i r s t treat t h e case where
and
and
E
is a simple algebra.
T(A)
can be regarded a s a l g e b r a s over t h e f i e l d
Z U 1 = F,
i.e. that
be t h e a l g e b r a i c c l o s u r e of
F
A
Mn(E)
F
n P
1.
Then c l e a r l y
TUE)= IAE,AEl
and
1
Because
dim[A,Al = d%IAE,AEl F
Z(A).
We may
is a c e n t r a l simple F-algebra.
so t h a t
AE = E @ A f o r some
Since f o r
x,y E A,
t h e r e f o r e assume t h a t Let
A
and d i m A = dim AE
F
E
135
,
then since dim[A,A] = d i m A - 1. If it were t r u e t h a t A = T ( A ) F F i s an E-subspace of AE w e would have AE = T [ A E ) , a c o n t r a d i c t i o n .
we i n f e r t h a t
T(A )
E
Hence A # T ( A 1 (A,$)
[A,A]
and t h e r e f o r e , by t h e above,
Ker$
and hence
proves the case where
T(A)
A
c o n t a i n s no nonzero l e f t i d e a l s of
H = A
1
Ai
< i < m,
A' 1
=
i s simple,
But then
A.
This
and write
A/J(A)
@...@Am
are pairwise orthogonal simple F-algebras.
Then,
T ( Az. )
=
[Ai,Ail,
by t h e s p e c i a l c a s e proved above, and t h u s
~ b v i o u s l y IA,AJ
+ J U ~c- T ( A )
and
T U ) / J ( A )5
T G l = T(A)/J(A) and so TlAl = IA,Al Assume t h a t
T
is a l e f t i d e a l of
contained i n
+ JU).
.
A
contained i n
T(A)
Tfl)
and hence i n
[z,x].
i s a l e f t i d e a l of
A'
~ 2 1 . Consequently,
is a symmetric a l g e b r a (Theorem 3 . 5 ( 1 1 1 ) ) .
semisimple, argument,
A
is a simple algebra.
Turning t o the g e n e r a l case, s e t
where t h e
Since
IA,Al.
=
J, E A* (Theorem 3.5(11)1.
is a symmetric a l g e b r a f o r some
C,
!?(A)
Iz,z]
c o n t a i n s no nonzero l e f t i d e a l s of
A'.
Then ( I + J ( A )1 / J ( A ) Since
is
Thus, by t h e previous Hence
I
5 JL 4 )
and
the r e q u i r e d a s s e r t i o n follows.
(ii) A s s u m e t h a t A
By ( i ) and (l),
JUl
Applying Lemma 3 . 9 ( i l ,
is a Frobenius algebra.
Thanks t o Theorem 3.6, w e have
1 i s t h e l a r g e s t l e f t i d e a l of
A
contained i n
(T(A)l)
we conclude t h a t
1
J ( A ) = L(T(A) Invoking (2) and ( 3 ) , it follows t h a t
1
= t(T(A) A)
(3)
C W T E R XI1
136
SocA =
1
r(J(A)I = T ( A ) A ,
a s required.
A
(iii) A s s u m e t h a t
i s a s p e t r i c algebra.
By (ii)and Lemma 3 . 9 ( i i ) / t o
prove (a) w e need only v e r i f y t h a t
Because
1 R(T(A) A )
A,
is a l e f t i d e a l of
we have
Consequently,
Property (b) i s a consequence of ( a ) and (ii), while ( d ) follows
proving ( a ) . from ( c ) .
W e a r e therefore l e f t t o v e r i f y ( c ) .
To prove ( c ) , w e may harmlessly assume t h a t
A
i s indecomposable, i n which
case it s u f f i c e s t o v e r i f y t h a t
J U ) = 0.
Suppose f i r s t t h a t i d e a l of
J ( A ) # 0.
ZU)
, so
Then
TU)'
19 S o d
Then
= ZL4)
and so
1
1 E SocA n Z ( A ) = T ( A )
1 T(A)
5 SocA
1
1 E ?"(A)
is a l o c a l algebra, we have
However, by ( a )
1 IT(A) 3
and thus
and thus
.
*
= Z(A)
.
.
But
TU)'
i s an
Next assume t h a t
Taking i n t o account t h a t
Z(A)
A CHARACTERIZATION OF ELEMENTS OF J ( F G )
as required.
137
.
We have completed the development of the general machinery, and we are confronted with the problem of characterizing the elements of J W G ) , is a field of characteristic p
where F
0.
Define a map
tr
:
FG
F
-+
called the truce, by
The following observation will enable us to take full advantage of the results so far obtained.
CFG,trl
3.11. LEMMA.
Cil
(ii) Let e
be an idemptent of FG
eFGe.
CeFGe,tr'l
Then
Proof.
y
=
Z y#
is a symmetric algebra and let tr'
be the restriction of
tr
to
is a symmetric algebra.
(i) It is clear that tr E HomCFG,F). F i n FG. Then we have
Fix x =
C x# SEG
and
SEG
trm) =
C x
=
$G
C y x -1 = tr Cyx) @Gg g
and for all g E G In particular, if t r ( F G x ) = 0 then x = 0 as required. Uil
Assume that y E e F G e
is such that tr' ( e F G e y ) = 0.
and
prwing that y = 0. For any subset
S o the lemma is true.
X of G , we write
x-1 =
{ s - l l x E XI,
x+ = c XE
x
x
Then y
=
eye = y e
CHAPTER 111
138
and denote by FX the F-linear span of X. Let C = 111, C 2 , 1
section of
G
...,Cr
be all p-regular classes of G .
By a p-regutar
associated Mith Ci, we understand the set
It is an easy consequence of the definition that
r with Si n S. = @ for i # j, each Si is a union of conjugacy (i) G = u S i=1 i 3 is a union of conjugacy classes of p-elements of G. classes of G and S (ii) An element C E Ck(G1
i s a p-regular section if and only if
C
is of p-
defect zero. Let X be a subset of G
3.12. LEMMA.
(i) tr(ux*) =
C u g E FG.
and let u =
SfG
c
and, in particular, tr CuSil =
Proof. By the definition of the product of elements in F G ,
Hence, by the definition of the trace map, tP(uX')
uX+
=
=
E
u -1 =
SEX $ 7
we have
C
u
6 x - 1 9'
Also
0 if and only i f for all t E G Cu
&X
=
tg-1
c
= o
u
FX-1 tg
as required. The following result will enable us to take full advantage of Theorem 3.10(iiil 3.13. LEMMA.
...,Sr
Let S1,S2,
(i) T V G I = { C
x g E FGI C x 9-i
be all p-regular sections of G . =
Then
0 for all i E {l , . . . , r } }
gEG l r (ii) T V G ) = C FS; i=1
Proof.
I*)
ti) We first observe that there exists n 2 1 such that
m gp
for all m
=
9P' Indeed, write / G I = pak with n E N such that pn
E
lhcd k).
Cp,k)
O(mod n ) = 1.
and all g E G
Because
(p,k) = 1, there exists
Replacing n by its multiple, if necessary,
CHARACTERIZATION OF ELEMENTS OF $@‘GI
A
we may assume that n 3 a.
Suppose that m
rn
Then
$
139
is divisible by
n and let g E G.
m
= 1
6, gp,.
and, since p m z lbcd k),
=
Thus
proving ( * ) .
m
x
x# E FG. Then, by ( * I , we may choose m 2 1 such that 9EG m for all g E G and such that x E TVG) if and only if $ E [FG,FG]. Let
=
8
=gpl
Wing to Lemma 2.2.1,
m
and thus x E T V G )
c
if and only if
2.2.2, that x E TWG)
$’“g E [FG,FGI. It follows, from Lemma SEG P’m if and only if C 2 = 0 for all i E (1 ,...,I-}. SEGi g
Since
the required assertion follows.
r (ii) By Lemma 3.12(i)
and (i),
c
FS;
5
1
i= 1 x = 9EG
E T(FG)’
regular section 5‘;’
and let a , b E Si.
.
Conversely, suppose that belong to the p -
Then a-1 and b-l
and hence, by (i), a-l-b-’
E !Z‘VG).
But then
tr(za-l-xb-l) = 0 which ensures that
proving (ii). We are now ready to prove the following important result. 3.14. THEOREM(Brauer(1955), Tsushima(1978al). Let F istic p > 0 and let S ,S2,...,Sr
Proof.
be a field of character-
be all p-regular sections of
By Theorem 3.1Otiii) (a),cb) ,
G.
Then
CHAPTER 1x1
140
r Invoking Lemma 3.13(ii), w e conclude that SocFG =
+
C PG*Si. i=1
It follows from
Proposition 1.6.30 and Theorem 3.8(iii) that
J(FGI
= =
r I ~ C S ~ ~ =I?( F G )c FG-S~) r i=l n I~(FG-s~) i=1
The desired conclusion is now a consequence of the equality k(FG*S;)
+
= R(Si).
’
We have now come to the demonstration for which the present section has been developed. 3.15. COROLLARY.
p
7
Let G be a finite group, let F be a field of characteristic
0 and let S1,S2,...,S r be all p-regular sections of G.
Proof.
Then
Apply Theorem 3.14 and Lemma 3,121ii) together with the fact that
Si I-+ Sii’is a permutation of the set ISl,. ..,S r 1 . 4. CONLON’S THEOREM, FONG‘S DIMENSION FORMULA ZLND RELATED RESULTS
Our main goal in this section is to investigate the decomposition of induced modules from normal subgroups and the dimensions of projective covers of irreducible FGmodules
.
study of S ( F G ) .
As a preparation for the proof of the main theorems, we prove a
We do this for use in the sequel where we will pursue our
number of important results which will be frequently applied at a later stage. Throughout, G denotes a finite group, field. Let
N
a normal subgroup of
G and F a
A l l modules are assumed to be finitely generated.
V be an FN-module and let g E G.
whose underlying space is
V
We define
to be the FN-module
and on which the elements n E N act according to
the rule
We say that two FN-modules V 1
g E G.
and
V 2 are C-oonjugate if
V
’V
for some
It is clear that G-conjugacy is an equivalence relation, so the set of
FN-modules is a disjoint-unionof G-conjugacy classes.
A straightforward
CONLON'S THEOREM
verification shows that V
is irreducible (completely reducible, indecomposable)
.
if and only if ' Y
is irreducible (completely reducible, indecomposable) consider the set H = I g E GIV
Given an FN-module V , subgroup of
G containing N.
be an irreducible FN-submodule of
U and the
VN.
Let H
Then there exists a positive integer e
$7 (i) VN cz e ( 'U @ for H
in G
... 8 g n U ) , 91
and
In particular, VN (ii) W
gn
U,,..,
W
denote, respectively, the
VN
isomorphic to
U.
such that
,...,gn}
is a left transversal
is completely reducible.
is an irreducible FH-module such that
(Here, of course, eU Proof.
V g W
and
e
means a direct sum of
First of all, it is clear that
module isomorphic to g U .
V =
Therefore each gU
is a sum of irreducible FN-modules gU. that "CyV)
Setting
W. = 2
Y
C
wV
for all x,y E G
gU, 1 g i
G
copies of
U).
C gU and that gU is an FNSfG is an irreducible FN-module. Thus
It follows, from the definition of
and so
X
U
yU if and only if zH= yH.
n , we deduce from Proposition 1.3.6, that
9fgp
VN
=
FJ
@
... CB wn
(W
=
and that
W.ze
z
gi i U
for some positive integer e i' Fix g E G , i E {1,2 and
V;
U are pairwise nonisomorphic irreducible FN-modules.
N
gV,
and
11 = g , , g 2
where
W "eU
VN
is a
be an irreducible FG-module and let U
of all submodules of
sum
Then H
is G-invariant.
V
Let V
4.1. THEOREM (Clifford(l937)).
gV}.
2
as the i n e r t i a g r o u p of
We shall refer to H
in case H = G we shall say that
inertia group of
141
s E H.
,...,n } ,
and write ggi = 9 . e 3
Then
gWi = g C g.hU =
%Hz
C g.8hU = W %H3 j
for some j E 11
,...,n}
W)
CHAPTER I11
142
and obviously z(yVi) = (q,')V the set { V action.
,. . .,Wn 1
i'
for all z,y E G .
transitively and H
Thus the group G
is the stabilizer of
W under this
It follows that dimWi = dimW and hence that all the ei
F Invoking Lemma 2.4.1(ii), Finally, because
v
acts on
are equal.
F
we also deduce that
W is an FH-module and V
1
8.
is irreducible, so is W and the result follows.
In order to make further progress, we need to introduce the notions of projective representations and twisted group algebras. Let F*
denote the multiplicative group of F and let Z2(G,F*) be the set
of all functions
which satisfy the following identities
Z2(G,F*)
We shall refer to the elements of define aB
as c o c y d e s .
Given
U,8 E
Z2(G,F*),
by the rule
It is then obvious that aB
is also a cocycle and that Z ' ( G , F * )
becomes an
abelian group. Let t : G
* F* be such that
t ( l )= 1
and let &t: G x G
-+
F*
be
defined by
We shall refer to B2CG,F*1
bt
as a cobounduq.
It is routine to verify that the set
of a l l coboundaries constitutes a subgroup of
Z2tG,F*).
The factor
group
H* (G,F*)
=
z 2 ( G , F * ) / B ~LG,F*)
is said to be the second cohomoZogy group of
G
over F*.
The elements of
CONLON ' S THEOREM
are called cohomology classes; any two cocycles contained in the same
H2CG,F*l
Given a E Z2(G,F*), we shall
cohomology class are said to be cohornologous. write
143
a
for the cohomology class containing a.
One can easily check that if the requirement (1) is dropped, then the resulting cohomology group will be isomoiphic to the one introduced. Let
V be a vector space over F.
projective representation of G
A
mapping
over F
p :G
d GL(V)
is called a
if there exists a mapping a
:G x G -
F*
such that
Thus an ordinary representation is a projective representation with a(Z,y) = 1 for all z,y E G.
If we identify G L ( Y )
with
GL(n,F),n = dimY, the resulting F map is called a projective matrix representation of G over F . As in the case of ordinary representations, we shall treat the terms "projective representation" and "projective matrix representation" as interchangeable. In view of the associativity of the multiplication in G, conditions (3) and (4) imply that a E Z2(G,F*).
shall often refer to TWO
p
To stress the dependence of
p
as an a-representation on the space
projective representations pi : G
+
GL(Yi),i = 1.2
V
on
and a ,
we
v. are said to be
linearly equivalent if there exists a vector space isomorphism f
:
V1-
V
such
that
for all g E G.
The projective representation
p
on the space
v
is called
irreducible if 0 and V are the only subspaces of V which are sent into themselves by all the transformations p(g),g be two a-representations.
defined by
Then the map
€
G.
Let pi : G -+
GL(Yi),i = 1,2,
CHAPTER I11
144
We warn the reader that unlike
is easily seen to be again an a-representation.
the situation for ordinary representations, there is no natrual way to define a sum of an a-representation and %-representation when a # B. p
on the space
An a-representation
V is completely reducible if for any subspace W invariant
under all transformations P ( g ) ,g E G, there exists another such subspace W' with
V = W 8 IJ'. over F can be interpreted as FG-modules.
G
Ordinary representations of
The same situation prevails for projective representations in which the role of the group algebra FG plays the twisted group algebra defined below. a Given a E Z2CG,F*), denote by F G the vector space over F with basis which is in one-to-one correspondence with G .
E GI
PG
in
Define multiplication
distributively using for all x , y E G r,-
xy
=
a&,y)G
a Then F G becomes an F-algebra to which we refer as the twisted group algebra G
over F.
Note that if a & , y ) = 1 for all z , y E G, then FaG
FG.
More
generally, we have:
4.2. LEMMA. Proof.
FG
P G
if and only if a a
Assume that F G say f
G ' F into F, so does ,
and so a
=
gt, where t
assume that a = 6t J, :
FG.
:
G
--t
for some t
is a coboundary.
Since FG admits an F-algebra homomorphism
FaG 4 F .
:
F* :
G
Then, for all x , y E G,
is defined by +
F*
P G + FG, which is the extension of
with +
t(g)
=
f(g).
tC1) = 1.
Conversely,
Then the map
t ( g ) g by F-linearity gives the
desired isomorphism. m
4.3. LEMMA.
Let A
unit group of A
be an F-algebra and let f
which satisfies
be a mapping of G
into the
145
CONLON'S THEOREM
Then the mapping f*
FaG
:
4
A
defined by
of F-algebras. Proof.
Since
p (LA
-
f* is the extension of g
vector space homomorphism.
To prove that f*
99
) =
f(g)
I-+
CXgf(g,
is a homomorphism
by F-linearity, f *
is a
preserves multiplication, it
suffices to check it on the basis elements g,g E G.
Since
the result follows. The next result shows that the study of a-representations is equivalent to the
a
study of F G-modules.
4.4. PROPOSITION. of G
There is a bijective correspondence between a-representations
a
and F G-modules.
This correspondence preserves sums and maps bijectively
linearly equivalent (irreducible, completely reducible) representations into isomorphic (Irreducible, completely reducible) modules. Proof.
Let
p
be an a-representation of
G on the space V.
Due to
Lemma 4.3, we can define a homomorphism f : FaG + EndCV) by setting and extending by linearity.
Thus V
F
a
becomes an F G-module by setting
(Xg
(CXgslu = EX p(g)u 9 a Conversely, given an F G-module define
so that
pCg1 =
p
f(5).
Then
V, and hence a homomorphism f
p(g) E
GLIV) since
is an a-representation on
dence between a-representations and A subspace
V.
9
V.
E
F.g
E G,u E V )
FaG --t. End(v), F is a unit of P G . Moreover, :
This sets up a bijective correspon-
FaG-modules.
W of V is invariant under all p(g)
PGsubmodule of
f(i) = p(g)
if and only i f
W is an
Therefore the correspondence preserves sums and maps bi-
jectively irreducible (completely reducible) representations into irreducible lcompletely reducible) modules.
CHAPTER 111
146
We next observe that an F-isomorphism f a
F G-isomorphism if and only if now that
=
if(V)
pi : G + GL(V .) , d = 1,2,
linearly equivalent to such that p 2 ( g ) f
p2(g)f(V) = fpl
=
(g)V
p2
:
a
Vl-
V2
of F G-modules is an
for all g E
fC$)
E Vl.
G,V
Assume
Then P1
are two a-representations.
is
if and only if there is an F-isomorphism f : V l +
fpl(g),
for all g E G.
or to gf
The latter is equivalent to for a l l
( V ) = f(sU),
V
.
g E G,V E V
Hence two
a-representations are linearly equivalent if and only if the corresponding modules
'
are isomorphic.
Given a E Z2(G/N,F*) , we denote by
infa the element of
Z2(G,F*) defined
by
a k infa induces a homomorphism H 2 (G/N,F*) -+
Then the map
the i n f k t i o n map.
FBG by n H
embed FN in 4.5.
n,
THEOREM CCliffordC1937)).
for all
v
Let
Then there exists w
FN-module.
V
Note that since (infcl)(x,y) TZ
E N,
=
H 2 (G,F*), called
1 for all x , y E N , where
B
= infa
we may
or B
=
be an absolutely irreducible G-invariant
= UG(V)
E Z2(G/N,F*)
such that for y = info,
can be extended to an FYG-module, i.e. there exists an FYG-module W
that WN
2
V.
Furthermore, if w
(infa)-'.
is a coboundary, then
V
such
can be extended to
an FG-module. Proof.
Let
I? be the representation of N afforded by V .
To prove the
first statement, it suffices b y Proposition 4.4) to exhibit y E Z2(G,F*), whose values are constant on the cosets of N , that PCn) =
r(n)
for all n E N .
a nonsingular matrix LCgl
By hypothesis, for each g E G
of
G such
there exists
such that
We may, of course, assume that L(1) = 1.
G containing 1 and set
and a Y-representation p
Let T
be a transversal for N
in
'
CONLON S THEOREM
147
Then, for all g E G , n E N ,
From Schur's lemma, we readily verify that a scalar
p ( x ) p ( y ) differs from
p ( q )
only by
:
with y ( x , y ) E F * .
Thus p
is a Y-representation of
Furthermore, since L ( 1 ) = 1, we have Given x,y E G
and n ,n 1
2
E N,
G
and hence Y E Z 2 ( G , F * ) .
p ( n ) = r ( n ) for all n E N .
we have
whence yLml , y n 2 ) = Y ( Z , Y ) . By the foregoing, we may write that w
t
:
G / N A F*
with p*
for all n E N .
tU )= 1.
(V)
E
inf CW) =
for some
W
E Z2(G/N,F*).
t(xN)t(yN)t(xyN)-'
Then, setting
P*(g) =
is an ordinary representation of
G
Assume
for some
tCgN)-lp(g),
it is
such that p * ( n ) = p ( n ) = r ( n )
This completes the proof of the theorem.
In view of Theorem 4.5,
G
=
is a coboundary, say W ( x i V , y N )
immediate that
W
y
Z*(G/N,F*).
the extendibility of
V
is governed by
For this reason we shall refer to
wG(V)
as the obstruction
t o the extension of V to FG. 4.6.
LEMMA.
Let A
be a finite-dimensional F-algebra, let V
be an absolutely
irreducible A-module, and let W = U @ V, where U is some finite-dimensional F For each a E A, let @aE EndV be defined by @ u ( v ) = u v , vector space over F .
v E
v.
(i) If
F 8 E EndW is such that for all a E A, 8(1 @ @a) = (1 @ F
$,)€I,
then
148
CHAPTER I11
8 = $
for some $ E EndU
81
F (ii) If 6 E G L ( W )
8‘l(1 @
= 1 @ ‘C-’@~T,
then 0 =
,...,um 1
Ci) Let { u
Proof.
are such that for all a E A ,
T E GLCV)
and
+8T
+
for some
E
GL(U).
Then, for all v E V,
be an F-basis for U.
we have
rn =
e(U.82)) 2
for some 0 . . E EndV.
”
a E A,
For a l l
c ~ . Q B3% ..(v) j=l 3
(7)
we have
F
m
Because the
{u.) 3
are linearly independent, we deduce that
or, equivalently, that 8 have
Ji
E EndV
A
g j z . = h ..*lV for some h . . 32
Then $ E G L ( U )
3%
and, by (71,
(it) Put y = 1 8 T.
. F.
Since
V
i s
absolutely irreducible, we
Define $ E EndU by
F
8 = $ 63 1, proving (i).
Then
and thus, by (i), 6Y-l =
II, 8 1
for some
JI E G L ( U 1 .
Hence
8= ($@l)y=+8
‘I
as we wished to show. Suppose that the map
defined by
p
i
: C .-,GLCV.) 2
,i = 1,2, is an a+-representation. Consider
149
CONLON'S THEOREM
p1 @ p 2
Then
is obviously an a a -representation.
We shall refer to
1 2
@
as the (inner) tansor product of the projective representations P I
and
In module-theoretic language, the (inner) tensor product can be defined Let V and W be FO"G and FBG-modules, respectively.
V
@
F"BG-module where
W is an
as
p2
pz.
follows.
Then the vector space
the action of the elements i,g E G, is defined by
F
and then extended to
V @ W and F '@G
by F-linearity.
F Let a E Z2CG/N,F*) an Finf(')G-mcdule
g,g
and let
V be an PCG/N)-module.
inflV) whose underlying space is V
4.7. THEOREM. CCliffordU937)).
as being infZated from
Let H
be the inertia group of V
tion to the extension of to FinfwH
(by Theorem 4.5.
V.
8, let w = w H ( I ' ) E Z2(H/N,F*)
V
is G-invariant.
and
T :G
=
infw,y = inf(w -+
-1
)
V
Then there exists an
U such that
By Theorem 4,1Ci), MN
that for all n E N,w E W, and v E
B
denote any extension of
such extension always exists).
we may therefore assume that M
put
be an obstruc-
Owing to Theorem 4.1Lii1, we may harmlessly assume that H = G
Proof. which case
be an irreducible constituent of MN.
to FH and let ext(V)
P-l IH/Nl -module
irreducible
V.
Let F be an algebraically closed field, let
be an irreducible FG-module and let V
of
on which the elements
G, act according to the rule:
E
In the future we shall refer to inf(V)
M
Then one can form
and S
=
=
in
is a direct sum of copies
W 8 V, where W is an F-space such
V,
ext(V).
If rl
:
N+
GL(V),p
GLG) are the respresentations afforded by V , M ,
:
G-
and S ,
GL(M)
C W T E R IIL
150
r e s p e c t i v e l y , then
and t h e r e f o r e
Thus, by Lemma 4 . 6 C i i )
, for
= qCg)w,W E W,
Setting
each
g E G
i t follows t h a t
M Furthermore, because
M
$(n) = 1
W
GL(W) such t h a t
i s an F%-module
such t h a t
W 8 extV
_y
W
is i r r e d u c i b l e ,
taking i n t o account t h a t
$Cg)
there exists
for all
must also be i r r e d u c i b l e .
n E N,
we deduce t h a t
Finally,
W
2
inf(U)
-1 f o r some i r r e d u c i b l e
CH/N)-module
U.
T h i s completes t h e proof of t h e
theorem. I n o r d e r t o apply Theorem 4.7, we need t h e following simple observation
4.8. LEMMA. order:
be an a l g e b r a i c a l l y closed f i e l d and l e t
G
be a group of
n.
If
(ki)
CY E
Z2(G,F*)
and
group a l l elements of which have o r d e r d i v i d i n g
rn is t h e order of t h e cohomology c l a s s of
01,
is cohomologous t o a cocycle of o r d e r m.
Ciikl
If
charF = p
Proof. CY E
F
H2(G,F*) i s a f i n i t e
(i)
a
Let
Z2(G,F*).
0,
then
p does not d i v i d e t h e order of
We f i r s t note t h a t (iii)is a consequence of C i i ) . Because
aLz,y)a(q,z) = a(y,z)a(t,ya),
we have
H 2 (G,F*). Assume t h a t
n. then
CONLON'S THEOREM
Setting t ( y ) =
151
it now follows that
6 G
Thus an = A t ,
proving that each element of
To prove that H2CG,F*)
H2CG,F*) has order dividing
is finite, it suffices to verify (ii).
71.
Indeed, if (ii)
is true, then H2(G,F*) can be regarded as a subset of all mappings from G to the group of n-th roots of H'(G,F*)
in F.
1
G
But the latter is finite, hence
is finite. a(z,y)" = t ( t C ) t ( y ) t ( x y ) - ' , x , y E G,
To prove (ii), write Because F
is algebraically closed, for any
that p ( r ) "
= t(3cI-l.
up=bp implies u=b.
for some t and hence
:
G+
c1 =
4.9. LEMMA.
3
=
a(6p),
3c
E G
0.
for some t
there exists
Ll(3c)
: G--,
E F*
F*.
such
we deduce that 8" = 1, as required.
(a-b)*
Then, for all a,b E F,
0.
Suppose now that
F*.
6p,
8
Setting
Assume that charF = p
= Up-bp,
so
E Z2(G,F*) is such that
Then a(z,y)P = p
~ p(y)p p p(xy)-P for some p
: ~4F*
as required. be an algebraically closed field of characteristic p > 0
Let F
and let a E Z2(G,F*). (i) G
X
Then G'F
2
FG under either of the following hypotheses:
is a p-group
(ii) For any prime
q # p,
the Sylow q-subgroups of
G are cyclic or general-
ized quaternion (Liil G
is a cyclic extension of a p-group
Proof. (if1 Let q
Ci)
Direct consequence of Lemmas 4.2 and 4.8.
be a prime and let S be a Sylow q-subgroup of
standard fact of cohomology theory Ce.g, see M Hall(1959)) of H 2 (G,F*)
is isomorphic to a subgroup of H 2 C S , F * ) .
quaternion,then H2CS,F*I = 1 we may assume that G
(see Karpilovsky (1985)1 .
is a cyclic q-group and q # p.
G.
It is a
that the q-component If S is generalized Hence, by Lemma 4.8,
Let g be a generator of
152
CHAPTER I11
G, say of order m. with pm
so that
=
G ' F F-basis of .
-m 9 = h * i far some h
Then
=
()I;)"
Thus FO"G
i.
F*.
E
Then the elements
Choose any
? . E l
- m-1 l,Llg, ..., (Vg)
-
c
F
form an
FG.
(iii) Direct consequence of (ii). 4.10. LEMMA. Let F be an algebraically closed field such that charF
IGI
and
c1
Then dimV divides the order of G. F By making a diagonal change of basis E G I of PG, we may
let V be an irreducible F G-module. Proof.
by any cohomologous cocycle.
replace CX
is of finite order, say m.
that a
F and let G* f(Eii)
=
E
=<Ez;lg
Thus, by Lemma 4.8 (ii), we may assume
Let
G, 1 G i G m>.
be a primitive m-th root of
E
Then the map f
:
G*+
G
1 in
defined by
g is a surjective homomorphism whose kernel i s a central subgroup
Furthermore, if then p* : G
p :G
GLCV)
-+
+
GL(V)
is an +representation of G
defined by
irreducible representation of
p*CEzgl
G*.
=
Eip(g)
Since charF
k
<E>.
afforded by ,'l
is easily seen to be an IG*l
G*FE> ' G ,
and
the
result follows by appealing to the following standard fact (see Curtis and Reiner (1981)1 :
If F
is algebraically closed, charF
1 IGl
and
Z is a central subgroup of
G, then the dimensions of irreducible FG-modules divide Let F be an algebraically closed field.
4.11. LEMMA.
G'F
i s a local ring if and only if
Proof.
Assume that G
Lemmas 4.2 and 4.8, p"lG Conversely, assume that
=
PH
3
0 and G
i s a p-group. 7
0.
Then, by
FG and hence, by Corollary 1.2, F"1G is a local ring.
a
P G is a local ring and assume that
G
has an element
It follows from Lemma 4.9 that for H=
,PH
Hence
F~HFx Fx proving that
p
Then, for any G # 1,
is a p-group and that charF = p
g of prime order p # charF. is semisimple.
charF
(G:Z).
... x F
has nontrivial idempotents.
P G , a contradiction.
(p times)
But then the same is true of
So the Lemma i s proved.
We are now ready to prove the following important result.
EFfl
CONLBNG'S THEOREM
153
Let
N
be a normal subgroup of
closed f i e l d and l e t
V
be an i r r e d u c i b l e FG-module.
4.12.
THEOREM.
(G:N),
assume t h a t
G/N
i s p-solvable.
G,
Then
If
dimV
F
be an a l g e b r a i c a l l y
charF = p > 0
(G:N)d,
divides
F
is t h e dimension of an i r r e d u c i b l e c o n s t i t u e n t of Proof.
let
divides
d
where
vN-
W e f i r s t show t h a t t h e r e s u l t i s t r u e under e i t h e r of t h e following
hypotheses:
y
(i) charF
(G:N)
charF = p > 0 and
lii)
G/N
i s a p-group
W be an i r r e d u c i b l e c o n s t i t u e n t of
Let
W,
and l e t
W E
VN,
H
let
be t h e i n e r t i a group of
Z2(H/N,F*) be an o b s t r u c t i o n t o t h e extension of
W
FH.
to
-1
By Theorem 4.7,
t h e r e e x i s t s an i r r e d u c i b l e
d i d = dim i n f (5') F F
Since
?
S
U€/N)-module
such t h a t
dimW = dim extlW1 , we have F F
and
dimV = (dims) (dimW) (G:H) F F F I n case l i )
d i d d i v i d e s (EI:N1, by Lemma 4.9, so dimV F -1 F i s a l o c a l r i n g , by Lemma 4.11. In case (ii), F@ W/N)
therefore
dimV
1G:N)dimW F Turning t o t h e g e n e r a l c a s e , we u s e induction on
r e s u l t i s t r u e f o r groups of lower o r d e r than charF = p
thesis. containing
Since
3
0 divides G/N
CG:N),
G.
\GI.
(G:N)dimW. F dimS = 1 and F
So assume t h a t t h e
By t h e above, w e may assume
i n which case
G/N
is p s o l v a b l e , by hypo-
i s p s o l v a b l e t h e r e e x i s t s a proper normal subgroup
N and such t h a t
i r r e d u c i b l e c o n s t i t u e n t of Then
Hence
.
again divides
F
that
divides
G/M VM
is either a
and l e t
p o r p'-group.
w
be an
S be an i r r e d u c i b l e c o n s t i t u e n t of
S is obviously an i r r e d u c i b l e c o n s t i t u e n t of dimV F
divides
dimW
divides
( G : M ) dimW F
and by induction hypothesis
F
Let
M of
(M:N) dimS F
vN.
By t h e foregoing,
wN.
G
CHAPTER 111
154
Hence d i d divides
F
(G:M) (M:N)dimS = (G:N)dimS F F
as asserted. 4.13. COROLLARY. (Dade(l968),Swan(1963)).
Let F be an algebraically closed
field of characteristic p > 0
be an abelian normal subgroup of a
p-solvable group G. Proof.
and let A
The the dimensions of irreducible FG-modules divide
(G:A).
Direct consequence of Theorem 4.12 and the fact that dimensions of
1.
all irreducible FA-modules are equal to
We next concentrate on decomposition of induced modules from normal subgroups.
4.14. THEOREM(Ward(1968),Willems(1976)). be an indecomposable FN-module, and let H
Let F
be an arbitrary field, let
be the inertia group of
V.
v
Suppose
that
where the ti)
Vi are indecomposable FH-modules.
fi fl @ fl @ .., @ f , =
that L$
2
(ii) If
1
J$
2
implies
Vi
where the
Then are indecomposable FG-modules such
V j'
Vi is irreducible for some i, then I$
Proof. (i) The equality
fl = fl @ ... @ f
induction is transitive and preserves direct sums.
and
Ir are irreducible.
follows from the fact that To prove that each
{
is
n
indecomposable, assume that V? = X 8 Y is a direct decomposition. is identifiable with a direct summand of Schmidt theorem, that Vi
form h @
V
with
t#)N
h E H.
'i
we may assume, by the K r u l l -
is a direct summand of XHI say
XH Next we observe that
I$)H,
Because
=
vi @ X'
(9)
is the direct sum of isomorphic FN-modules of the Applying the Krull-Schmidt theorem, we obtain from (8)
CONLON'S THEOREM
155
Thus, for all g E G, there i s an FN-isomorphism
for some positive integer mi,
g 0 Vi
I
mi(g Q V)
(1 ,G
i ,G
s)
which in turn yields
where
T is a transversal for H in G containing 1.
Restricting (9) to N ,
However X
we see by (10) that
is an FG-module, so
proving that for all
t
T , t 8 V appears as a direct summand of XN with
E
multiplicity at least mi.
t
@
V and
t 28
Taking into account that for distinct t ,t 1
2
E
T,
G V are nonisomorphic FN-modules, we infer from Cii) that (Vi)N
is isomorphic to a direct summand of
XN.
Thus
ViG
d i m 8 G dimX and therefore
FZ
F
is indeed indecomposable. We are left to prove that contradiction that direct summand of
<
J$
($IH,
k$
but
(
implies
Vi $ V
j*
Vi
Since
V
i*
Assume by way of
vi is identifiable with a
we may write
V'
i
for some FH-module V:
V!. By the Krull-Schmidt theorem, Vi is iso3 Hence (Vi)N is isomorphic to an FNmorphic to an FH-direct summand of V' and
i'
(V!) Invoking (10) we infer that V 3 " and that FN-direct summand of I V ! ) 3 N direct summand of
is isomorphic to an
n
But, by (ll), the multiplicity of is exactly m
j'
1' as an indecomposable direct summand of
This contradiction proves the desired assertion.
(@)
jfl
156
CHaPTER I11
i E {l,...,~}
(if) Assume that there is an by Theorem 4.1, ible.
Vi
(Vi)N
such that V2
Thus, by (101, V
is completely reducible.
Let W be an irreducible factor module of
WH.
is isomorphic to an FH-submodule of
FN-submodule of
WN and so, by (lo), WN
6.
( V z ) N is isomorphic to an
Hence
contains a submodule X
P
m.V.
It
m Z ( g 49 V ) .
Then the FN-modules t 8 V , t E T ,
H in G.
be a transversal for
is irreduc-
Owing to Theorem 2.4.10,
contains the FN-submodule gX
follows that for all g E G,WN Let T
is irreducible. Then,
are irreducible and nonisomorphic and they all occur as direct summands of with multiplicity at least m
wN
Applying (111, we conclude that
3'
dimW > d i m E
F and hence that L$ = W
.
F"
is irreducible.
Next we exhibit a relationship between Idempotents In a ring R, and idem-
patents in a factor ring R/I, where I is a nil ideal. denotes the image of P E R 4.15. LEMMA.
R'
p
= R/I.
I be a nil ideal of a ring R.
Let
(I) Each fdempotent
-e =
in
In what follows
E
can be lifted to an idempotent e E 8 ,
E
that is,
E.
(ii) If e
and f
and only if
eii
(iii) Let I= potents in
R'.
fR as right R-modules if
then eR
are idempotents of R,
as right R-modules. E
+
... +
En
be a decomposition of
-1
into orthogonal idem-
...,e
Then there exist orthogonal idempotents e l , e 2 ,
E
R
=
E.
such
that 1 = e
Moreover,
+
... + en
i is primitive if and only if ei
E
Proof. (i) Given an idempotent u-u2 E
and
I and hence
( U - U ~= ~0 ~
E
of
E,
for some m
ei
= E
i
is also. choose u E R
a 1.
We have
with
Then
157
CONLON'S THEOREM
rn
Note that on the right each term after the first rn while the first rn the first rn u(1-u) E
terms, then 1
-
=
-
=
u =
ba = f.
to f.
and f
=
denotes the
sum
is a polynomial in u .
of
Now
a,b E R :
eR-
2
Thus e
fR
=
(e-'e)e
where z E e r e .
Because
' = zz ' = z ' z .
a h (e-z") = e .
Next write b a 2
2
a
f-y,
=
y
2
YE
ZE
a ,b
=
1
(f-y)2
and since y
Because a 2 b 2 = e
e,
=
and
b
=
=
e-l(fi)
b, =
to b
fbe
so
=
b,
e - l ( f ) b = a b , and
satisfy the condition above, then
with inverse y -
ay.
satisfy conditions equivalent to this
and a , b E R
2
=
=
and b a 2
are such that a Then a b
e a f , b1 = fbe.
1
eaf, 1
= e-z
,
=
Z+Z"
= ZZ"
=
Z"Z
and hence
b (e-z"), then
then y E fIf..
f-y = Hence y 2
e-l@)
Setting z" = ez ' e , we obtain
=
e a f , b = j%e,ab
=
i s nilpotent, 1-2 has an inverse 1-2' so that
2
NOW put
1
to f R
fbod I ) , and set a
e , ba
a
satisfy the same conditions for the isomorphism ZH'FR,
Now suppose that
fbe,ab
and
=
fR
We first show that eR
fR is an isomorphism that maps e
and let a,b E R
z,z
Then
is a required idempotent.
b and @ ( e ) e= @ ( e l , hence be
=
a.
such that
is a homomorphism from eR
FE.
u (mod I),
are idempotents of R.
Conversely, if a
ba.
isomorphism.
1
fb
Then
Assume that eR
z+z
2 m ~ ~ ~ l ( l +- u )
Assume first that 0
similarly f
E
where g
m
and similarly eaf =
b
Hence, if e
Because e ( 1 - e ) = e ( 1 - u ) g = 0 , e
E.
if and only if there exist
so
.
...
+
u2m
(ii) Assume that e
bx
rn e+(l-u) g ,
,
I, so
that is, e
3~ t+
u
terms are divisible by
e
and a
rn
is dfvisfble By (1-u)
1
(b a
Since
2
f2-fy-yf+y2
=
2
)'
=
b ea 2
2
=
b a
2 2
,
we have
f-2y+y2
is nilpotent, we find that y = 0 and so b 2 a 2 = f. 2
=
f
imply a
= ea
f , b 2 = fb,e,
the desired
assertion is proved. (iii) Assume that e
and f
are idempotents of R
first show that there exists an idempotent g
such that
such that
=
7
z F = F z = 0.
We
and eg
0.
=
ge
=
158
CWTER 1x1
To t h i s end, note t h a t
h2 = h,he = 0 ,
Then
is n i l p o t e n t and so
fe
and
7.
=
is a u n i t .
1-fe
g = (1-e)h.
Now p u t
Then
proving t h e a u x i l i a r y a s s e r t i o n . W e now show t h a t i f
<#
j,
1
,...,en
For
n.
3 # j,<,j
e
0
for
I n o t h e r words, we may suppose t h a t
... + en,
+
e
then
= 0
i s an idempotent such t h a t
Invoking t h e a u x i l i a r y a s s e r t i o n , t h e r e e x i s t s an idempotent
1
c
1
u
e
1
such t h a t
-
= 1,
1
Thus
1-u
so
ei =
1
i
n.
e # 0,
-.
only i n t h e case
= e
+e
1
u
If
z
=
2
x+y
e = 0 or
(?ding t o l i i ) ,
e
= 0.
+. . .+e n
= 1.
e E 8 i s a p r i m i t i v e idempotent. f o r some orthogonal idempotents
e 1 , e 2E R
then t h e r e e x i s t orthogonal idempotents
2
Therefore f o r u=e
ts a n i l p o t e n t idempotent, and t h u s
we have
and hence with
1
a r e pairwise orthog-
er,e
E ~ 1 ,
To prove t h e f i n a l a s s e r t i o n , assume t h a t
e # 0,
,...,en 2
e'
Hence, by (i) and t h e above, t h e r e e x i s t pairwise orthogonal
,...,e
I
ee' = e r e = 0.
and
1
for
0.
=
e
e.e
n
=
er= 2
5,
=
Put
idempotents
x,y E
ere' z'j
for a l l
1.
onal idempotents.
Because
and
0
3
such t h a t
we have
3=5
such t h a t
el
=
1 t h e r e is nothing t o prove, so we may suppose t h a t
11 =
and use induction on
-e -e = -e e-
-eiej
a r e idempotents such t h a t
then t h e r e e x i s t idempotents
i # j,
all
e
Put
eR a (e +e ) R , 1
Thus e i t h e r
2
x
= 0
with
e
= x,;
= y
which i s p o s s i b l e or
y = 0
and so
e
is p r i m i t i v e . Conversely, suppose t h a t
e = u+V, either
where
u=0
primitive.
a r e orthogonal idempotents,
u,V or
i s primitive.
= 0.
Hence e i t h e r
u = 0
e # 0 and so e # 0. I f - - - - - - - and so then e = u + V , u V = v u
Then
or
V = 0,
proving t h a t
e
is
So t h e lemma i s v e r i f i e d .
I t i s now an easy matter t o prove t h e following u s e f u l observation.
4.16. PROPOSITION.
Let
admit a decomposition
I
be a n i l i d e a l of a r i n g
R
and l e t t h e r i n g
z=R/I
159
COPILON'S TmOREM
into finitely many indecomposable right ideals.
Then there exists a decompo-
R,
sition of
R
=
... @I,
Il @ I2 @
into indecomposable right ideals such that (i) L .
i s the image of
(ii) L.
~i
L
Proof. with
1
Ii under the natural map R
as right kmodules if and only if Ii
2
I
as right R-modules.
i
By assumption, there exist orthogonal primitive idempotents
+
= E 1
... + E~
and L
=
ciE.
Thanks to Lemma 4.15(iii),
orthogonal primitive idempotents e , . . . r e E R with
n
1
1
e 1
Setting
--f
Ii
=
+
and
E
f
eiR, we have the decomposition
into indecomposable rtght *deals of
established.
there exist
... + en.
R = II @ I2 8
(i) holds.
ei =
ci E 8
R.
... @ I n
Since
72. =
e .z = 2
E
.E =
2
Property (ii) being a consequence of Lemma 4.15(iii),
Lj, property the result is
9
We next exhibit a general fact which provides a link between a module and its ring of endomorphisms.
Let R
be a ring and let V be a left R-module.
For
any subset S of EndCV), let SV be the set of all finite sums Csi(ui),8 E S i R vi E V . Then, clearly, SV is an R-submodule of V . Observe also that if I is a principal right ideal of
End(V)
generated by
then I V = $(V).
R 4.17. LEMMA.
Let R be a ring, let
admit a decomposition
V be a left R-module, and let End(V) R
CHAPTER I11
160
Then
into finitely many indecomposable right ideals. V
lil
=
... @ InV
1 1 V 8I V 8 2
(ii)1.V
z modules.
proof.
is a decomposition into indecomposable submodules.
as left R-modules if and only if I.
I,V 3
z
put E
ti)
EndCV)
=
I
as right End(V)R
j
and write
R + $
l = $
i
1
Then $ ? = $., 2
2
$.$.
=
2 3
of the idempotents $
j , Ti
0, i
2
=
c + ~E rii
@n
and since Ii is indecomposable, each
$iEl
If v E V ,
i s primitive.
i
...
then v
=
$ Iv)
i
... +
$),(V),
so
v= Assume that v
1
,t,
2
...,v
IV i I 2
1
X
and 1
we denote By
T
i
$
Next put
some 8-modules X E
2
E
... iInV
V satisfy $ 1 (vl)
E
$
taking the images of both stdes under Hence V = 8 I.V.
v i
.
1
Then
X
i'
=
... + $n (vn l
we infer that each
i'
$
=
i$)2Cv2)i
= 0.
$i(t,i)
is 0 .
and assume that I V = X @ X
i'
1
and X 1
Then,
are direct summands of
for
2
V and
2
X2.
the canonical projection V -
orthogonal idempotents such that for all
t,
E
IVlv
Then
and 71
71
are
L
Because
= 71 ( U ) i I T ( V ) . 1
IV
= $(V)
we must also have
therefore either IT
1
X
= 2
2,
or IT
module via
I
j
@ = $(Vl,$ E
Now the canonical map
$ ( v ) for all v E
must be
0.
But
E
Thus $
= 71 i 7 1 1 2
.
2
1
= O
and
or
is indecomposable.
as right E-modules.
E,v
IT.
X. = n.(V), so X 2
2
0, which proves that TV
lii) Assume that Ti
=
V.
Then E 8 V E
We may view
V
as a left E-
becomes a left E-module via
161
CONLON’S THEOREM
Since Ii
is an isomorphism of E-modules.
i s a
direct summand of E ,
But this image is identify Ti 8 V with its image in E 8 V . E E I . 8 V 1 I . V which in turn implies that I . V 2 I .V. Z E 2 2 3 Now assume that 8 : I i V 4 I . V is an R-isomorphism and fix 3 @ E I i , we have Im@ 5 I i V , so Im(8@) 5 I . V . Because I . V = 3 3 deduce that (€I@) IU) = @.CV ) for some 2, E v. Thus 3
IiV,
we may
so
u E V. @.CV), 3
Given we
1
and so
If we now define
:
I;
+
ism of right E-modules. pI$l = OW’$
defined by of
h,
A
Ij
by
A(@)
=
84,
then
is obviously a homomorph-
similar argument shows that the mapping
is also an E-homomorphism.
Because p
p : I
j
+
Ii
is the inverse
the result follows. Let K
4.18. LEMMA.
be a local ring.
Then all (finitely generated) projective
R-modules are free. Let V
Proof. U
,..., n 2)
be a finitely generated projective R-module, and let
be a minimal generating set for V .
Suppose that we are given a
relation
r 1u 1 We claim that all such
r . E J(R). 2
+ . . . + r nun
(ri E R)
= O
Indeed, if ri & J(R),
then it is a unit in
as an R-linear sum of the U ! s . 3 This, however, is impossible since otherwise the generator ui can be deleted. R,
and we can solve for the corresponding U i
The claim is therefore substantiated. Let
W be a free R-module freely generated by w 1 ,...,Wn and map W onto V
by sending wi
to u . .
contained in J(R)W. with
V’
V.
By the foregoing, the kernel of this map, say S By hypothesis, V
Since S
P
W/V’
is projective so that we have
it must be finitely generated.
is
W=S
Moreover,
@
v’
CElZlPTER I11
162
and hence the inclusion S C_ J(R)W
forces S = J ( R ) S .
Invoking Nakayama's lemma
(Proposition 1.6.151, we conclude that S = 0 and hence that V lemma is proved.
2
bJ.
So the
8
We now analyze the indecomposable components of the induced module
V is an indecomposable FN-module.
fl,
where
The following preliminary observation will
clear our path Let H be a subgroup of G
4.19. LEMHR.
(1) The map
-
I End (v)
I
and let V
End
be an FH-module.
(fl)
FG
FH $ e l @ $
is an injective homomorphism of F-algebras whose image consists of all $
such that $Cl 8 Vl Ui1
If g E I ( H )
518 V and if
18 V
G
'+
g 8V
uniquely extended to an FG-isomorphism Proof.
E
is an FH-isomorphism, then $
End(fl) FG can be
@+ F.
li) It is clear that the given map is an injective homomorphism of $
F-algebras and that for a l l
E
EndCV) , 1 8 $
sends 1 8 V
FH Suppose that $ E End(@)
is such that $(l 8 V)
into 1 8 V .
5 1 8 V.
Then there exists
FG $
E
EndCV)
such that $U 8 V l = 1 8 $ C V )
for all tr
Since $
E V.
is
FH uniquely determined by its restriction to
1 8 V, we have $ = 1 8 $,
as
required. Cii) By a universal characterization of induced modules (Proposition 2.4.81, can be uniquely extended to an FG-homomorphism $
x E GI
Because
:
VG+
VG
.
For an arbitrary
we have
fl
is the sum of all vector spaces on the right, $
Taking into account that
fi
4,
is surjective.
is finite-dimensional over F, we deduce that J,
163
CONLON'S THEOREM
is an isomorphism.
1
-
With the aid of the lemma above we shall identify End(Vl and any FN-isomorphism 1 8 V
End(fl)
FG
f l 4 fl.
isomorphism V
with its image in
FN with its extension to an FG-
g @ V
We now fix the following assumptions and notation:
is a G-invariant FN-module
E = EndCfl), A = End(V)
FN
FG Q
= G/N
{txlxE
Ql
-
N in G with t .
is a transversal for
For each x E Q,@,
:
1@ V
send
1 8 V
1 and 3: = t-lN. X
t x @ V is an FN-isomorphism with
by our convention and by Lemma 4.19(ii), each For each x E Q,Ex
=
@x
@
=
1
(thus
is a unit of E)
is the F-subspace of E consisting of all homomorphisms that
into tx 8 V
(hence E- = A ) . With the notation above, the ring E
4.20. THEOREM. lConlon C196411.
satsifies
the following properties. (iE ) =
@E fiQ
and
ki) For all x,y
E
Ex
= A@ = X
@xA
for all x E Q
Q,@x@y@Z EA
(fii) J(A)*E= E*J(A) f J ( E ) and, in particular, J(A)E
is a nilpotent ideal of
E. (iv) Assume that V
is indecomposable, F
algebraically closed (hence A / J ( A ) %F,
by Lemma 1.7.4Cii)). and let @(x,y) denote the image of
B E Z2C&,F*) and E/J(A1E Proof.
@ @
4-l
XYW
in F*.
Then
FBQ.
X yx
First assume that
= 0
with
yx E Ex.
Then, for all
X=Q
v E V,g E GI
we have
vc Thus yx(g 8 U ) = 0 for a l l
v
E
=
@gtxQ
v
&Q
V , g E G, and so yx = 0.
Next assume that
164
CHAPTER I11
fEE.
Let
T
2
be the projection of
G V onto t x 8 V.
Then nx
is an FN-
homomorphism such that
Let g
:
to 1 8
v.
X
1@ V
-
tx 8 V be the FN-homomorphism that is the restriction of
E
E such that fx(l Q u1
Then, for all X E Q, we have f E Ex
proving that f = If
xx
$
4.
Then
f4i1
Therefore f4-l E A , X
( i t 1 FEx x , y E Q.
gx(l 63
5 Ex.
so f E
A$x
then
Because for all
JI
@ilfE A
E V
E
f E Ex, say f
and so
A,$(tx@ V ) = $ $
to t x @ V and hence 1 8 V
5tx8 V ,
for some
X
to
1 @ V.
and Ci) is established.
Our choice of the transversal implies that t t t
-' N
Y x XY
and hence
t t Q V = t Y X
It therefore follows that
2)
.
If f E Ex,
Conversely, assume that sends t x @ V
for all
2))
and by (12) we also have
5
Thts proves that Ex = $,A.
we have A$x C_ E,.
=
and hence that E = @Ex
By the definition of A,4JX4 $xA.
Tx"f
By a universal characterization of unduced modules (Proposition
2.4.81, there exists fx
f
= 1
C T
6Q
;cy
Qv
E
CONLON'S THEOREM
Since A
n 2- 1.
is a finite-dimensional algebra over F , [J(A)Eln = J(AInE
Hence
(iv) Given a subset X of E ,
Wx
165
= J ( A )$z(x E &)
, then Vx
=
0,
let Ex
-X
we have J ( A ) n
=
0
for some
proving that J ( A ) E 5 J ( E ) . denote its image in E / J ( A ) E .
If
and
J ( A ) E = 8 J ( A ) A @ , = 8 'J, SQ SQ
Because E
=
8 E it follows that x x
E= NOW A = F @
-
J U ) , so A = A
that
9x9y9w
=
-
Identifying A
for all
X
E
F
B(x,y)+Z.
63
ZTx
&Q
+ J(A)E= F + J U ) E
-
is a central subring of E there exists z E J ( A )
such
Thus
with its image in F ,
and x,y E Q.
implies that B E Z 2 ( Q , F * ) As
Ex =
By hypothesis, given x,y E Q
isomorphic to F . -1
8 &Q
we therefore derive
-
The associativity of multiplication in E
-
and hence that E
1
B
F Q,
now
thus completing the proof..
an application of the theorem above, we shall establish two results.
first, pertaining to arbitrary fields, is of a general nature.
The
The second, valid
for algebraically closed fields, gives a complete description of the indecomposable components of
fl
B
exclusively in terms of the twisted group algebra F &
con-
structed above.
4.21. THEOREM.
Let F
be an artibrary field, let V be an indecomposable G-
n invariant FN-module and let E = End(fl), A = End(V1. FG FN
Let E = @ li be a i= 1
166
CHWTER I11
decomposition into indecomposable right ideals. (i)
fl =
n @ I.V Z i=1
is a decomposition into indecomposable FG-modules
(ii) I. z I as right E-modules if and only if I.V I .V as FG-modules z j 2 3 (iii) Each Ti is a free A-module and d i d .V = (A-rankI.) (dim Vl EZ ' F Proof. (i) Owing to Lemma 4.17(i),
vc
rivG
=
i=1
is a decomposition into indecomposable submodules.
Since
v 5 F,
n
have IiV C_ IiVb. To prove the opposite inclusion, note that Since I i f i = Q(@)
idempotent @ E E . the form g @ V , Let x
9
E E
and since Ve
it suffices to verify that $ ( g @ Y )
be defined by
x (v) 9
= gU,u E
V.
= $S
for some
is a sum of subspaces of
5 TiV @x
Then
xi
we obviously
9
E Ii
for a l l
gE
G.
and for all U E V,
we have
$cg @ U ) and hence l i V = I i F ,
= $ g W ) = ($32 1
9
( v ) E T{V
proving (i).
(ti) Direct consequence of Lemma 4.17Uil (tii) Since
V is indecomposable, A is a local ring (Lemma 1.7.41.
Lemma 4.18, any projectfve A-module is free. A-module.
Hence Ii
By Theorem 4.20(i),
E
Thus, by is a free
is a projective A-module and therefore is free.
We now claim that the map
is an F-isomorphism. (G:8),
so
Indeed, by Theorem 4.20(1),
dim(E @ V ) = (dim V ) (G:N) = dimfl. F A F F
E
is a free A-module of rank
The map above is surjective,
since
and since
fi
is a direct sum of all such t @ V. 2
This substantiates our claim.
CONLON'S THEOREM
167
Ti is a direct summand of E, Ti @ V A image in E 8 V and so the induced map A Because
Ii @
v-
A Since dimU. 8
is injective.
V)
vc and since I.V
(A-ranM.)(dim V )
=
F 'A image of li 8 V,
i s identifiable with its
'
is the
F
the result follows.
A 4.22. THEOREM (Conlon 11964)).
Let F be an algebraically closed field, let
be an indecomposable G-invariant FN-module and let
E/J(A)E
FB(G/N),
where E = End/II(;) and A
=
End(V)
FG such
B always exists).
B E Z*(G/N,F*)
V
be such that
(by Theorem 4.2O(ivlI
FN
Let
FB CG/N)
=
L @
... d Ln
be a decomposition into indecomposable right ideals and let
E = I @I 1
.
2
@...ern
be the corresponding lifted decomposition of E
(see Proposition 4.16Ci)).
Then
the following properties hold:
n
(i)
II(; = d IiV
is a decomposition into indecomposable FG-modules
i=1
(ii) L. 2 L
z
(iiil
i
dimIiV
F
as =
FB CG/N)-modules if and only if IiV
P
1.V as FGmodules 3
(didi) (dim V)
F
F
Proof. Properties ti) and U i l follow from Theorem 4.21Cil and (ii) and Owing to Theorem 4.21(iii), to prove (iii) it suffices to
Proposition 4.16W1, verify that for I = I
i A-rank I = d M i
F The latter will follow provided we show that for n dimI(I+J(A)E)/JCA)E]
=
=
A-rank I,
n
F To this end, choose an A-basis { d
{ d l ,...,dn,dn+l,...,am} of E
,...,dn}
of I and extend it to an A-basis
(this is always possible since all direct
168
C W T E R I11
suuunands of E
and since A
=
are free A-modules).
Then
J(A) 0 F, we conclude that each element in
.
a unique F-linear combination of the images dl,...,d This proves (13) and hence the result.
in
11
f
J u l ) E ] / J C A ) E is
I1 + J ( A ) E l / J ( A l E .
We now apply the foregoing to prove the following fundamental result. 4.23. THEOREM.(Green(1959), Conlon(l964),Willems(l976)).
be an algebra-
be an indecomposable FN-module, where N
ically closed field, let V normal subgroup of G
Let F
and let H be the inertia group of
v.
is a proper
Then the follow-
ing properties are equivalent: n
v"
(il (iil
is indecomposable
8
is indecomposable
(iii) charF = p
0 and H/N
3
In particular, if charF
=
p
3
is a p-group
0 and G/N
is a p-group, then
fi
is indecom-
psable. Proof. That ti) is equivalent to (ii) is a consequence of Theorem 4.14.
It
therefore suffices to verify that (i) is equivalent to (iii) for the case H = G. By Theorem 4.22, idempotents.
fi
is indecomposable if and only if FB (G/N)
has no nontrivial
The latter is equivalent to the requirement that FB(G/N)
is local.
.
Since G/N # 1, by hypothesis, the desired conclusion follows by virtue of Lemma 4.11.
We now turn our attention to projective covers of irreducible FG-modules. 4.24. LEMMA.
Let
H be a subgroup of G and let V and W be FH and FG-
modules respectivley.
Then
(il P(fi1 i s isomorphic to a direct summand of
( i i l P(W I H
is isomorphic to a direct summand of
G
P(V)
PCW)H.
CONLON ' S THEOREM
,
(i) By the definition of PCVl
Proof.
P(F)
of FG-modules. Hence, by Lemma 1.9.2(ii), P(P(V)G)
.
R
there is an FH-epimorphism P(V)+
we infer that there is an epimorphism P ( V ) G d
Invoking Proposition 2.4.4(i),
summand of
169
fl
is isomorphic to a direct p ( V ) G is projective.
But by Theorem 2.5.9cii)
Hence
PPCVIG1 = P ( V I G and the required assertion follows. (ii) Since there is a homomorphism of PCW)
W ,P(W H
H
onto W
is isomorphic to a direct summand of
)
and hence of P(W),
onto
P(P(W)H). Since P(W) is Hence P(WH) is iso-
projective, so is PCW)H, by virtue of Theorem 2.5.9(1). morphic to a direct summand of P ( W H . Let N
4.25. LEMMA.
be a normal subgroup of
and let V
G
and
W be irre-
ducible FN and FGmodules, respectively. (i) If
(G:N)
=
pn,
characteristic p ,
p
a prime, and F
is an algebraically closed field of
then
PCF) = P ( V I G
= PtU)
where
U is an irreducible FG-module such that V is a constituent of U N .
tiil
If
G : N ) # 0 In F, then P ( W N )
P(mN.
P(Y)
(I) Owing to Lemma 1.9.1(iv),
Proof.
is an indecomposable module.
"
Hence, by Theorem 4.23, infer that P(fi) 1PLY)
PCV)' G
.
is also indecomposable. Since
v
is isomorphic to a submodule of
Theorem 2.4.10 implies that there is an FGepimorphism PCU)
1.9.2Cii),
Applying Lemma 4.24, we
fi+
is isomorphic to a direct summand of P ( f i ) .
U.
uN,
Thus, by Lemma Since by the
n
above PCPl
is indecomposable, Ci) is established.
(ii) Consider FG
as an FN-module.
Then, by Proposition 1.8cii). we have
Write
FG/J(FG)
=
with nonisomorphic irreducible FG-modules
8 diVi z
Vi.
Then, by Theorem 1.9.3 (iii),
(15)
CHAPTER I11
170
FG
=
? diPCVi) 2
and thus
Bearing in mind that FG is a free (hence, projective) FN-module, we also have
(by (1411 (by (15) and Lemma l.g.l(iii))
Thus the desired conclusion follows from (16) and Lemma 4.24tii). We have now come to the final result of this section. 4.26.
THEOREM. (Fong's dimension formula, FongC1961)).
group, let F
Since G
subgroup of C .
is a p'-group.
1 GI .
Let N
is p-solvable, either
and let V
denote a maximal normal
(G/N( = pn
for some
n
1 or
By Clifford's theorem, there exists a decomposition
with nonisomorphic irreducible FN-modules Vi Theorem 4.12, e8
divides
,
we have
F
p-solvable
Then
Proof. We argue by induction on
d f m V = e8 dfmV
a
be an algebraically closed field of characteristic p ,
be an irredncible FG-module.
G/N
Let G be
)G/NI.
of equal dimension.
Assume first that
IG/NI
=
p".
F'
On the other hand, by Lemma 4.25(i),
P(V)
2
P(f)
2
G
P(vl)
.
Hence
Moreover, by Because
A CHAR~CTERIZATIONOF ELEMENTS OF
=
J(FG)
171
IG/NI IN1 (dimV ) b y induction hypothesis) p F'p'
as required. Now suppose th t G/N
is a p'-group.
By Lemma 4.25(ii),
P(V),
P(V,)
nd
thus
so
(by induction hypothesis) (since esllG/NI
as required.
I N ( p = IGlp)
and since
'
5. A CHARACTERIZATION OF ELEMENTS OF
J(FG)
:
G IS p - SOLVABLE
Throughout this section, F denotes a field of characteristic p > 0, G group, and c
the sum of all p-elements of
G
(including 1).
Our
a finite
aim is to
discover necessary and sufficient conditions under which
In particular, it will be shown that J(FG) Given an FG-module
V, we write
xy
=
Ilk?) in case G is p-solvable.
for the character of
G afforded by
V
and
set
If
H is a subgroup of G, then lH denotes the trivial FH-module.
Define the
CHAPTER XI1
172
F-linear map
T
FG
F C
Ezgtf
t
and put KerT Finally, if
= It E
V
5.1. LEMMA. (i) KerT
(ii) If S E Syl (GI
FG.
=
k(c)
and
W.
V
T can be brought into argument.
=
(lS)
G , then Xv(xl
have TIDI
c
t
=
Kerx
for all g E G
g E FG, then tr(g-'r)
(Lemma 3.12Ci) 1.
= D
Z
=
XY
Hence, if g
Hence T(gcc)
E
G.
g
for all g E G Since G-l
P
= tP(g(zC1)
=
and
G
P'
we
and so T ( g x ) = o
Y be left transversals for N(S) in G and S in N(S), G G
respectively, and for a given g E G ,
Then
for all
if and only if zc = 0.
Let X and
CiiI
SIT(z)
V.
0 if and only if tr(gz) = 0 for all g = tP(3CC)
:
G
(i) If 2 =
D =
is an ideal of FG.
IIv(S)
=
SfG
hence
for the
The following simple obser-
and, in particular, KerT
In particular KerT
Proof.
0 for all g E GI
as a composition factor of
vation illustrates how the map
P
=
is irreducible, we write n(V,W)
V
V,W are FG-modules and
multiplicity of
D E
FGITCgcc)
set
is a left transversal for S
in G
and
is not a p-element, then by (1) XV(g)
Let g be a p-element.
Then g E P
=
T(g) = 0 .
if and only if g
NCP), for all G
P E Syl (G), and therefore M = {P E Syl (GI I# = PI. We claim that P P if sustained, it will follow from (1) that X ( g ) = IrJ(S): S I T ( g ) , l(modp); IMl V G
A CHP\RACTERIZATION OF ELEMENTS OF
T(g) = 1,
since
and hence t h e r e s u l t .
(GI I
lSyl
1
P
5.2.
[MI
< g > a c t s on S y l ( G ) P
d u c i b l e FG-module.
is a splitting field for
F
by
Hence
'
(modp) , as required.
Assume t h a t
LEMMA.
173
i s e x a c t l y t h e number of o r b i t s of l e n g t h 1.
lMl
conjugation and
The p-group
J(FG)
FG
V
and l e t
be an i r r e -
Then
KerXv = ann(V)
Proof.
rX ( V )
n
Let
= dimV
x E FG,
and f o r each
let
F
End(V1 with M n ( F ) , F homomorphism FG 3 I V ~ ( F ) of F-algebras given by x = 33,z) E
V.
Identifying
Tr(r
only v e r i f y t h a t
r = o,
r
then
X
X
rx
=
r r
=
) =
=
9 2 g E G.
0 for a l l
=
0 for all
and l e t
V 1
,...,V
J ( F G ) = Kerx Proof.
1 < i
Then
A E Mn(F1.
Assume t h a t
F
Put
x E
n
KerXi,
Tr(r
Kerx
=
0.
gx
Tr(r r ) 9 x
Hence we need
rX
k
n(Vi,V)
for a l l
and denote by
Then, by t h e d e f i n i t i o n of
then f o r a l l
3
Conversely, assume
0.
If
g E G
and hence
Since t h e l a t t e r i s p o s s i b l e o n l y i n t h e c a s e
is a s p l i t t i n g f i e l d f o r
= n(V$,V)
V -
=
Conversely, assume t h a t
0 for a l l
=
xi X
g E G,
Phi
proving t h a t
Tx.
i f and only i f
and we claim t h a t
If
-
FG, l e t
V
be a l l nonisomorphic i r r e d u c i b l e FG-modules.
ri
s.
G
E End(V) be defined by " F we o b t a i n a s u r j e c t i v e
'
i f and only i f p
V
g
0 and hence
0, t h e r e s u l t follows.
5 . 3 . LEMMA.
Vi,
gx
0 for a l l
TP(r ) = gx
Tr(A*r 1
g+
r
n KerXi x E KeryV.
Then
V'
be an FG-module Then
i E 11,...,s } .
t h e c h a r a c t e r of
G
afforded by
CHAPTER I11
174
C r .x .(gx1 2 2 PXri
Let E
= e
1
1
+
... ,E
J(FG),
=
es
+
for all g E G
= 0
J(FG) be all block idempotents of FG/J(FG)
Vi, if necessary, we may assume that
By renumbering the
Vi
identity transformation on
(3)
and annihilated all V j
Ei
.
acts as the
with j # i.
Hence, for
x E FG,
any
if i = j (4)
if
z
Now (3) implies that
r.X.(e .(gr))= 0, for a l l
g E G.
i # j
Invoking ( 4 ) , we
vkr.
I .I
deduce that X i ( g s ) = 0
gor all g E G
n KerXi, proving
xE
and all i
such that p
k
ri.
Hence
(2).
P k i Applying (2) and Lemma 5.2, we derive
S
Since J(FG1 = n ann(Vi)
and since for any proper subset X
of
{ V l,...,vsl
i=1
J(FG)
C
n ann(V1, the result follows.
V€X 5.4.
LEMMA.
Let F be a splitting field for FG, let V be an irreducible FG-
module and let S E Syl (GI.
P
(i) n(V,FG) = dimP(V)
F
G
(ii) dimP(V1 = ISln(V,(lS1 1 F Proof. (i) Let e be a primitive idempotent of FG such that
V
1
FGe/J(FG)e.
Theorem 3.6(ii),
Then, by Proposition 1.10.16 (iii), n(V,FG) = dimeFG. But, by F Since P t V ) FGe, (i) is established. dimeFG = dimFGe.
F (iil is
Since lS
IS!.
F
is the only irreducible FS-module, the composition length of FS
Let
O = X C X 0
be a composition series of FS. we have
1
C . . . C X
Is1
=FS
Then, by Lemma 2.4.2(11 and Proposition 2.4.3(1),
J(FG)
A CHARRCTERIZATION OF ELEMENTS OF
0
=
XoG C X G C
... C X G I S I =
(FS1 G
2
175
FG,
where, by Proposition 2.4.4(iii1,
Invoking (i), we therefore derive
as asserted. As
a final step in the preparation for the proof of the main result, we
establish the following property due to Okuyama (1980). 5.5.
Let F
LEMMA.
r potent of FG sections of Proof.
t r ( e z ) = 0.
and let Z
G.
=
P'
C FS;
=
,
where S1,
... ,Sr
be a primitive idemare all p-regular
ez = 0 if and only if t r c e z ) = 0 . P' ' 0, then clearly t r ( e z ) = 0. Conversely, assume that
Then, for any Z
If ez
let e
be a splitting field for FG,
E
Z
Given y E FG, we have e y = eye
+ (eey - eye)
and so
ezFG C e z F G e + [FG,PG] Because F is a splitting field for G, e F G e / e J ( F G ) e
eFGe
=
Fe
+
2
F
and so
eJ (FG) e
Invoking Theorem 3.14, it follows that
ezFG C - Fez and hence that t r ( e z F G 1
=
+ e J ( F G ) z e + [FG,FGl = F e z + IFG,FGl 0.
Thus
ez = 0 as asserted.
We have now come to the demonstration for which this section has been developed. 5 . 6 . THEOREM (Brockhaus (19851).
Vl,...,V
r
Let F
be a splitting field for FG, let
be all nonisomorphic irreducible FG-modules and let S E Syl ( G ) .
Then the following conditions are equivalent: (i) J(FG1 = L(c1
P
CHAPTER I11
176
(iil
SocFG
(iiil p
j'
FGc dimP ( Vi)
=
for a l l
iE
{l,.
.. ,r}
Is1 (iv) cP(vi) # (v) cP
(vi) tr(ec1
o vi
for all i E {I,.. for all i E {l,
.,PI
...,r}
# 0 for every primitive idempotent e of FG
Proof. (i) * (ii): Since c E Z ( F G 1 , J(FG)
=
&(el, then by Proposition
Theorem 3.6(iii),
SocFG = FGc.
we have
R(cl
=
R(FGc).
SocFG = r(R(FGc) and hence, by
1.6.29,
Conversely, assume that SocFG =
Proposition 1.6.30 and Theorem 3.8(iii),
If
JtFG) = LCFGC)
FGC.
Then, by
Thus (1) is
= k(C).
equivalent to (ii). (i) * (iii): It is a consequence of Lemmas S.l(fi) and 5.3 that J(FG) = KerT if and only if p
1 n(Vi' (lS)G )
for all i E {l,
... ,r}
Hence, by Lemma S.l(i) ,
and so the desired conclusion follows by virtue of Lemma5.4.(ii). (ii) * (iv) and
(ii) * (v): Let FG = FGe @
is an indecomposable left ideal and FGce
Vi
... @
FGei/J(FG)ei, 1
FGet where each FGei
i
P.
Then
. C F G ez. n FGc
z-
C FGe. n SocFG 2
(by Theorem 3.14)
= Soc CFGe .) 2
t t Since Soc(FG) = CB soc(FGei) and FG" = FGcei , it follows that (ii) is i=1 i=1 equivalent to soc
=
FGce
i
But, by Theorem 3 . 8 ( i i i ) ,
Soc(P(vi))
vi
i
for all
{l,
E
...,PI,
hence t h e
assertion. (iv)
0
idempotent
(vi):
e
BY Lemma 5.5,
of
FG.
ce # 0 f o r any p r i m i t i v e
(vi) is equivalent t o
Since t h e l a t t e r i s obviously e q u i v a l e n t t o ( i v ) , t h e
r e s u l t follows. 5.7. COROLLARY
a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
p
J(FG)
=
be an
be a p-solvable group.
Then
and
J(FG)
=
Let
R(c)
By P r o p o s i t i o n 1.6.29 and Theorem 3 . 6 ( i i i ) , it s u f f i c e s t o v e r i f y t h a t
R(c).
i c a l l y closed. 5.8.
G
and l e t
SocFG = FGc
Proof.
F
(Formanek and Snider (19771, Tsushima (1971a)).
COROLLARY.
.
F
By Corollary 1.18, we may harmlessly assume t h a t
Now apply Theorems 5.6 and 4.26. Let
F
be an a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
G
a p-solvable group and l e t
P
be t h e s e t of a l l p-elements of
i s algebra-
p,
G
let
G
be
including 1.
Then
Proof.
D i r e c t consequence of Lemma 3 . 1 2 ( i i ) and Corollary 5.7.
6. A CHARACTERIZATION OF ELEMENTS OF
F
Throughout t h i s s e c t i o n ,
G
a f i n i t e group and
c
.
J ( z (FG)
denotes an a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
t h e sum of a l l p-elements of
a consequence of Theorem 3.14 t h a t
J ( Z V G )1
=
Z (FGI
G
(including 1).
R (St), where
i= 1 a r e a l l p-regular s e c t i o n s of
if that
G
i s p-solvable.
.
G.
S1,.
p > 0, It is
.. ,S
Furthermore, by Corollary 5.7,
The aim of t h i s s e c t i o n i s t o prove t h e remarkable f a c t
CHAPTER I11
178
G.
f o r an a r b i t r a r y f i n i t e group
6.1. LEMMA.
A
algebra
e
Let
over a f i e l d .
A
p o t e n t s of
I
and of
of
A.
Then, f o r a l l
e
i n t o p r i m i t i v e idem-
a r e t h e same.
,
1 ei i=1
e =
Let
of a finite-dimensional
Then t h e decompositions of
n Proof.
I
be any idempotent i n an i d e a l
e
where t h e
a r e orthogonal p r i m i t i v e i d e m p t e n t s
i
j E 11 ,..., n } , n ee j
e
so t h a t 6.2.
E
j
I,
c eiej
=
j
ej
a s required.
THEOREM (Iizuka-Watanabe (1973)).
Proof.
= e2 =
.
(Kiilshammer (1981a)
R(FG)
I t i s obvious t h a t
J(Z(FG))
=
z(FG)
fl R ( C )
Put
i s an F-subspace of
FG.
X
Denote by
sets of a l l p-regular elements (excluding 1) and a l l p-elements respectively.
Since f o r a l l
y E Y,y-l
5
R(FG1
exist
a
Y,
then
a
F(y-1).
E
(including l),
+ Z F(y-1)
C F3:
y=Y
To prove t h e reverse containment, w e need only v e r i f y t h a t i f
5
the
Is n i l p o t e n t , we have
fix
Suppa
Y
and
But i f
Suppa
C_ Y and a
a
E RVG)
E R(FG),
with
then t h e r e
6 Y Y
E F
such t h a t
u = C a (y-1) + C a 9YEY y and
C a YEY
E R(FG),
which i s p o s s i b l e only i n t h e case
I t follows from (1) t h a t
tr(a R(FG))
ac€X,y E Y ,
c
c ~ ~ ( y - 1 ) . This
YEY
proves t h a t
that
a =
= 0
we have
Fc
5 R ( F G )1 .
f o r some
tr(&
To prove t h e r e v e r s e containment, assume
a = C a#
= 0
E FG. Then, by (11, f o r a l l SEG and t r k z y ) = tra. Thus, f o r a l l
A CH~ACTERIZATIONOF ELEMENTS OF
x E X,y E Y , a -1
=
0 and a
= u
Y
X
-1’
Fc We now claim that R ( F G )
J(z(pG1)
This proves that a
=
=
a
179
c and hence that
1 R(FG)
(2)
does not contain blocks of FG.
Let e
=
Ze g g
By Theorem 2.2.4, there exists
a
p-regular element g E G
such that e
g
n hypothesis, tr(eg-’
5R t F G ) .
and assume, by way of contradiction, that F G e
block idempotent of FG
0 for all sufficiently large n.
=
be a
# 0.
By
We may choose n
a contradiction. It is now an easy matter to complete the proof.
J(Z(FG))
1.10.6,
=
Z(FG) n J(FG)
Indeed, by Proposition
and hence, by Theorem 3.14,
n Z(FG1
To prove the reverse containment, it suffices to show that t ( e ) nilpotent ideal of
Z(FG)
.
Assume the contrary.
Then R(c)
n ZtFG)
primitive idempotent, say e, by virtue of Corollary 1.6.38. is a block idempotent of FG
and so
PGe
5 Rlc).
is a contains a
By Lemma 6.1,
e
However, by (2) and Theorem
3.6(i),
which implies that R(FG]
where
G
P
contains the block
is the set of all p-elements of
G
FGe,
a contradiction.
H
(including 1).
Proof. Direct consequence of Theorem 6.2 and Lemma 3.12Cii). For later use, we next establish the following result due to Reynolds (1972) Let S 1 ,s 2 ,...,SP
6.4. THEOREM.
r
Z
P’
=
x
FS;
properties:
.
Then Z
P’
be all p-regular sections of
is an ideal of
Z(FG)
G
and let
which satisfies the following
180
C W T E R I11
of d e f e c t zero. if
Z
The p r i m i t i v e idempotents of
(il
B
(ii)
where
FG
a r e t h e same a s block idempotents of
P'
B
I n p a r t i c u l a r , a block
of
FG
i s of zero d e f e c t i f and only
i s a simple F-algebra. =
P' Bl,
z ( B ~ )8
...,I3
.. . @ z ( B ~ )
are all blocks of
21
is a s p l i t t i n g f i e l d f o r
FG,
then
FG
of d e f e c t zero.
dimZ2 F P'
F
I n particular, if
i s equal t o t h e number of blocks of
FG of d e f e c t zero. Proof.
Z
That
P'
and Lemma 3.13 ( i i l
.
(i) Assume t h a t
e
ZPG)
i s an i d e a l of
i s a p r i m i t i v e idempotent of
i s a block idempotent of
FG.
Because
suppe
follows t h a t each p-regular s e c t i o n i n l a t t e r implies t h a t
e
Thus
i s a consequence of Theorem 3 . l O ( i i i l
suppe
G, i t
G.
The
i s a union of conjugacy c l a s s e s of p-defect zero.
e
i s a block idernpotent of d e f e c t zero.
i s a union of conjugacy c l a s s e s of p-defect zero.
Suppe
of
i s a block idempotent of d e f e c t zero.
Conversely, suppose t h a t
we have
e E Z P"
Hence, by t h e foregoing, But
B = B(e)
J ( B ) = 0 i f and only i f
Theorem 3 . l O ( i i i ) , Soc(FG1 n Z(FG) = 2
P'
i s a block of
e E Soc(FG) n Z(FG).
and thus
'
Because these c l a s s e s
P"
To prove t h e f i n a l statement, assume t h a t
J ( B ) = J(FG)e,
J(B1 = 0
J ( B ) = 0 i f and only i f
J ( B ) = 0 if and only i f
B
Then
e E Z
a r e a l s o p-regular s e c t i o n s , we deduce t h a t
zero.
c o n s i s t s of p'-elements
i s a conjugacy c l a s s of
Suppe
e
Then, by Lemma 6.1,
P"
FG.
Since
By
i f and only i f
B
i s of d e f e c t
i s a simple F-algebra, hence t h e a s s e r -
tion.
Ciil
By Theorem 3.10Ciii) and Lemma 3 . 1 3 ( i i ) ,
where
B
B 1 ,...,Br
,...,Bp
FG
a r e a l l blocks of
Assume t h a t
FG/J(FG1,
a r e a l l blocks of
F
FG
which a r e simple F-algebras.
of d e f e c t zero.
is a s p l i t t i n g f i e l d f o r
i t follows t h a t
Bi
2
By ( i ) ,
Mni(F)
FG.
f o r some
Since each
Bi
i s a block of
ni 3 1, 1 6 i 6 r .
Thus
A CHARACTERIZATION OF ELEMENTS OF
J
(Z(FG11
.
i' C r , and t h e required a s s e r t i o n follows.
dirnZ(B.1 = 1, 1 C
F ' Z;,
We c l o s e by providing circumstances under which
c J ( Z ( F G ) 1. P' -
P
Let
LEMMA.
= 0
and hence
For t h i s w e need t h e following two simple observations.
2
6.5.
181
G.
be a normal p-subgroup of
Then t h e k e r n e l of t h e
Brauer homomorphism
is a nilpotent ideal. Proof.
cE
Let
CR(G)
c(P) = @
G is nilpotent.
C+
it s u f f i c e s t o v e r i f y t h a t
2.2.3,
cn
be such t h a t
.
Invoking Theorem
FG'I(P)
Since
is a
n i l p o t e n t i d e a l of FG (Proposition l . l ( i ) )t,h e l a t t e r w i l l follow provided we C+ E F G . T ( P ) .
show t h a t
P
The group
0 ,
C fl C(P1 =
Since
t h e s i z e of each o r b i t under t h i s a c t i o n i s d i v i s i b l e by
G
p.
x
But f o r a l l
C'
Hence
E FG.T(P)
E P
w e have
and t h e r e s u l t follows. C E CR(G),
which is t h e F - l i n e a r span of
6(C)
c,
g
and
I n what follows, given
6(X1
C.
a c t s by conjugation a s a permutation group of t h e set
,'X
I[Cl
we w r i t e
X
where
f o r t h e i d e a l of
ranges a l l elements i n
Z(FG)
CR(G)
with
(see Lemma 2.2.5).
n
Ll
6.6. LEMMA. tively,
P be a normal p-subgroup of
Let
Ck+l,...,C
t 1 be a l l conjugacy c l a s s e s of
( r e s p e c t i v e l y , a l l conjugacy c l a s s e s of proper subgroups of
B
t h e F-linear
(i) A
lii)
P)
span of
.
Denote by
A
.. , C i .
Ci+l,.
is a subalgebra of
C E { C I,...,C
G
Z ( F G ) ,B
k 1, T I C ] = A @ B
G
and l e t
G
C
1
,...,Ck
with d e f e c t group
t h e F - l i n e a r span of
..., C i
,C:
and by
Then
is a n i l p o t e n t i d e a l of
Z(FG)
and f o r any
( d i r e c t sum of F-spaces)
P.
FG
contain
d i s p , then any block idempotent of
FG
of d e f e c t
P
P
whose d e f e c t groups a r e conjugate t o
The d e f e c t groups of any block idempotent of
o r d e r of
(respec-
If the
d
lies in
A.
CHAPTER 1x1
182
z',j e {l,
(i) Fix
Proof.
++
C.C. = 2 3
where X r E F , 1 G r
x
c+ +
1 1
t.
Q
..., k j .
we have
... + hkc+k + xk+lc;+l + ... + XtCL
Let
be the Brauer homomorphism.
Since
C C P ) , 1 G r G k, and so n(C,l
Cr
By Lemma 2.2.5,
C1,C2,
...,Ck
= C :
for all r E 11
have P as their defect group,
,...,k}.
On the other
G
hand, since the defect groups of of P ,
Thus
are conjugate to proper subgroups
Ck+l,...,Ct
we have
+
n ( Ck+l) =
... =
proving that A
n ( C + ) = 0 and therefore t
i s a subalgebra.
By the definition of I [ C ] , I [ C ] = A @ B also have B
=
Kern n T [ C ] and so B
nilpotent (Lemma 6 . 5 ) ,
B
as F-spaces.
is an ideal.
e.
Let C E CR(G)
and so, by Lemma 6.5,
Now assume that e group of
is
D as a defect group.
By
Suppe is a union of conjugacy classes whose defect groups are
conjugate to subgroups of D. C n C(P) = 0 G P C D.
Finally, since Kern
we
is a nilpotent ideal,
(ii) Let e be a block idempotent of FG with Proposition 2.2.6,
By Theorem 2.2.3,
e
with
Then, by the above, P
Hence, by (i) and Proposition 2 . 2 . 6 ( 1 ) ,
and the result follows.
=
5 Suppe.
If P
is nilpotent which is impossible.
is of defect d.
n But there is an integer n with bp
C
0.
Hence
D then Thus
i s a defect
e = a + b with a E A,b
EB.
183
FROBENIUS GROUPS
6.7.
COROLLARY. F
Z
P'
=
C FS; i=1
.
,...,sr
s1 ,s2
Let
G
If
has a n o n t r i v i a l normal p-subgroup,
,z; Z
and, i n p a r t i c u l a r , Proof.
and l e t
G
be a l l p-regular s e c t i o n s of then
= 0
J ( Z (FG) 1 .
P'
Apply Lemma 6 . 6 ( i i ) and Theorem 6.4Cii1.
7. FROBENIUS GROUPS. A t r a n s i t i v e permutation group
G
i n which o n l y t h e i d e n t i t y f i x e s more than one
l e t t e r , b u t t h e subgroup f i x i n g a l e t t e r i s n o n t r i v i a l i s c a l l e d a Frobenius group. I n what follows w e s h a l l use t h e following two standard f a c t s f o r t h e proof of which w e r e f e r t o Gorenstein (1968). (Frobenius).
(1)
G
Let
f i x i n g a letter.
be a Frobenius group and l e t
G
Then t h e subset o f
H
be the subgroup
c o n s i s t i n g of t h e i d e n t i t y t o g e t h e r with
N
those elements which f i x no l e t t e r s forms a normal subgroup
G
of
of o r d e r
1G:HI (21
(Thompson).
L e t a group
G
admit an automorphism of prime o r d e r which
f i x e s only t h e i d e n t i t y element o f
G.
Then
W e s h a l l r e f e r t o t h e normal subgroup
N
G
is nilpotent
and t h e subgroup
H
as t h e Frobenius kernel and Frobenhm complement, r e s p e c t i v e l y . s e c t i o n i s t o reduce t h e c a l c u l a t i o n of
J(FG1
t o t h a t of
G i n (1)
of
Our a i m i n t h i s
J(FH).
W e s t a r t by
recording some group-theoretic r e s u l t s . LEMMA.
Let
G
(i) G = NH
with
N
Cii1
h # 1 in
7.1.
be a Frobenius group with complement
H
and k e r n e l
N.
Then
Every
fl H =
H
1 induces by conjugation an automorphism of
f i x e s only t h e i d e n t i t y element of
Ciiil
IHI
(iv)
N
Cvl
If
(vil
IN[
divides
N
-1
is nilpotent
g e G-Nl
For every
2
then
g
lies i n a conjugate of
# 1 in N, CGbl
5N
H
N
which
IIL
CWTER
184
(i) BY the definition of N ,
proof.
IGI = I N H I .
by (1) and hence
Thus G = NH
is transitive and H
G
(ii) Since
N nH
we have
=
nh
=
n
is the subgroup fixing a letter, our
themselves as a transversal for H
for some h # 1 in B
fixes the coset H n
and n # 1 in N .
as well as the coset H .
and n # 1 in N ,
of rn
the set
rX
=
But, by definition of a Frobenius Hence n
rY
rn
{nh I H E H
=
or
r
X
n
rY
=
.
0
h
1
# n for any
must consist of
in N
Hence
liv-{l}
with
I
x # 1 and
is a multiple
and (iii) follows.
(iv) Since H # 1, we may choose an element h E H
h
Suppose that
Then it is immediate that h
But clearly for x,@
distinct elements of N .
y # 1, we have either
can take
proving ( i i ) .
(iii) For a fixed n # 1 in N ,
rn = IHI
, we
By ( i l
in G .
group, only the identity fixes more than one letter.
h # 1 in H
IN1 = 1G:HI
But
as required.
representation is equivalent to that on the cosets of H . the elements of N
1.
of prime order.
By (ii),
induces an automorphism of N which fixes only the identity element of
Hence, by (21,
N is nilpotent. then g
(v) If g E G - N ,
fixes a letter.
simply the subgroups fixing a letter. Ivi) By (v), N
LEMMA.
in G.
Let H
Frobenius group with complement H
or equivalently, h E =
for all g E G - H ,
H n /?
1 for
=
i2
Thus G 2.
N,
as required.
1 for all g E G - H.
and H
H
be a complete set of cosets of
with
i
=
.
is a Probenius
then by transitivity some
i > 1.
Thus G
if and only if no element of H - {l)
However for h # 1 in H,Hgi
> 1.
Then G
G.
G fixes two cosets of H ,
fixes one of the cosets H g
are
G which lie in no con-
for x # 1 i n
@
q = 1, 1 G i G n,
G
If an element of
only if H n Hgi
5N
if and only if H n
H
lies in a conjugate of H .
be a nontrivial subgroup of
Let H g i , g i
element of H - { l )
any H g i , i
Hence g
We conclude that C (XI G
group with complement H Proof.
But the conjugates of
consists precisely of the elements of
jugate of H - 11). 7.2.
N.
Hg.h 2
is a fixes
if and only if gihgil
E H
have the given properties if and
Since the latter is equivalent to H n H g = 1
the result follows.
185
FROBENIUS GROUPS
Turntng our attention to S(FG1,
we next establish the followtng elementary
result. Let F be an arbitrary field of characteristic p
7.3. LEMMA.
normal p'-subgroup of a group G
H such that G
NH and N n H
=
and let e = 1,
=
1BI-l C n. TEN
0,
7
If G
let N
be a
has a subgroup
then
(i) J(FG)e = J(FH)e dimJ(FHle = dimJ(FU1
liil
F
F
Proof. The element e J(FGe)
=
J(FG)e
by Proposition 1.6.36.
FH
FGe zz F(G/N)
is obviously a central idempotent of
-
is an isomorphism of F-algebras.
7.4.
LEMMA.
On the other hand, by Proposition 2.1.9,
FU -+
FGe
z++ ze
Hence J(FGe)
=
J(FH)e
1
IGl
and
. Let F be a field such that
splitting field for FG.
xi
FG-modules and by
Denote by
charF
...,Vr
Vl,V2,
and such that F
FG.
is a
all nonisomorphic irreducible
the character of FG afforded by
are all distinct block idempotents of F-basis for
Hence
Thus
I as required.
FG.
i
r.
Then
,... ,er}
is an
Vi, 1
In particular, {e
2 (FG).
Proof. Because FG
is semisimple (Maschke's theorem) and F is a splitting
field for FG, there exists exactly r
block idempotents, say
2,
,...,U r ,
of
FG. Thus we have: (a) FG = FGV 0 (b) FGvi
(uivj
Mn ( F ) , where n
i (c)
... @ FGu, =
i
= 0
for i # j ,
dimV
F
The irreducible constituents of the FG-module FGvi
are isomorphic to i'
CHAPTER 111
186
and
occurs with multiplicity n
Vi
Let p
2" FG,
be the regular character of
i.e. the character afforded by
FG.
It
follows from (a) - (c) that
(FGu .l = 0 (1 S i,j r,i # i l 3 Let x be an arbitrary element of FG. (el
i
xi
E
such that x
FGvi
mi
that xi =
(1 4
=
x
+
i < PI.
By (a1 there is a unique element
... + X r .
Because
viuj
= 0 for
Hence (e) implies if j = l,2r...,r
X.(Xv.)
3
-L
=
x.(x.) 3
%
i # j , we see
and j #
i
=
if j = i
In view of (a), this gives
P(rni1
=
Xi(l)Xi(X)
We now fix
i
E {l,
(fl
x
9
...,PI
is the coefficient of
for all x E F G , 1 G and write V
=
1 when g - l V i
i
c x g for some x E F . Evidently, &Gg 9 is written as a linear combination of
elements of G :
g Applying
P
-1
uz = x '1 + 9
...
and using the equality PCCX g ) = x / G I ,
9
we obtain
Hence, by (fl , we derive
and therefore
Thus
vi
= ei
Let G
for all i E {l,
...
and the result follows.
be a group of linear transformations of a vector space V .
fixed-point space of V
(relative to GI
By the
we understand the set of those V E V
187
FROBENIUS GROUPS
for which gW1
= t,
for all g E G.
A weaker verston of the following result
would be sufficient for the proof of the main theorem of this section.
However,
the more general result will be needed later. Let G be a group of linear transformations of a vector space V
7.5. LEMMA.
over a field F. element of
Assume that X
Y are two bases for V such that each
and
induces a permutation of the sets X
G
have the same number of G-orbits.
Furthermore, if
g E G, the number of fixed elements of X Proof. Let X ,X 1
2
,...,Xn
respectively, and let W sitively on each X i ,
Yl,
and
and
Y.
charF
=
Then X
Y.
be all G-orbits of
be the fixed-point space of
V.
Y
0, then for each
is equal to that of
...,Ym
and
X and Y,
Since G
acts tran-
it is immediate that the elements
E x "'Xi
constitute a basis for proving that n Fix g E G mation g
=
W.
= dimW
and, by the same argument, m=dimW
m.
and denote by A
with respect to X
similar and so
n
Thus
trA= t r B
and B
and
Y,
the matrices of the linear transforThen A
respectively.
Now assume that charF = 0 .
elements of
X
and
Y, trA and trB
elements of
X
and
Y.
and
Since g
B
are
permutes the
are, respectively, the numbers of fixed
So the lemma is true.
The next observation is crucial.
7.6. LEMMA.
Let N
be a normal subgroup of G, let F be an algebraically
closed field such that
charF
) IN1 and let H be
(i) The number of conjugacy classes of
G
a subgroup of
contained in N
number of H-conjugacy classes of block idempotents of (iil For each h E H,
(iii) If G
FN. with
(i) Let
x
be the character of N
V
2
h
V
with C = hCh-'.
is a Frobenius group with complement H and kernel N,
is irreducible for any irreducible FN-module V Proof.
is equal to the
the number of irreducible FN-modules V
equals the number of conjugacy classes C of N
G with G=NH.
with
V
then
?
lN.
afforded by an irreducible FN-
CHAPTER I11
188
V,
module
If
gx
n
N.
E
and let
is the character of N
for all
Hence
Let C
{1},C2,...,C,
=
exists exactly
x
and if 1Q
gV, then ’X(n) = X(g-’ng)
afforded by
?1
be all conjugacy classes of N .
nonisomorphic irreducible FN-modules, say
is the character of
i
of linear transformations of
X
=
Z(FN1.
{C+
V
=
lNlV2,-..,V,
FN afforded by Vi, then the elements e
i < r , form an F-basis for Z(FN).
Then X
By Lemma 7.4, there
,...}:C,
Now H
,
X i
acts by conjugation as a group
Put and
Y
=
,...,e$1
{eXl
Y are F-bases of ZWN) and each element of H induces a
and
permutation of X and Y.
Y are conjugacy
Since the H-orBits of X and
G contained in N and 8-conjugacy classes of block idempotents of
classes of
FN, respectively, the assertion follows by appealing to Lemma 7.5.
x
(ii) Let
be the character of N
h
Then, for any h E H , V Since charF
k
= 0,
h
x= x
V if and only if
IN1 , the number of X
teristic of F charF
1:
V.
afforded by an irreducible FN-module
with
x
=
h
-1
or if and only if he,,h
=e
A
x
is independent of the charac-
(see Curtis and Reiner (1981,p.420)).
Hence we may assume that
in which case (ii) follows by virtue of Lemma 7.5. and let W
(iii) Let V be a nontrivial irreducible FN-module
be an irre-
n
ducible factor module of submodule of 4.l[ii),
h~
9
v
W
1
WN.
x # l,n
G
V
for all h # 1
E N
Then, by Theorem 2.4.10,
Hence if the inertia group of
VG and
Let C E CRU)
v“.
is irreducible.
v
is
V
is isomorphic to a
N,
then by Theorem
We are therefore left to verify that
in H.
be such that C = hCh-’ and -1 and hence with with hxh-’ = nxn
.
X
C # 1.
Then there exists
n-lh E CG(x). But by Lemma
FROBENIUS GROUPS
7.l(vi),
5N ,
CG(xl
and thus h
so n-lh E N
.
fixes no nonidentity conjugacy class of H required.
=
189
Thus h
1, a contradiction.
and hence, by (iil, hI'
v,
?
as
We have now accumulated all the information necessary to prove the following result. Let G
7.7. THEOREM. (Motose (1974a)l.
and let F be an arbitrary field of characteristic p > 0.
and kernel N ti)
If p
If p
IN1
divides
P is a Sylow p-subgroup of N
and
(and hence of
G,
then J(FG1 = FG-ICPI
by Lemma 7.l(iii)),
(iil
be a Frobenius group with complement H
does not divide
I N ! , then
J(FG) = J ( F H ) e and dimJ(FG) F
=
dimJ(FflH)
F
where
Proof.
(il
Assume that p
By Lemma 7.1 (ivl, N
of N .
divides
IN\ and that P
is a Sylow p-subgroup
is nilpotent and hence P 4 N .
P 4 G and
Thus
the result follows by virtue of Proposition l.l(iiil. (ii)
Suppose that p
that F
does not divide
is algebraically closed.
Hence FN(1-el
Note that FN = FNe @3 FN(1-el
=
0
be a field of characteristic p P
and kernel Iv.
4
(FN (1-el 1 G
and therefore J(FG) = J(FGe1.
7.8. COROLLARY (Wallace (195811.
Then
FNe
1
1N'
Invoking
we deduce that
FGC1-el
J V G ] (1-el
and
is a direct sum of nontrivial irreducible FN-modules.
Lemma 7.6 and Proposition 2.4.3(11,
Thus
By Corollary 1.18, we may assume
Let P and l e t
is completely reducible
Now apply Lemma 7.3.
be a Sylow p-subgroup o f
G,
let F
G be a Frobenius group with complement
CHAPTER I11
190
where
In particular, dimJ(FG) = IPI
-1
F Proof.
Apply Theorem 7.7 for H = P
and Corollary 1.2.
We close this section by considering the following two examples. 7.9. EXAMPLE.
p = 2.
Let C = S
Then G
G =
i.e.
3'
=
l , b 2 = l,bab-I = a'>,
and let
is a Frobenius group with complement P = < b > and kernel
N =. The element e in this case is given by e
1 -(l+a+a2) 3
=
=
(l+a+a2).
Since I(P) = { A (b+l)I A E F) , Corollary 7.8 implies that J ( F G ) = {hCl+a+a2)(l+b) Ih E F)
Ix E
= {X(1+u+a2+b+ab+a2bI
Let G
7.10. EXAMPLE.
cbc-l
=
=
G =
A ,i.e.
ab> and let p = 3.
Then G
P = < c > and kernel N =.
=
e3
=
l,ub=ba,eae-l=b,
is a Probenius group with complement
The element e
1
b2
F}
e = -(l+a+b+ab) 4
=
in this case is given by
l+a+b+ab
Hence, by Corollary 7.8,
7.11. REMARK.
Let G be a Frobenius group with complement P E Syl ( G I .
P
According to Burnside (1955,p.335), Hence, by Zassenhaus C1949,p.119), (i) If p # 2 , (ii) If
p = 2,
then P then P
P
has no noncyclic subgroup of order p2.
we have
is cyclic
is either cyclic or a generalized quaternion group.
'
UPPER AND LOWER BOUNDS FOR dimJ(FG)
191
F 8 . UPPER AND LOWER BOUNDS FOR
dimJ(FG)
F Throughout this section, G denotes a finite group, F an arbitrary field of characteristic p > 0, lG or
lFG
the trivial FG-module and u = dimP(lG)
F Our aim is to provide an upper and lower bounds for dimJ(FG1.
It turns out that
F is of maximal F-dimension if and only if the Sylow p-subgroup P
J(FG1
normal, while S(FG1
is of minimal F-dimension if and only if P
of
G is
is disjoint.
Some of the results presented here were originally stated
from its conjugates. for splitting fields.
The following simple observation will eliminate that
assumption. 8.1. LEMMA.
Proof.
Let L be a field extension of
Put
V
=
1FG
.
Then V
Proposition 1.19 and Corollary 1.18. 8.2.
LEMMA.
Let P
= lFL
L
F.
Then
and the result follows by virtue of
H
be a Sylowp-subgroup of
G
and let V be a projective FG-
IPI divides did. F Proof. By Corollary 1.2, PP is a local ring.
module.
module,
Then
V p is a projective module.
8 . 3 . LEMMA.
Let V
V.
is a free FP-
NOW apply Lemma 4.18.
Then P(V1
be an irreducible FG-module.
a direct summand of P(lG) 8
Since FG
is isomorphic to
In particular,
F dimP(V1
u dimV
F Proof.
BY Proposition 2.5.7,
F
P(1 1 8 V
is a projective FG-module.
On the
G F other hand, by Theorem 3.8(1111,
1 c P(lG). G-
Hence
n
...,Vn.
for some irreducible FG-modules V1,
It follows that
V is isomorphic
to
CHaPTER ILI
192
a submodule of
n
n
P(Vi)) = Q soc(P(Vi)) i=1 i= 1
SOC( @
But, Theorem 3.8(iii),
i
SocP(V.)
E c1,. . .,n}, and thus P ( V )
2
2
for all 2.
V.
P(VJ
Hence
V.
V
is a direct summand of
for some
P(lG) 8 V
F
.
We are now ready to prove the following important result. U
be a group of order p rn
Let G
8.4. THEOREM.
(i) (Wallace (1958))
dimJ(FG)
F
(p,rn) = 1.
where
Then
> pa - 1
(ii) (Brauer and Nesbitt (1941)) dimS(FG) 4 I G (
M
-
F Proof. e
,...,e21
... @ FGe,
FG = FGe @
(i) Write
of PG with FGe /J(FG)e
5 J(FG),
Since J(FG)el
Then, by Lemma 8.2,
= dimFGel - 1
dimJ (FGIe
F
lG.
2
1
for some primitive idempotents
'
> pa - 1
F
the required assertion follows.
(iil By Lemma 8.1 and Corollary 1.18, we may assume that F is algebraically closed.
Let V ,V , . . . l V 2
-
be all nonisomorphic irreducible FG-modules.
n
Then
n dimFG/J(FG)
=
F =
C (dimVi) i=1 F
(by Corollary 1 10.17)
p i=l(dimV
l n 3; C (dimVil (.dimP(Vil i=1
F
F
=IGI
(by Lemma 8.2)
(by Corollary 1.10.17)
and the result follows. We next provide necessary and sufficient conditions for J(FG) to be of maximal P-dimension.
The lemma which follows disposes of some of the technical-
ities. 8.5. LEMMA.
Let P
be a p-group acting on a p'-group
izes each conjugacy class of
Q.
P ) I CI
I
stabil-
Then P acts trivially.
Proof. Let C be a conjugacy class of Q. Since
Q such that P
there exists c E C
By assumption P acts on C.
such that c E
c P).
l?
Hence
UPPER AND LOWER BOUNDS FOR
dimJ(FG) F
193
and t h e r e f o r e
Q
u I C Q ( P)P
=
SfQ The d e s i r e d a s s e r t i o n i s now a consequence of t h e following g e n e r a l observation: If
H
is a subgroup of
G
Indeed, assume t h a t
(H n
ha1
< (HI.
G =
with
G # H.
Then
a contradiction.
IHIIG:N(H)I
<
Let
G
G.
-
1G:PI
then
I
P-solvable groups of smaller o r d e r .
N
Case 1:
of
G/N
and hence
IGI
Then
i s normal.
u = [PI where
F
P i s a Sylow
Thus we a r e l e f t t o v e r i f y t h a t
dimJ(FG) = IGI - ] G : P I , F Assume t h a t dimJ(FG) = [GI - G:PI. F case [GI = 1 being obvious, assume t h a t
subgroup
g
G
Hence, by Theorem 4 . 2 6 ,
dimJ(FG) = I G l F
P d G,
g E G
f o r some
By Lemma 8.1, w e may harmlessly assume t h a t
(Brockhaus (19821).
is a l g e b r a i c a l l y closed.
If
# H
be a p-solvable group.
G
i f and only i f t h e Sylow p-subgroup of
p-subgroup of
@
So t h e lemma i s t r u e .
THEOREM (Wallace (1961) 1.
Proof.
G = H.
then
But then
[GI
8.6.
ha,
U
sEG
G
such t h a t e i t h e r
i s a p’-group.
i f and only i f
by v i r t u e of Corollary 1.4. W e argue by induction on
Then
IGI.
The
/GI > 1 and t h a t t h e r e s u l t holds f o r
Because
G/N
P4G
G i s p-solvable, t h e r e i s a normal
i s a p‘-group or
P E Syl(N)
P
dimJ(FN) = IG:NI-’dimJ(FG) F F
G/N
is of order
p.
and
(by P r o p o s i t i o n 1 . 8 ( i i ) )
CHAPTER I11
194
P 4N
Thus, by induction,
IG/NI
Case 2: FN n JVG)
p
=
and
J(FN1 = 0
P 4 G.
and t h e r e f o r e
p ) INI.
N = 0 ,(GI
Then
P
and
G
=
NP
.
Since
(by Maschke's theorem) t h e map
i s an i n j e c t i v e homomorphism of F-algebras.
@
Furthermore,
i s an isomorphism
since dim(FG/J(FG)) = IGI F Define
$
:
FG+
FN
morphism such t h a t of
N
g E P.
and
and hence
dimJ(FG) = IN1 F
$(z) = $-'(s+JCFG)).
by
$
-
Then
FN.
i s t h e i d e n t i t y map on
$
Let
is an F-algebra homobe a conjugacy c l a s s
C
Then
g-lCg = C.
Tnvoking Lemma 8.5, w e deduce t h a t
N
= C(P1
and hence
N that
P d G. Case 3 :
IG:NI = p
and
p i 111.
and t h e r e f o r e , by Theorem 8 . 4 C i i 1 , induction, t h a t deduce t h a t
Q4N
Let
Q E Syl U). Then P
dimJ(FN) = F and hence t h a t Q 4 G.
IN1 - 1 N : Q I .
I t follows, by
We now apply Corollary 1.5 t o
UPPER AND LOWER BOUNDS FOR dimJ(FG)
195
F and the result i s established.
Remark.
The result above holds f o r an arbitrary group G.
This was proved
by Brockhaus (1985) whose argument relies on classification theorem of finite simple groups. We now head towards providing circumstances under which J(FG)
is of minimal
F-dimention. By the principal bZock of FG we understand the block which contains the trivial FC-module.
Let
V be an FG-module.
By the k e r n e l of
KerV, we understand the kernel of the representation of Assume that B
is a block of FG.
V,
written
G afforded by V.
Following Brauer (19711, we define the k e r n e l
of B, written KerB, by KerB = n KerV
EX where
X is the set of all irreducible FG-modules in B .
The following results
?re extracted from a work of Michler (1973). 8.7.
LEMMA.
Let B
=
B(e1
be a block o f KerB
Proof.
We may write
e
=
FG.
Then
{ g E GI (1-gle E J ( F G ) e )
in the form e = e
i
... + es,
where e , . . - , e 1
are orthogonal primitive idempotents of FG, and where
Vi
= FGei/J (FG)ei
are all nonisomorphic irreducible FG-modules in B . If g E KerB, then G acts as the identity operator on each
Vi and so
(1 - g ) ei E J (FG)ei
for all
E {l,...,~}.
Thus
(l-gle E J(FG)e.
Conversely, assume that g E G
satisfies
( l - g l e E J(FG)e.
Then
CHAPTER 111
196
since S ( F G I V
=
i
0 and
V . =
evi
for all V i
E
Vi,
1 C
r.
This proves the
lemma. 8.8.
g
E
Let B = B ( e ) be a block of
LEMMA.
G
FG.
Then the following properties of
are equivalent:
(i) ge (ii) g
=
e
is a p-regular element of KerB
Proof.
Suppose first that ge = e .
Assume, by way of contradiction, that g
Replacing g by its suitable power, if necessary we may
is not p-regular.
Let D denote the cyclic group generated by
assume that g has order p.
g
t and write e
c
=
higi
with
hi
E
F
Then e = ge
and t = \ G I .
implies that
i=1
D acts fixed-point free on fg.11 z
i
t}
by left multiplication.
Hence we
may assume that
k
e =
c h,(g+g2+ j=1 J
... + 8-l+ d h j
Because e = e 2 and e
where k = t/p.
Y
=
k
e = e2 =
c h.(ge+g2e + j=1 J
gSe
I
for all s
a
1,
it follows that
... + $-le+3Pe1gj
k
a contradiction.
=
C pX .g .e j=1 J J
=
0.
Consequently (il implies (ii)
Now assume that
.
g is a pregular element of KerB.
Since J ( F Q
is nil-
potent, it follows from Lemma 8.7 that
for some integer k 2 0. g =
gpkucru,
proving (il.
If r
for some integers u
is the order of and V .
So the lemma is proved,
g,
then
(pk ,r)
= 1
But then (2) ensures that
and hence
dimJ(FG) F
UPPER AND LOWER BOUNDS FOR
n o m t p-comptement,
called the
G
wise,
is s a i d t o be p-nczpotdnt i f
G
A group
G
i s p-nilpotent i f
=
G h a s a normal p'-subgroup
G/N
such t h a t
0 (G). P'tP
197
i s a p-group.
N,
Expressed o t h e r -
We have now accumulated a l l t h e
information necessary t o e s t a b l i s h t h e following two u s e f u l r e s u l t s .
8.9.
PROPOSITION.
(il
N
(iil
i s a p - n i l p o t e n t normal subgroup of = 0 (G/O
N / O ,(A']
P
P
( i ) Let
Proof.
X
Lemma 8.8,
P
c
f = [XI-'
r.
X€X .~ such t h a t F G f
M/X
P
FG
N = KerB.
and l e t
G
r(N)) be t h e s e t of a l l p-regular elements of
X
i s m u l t i p l i c a t i v e l y closed.
(ii) By ( i ) , X = 0
that
B = B ( e ) be a block of
Let
(N)
and
N/X
Hence
G/X.
G/X.
Set
f is a c e n t r a l idempotent of
Applying Lennna 8.8,
i s a normal p-subgroup of
Then, by
must be p - n i l p o t e n t .
is a normal p-subgroup of
Then, by Proposition 2.1.9,
P(G/X).
N
N.
e
we a l s o have
=
f e = ef.
FG Assume
Then, by P r o p o s i t i o n l . l ( i . 1 ,
and t h e r e f o r e
( l - m l e = (1-mlfe
proving t h a t
M
5N ,
(i)
B = B(e)
E GIge = e ) = 0
P'
(ii) K e r B = 0
P'IP
Proof.
J(FG)fe = S(FG)e,
be t h e p r i n c i p a l block of
( i ) Assume t h a t
But i f
V
FG.
E
M
.
FG.
(G)
i s a c e n t r a l idempotent of
p o t e n t s of
rn
(G1
g E 0
(GI
and p u t
P' f
for a l l
So t h e p r o p o s i t i o n i s t r u e .
by appealing t o Lemma 8.7.
Let
8-10. PROPOSITION.
E
FGe
Since
FG,
so
ef
2,
= 10 ,(GI 1-l Ex Then P S O t(G) e ( 1 - f ) a r e a l s o c e g t r a l idem-
i s indecomposable, e i t h e r
i s t h e t r i v i a l FG-module, then
( e f l u = e ( f 2 , ) = e2, =
and
f
and t h u s
e f # 0.
fV =
2,
Hence
fg = f , w e conclude t h a t eg = efg = e f = e ,
ef = 0 or 2,
E V,
e(1-f) = 0
and
for a l l
e ( 1 - f ) = 0. so
e = ef.
Since
198
CHAPTER I11
proving that 0
P'
(GI
{ g E G ( g e = e1
Conversely, assume that ge = e p-regular element of KerB.
Then, by Lemma 8.8, g is a
for some g E G.
Applying Proposition E.lO(i), we therefore derive
as required. (ii) Owing to Proposition 8.10(ii), it suffices to verify that 0
P
The latter will follow provided we show that 0
g
0
P
.
(GI.
asserted.
P
Then, by (i), ge
=
(G)
5 op
.
g
6
P
So assume that
(KerBI
and thus, by Lemma 8 . 8 ,
e
(GI.
(KerB)= 0
oPf(KerB),
as
We are now ready to prove the following result. Let P
8.11. THEOREM (Wallace (1958)).
if and only if either G = P
Proof.
or G
be a Sylow p-subgroup of G.
Then
is a Frobenius group with complement P.
Owing to Corollaries 7.8 and 1.2, we need only verify the "only if"
and dimJ(FG1 = IPI - 1. By Corollary 1.18, there F is no loss of generality in assuming that F is algebraically closed. Write
part.
So assume that
G # P
FG where e ,...,e
=
FGe @
...
@ FGen
are mutually orthogonal primitive idempotents of FG
P
1 By (1) dimS(FG)el 2 IPI G' F assumption on dimJCFG) forces
FGe /J(FG)e
-
1.
and
Since J(FGle c J(FG), our 1 -
1
F J(FG)e2 = S(FG)e =
... = J(FG)en = 0
It follows from ( 3 ) that each FGei, i E {2,...,n},
is an irreducible module.
Furthermore, by Proposition 1.9.4, PGei $ lG for all that by Proposition 1.10.16(iii) a composition factor of FG lG.
is dimFGe
Thus e
i
E {2,...,n}.
Note also
and Theorem 3.6(ii), the multiplicity of
F are isomorphic to
(3)
1
.
lG
Hence all composition factors of
is linked to none of the e
i'
2
ci
as FLkl
n, which
UPPER FWD LOWER BOUNDS FOR dimJ(FG)
199
F i s the principal block.
implies, by Proposition 1.10.14, that FGe
Since
lG
is the only irreducible module belonging to FGel, we conclude from Proposition that G
8.10(ii)
is p-nilpotent.
Setting N = 0 ,(GI
P
G
we have
FN where
V 1
,...,I)
=
=
NP.
FNV @
Write
... 8 FNVs
are mutually orthogonal primitive idempotents of FN.
FG
=
FGv 8
.. @ Fms
,
and, by Proposition 2.4.3(i) and Theorem 4.23, each FGUi Hence, by the Krull-Scmidt theorem, s = n
i
,...,n ) ,
E I2
(81,
each h # 1 in P
x
8,
h
V.
fixes no conjugacy class of N
E if, C(x1 5 N. G Suppose that g E G-N.
g 8 g-'
2
n,
after
is irreducible for all
V $ lN
contains V with multiplicity 1.
Clifford theorem, for each h # 1 in P , V F
x
(FNV .lG
is irreducible for any irreducible FN-module
Hence, by Theorem 2.4.10,
1 #
.*
2
i
by virtue of (31, it follows that
fl
any
is indecomposable.
and FOvi P FGei, 1 Q
Since FCU
possible renumbering of the FGVi.
Then
Thus, by
By Lemma 7.6, we conclude that distinct from 1.
Hence for
3c
=
8
x
__h
Then
8
which is impossible unless
group of G.
E N
for some positive integer A;
but
i s in the centralizer of the p-regular element
and therefore g
Sylow p-subgroup.
8 x
= 1.
Hence g
is an element of some
Therefore an element either lies in
N or in a Sylow p-sub-
Hence the number of elements, excluding 1, in Sylow p-subgroups is
exactly m(pa-l),
where
for all g E G-P.
IGl
0.
= p
m , (p,m)
Hence, by Lemma 7.2,
= 1.
The latter implies that Pn@= 1
G is a Frobenius group with complement
P. As
a preparation for the proof of the final result, we record the following
fact. 8.12. LEMMA.
Let K
be a finite field.
Then each element k
of
K
can be
200
C W T E R I11
k
w r i t t e n i n t h e form Proof.
K
of
L
x'+y'
=
The r e s u l t is t r i v i a l i f
is a square.
Now assume t h a t
i s a proper subgroup of
k E K.
kk
If
k E L
=
K*
kR
=
kL
C ' 0
s i n c e i n t h i s case each element
charK # 2.
K*
k E L
k = c 2 k0 , c E K*,
K.
E
charK = 2 ,
and s i n c e
x2+u2 f o r some
i n t h e form
x,y
f o r some
L = {a21a E K*l.
Set
K*
is cyclic,
and some
=
x,y E K ,
LUkL
we obtain =
(cx)' +
Assume, by way of c o n t r a d i c t i o n , t h a t f o r a l l
square.
Then
K
( K 11 =
.
But
U L
i s a s u b f i e l d of
91( q ,
s contridiction.
= (0)
8.13. THEOREM (Brauer and N e s b i t t (1941)). a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
p.
f o r some
then w r i t i n g any
Hence it s u f f i c e s t o prove t h a t a t l e a s t one element of t h e form nonsquare.
Then
Let
K.
x,y E K , x2+y2 i s a
1x1
If
is a
X2+y2
=
G = SL(2,p)
q,
then we have
and l e t
F be an
Then
F Proof.
By Corollary 1.18, we may assume t h a t
F
i s a l g e b r a i c a l l y closed.
For t h e sake of c l a r i t y , we d i v i d e t h e proof i n t o t h r e e s t e p s . Step 1.
Here we prove t h a t t h e number of p-regular c k s s e s of
We denote by 2'
P
element of 2'
P
t h e f i e l d of
p
elements.
conjugate elements of
G
I
and
-I, where
equal t o
p.
and t h a t a l l pThis i s so s i n c e
have t h e same t r a c e ,
I t i s c l e a r t h a t t h e r e e x i s t p-regular elements i n
namely
G
element i n
with t h e same t r a c e a r e conjugate.
G
is
I t s u f f i c e s t o show t h a t every
i s t h e t r a c e of some p-regular
r e g u l a r elements of
G
I
i s t h e i d e n t i t y matrix.
G
with t r a c e If
1 €IF
P
2
and
and
1 #
-2,
*
Put
Then
T(X1 E G
and
Assume t h e c o n t r a r y ,
tr(r(All
= A.
We show t h a t
Then t h e p-component
r(xIp
T(X)
i s a p-regular element.
of
i s conjugate t o an
2,
UPPER AND LOWER BOUNDS FOR
dirrJ(FG1 F
201
element
for some a # 0 ,
and
T(x1
is conjugate to an element in the centralizer of A in G consists of all elements of the form
However the centralizer of A
and all these elements have trace f2, contrary to the assumption
# *2.
Thus
T ( h ) is a p-regular element. Let C be a p-regular element of
fcx,
=
is the characteristic polynomial of the eigenvalues 1,l of the form easily that
or
G with trace 1.
x2 -AX +
=
fT.
f2, f ( X ]
=
=
(1flf
Thus C is conjugate in GL(2,p)
-1,-1.
fA ( A as above], and C2’ C
1
If h
C.
Then
=
I.
Since
and
to a matrix
C is p-regular, it follows
If h # f2, then C is not a scalar multiple of
there is a vector V . such that
and
V
V
2
= V
1
C has
I, so
C are linearly independent.
Invoking the Cayley-Hamilton theorem, we derive
Hence
v
c
=
v
c2 = -2, +xu 1 2
and therefore C is conjugate in GL(2,p)
to T ( h ) .
We are now left to verify that C is conjugate to
T(A)
in G.
To this
end, write
for some D E GL(2,p). if accomplished, then
We wish to find E E CGLC2
,PI
( T ( h ) ) such that detDE=l.
CHAPTER I11
202
and DE E G.
cGLc2,p,(r ( A ) )
Thus we need only verify that the determinants of
take all values in
P*. P
An elementary calculation shows that
Consequently we have to prove that the quadratic form x2+Axy+y2 represents all values in P
If
P'
This is obvious for p
2.
=
If p # 2,
then
1 1-2' is not a square, it is trivial that this form represents all values
Otherwise, 1 -+A' is a nonzero square since 1 # f2. Hence the form P' is equivalent to x2+y2 and represents all values, by virtue of Lemma 8.12. in F
S t e p 2.
Here we construct i r r e d u c i b Z e FG-moduZes Vm dimV Frn
=
(0 C rn C p
- 1I
for which
rn + 1
It will then follow, from Step 1 and Theorem 2.3.2, that these
Vm are all non-
isomorphic irreducible FG-modules.
rn denote the F-space of homogeneous polynomials of degree rn in the independent variables z and y . If A = ( a . .I E G, we put V
Let
w
(xiyWi)~ Then
=
(a x + a 11
12
y ) i ( a x + a Y J m-i 21
22
Vm becomes a (right) FG-module of F-dimension r n + l .
Vm
is irreducible for 0 G m G p-1.
V,
and choose
where a
n
# 0 and n
< rn.
So assume that W
Given 1-I E P D ,
we put
We next prove that
is nonzero submodule of
UPPER AND LOWER BOUNDS FOR
dimJ(FG)
203
P Then there exist polynomials f j , 0
n, with fo
J'
=
and f n
f
=
anym such
that
Note that
P
Bearing in mind that in
P P-1 lli
=
)I=l
1
-1 if p-1 divides i
0
otherwise
I
we derive
-
c I
fj
p - 1 j-i
for l G i G p - 2 ,
for i = 0,p-1 = n=m
-fo-fp-l Because
f
=
f E W , we infer that fi E V for all i
ticular, since fn = any
rn
,
we
have ym E W.
A
l
E (1
,...,n j .
In par-
similar argument shows that
W
contains the element
for 0 < j Q m < p - 1 and 0 < j < p - I = rn for rn = p-l,j=O But yP-l E W has already been shown in the case m = p-1. for all j E {O,l,...,m}. irreducible for all rn E
Step 3.
The conclusion is that
{o,l,..
Hence Jym-'
W = Vrn and therefore
E
vm
W is
., p - l } .
Completion of the proof.
By Step 2 , the dimensions of irreducible FG-modules are
l I 2 ,. . . , p .
Invoking
CHAPTER I11
204
Corollary 1.10.17, we deduce that
thus completing the proof. 9. A CHARACTERIZATION OF
dimJ(Z(FG1)
AND ITS APPLICATIONS,
F Throughout this section, G characteristic p
0.
As
conjugacy classes of G
denotes a finite group and F usual, we write
and the exponent of
and exp(G1
for the set of
G, respectively.
Given a subset X
set of all nonnegative integers.
FX
CR(G)
an arbitrary field of
of
G
no
Let
nEno
and
be the we put
F-linear span of X
= the
n
Ix E F G l P
Tn(FG) =
X+
=
1x 6 X
We denote by
c
is empty, then by definition 'X
=
rP
C.
We
=
0)
lx E XI
the sum of all p-elements of G (including 1).
Recall that a defect group of g E
IFG,FGl]
n
n
xp
(if X
E
write Def ( C )
P
C E CR(G)
is a Sylow p-subgroup of
for the set of all defect groups of
denotes the set of all p-regular elements of G. number dimJ(Z(FG11
Our
c.
cG(gl with
Finally, G
P'
aim is to characterize the
in terms of the rank of a matrix associated with G.
A
F number of important applications is also provided. Tn (FG)
and Tn (FG)'
9.1. LEMMA.
(i) Tn(FGI
(ii) Tn(FG)'
As a first step, we describe
explicitely.
For any n Eno, the following properties hold is a
ZVGl-module and
is an ideal of
Z(FGl
and
205
and so yz E T n ( F G ) , Now fix x = Ex#
proving that T (FG) is a E
FG.
Because
gn it follows that
5
Z(FG)-module.
(mod[FG,FG])
E Tn(FG)
C$"gn E
if and only if
IFG,FGI .
n
assertion is now a consequence of the equality 1 2 = (Ex 1'
9
Z(FG)
(ii) Owing to Lemma 3.9,
=
[FG,FGlI .
The required
n and Lemma 2 . 2 . 2 .
9
Since Tn~(FG)2 [FG,FGI, we have
Tn(FGll C - IFG,FGJ1= Z(FG) But, by (i), Tn (FG) is a Let
Z(FG)-module, so T (FG)1 is an ideal of
n
c
E
CR(G1
and put
X
=
C1Ip
n Then I-' = (C-l)l/pn
.
Z(FG).
and, by (i), for
x E T ( F G ) , we have t r k X
+)
x
C
=
proving that 'X
=
0
g
$,TI
5 Tn(FG1 1 .
To prove the opposite containment, suppose that x = cZgg tr(q4) = 0 for all y E T,(FG).
a-l,b-l E
(c-')l/pn
In particular, if a , b E C1'pn
and thus, by (i ,) tP(X(a-'-b-'I
xa
=
Finally assume that g 9 C1/pn
tr(m-'i
=
9.2. LEMMA.
) =
tr(xb-li = x
for all C E Ck(G).
C E Ci(G) and hence, by (i), g -1 e Tn(FGl .
as asserted.
Tn(FG)'
0.
.
Then
then
Hence
b Then g-l 9 C1Ipn
for all
Thus
. Let D
be a defect group of
C E CR(G1,
let
x
E FG
and let
206
CHAPTER 111
(iil
Under t h e assumption t h a t
(a)
If
C
i s p - s i n g u l a r , then
(bl
If
C
is p-regular,
i i s t h e inverse of
(il
Proof.
q
then
C1lq =
0
and
tr(xqC+) = 0
C1lq is a p-regular
s e c t i o n containing
modulo t h e o r d e r of t h e elements i n
x =
Write
q 2 e x p ( D ) , t h e following hold:
E x g,
E
I
F,
.'6
where
Ci,
Moreover,
so t h a t
+Gg
Since
IFG,FGl
is a
Z(FG1 -module, we have
tr(IFG,FGl1
Taking i n t o account t h a t
=
0,
it follows t h a t
a s desired.
( i i l (a) and that (bl
g
P C
Since
E CG(gql.
i s p-regular.
Suppose t h a t
(xi)q = x.
c
x,y
E
g P"
.
Then t h e r e e x i s t s
q 2 exp(D1,
g E C1/'
so t h a t
g4 E C
( g 4 ) p = (gPlq = I, proving
we must have
i s p-regular.
Then
g E C1Iq.
Then
C1/q,(x '1' P
a r e a l s o conjugate.
conjugate t o
# 0
The d e s i r e d conclusion now follows by v i r t u e of ( i ) .
Assume t h a t
hence f o r any
ypl
C1"
Suppose t h a t
say
and
ci c- cl/'
-1
gp,y.
2 E
C,
E C and gq = (gp)f?(gpr)' = (g P ( y P r l 4 a r e conjugate. Thus r and
Conversely, suppose t h a t
xpr = y
s i n c e f o r any
Then
r
E G
P' is such t h a t x
P'
is
207
D is
and
CIIq
proving t h a t
C
Since
(81'Iq
D.
subgroup of
,I.
Thus
(z I q
xq
1 and so
=
P
8
i s a p-regular conjugacy class of
i s a p-regular s e c t i o n .
(811/q. Then
XE
G P
C,
is a p-regular s e c t i o n .
i s p-regular,
t h e foregoing, and
C (g
a Sylow p-subgroup of
Suppose t h a t
X i s a p-singular c l a s s and b ( X ) tr (xqX1
Thus, by ( a ) ,
=
X
G E
and so, by
CR(G) , X # C
i s conjugate t o a
0 and t h e r e f o r e
The r e q u i r e d a s s e r t i o n now follows by appealing t o (i).
,
C E CE(G)
For any
C(p1 E CR(G)
define
by
W e a r e now ready t o prove t h e following important r e s u l t .
9 . 3 . THEOREM U<;ilshammer (198211.
n EDo
a s a d e f e c t group and l e t C,,
...,Cm
(i) I f
(ii)
a l l p-regular
z E Z(PG)
If
C
Let
C
be a conjugacy c l a s s of
conjugacy c l a s s e s of
then
i s p-regular,
q
be such t h a t
C ( p ) = K(p) then f o r any
f o r any
=
p
n
a
exp(D).
G with D Denote by
G.
K E CR(G)
z E Z(FG)
with
there e x i s t
K
5 SuppZqc+
1 ,...,Am
E
F
such t h a t
(c)
zqC+ =
Proof.
m X A%.' (il
G
Let
be a counter-example of minimal o r d e r and w r i t e
(A
Then t h e r e e x i s t s Let
II : Z ( F G ) +
g
G with
E
A
9
h = g P"
# 0 and such t h a t f o r
Z ( F C ( h ) ) be t h e Brauer homomorphism.
Then
G
xg
= t r ( g - l v ( z g ~ + ~=) t r ( g - l n ( z ) q n ( C + ) )
+o
h
4
C(p).
$7
E
F)
CHAPTER I11
208
C n C(h) # @
and hence
.
Let
C and f o r each 2 E
contained i n Since each
..,Cr be {l, ...,P I ,
cl,.
G
Di i s G-conjugate t o a subgroup
i E 11 ,...,PI.
D,Q
K
gh-l
E
>
exp(Di)
‘i
for a l l
E
11,. .. ,rI.
h 9
Because
ci
( p ),
.
C ( k ) i s a l s o a counter example t o (i) Hence h E Z ( G ) and so for G C!L(G) ,Kh-l E C R ( G ) , Ck-’ i s p-singular with D a s a d e f e c t group and
i s p-regular.
y E G.
G
BY (11, w e may w r i t e
t h e group all
C(h)
Di be a d e f e c t group of
let
of
tr ( 9 - h ( z ) q 2~ + )= o f o r a s u i t a b l e i
so t h a t
a l l conjugacy c l a s s e s of
Since
hg-I
y q = k7-l
i s p-regular, we have
f o r some
Applying Lemma 9 . l ( i i ) , we t h e r e f o r e deduce t h a t
a contradiction.
Cm+l, Cm+2,...,C
(ii) Denote by 2
E Z(FG).
21
Owing t o Lemma 9 . l ( i i l , t h e r e e x i s t
W e may harmlessly assume t h a t each
g E C;Iq.
Then
hi
=
But, by ( i ) Suppzqc+
all
G and l e t
a l l p-singular c l a s s e s of
CIIq
1
,...,h r E F
i s nonempty.
i
tr(g-lz(C1/ql+l
h
Fix
i
such t h a t
E
... ,r}
11,
and
and hence, by Lemma 9 . 2 ( i i ) b ,
i s a union of some p-regular c l a s s e s .
Thus
1. = 0
for
i E {m+l,-.., r } , proving (a) and ( c ) . Since
8
is a p-regular c l a s s of
G
with d e f e c t group
D,
it follows from
(c) t h a t
g E C;jq.
Then
gq E C . C ( $ ) l / q 2 -
and
Applying Lemma 9 . 2 ( i i ) (bl , w e d e r i v e
209
as asserted. For future use we next quote the following standard result. THEOREM (Wedderburn-Malcevl.
9.4.
field K .
If A / J ( A )
Let A
be a finite-dimensional algebra over a
is separable, then
(i) There exists a semisimple subalgebra A ( A ) A =
(it) If A ( A )
and
1
A
(A)
of A
such that
nu) @J(A)
are subalgebras of A
(direct sum of K-spaces) with A = A . ( A ) @ J ( A ) , i = 1,2, z
then
for some 3: E J ( A l . Proof.
See Curtis and Reiner (1962).
The following useful observation will be derived with the aid of the result above. 9.5.
LEMMA.
(i) dimZ(FG)/J(Z(FG))
= dimZ(FGlc.
splitting field for Z(FG), FG
In particular, if F is a
F
F
then dimZCFGlc is equal to the number of blocks of
F
.
(ii) There exists a unique F-subalgebra
A(Z (FG)
Z(5'G) = A ( Z ( F G ) ) @
Proof. z E Z(FG).
(i) Let 4 : Z ( F G ) Then 4
.-+
of
Z (FG) such that
J(Z(FG))
Z ( F G ) c be defined by
i s a surjective 5'-linear map with
(direct sum of F-spaces)
$ ( z ) = zc
for all
Ker4 = (l(c) n Z(FG1
.
Now apply Theorem 6.2. (ii)
Owing to Corollary 1.18,
Z(FG)/J(Z(FG)) is a separable F-algebra.
desired conclusion now follows by appealing to Theorem 9.4. From now on, the maps:
The
2 10
CHAPTER I11
designate t h e n a t u r a l p r o j e c t i o n s .
A(Z(FG1)
The following r e s u l t i l l u s t r a t e s how t h e algebra
can be brought
i n t o argument. (i) 6 ( z ) = n ( z c )
9.6. LEMMA. (ii) A ( Z
z E Z(FGI
(FG)) contains a l l c e n t r a l idempotents of
Proof.
FG.
n
( b ) , (c) for a sufficiently large
( i ) Invoking Theorem 9 . 3 ( i i ) ,
C = ill,
and
for a l l
we o b t a i n
z E Z(FG)
(3)
Frect sum
A(Z(FG)) i s a commutative separable F-a qebra, it must be a
Since
of separable f i e l d extensions of
E = F(y)
may w r i t e
f o r some
n n(zc)
T(zcl = 6 ( z ) (ii)
for a l l
e
Let
y E A(Z(FGI1.
z E A(Z(FG)).
for a l l
= z
Assume t h a t
E
2
E/F
i s one of them.
Then we
E/F
Moreover, t h e s e p a r a b i l i t y of
and so, by ( 3 ) , t h a t
E = F(yp J
ensures t h a t
F.
lT(ZC)
for a l l
= 2
Z
E E.
Hence,
Applying Theorem 6.2, w e i n f e r t h a t
Z(FGJ , as a s s e r t e d .
denote a c e n t r a l idempotent of
FG.
Invoking (3) together with (i),
we d e r i v e
a s required.
.
For any p-subgroup C E
CR(G1
,
d(C)
with
D
5 D.
of
G,
let
rDCG) be
t h e F-linear span of a l l
Note t h a t by Lemma 2 . 2 . 5 ,
I,(G)
i s an i d e a l of
G
.
Z (FG)
9.7. LEMMA.
FG
idempotents of (i)
{ e l ,...,e
D be a p-subgroup of
Let
1
with a subgroup of
G D
and l e t
...,en
el,
+,
C
be a l l block
a s a d e f e c t group.
i s t h e s e t of a l l p r i m i t i v e idempotents of t h e i d e a l
VGl)
did
A CHARACTERIZATION OF
211
F By Lemma 6.1, e 9.6(ii),
e E {e
e
is a primitive idempotent of a(Z(FG)).
is a block idempotent of FG.
e
i i
lies in I D ( G ) n A ( Z ( F G ) ) .
i
E TD(G)
ID(G) n A(Z(FG)).
=
TDC0 n A ( Z ( F G ) )
Bearing in mind that
of fields and that I is an ideal of
.
Since n(TD(G))
virtue of (i)
and so, by Lemma 9.6(ii), each
Again, applying Lemma 6.1, we conclude that each
is in fact a primitive idempotent of
(ii) Let I
On
Because e E T D ( G ) , we also have
,...,e I .
Conversely, by hypothesis, each e
e
Therefore, by Lemma
.
A(Z(FGI
is a direct sum
n ( z ( F G ) ) , the second equality is true by
5T D ( G ) ,
Lemma 6.l(i) implies that
the other hand, we have
I C_ 6 (ID(Gl1 .
proving that
.
A
Let D denote a p-subgroup o f all C'
with
Because
?D ( G )
ideal of
such that D
C E CR(G)
We write ID ( G )
Denote by
a p-regular class of
V,(G)
with L
c
D,
we see that Y D ( G )
the F-linear span of all 'C',
C.
where
is an C
is
G with defect group D.
Let D be a p-subgroup of
idempotents of FG
for the F-linear span of
contains properly a defect group of
is the sum of all T,(G)
Z(FG1.
9.8. LEMMA,
G.
with defect group D.
G and let eD be the
sum
of all block
Then
dimIZ ( F G ) eo/J (2( F G ) eDl =
F
(By convention, e D = 0 Proof.
Let X
and
i f there are no blocks of
Y
FG
with defect group D).
be the left and right hand side spaces, respectively.
By Lemma 9.6(ii), e D E A ( Z ( F G ) )
and thus
CHAPTER I11
212
proving that dim X = dim b(Z(FG))eD.
On the other hand, by Lemma 9.7,
F
F
Bearing in mind that
we infer that
Invoking the argument employed in the proof of Lemma 9.7(iil, we have
Thus the right hand side of (4) is the F-linear span of the elements
where C E CR(G)
is such that D
is a defect group of
c.
6(c*) + T D ( G ) ,
Owing to Lemmas
6 ( C + ) E FGp, n TD(G) and hence
9.6(i] and 9.7(iilI
6(c+) e zx;6(c)
(mod ?,CG))
i
for some p-regular classes Ci
with defect group D
and some 1 . E F z
.
This
implies that
and thus dim X = dim F F
Y
Finally note that if there are no blocks of FG with defect group D, then either
6(1D (GI1
FG with e
E
=
0 or
IDGl.
6(1D (G))
=
In both cases we have dimY = 0, as required.
F 9.9. LEMMA.
Let Q
be a p-subgroup of
G,
of
let S be a p-regular section of
and let C be a unique p-regular class of G (i) The number r
.
6(lD(Gl)e for some block idempotent e
of Sylow p-subgroups of G
G
contained in S. containing Q
is congruent to
1
d i d ( 2 (FG) )
A CHARACTERIZATION OF
213
F
p
modulo
(iil
S+ =
E Syl
P
E S y l ( C (gl I
gp P G ( i ) The group & a c t s on t h e s e t Syl (Gl
Proof.
P
+
c
c
sfc
by conjugation.
P
(G),P
3&
{PI
i f and only i f
i s a &-orbit.
p,
n o n t r i v i a l & - o r b i t i s d i v i s i b l e by
Since t h e s i z e of each
we have
lSyl (Gl
On t h e o t h e r hand, by Sylow theorem
Given
P
I
lbod p )
.
Thus
r
1 (mod p)
,
a s asserted. (ii) Fix
x
E S
r
and denote by
t h e number of Sylow p-subgroups of X
containing
(iil.
<xP >.
Then
r
By ( i l ,
9-10. LEMMA.
r e g u l a r class of given
E
C , K E CR(G)
i s t h e c o e f f i c i e n t of
3:
contained i n
,
put
S,
Fix
1.
G
Co
and l e t
=
s E Co, P E Syl (C (s1)
P
G
C F C R (Gl
Then
X(C,KI
i s an i n t e g e r such t h a t
I C l p , a KCs
Proof.
Let
co
E C E
CR(G)
.
=
X(C,K)*lF
Invoking Lemma 9.9 ( i i l
On t h e o t h e r hand, a n easy c a l c u l a t i o n shows t h a t
‘1
Since f o r
be a unique
and w r i t e
K+S+
P
t h e r e q u i r e d a s s e r t i o n follows.
be a p-regular s e c t i o n of
G
G
i n t h e r i g h t hand s i d e of
and so t h e above c o e f f i c i e n t is
P E Syl (C (g1 1 ,gy E S P G
S
Let
X
l b o d pl
X
g E C and y
any
r
c (x
, we
derive
and, f o r
’ p-
CHAPTER I11
214
Because the number above is an integer and conclude that h ( C , K )
as required.
lSyl I C (s) 1
P G
I
1 (mod p) , we
is an integer such that
.
We have now come to the demonstration for which this section has been developed.
The following result for the case D 4 G
9.11. THEOREM (Kulshammer (1984b)). of
Let
is due to Robinson (1983).
c1 ,C2 ,...,Crn
be all p-regular classes
G having the same defect group D, let eD be the sum of all block idem-
potents of FG with defect group D and let P Define the matrix MD
=
M..)E M 23
m
OF 1
P
be a Sylow p-subgroup of
G.
as follows:
a and pd are the orders of P and D, respectively (by Lemma 9.10
where p
+
applied for S = c,K = C
j
and C = Ci, the number in parenthesis is an integer).
The? dim2 (FG)e 0 / 3 ( 2 (FG)1 eD = rank
F
MD
(By convention, eD = 0 if there are no blocks of FG with defect group D)
Proof. tives in
Then
Let gl,g2,...,gp
G and put
be a full set of (P,P)-double coset representa-
A
CHARACTERIZATION OF
dimJ(Z(FG))
215
F
N i the transpose of No.
and we denote by
11,...,m} and denote by
Fix j E
6(c+) E 3
Lemma 9.7(ii),
rD(G)and, by
+
Applying Lemma 9.10 for S
= c,K = C
j
ci.. 23
6(C:)
Lemma 9.6(i),
C = Ci,
and
3
=
+
Ci
in C.C. 3 a(C+c). Thus 3
the coefficient of
By
we also have
We now claim that the result will follow provided we show that P
(CiIp'aij = Indeed, put M'
W
Is
=
(a..) ,U $3
i
=
the F-linear span o f
linear span that dimV
of
=
F
,um 1 .
{Vl,.,.
rn C a C?
i=l ij - z
{Vl
+ ?,(GI Because
and
c nkinkj k=l FI
r...,t(rn
=
[s(VD(G) + ?D(G)I/?D(G).
+ YD(G)1.
V C_ VDlg(G) and
d i d = rankMr. By ( 6 ) and (7)
(71
-note by
VD(G) n
fDD(G)
By ( 5 ) , V
the F-
= 0,
we also have
F t rankMD = rankM' = rankN$b
Applying Lemma 9.8, we conclude that
t
dim2 (FGIeD/J (2 ( F G ) e D ) = rankNDND = rankMD
as claimed. By the foregoing, we are left to verify ( 7 ) .
XP
5 PgP
so that XP
=
YgP for some y E P.
To this end, assume that Then
we see
216
CHAPTER I11
-1
x ax
E gkPgk
-1
( g k P ) P = gkP
( a r b )(Icty) = (a,bl
Note a l s o t h a t
and
y-lby E gkPgk
i f and only i f
y E C(b ) n P n g k P g k - l = C(b) G P Thus t h e s i z e
and hence
k
of t h e o r b i t of
(a,b)
i s given by
-1
(gkPIP = gkP
A
CHARACTERIZATION OF
d i d (Z(FG))
217
F
Thus in this case we have
R
=
1 and
Applying (91, we therefore conclude that
( c ~ I ~ = , k=c~1nkinkj ~ ~ proving (7) and hence the result. I.t is now an easy matter toprovidea characterization of
dimJ(Z(FG1).
F 9.12. COROLLARY.
Let
21
be the number of conjugacy classes of
G and let X
be the set of all nonisomorphic defect groups of p-regular classes of each D E X, let the matrix
MD E M
be defined as in Theorem 9.11.
[F )
m P
dimJ(Z(FG1 I
=
r
-
F Proof.
Our choice of
groups of blocks of FG.
G.
For Then
C rankMD
E X
X ensures that it contains all nonisomorphic defect Hence Z(FG) = @ Z(FG)eD
EX where
eD
is the sum of all block idempotents with defect group
if there are no such idempotents.
D and eD
Accordingly,
Taking dimensions of both sides and applying Theorem 9.11, we derive
r - dimJ(Z (FG)I F
=
C rankMD
,
E X
whence the result. We close this section by providing some applications of Theorem 9.11.
=
0
218
CHAPTER I11
9.13. COROLLARY [Green (19681).
D
Let
t E CG ( D )
Then t h e r e e x i s t s a p-regular element
G
be a d e f e c t group of a block of
FG.
Q
and a Sylow p-subgroup
of
such t h a t
Proof.
We keep t h e n o t a t i o n of Theorem 9.11.
t
i t follows from ( 8 ) t h a t rankNDND# 0.
FG,
a block of
regular c l a s s
C
G
of
D
Taking i n t o account t h a t
is a d e f e c t group of
P
Setting
t
.
z-ldx
=
a s asserted.
n
Q = x-lPx,
and
d E gP, we have
Since
P G
dpd-l
Hence t h e r e e x i s t s a p-
D, a n element g E G and d E gP n C
with d e f e c t group
P n 9Pg-l E Syl (C (dl 1 .
such t h a t
D i s a d e f e c t group of
Since
= XDX
gPg-’
=
dpd-l.
C, we conclude t h a t f o r some
-1
it follows t h a t
t
x
E G
C (D) and D = Q n t Q t l ,
E
G
I n o r d e r t o provide f u r t h e r information about i n v e s t i g a t e t h e rank of t h e matrix
MD
dimJ(Z(FG1 it is d e s i r a b l e t o F defined i n Theorem 9.11. In general,
t h i s i s a formidable t a s k and no p r e c i s e formula f o r t h e rank of
G has been discovered so f a r .
t h e group s t r u c t u r e of
i n terms of
MD
However, it i s q u i t e
p o s s i b l e t o provide some important u s e f u l information, which w e p r e s e n t below. A conjugacy c l a s s
is s a i d t o be nilpotent i f
G
Let
LEMMA.
rnD
d e f e c t group
D
contained i n
0 ,(GI
D. If
Then
:C
A(Z(FG))
a nilpotent
P
be t h e number of nonnflpotent p-regular c l a s s e s of
and l e t
.tD be
t h e number of conjugacy c l a s s e s of
and have d e f e c t group
RD Proof.
C+ is
Z (FG) .
element of 9.14.
C of
Let
C1,Cp,
...,Crn
D.
G rankMD G
i s semisimple, so
i s a n i l p o t e n t element of
&(c,+)= 0.
which a r e
mD
be a l l p-regular c l a s s e s of
6(C?)
with
Then
[6(VD(G)) + ?D (G)l/iD(G) is t h e F-linear span of
i s n i l p o t e n t , then
G
G
Hence
G
with d e f e c t group
6(Ci) +iD(G),1GiGrn h(Z(FG)).
But
A CHARACPERIZATION OF
dtmd (2(,?GI 1
219
F
and therefore, by Lemma 9.8 and Theorem 9.11, rankMD < m D . Let C ,C 1
,...,CII ,II
2
=
LD, be all conjugacy classes of G contained in For each i E 11 ,... ,R},
(G) and having defect group D.
0
P
Assume that I is conjugate to y Since x , y E 0 ,(G), x-'y
y
i'
is conjugate to y
Thus I y = 1,
is p-regular.
P
j
-1
fix yi E Ci.
and z-lyEPESyl (GI.
P
so
I = y.
This
proves that
Hence mij
=
O for
m.. # 0 as ZZ
so
ID1
-
i#j
=
N~~ m
ICG(yil
Ip.
ii'
1
< i < R,
I s
Hence rankMD 2 R,
the residue (modp) of
as asserted.
We are now ready to provide the following application of Corollary 9.12. 9.15. THEOREM. r r
r
Define the numbers
P,Tl,
and
T
as follows:
G
=
the number of conjugacy classes of
=
the number of nonnilpotent p-regular classes of G
=
the number of conjugacy classes of
1
G contained in 0 ,(GI
P
2
Then r
-
< did(Z(FG1) < r - r 2 .
r '
F Proof. We use the notation of Corollary 9.12 and Lemma 9.14.
It is a
consequence of Lemma 9.14 that C
R <
E X D
C rankMD
E X
C mD E X
Applying Corollary 9.12, we therefore deduce that
r
-
C
#X But
c mD
mD
dimJ(Z(FG)) F
IID
E X
is the number of nonnilpotent p-regular classes of
G, while
E X C RD
E X
is the number of conjugacy classes of
G contained in 0 ,(GI.
P
Thus
CHAPTER LIL
2 20
r
-r
r
F
- r2
as required.
G be p-nilpotent.
9.16. COROLLARY. Let
is equal to the
Then dimJ(Z(FG1
F G.
number of p-singular classes of Since G
Proof. in
0
P
is p-nilpotent, all p-regular classes of
r(G) and therefore are nonnilpotent.
Hence r
= 1
G
are contained
and r
r 2
- r1
is the
G. Now apply Theorem 9.15.
number of p-singular classes of
To provide further applications of Theorem 9.11, we need to introduce the The following preliminary observations will clear our
Brauer correspondence. path. 9.17. LEMMA. Let P
be a p-subgroup of G
and let H = N ( P )
G C E Ck(G1
(i) If
then C n C(P1
6(C) = P
with
H
is a conjugacy class of
G with defect group is a conjugacy class of H
(ii) If L
C E Ck(G)
such that L
=
with defect group P, then there exists
C n C(P) and 61C) = P.
G Proof.
(il
There is an element g E C
such that P
is a Sylow p-subgroup
Hence g E C n C(P). Tf g' E C n C(P), then there is an z E G G G G Because g' E C(P1, P is a Sylow p-subgroup of such that zgz-' = g ' .
of
C(g).
G
c
(g'l
yG3-l
so
=
But z€'z-' is one also and so there is a YECkg') with
zC(g1z-l.
=
zPxG1., hence Y-lz E H.
that g'
is conjugate to g
Since y
in H.
commutes with g ' ,
Finally, because c ( P )
element of H which is H-conjugate to g ,
we have
a u(P1
any
G
G lies in C 'lC(P).
G (ii) By assumption, there is an element h subgroup of the centralizer of h
in
H.
such that P
in L Thus
h
is a Sylow p-
5C n
E C(P) and so L
G where
C
i s
the conjugacy class of
a defect group of
C.
G containing h .
C(P),
G We next show that P is
It will then follow, by (i), that L = C n C(P) and hence
G the proof will be complete.
Let Q
denote a Sylow p-subgroup of
C ( h ) with
G P
CQ;
then Q
is a defect group of
C.
If Q # P , choose a subgroup P, of
dimJ(2 ( F G )
A CHARACTERIZATION OF
221
F
(P :P] = p .
Q such t h a t P
fact that
Pc - N(P1
Then
1
L.
i s a d e f e c t group of
9.18. LEMMA.
=
6' and P c C(h1,
Q
Hence
P be a p-subgroup of
Let
c o n t r a r y t o the
1 - H
G
=
let
G,
P and t h e r e s u l t follows.
H
=
#(PI
and l e t
G
be t h e Brauer homomorphism. p o t e n t s of
FG
having
P
Then
71
induces an i n j e c t i o n of t h e block idem-
a s a d e f e c t group i n t o nonzero idempotents of e
Moreover, f o r any block idempotent
FG with d e f e c t group P, SuppT((e)
of
P.
i s a union of H-conjugacy c l a s s e s having d e f e c t group Proof.
Let
e
and ( b ) of Proposition 2.2.6
C
hi
classes
Cl,-..,Ct
=
x 1 c+1 +
CRIG)
in
Owing t o
which s a t i s f y c o n d i t i o n s ( a )
\PI.
... +
Hence, i n t h e n o t a t i o n of P r o p o s i t i o n 2.2.6,
have d e f e c t group
have o r d e r less than
I t follows t h a t
F.
in
,...,Cr
P.
with d e f e c t group
and such t h a t
e with some nonzero
FG
be a block idempotent of there exist
P r o p o s i t i o n 2.2.6,
Z(FH).
P,
Ct+l'
while d e f e c t groups of
the
- . ,cr *
Thus
k
?T(C~I = 0,t
P,
n ( e ) = n(C+)
and t h e r e f o r e
+
... + T ( C+t ) .
Now
and, by Lemma 9.17(1), the of
H with d e f e c t group Assume t h a t
and l e t
e'
C. n C(P1,l G
Z
P.
G
Hence
i
n(e)
t, are d i s t i n c t conjugacy classes s a t i s f i e s t h e r e q u i r e d property.
i s another block idempotent of
*(el = n ( e '1.
Then
Supp(e-e 'I
y(e-e ') = 0
and so
Z(FG)
a s s o c i a t e d with
y(e1 = y f e '1.
But i f
with d e f e c t group
P,
is a union of conjugacy classes whose
d e f e c t groups are conjugate t o proper subgroups of i b l e r e p r e s e n t a t i o n of
FG
e
P.
Let
y
be t h e irreduc-
e.
By P r o p o s i t i o n 2 . 2 . 6 ( i i ) ,
and
e'
r e a l l y were d i s t i n c t ,
CHAPTER I11
222
then y [ e )
=
Let A
LEMMA.
9.19.
:A
-
:
rl(Al---t
r2
r2 (A)
r
f
r 1 (el
such that
Let p : A+
B
for all i , j E {l,
r
1
(a) = C-lY =
C-
is equivalent to
be a block idempotent of A
u +u
associated with
...,n } ,
fCcl
y1
=
n
and there 2
(a)C
for all a E A
1
xC is an isomorphism of e
the block
Then
1.
=
1
are block idempotents of
presentation of B
r
r .
So the lemma is
be a homomorphism of commutative algebras over a
p(e) =
where the ui
r
Then
To prove the converse, denote by
.
let e
such that
defined by
and so, by Proposition 1.10.8(vl,
LEMMA.
.
1
:
idempotent of A
r
is equivalent to
r l (A)-+ r2(AJ K-algebras such that r = fr . 1
Hence the map
Then
which renders commutative the following diagram
1
field K ,
and let
if and only if there exists a K-algebra isomorphism
exists a nonsingular a x n -matrix C
9.20.
and the result follows.
be a commutative algebra over a field K
Proof. Assume that
proved.
e = e'
Aence
( K ) , i = 1,2, be two irreducible representations of A .
Mni
is equivalent to
f
y(e'1 = 0.
1 and
u
2
+ B.
and let
... + u Denote by
xi
i and set K i -- (xip)
the irreducible re(A), 1
i 6 n.
there exists a K-algebra isomorphism & . . : K
w
j
---+
Then, Ki
A
CHARACTERIZATION OF
d i d (2( F G ) 1 F
223
which renders commutative the following diagram
Proof.
Because
Xi(&
Fix i , j E 11 ,...,n } .
Owing to Proposition l.lO.E(v),
is a finite field extension of K
conclude that K .
and since K 2. c 1?.. ( B ) ,
is also a finite field extension of K.
by its image in the regular representation of Ki, irreducible representation of A , 1 it follows that l . p
and
x.p
3
i
n,
we
Thus, replacing Ki
we may assume that A2. p
is an
By (10) and Proposition 1.10.8[v) ,
are equivalent.
The desired conclusion is now a
consequence of Lemma 9.19. We are now ready to prove the following important result.
9.21. THEOREM (Brauer correspondence). H
be the normalizer of P
in G.
Let P
be a p-subgroup of
G and let
Then the Brauer homomorphism
determines a bijective correspondence between the block idempotents of FG
which
have P
as one of their defect groups and the block idempotents of FH which
have P
as their unique defect group.
Proof.
Step 1. P.
For the sake of clarity, we shall divide the proof into two steps.
Here we prove that
?r(el is a block idempotent of
FH
with defect group
224
CHAPTER I11
Owing to Lemma 9.18, it suffices to show that n(e1 potent of Z(FG).
is a primitive idem-
Assume the contrary, say,
~ ( e =) e
+
1
e
2
+
... + e t '
t > 1, where the ei are orthogonal primitive idempotents of Z(FH1. by
yi
the irreducible representation of
Z(FH1 associated with e i and by Fi
the image of Z(FG) under the homomorphism yio7T, 1 < there exist F-isomorphisms Qij
:
F
j
Denote
i G t.
By Lemma 9.20,
Fi which render commutative the following
-+
diagram
Owing to Lema 6.6(ii), the defect groups of each ei
k E {1,2,
...,t1
is such that the defect groups of
Then, by Proposition 2.2.6(ii), respect to FH all the ei Let L
.
Assume that
ek contain P properly.
Yk(AI = 0 where A
It follows that
contain P.
is as in Lemma 6.6 with
(yk-71) ( e ) = 0, which is impossible.
Hence
are in A .
,...,Lr
denote all conjugacy classes of H with defect group P.
Then, by Lemma 9.17(iil, there exist conjugacy classes C 1 defect group P
and with Li = Ci n C ( P ) , 1 G
i G r.
,...,Cr
of
G with
Thus
P
for some
x.. 23
E
F.
Because t > 1, the commutativity of the diagram above yields
dimJ(Z ( F G ) ) F
A CHARACTERIZATION OF
r 0
(e 1 = y
= y 2
1
T( 2
+
Z X..C.l j = l 1-3 3
e y n(
=
225
1
21
r
y (e 1
= 8
C h.C.1 jZ1 3 3
21
I
1
= 1,
Thus Tr(e) i s p r i m i t i v e . f be a block idempotent of
a contradiction.
Let
Step 2.
we are l e f t t o v e r i f y t h a t
L e m 9.18, FG
w i t h d e f e c t group
FH
n(el = f
By
P.
for some block idernpotent
e
of
w i t h d e f e c t group P. f
Owing t o Lemma 6 . 6 ( i i ) ,
Z(FH)
r e p r e s e n t a t i o n of
L
conjugacy c l a s s some
C E Ck(G)
yn
and t h e r e f o r e Let
e
H
of
+
# 0, w e have
.. . + A
Cs
+ (yn) (cs) =
~
P
0
CG(Pl # 0.
c o n t a i n subgroups conjugate t o
contain
i s of t h e form
n ( Z ( F G ) 1.
FG
Y C f , = 1.
Therefore, by S t e p 1,
a s s o c i a t e d with
n(e) = f ,
9.22. L E W .
for
yT
# 0
yon.
Write
.
+C A~ ~ + ~ +c .~. + X~C: ~ (YW ( C i + k ) = 0 f o r a l l
Then
s
f o r some
E 11
,...,t ) .
Because Cs,
Hence t h e d e f e c t groups of
P.
But i f t h e d e f e c t groups of
Thus t h e d e f e c t groups of
n(el
C n C(P)
Hence'
Z(FG).
p r o p e r l y , then by Proposition 2.2.6 and Lemma 6.5,
contradicts
have
1
Ci a s i n Proposition 2.2.6.
k E 11 ,...,~ t ] while ,
e,
P
i s contained i n
denote t h e block idempotent of
1
n(C:)
denote t h e i r r e d u c i b l e
Since by Lemma 9.17 every
i s an i r r e d u c i b l e r e p r e s e n t a t i o n of
e = A C+ with t h e
y
Let
f.
with d e f e c t group
6 (C) = P , A
with
A.
lies in
a s s o c i a t e d with
is p r i m i t i v e .
e
Y(A)
and so of
e
0 which
=
a r e conjugate t o
But then s i n c e
were t o
p.
(Yn) ( e ) = Y ( f ) ,
we
t h u s completing t h e proof.
Let
D be a subgroup of
o r d e r of Sylow p-subgroups of
G.
G
Then
of o r d e r
pd
and l e t
p
U
be t h e
N(D)/D has a normal pcomplement under G
e i t h e r of t h e following hypotheses
(il
d = a-1
and
(ii) d = a-2,
Proof.
G Let
i s t h e smallest prime d i v i s o r of
i s of odd o r d e r and
p
izer i n
G,
then
i s t h e smallest prime d i v i s o r of
P denote a Sylow p-subgroup of
Burnside (Burnside (1955)) a s s e r t s t h a t i f
\GI
P
G.
G.
A c l a s s i c a l theorem of
l i e s i n t h e c e n t r e of i t s normal-
G h a s a normal p-complement.
W e claim t h a t
CHAPTER I11
226
is abelian and q
If P
1
for any prime 9 # p
IAutPl
dividing
IGl,
G has a normal p-complement.
then
Indeed, assume that g E N(P)
(11)
and let < g > = < g
> x . .. x
< g n > be the
G
decomposition of
into direct product of primary components.
Owing to
Burnside's theorem mentioned above, it suffices to verify that g E C ( P ) clearly assume that p
3
1
Since conjugation by
Il.
1
the assumption that q
Autk',
.
We may
G
IAutPI
gi
induces a homomorphism
ensures that the image of gi
is
the identity automorphism, as claimed. Then N ( D ) / D
Suppose that (i) holds.
has a Sylow p-subgroup S
of order
G
p.
Because
IAutSI
=
p-1
and p
follows from (11) that Y C D ) / D
is the smallest prime divisor of
IGI,
it
has a normal p-complement.
G
Then N ( D I / D
Now assume that (iil holds.
has a Sylow p-subgroup S
of
n
CI
order p
p2. By the foregoing, we may assume that IS( = p2 in which case
or
p (p-1)
if
s
(p-1) 2p(p+ll
if
s
Ep2
IAuGI =
Let q
denote a prime divisor of
order, q
7
p+l
ICl
zP
distinct from p.
1
and so in both cases q
IAutSl.
x
E P
Since G
is of odd
Hence, by (ii), N ( D ) / D
has
G
a normal p-complement, as required. We are now ready to provide an important application of Theorem 9.11.
Part
(ii)d of the following result is due to Brauer and Nesbitt (19411, while the rest generalizes theorems due to Gow ( 1 9 7 8 1 , Kawada (1966) and Michler (1972).
9.23. THEOREM. Let F
be a field of characteristic p > 0 which is a splitting
field for Z(FG), let D orders of D
be a p-subgroup of G
and Sylow p-subgroups of
G,
and let pa
respectively.
and pa
denote the
Define the numbers
nD ,mD ,6D , and LD by: nD = the number of blocks of G with defect group D
mD
= the number of nonnilpotent p-regular classes of
G with defect group D
sD = the number of p-regular classes of G with defect group D ED
= the number of conjugacy classes of
G contained in 0 ,(GI
P
and having
227
dimJ(Z (FG))
A CHARACTERIZATION OF
F defect group D.
Then
G nD G mD equality nD =
is true under either of the following hypotheses:
SD
G is p-nilpotent DC(D)/D
is p-nilpotent
G
N ( D ) / D is p-nilpotent G D
G
is a Sylow p-subgroup of
The Sylow p-subgroups of
d = a-1
and p
d = a-2,
G are metacyclic and (p2-l,lGI) = 1
is the smallest prime divisor of
i s of odd order and p
G
IGl
is the smallest prime divisor of
IGI. Proof. (i) We keep the notation of Theorem 9.11.
If e
is a block idem-
potent of FC, then Z(FGIe/J(ZCFG) since F is a splitting field for Z(FG1.
and so, by Theorem 9.11,
nD = rankMD.
e
4
F
Hence
The desired assertion is now a conse-
quence of Lemma 9.14. (ii) If G Lemma 9.22,
is p-nilpotent, then S = D
RD
= mD
and so, by (i), nD =
each of the conditions (f) , (g) implies (c)
group of N ( D ) / D ,
G
By
SD.
.
Since DCCD) is a subG Moreover, (e) implies (a) (see Huppert
(c] implies (bl.
Since (d) obviously implies Ic) , we are left to verify (b).
(1967) ,p.437).
Suppose that D C ( D ) / D
i s p-nilpotent.
If the result is true for N ( D ) ,
G
G the corresponding result is true for G.
then by Lemma 9.17 and Theorem 9.21,
Thus we may harmlessly assume that D 4 G. Let Ci,C z E Ci,y E C
j
be p-regular classes of
j
with x-ly E P E Syl (GI.
and D E Syl ( C ( y ) )
P G
G with defect group D and let Because D 4 G we have
P
D E Syl (Ch)1
P G
so that r , y E C L D ) .
G
Then r - l D
and y D
are p-regular
CHAPTER I11
228
Thus X.y-'D
elements of the p-nilpotent group D C ( D ) /D. However, x-lyD
is also a p-element, so x
It follows that if = xd
with zd = dx
Hence r - l y = 1 and so x
.
conclude that rankM 2 s asserted.
=
.
y.
is a p-regular element.
-1
y E D, say x-ly = d with d E D.
which is only possible in the case d
=
1.
By repeating the argument of Lemma 9.14, we rankMD = nD, so by (i), nD = SD, as
But
10. MORITA'S THEOREM.
denotes a normal subgroup of a finite group G, F
Throughout this section, N
a field of characteristic p > 0 and
Our aim is to provide an important decomposition theorem due to Morita which will be frequently applied in the investigation of J V G )
.
We start by exhibiting some connections between the blocks of FG Assume that f
10.1. LEMMA.
(i) For all g E G, gfg-l ducible representation of
and FN.
is a block idempotent of FN. If y
is a block idempotent of FN.
z (FN)
associated with f, then
is the irreducible representation of
9
is the irre-
defined by
Z (FN) associated with gfg-l.
(ii) The set
is a subgroup of G tiii)
If 2'
containing N.
is a left transversal for G(f)
all distinct conjugates of
f and f*
=
in G,
C tft-l
then the tft-',t
is an idempotent of
E
T , are
Z*(FN)
ET which is independent of the choice of T. (iv) Suppf* Proof. of F N .
is a union of some p-regular classes of
G
contained in N.
(i) Follows from the fact that conjugation by
g is an automorphism
229
MORTTA'S THEOREM
(ii
Straightforward
(iii) It is obvious that the tft-',t E T, are all distinct conjugates of Hence f* is an idempotent of FN g E G
and T
=
1
c
gf*g-l =
T.
which is independent of the choice of
is also a left transversal for GCf,
gT, then T I
c
(gt)f(gt)-l =
f.
in G
If
and
zfz-l = f*
ET
This proves that f* E
ST 1 Z (FG) and hence that f* E Z* ( F N ) .
(iv) Apply Theorem 2.3.4.
.
Let f be a block idempotent of F N . group
covers
G(f)
as the inertia group of f.
f (or B
=
B ( e ) covers
b
=
b Cfl)
In what follows we shall refer to the We say that a block idempotent e E FG
e
if
occurs in the decomposition of
f* into the sum of block idempotents of FG. Observe that conjugation by block idempotents of FN
g E G induces an action of G on the sets of all
and all blocks of F N ,
the elements in the G-orbit of f. G-conjugacy class of f.
and that f* is just the sum of
We shall refer to the G-orbit of
Similarly, the G-orbit of a block b = b ( f )
f as the will be
called the G-cow'ugacy class of b. Assume that b = b ( f ) is a block of FN
Then G(b) = G(f)
and we refer to
G(b)
as
and put
the inertia group of t h e block
b.
(it) The mapping f k f* induces a bijective correspondence
10.2. LEMMA.
Z(FN) and all block
between the G-conjugacy classes of block idempotents of idempotents of
Z*(FN).
(ii) Every block idempotent of FG covers exactly one G-conjugacy class of block idempotents of
FN.
(iii) A block
B = B ( e ) of
FG
covers the block b = b ( f ) of FN
if and only
if ef # 0. Proof.
(i) Let f be a block idempotent of
f* is an idempotent of Z * ( F N ) .
ZCFN)
Assume that f* =
.
UW,
By Lemma 10.1 (iii), where
u
and
V
are
230
CHAPTER I11
nonzero orthogonal idempotents of
Z*(FN).
block idempotents ui,uj
Sfnce
n, 1
i
in F N , 1
u,V E
< j <m,
Z(FN1,
there exist
such that
Thus
m
n c u
f"= Since
c uj i + j=1
is a unique sum of block idempotents of F N
we may assume that u
=
1
a contradiction.
=
zu
=
gZIFN)g-'u
=
Z(FN)u
for a suitable z e Z ( F N ) .
Thus f* is a block idempotent of
Now assume that x
+ f, +
... + f,
But f* = u+u,
Z*(FN).
a left transversal for
for all j E { 1 , 2 ,
...,t}.
the uniqueness of
(*)
Because every g . f g
GCfl
-1
~j
(*)
in G.
Put f
=
f
Then
is a block idempotent of F N ,
implies that
9jf9j for all j E {l,Z,
Then
,
where the fi are uniquely determined block idempotents of F N .
g ,...,gt
so
Z*(FN).
is an arbitrary block idempotent of
x = f
and denote by
.
we therefore deduce that
gfg-l E gZ(FN)ug-' Hence f*
is one of them,
Because u u = u , we have f = u l u E Z ( F N ) U
f.
Bearing in mind that u E Z ( F G ) ,
for all g E G .
and f
...,t}.
-1
E
{fl .,fJ t * .
Hence
x = f*+y for some idempotent y
in
ZCFN)
such that yf* = f*y = 0.
Since x and f*
MORITA'S THEOREM
Z* (FNI,
are block idempotents of
we infer that
Finally, since for any two block idempotents f and f
and only if f
2
Z* (FN).
of FG.
and hence that
and f
of
=
=
f';
z ( F N ) , f*
f*. if
are all block idempotents
k E {1,2,..
. ,n}
such that e
as the sum of block idempotents of FG.
.
if and only if f* = f;
occurs in
This proves
of F N , e
and that, for any block idempotent f
covers f k
x
Owing to (i), we can find block idem-
Thus there exists a unique
the decomposition of f; that e
= 0
of FN such that f ; , c , . . . , f ;
potents f ,f ,...,fn of
y
are G-conjugate, (i) follows.
(ii) Fix a block idempotent e 1
231
Because the latter is equivalent to f
covers f
fk
and
being
Gconjugate, the required assertion follows. (iii) It is clear that ef # 0 if and only if ef* # 0. if and only if e
covers f, the result follows.
Let A
10.3. LEMMA.
be a finite-dimensional local commutative algebra over F .
Then there exists a subfield E
2F
of A
such that
A = E@ J U )
Proof. Observe that if A hence E
3.F .
Since the latter holds
=
(direct sum of F-spaces).
E @ J U ) , then E
A/J(A)
is an F-algebra and
We divide the rest of the proof into two steps.
Step 1. Here we establish the case where J(AI2 = 0 . Put Ap = { $ l a E A }
Therefore, for all a E A , a
apb = 1 for some b E A , A.
A = E @ J(Al,
Ap C - E,
2
If f ( X ) with
2
and
are units of A
then b = ( u - ' ) ~ E A p
Let E denote a maximal subfield of A
E n J U )= 0.
charE = p ,
containing A p .
assume by way of contradiction that
E
and so
2E
E.
Since
(r,p)
we obtain
=
PU
= A/J(A).
is a coefficient of
z.
If
To prove that
a = a + J(A) 4
1, 1 = sr+tp for some t , s E
0.
Clearly
Consider the polynomial f ( X I
Now
2#
is a subfield of
is not irreducible, then it has an irreducible factor because a 4 E.
since J ( A j 2 = 0 .
whenever
and hence A p
the image of E in
We denote by
1 C r < p,
r a E E.
then a 9 J ( A I 2 ,
and note that if up # 0
E.
Since
9-2 E
E[Xl.
(X-aIr
E[Xl
=
(X-a)
E
r
and so
Bearing in mind that
CHAPTER I11
232
a = sra
contrary to our choice of a. is a subfield of
+
Thus
tpa = s(ra) e E ,
f(X) is irreducible, in which case EIal
A properly containing E , a contradiction.
Completion of the proof.
Step 2 .
Assume the contrary and denote by A
a counter-example of minimal F-dimension.
By Step 1, J(A)'
# 0 and so A l = A/J(A)*
Clearly, J(A
J(A)/J(A)'
) =
has smaller dimension than A .
and so J(A1)' = 0.
Because A
is also local,
it follows from Step 1 that
of A l
for a suitable subfield E some F-subalgebra B
Since B
of A .
containing F.
of B
= B/J(AI'
for
Then, by (11, we have
is a local proper F-subalgebra of
subfield E
Write E
A with J(B1
= J(A)
',
there is a
for which B = E ED J(A1'
(3)
Applying ( 2 ) and ( 3 1 , we therefore derive A=E+J(A) and
a contradiction.
10.4. LEMMA. and let y ZCFG)
e
and
Let e and yf
ZLFN).
and f
be block idempotents of FG
and FN, respectively,
denote the corresponding irreducible representations of Then e
covers f
if and only if there exists an F-algebra
isomorphism IT : y
f
U *(FN))
3
ye (2" (FN)1
MORITA‘S THEOREM
233
such that
is algebraically closed, then e
In particular, if F
y (x)
f
Proof.
e
=
e ,e 1
2
Assume that e
,...,er
in FG
covers f .
Jacobson radical of local algebra over F
if and only if
for all x E Z*(FN)
Ye&)
Then there are block idempotents
such that
f*=e and, by Lemma lO.Z(i),
=
covers f
f*
1
+ e
2
+...+
e
r
is a block idempotent of
Z*(F”N).
Then
z*(FN)f*
Z*(FN).
Denote by
J the
is a finite-dimensional commutative
with Jacobson radical J
=
1
Jf*.
Applying Lemma 10.3, we
have
for some subfield E
and a unique z E J
there is a unique y E E
Thus, given x E Z*(FN),
containing F.
Z*(FIVlf*
of
with 1
xf* = y + z Because f*
is the identity element of
Z*(FN)f*
(4)
and since f*e = e ,
from (4) that
xe + ye + ze Since z E J
J(Z(FG11,
w e have
0 =y
e (2)
and thus
=
ye(z)ye(e) = ye(ze)
it follows
CHAPTER I11
234
of ye
NOW the image of the restriction Y ' extension of F
F
of FN,
Hence the restriction y
(Proposition 1.10.8(iv)).
occurs in the decomposition of $*
Of
where
onto y (Z* (FN)1
deduce that
TI =
y6-l
TI
i
E 11
i
,...,k},
Invoking (5) and (61, we therefore
f' is a central fdempotent of FG.
+...+
then y (e.1 = 0 for all i e z
0 = Y , ( f * ) = ny ( f * ) =
f
Hence e
10.5. LEMMA.
covers f
Hence
If e # ei
for all
by Proposition 1.10.8(v).
TIy
f
Owing
ek
are suitable block idempotents of FG.
a contradiction.
is
to
satisfies the hypothesis of the theorem.
p = e 1 + e2 where the e
Yf
is a required isomorphism.
Conversely, suppose that to Lemma lO.l(iii),
Thus
such that
f
x and y are chosen as in (4).
to
e
into the sum of block idempotents
it follows similarly from (4) that the restriction 6
an F-algebra isomorphism from E
of y
onto y e ( Z * ( F N ) ) , by virtue of ( 5 ) .
is an F-algebra isomorphism of E
Because f
is a finite field
to Z*(FN)
(fl =
T(1)
Thus
= 1,
and the result follows. m
Let L be a conjugacy class of N
containing x.
Then
N ( L ) = Ch)N
G Proof. Assume that with
-1 -1
g n mg = n
-1
1
normal.
n E N.
h E NI
xn 1
.
g E NU]. G Hence
Then, for any
"9";'
Conversely, suppose that Then
-1 y
-1
= ng
-1
xgn
G
=x
g E
n
E CG(x) and
E
N, there exists n E
g E CG(")N
since
CGCz)N1 say g = yn with y
and so
-1
-1
g xg = n ma.
E
N
is
C(z) and G
Hence, for all
MORITA'S THEOREM
g E N(L).
proving t h a t
235
So t h e lemma i s t r u e .
G The following r e s u l t i s e x t r a c t e d from a work of Michler(1972). 10.6. PROPOSITION.
,...,e
e
potents
i E { l , ...,s l r Di
(i) For all
i
(ii) For a l l
D
(iv)
N
C
,...,Cr
Civ)).
(i)
yi
Denote by
ei, 1 4 i
s.
Q
k E {l,.. . , r } . f.
1J. (Z* ( F N )
Because
f*
Write
D.
=
y l ( f * l = y Ce ) = 1,
h C+ 1 1
Let
p
GCf)
G
ZCFG) a s s o c i 0 # hi E F
where
contained i n
it follows t h a t
y
i
N
(ck+)
be t h e i r r e d u c i b l e r e p r e s e n t a t i o n of
yi ( Z * (FN1 1
C i E Z*(FN),
yi(Ck+) # 0
# 0 f o r some
Z(FN)
associated from
Ti
such t h a t
it follows t h a t
for a l l
2
71
+
U(Ck)
E 11,
-1
= 1T 7 1 .
1 %
...,s].
+
Yi(Ck)
Thus, by P r o p o s i t i o n 2 . 2 . 6 ( i v ) ,
i s conjugate t o a subgroup of a d e f e c t group, say
D, of
Ck.
Taking
i n t o account t h a t
f* =
and t h a t
Ck
and
(see Lemma 10.1
Then, by Lemma 10.4, t h e r e e x i s t F-algebra isomorphisms onto
t o be a
of p'-index.
... + h,C:
+
D
and choose
1
+
each
ZCFNl
t h e i r r e d u c i b l e r e p r e s e n t a t i o n of
0 # y (Ck) =
and t h u s
f
are s u i t a b l e p-regular c l a s s e s of Because
G(f)
i s conjugate t o a subgroup of
i s a subgroup of
1
with
Dl
Di
DN
Then
be of maximal d e f e c t and l e t
Then
is a s p l i t t i n g f i e l d f o r
GCfI.
Proof. a t e d with
F
e
let
Among t h e block idem-
i s conjugate t o a subgroup of
i s a d e f e c t group of
Assume t h a t
subgroup of
...,s},
E {l,
f,
covering
ei, 1< i Gs.
of
FN.
be a block idempotent of
FG
of
Di be a d e f e c t group
(iii)
f
Let
5 S u ~ p f * ~w e
x 1 c+1
+
... + AP cT+ = e
must have
1
Ck C_ suppej
+
... +
es
f o r some
j E 11,
... , s } .
236
CHAPTER I11
Therefore D is conjugate to a subgroup of D
j
conjugate to D
j-
Since D
and hence, by the above, D
is of maximal order among the D 1
1
,...,D8,
is
we infer
is conjugate to D, proving ti).
that D 1
(ii) Fix x E Ck
together with a complete system g ,...,gn
and choose xi j E CG(x) such that
of the double cosets G ( f ) y C G ( z )
in G
is a right transversal for G ( f )
in G(f)giCG(x).
Assume that g
E
of representatives
N is such that gi-1ggi
=
x for some i E 11,...,nl, and
n
n
c
r.1
-1 is the coefficient of x in f*. Because x E Suppf*, gzw< there is at least one integer t E (1,. ,n) such that p 1 rt. Since and so
2
..
r = t
it follows that G(f)
(iii) Let x
IGlf) I
W gtCG (x)gill
IGCf)
contains a Sylow p-subgroup of
account that g-l xgCSt E Ck, But, by (i), each Di
I G(f)gtCG(x)I - I G
I
Cc(gt-1xgtl.
Taking into
we infer that D is conjugate to a subgroup of C ( f ) .
is conjugate to a subgroup of
be as in (ii) and let L
D, proving (ii).
be the N-conjugacy class containing x.
rn
-1 Then Ck = U ti L t i where { t i l l G i G m3 is a right transversal for NGK) in i=1 G. Denote by l.l the irreducible representation of Z(FN) associated with f.
Owing to Lemma 10.4, there exists an F-algebra isomorphism ylCT),
where T = Z*(FN),
Because
+ yI(Ck) #
such that W(y1 = y (y)
0, it follows that
71
from p ( T )
for all y E T.
onto
237
MORITA' THEOREM
(7)
and hence t h a t
o
p(t-lL+t ) #
4
I t i s obvious that
t-lL+t
normal subgroup of
C
4
f o r some
q
E
11, ... ,m)
(8)
i s a c e n t r a l element of FN whose support i s t h e Nq q conjugacy c l a s s containing t - l x t Since t-lxt E C k , w e may assume t h a t Dl 4 4' 9 4 is a sylow p-subgroup o f c (t-lxt 1 . NOW N n c (t-lxt = c ( t - l t ) is a 4 G 4 4 N q 4 G q
(t-'xt4I,
(t-lxt 1. q 4 containing t-'xtq. 4
group of
L
N
nN
D
Thus
C
f
f
Because t h e d e f e c t groups of
I D 1 n N1 (iv)
D
n N.
f.
Thus
L1
and
gfg-l
i s N-conjugate t o a subgroup of
is t h e sum of t h e conjugates of
l6(fl
Since
For a given
I.
D nN
Hence
f E ZIFN) , we have
g E G , gfg-'
is a Sylow p-sub-
i s a d e f e c t group of t h e N-conjugacy
I t now follows, from (81 and Proposition 2.2.6 ( i v )
of
nN
D
so by Sylow theorem
G q
NOW
, that L
c Ck
5 Suppf*
f o r some
6 (f)
and
f*
g E G.
a r e of t h e same o r d e r , w e have
6(f),
i s N-conjugate t o
N C_ GCf)
a d e f e c t group
1 -
5 Suppgfg-I
class
and so
is a subgroup of
D N
i s a block idempotent of
FN
proving ( i i i ) .
GCf).
whose a s s o c i a t e d i r r e -
d u c i b l e r e p r e s e n t a t i o n i s given by.
Invoking ( 7 ) , we d e r i v e
1:= 1 Assume that
g E G(f).
Z (FN)
s e n t a t i o n s of Let
cosets
Then
{ w j E GI1
G(f)wN(L)
and hence
j in
<s} G.
and
llg
a r e e q u i v a l e n t one dimensional r e p r e s -
p
pg = p.
be a f u l l system of r e p r e s e n t a t i v e s of t h e double Choose
tij E GLf)
such t h a t
{tijUilj=l,...,n.1
G N(LI G
a l e f t transversal f o r
in
i= 1
Then
G
m t.
cu
G(f)wiN(L).
s
'(L+] =
c
ni t . . w s w ' ( L + ) = c nip ;(L+) I: 1-I
i=1 j=1
i=1
,
is
238
CHAPTER I11
t.. w
t..w. p
=
(p
=
)’
wi
for all
? . i
Invoking ( 9 ) , we infer that
E {l,..,,ni}.
J’
and hence that there is an integer d E (1,.
. .,s}
nd.
such that p
Because
it follows that
p ) (G(f) By Lemma 10.5,
:
t G ( f l n iVG(wdLwil))(
NG(L)dLwd-l) = CG(”dxwdl)N,
a Sylow p-subgroup of GCf).
D and D
p-subgroup of
contains
Thus
dividing n.
are conjugate, it follows that D
cG(Wd 3 : W d-1 I .
cG (wd Xw-l)N d
so by (10)
where v ( n ) denotes the highest power of p by ti),
(10)
Because
x E ck and,
is G-conjugate to a Sylow
The conclusion is that
proving (ivl, by applying (11) and (12).
.
At this point we begin to attack Morita’s theorem directly. 10.7. LEMMA. of
Let S
be a ring, let 1 = e
1
+
e
2
+
... + e n
1 into a sum of orthogonal idempotents, and let R = e Se 1
is a subgroup of the unit group of S sitively by conjugation. Proof.
since G
able gi E G , 1
i
Then S
be a decomposition 1
.
Assume
which permutes the set { e l , .
..,e n 1
that G tran-
M ( R ) as rings and R-modules. n
-1 acts transitively, we may write ei = gi e l g i for a suit-
n, with g l
= 1.
Given i,j e 11,...,n3,
set
MORITA’S THEOREM
-1 Then g-’ e , g j # gk e , g k
and fix k # j .
239
so that
( g-1 . e g .) ( g i l e g ) = 3 1 3 i k
o
Thus
and
= e it By definition, eii
ei
=
and therefore
1 = e
11
+
e 22
+
... + Qnn.
The desired
assertion is now a consequence of Proposition 1.5.1. 10.8. LEMMA.
Let
{e
,...,e 1
be a full set of representatives of the G-con-
ei,
be the sum of all G-conjugates of
For each i E {l,.
FN.
jugacy classes o f block idempotents of
let G . = G(e.1
.. , s ) ,
and let ni
let e? 2.
( G : Gi).
=
Then (i) FGe;
p
M, (FGiei) i
(1 Q i
proof. By Lemma 10.1, each e? account that
1
=
e*
... + e*
+
is a central idempotent of
and that e?e? 2 3
=
FG.
0 for i # j ,
Q s)
Taking into
we see that (ii)
is a consequence of (i). To prove (i), fix a block idempotent e
{e = u
1
,...,un
be the G-orbit of e .
also that S = FGe* group of units the set
{*I
and put
By hypothesis,
Because
e* =
G*
=
G(e).
It is obvious that n = (G : G * ) .
is a ring with identity element e * ,
-G = Ge*.
,.. . , V n 1 .
of FN
V
+
-G
and S
Let Note
contains the
acts transitively by conjugation on
... + on
is a decomposition of the
CHAPTER III
240
identity element of S into a sum of orthogonal idempotents of
S, Lemma 10.7
tells us that
FGe* = S Furthermore, since e*e = e ,
Mn(eSel
we have
eSe
=
e (FGe*)e
and it remains to prove that eFGe
=
FG*e.
ege
=
g(g-'eg)e
= 0.
eFGe
is the F-linear span of the set {egelg E G}.
To this end, note that eFGe
If g E G*, then certainly ege
=
=
Hence eFGe
On the other hand, if g 9 G*, then
ge. FG*e
=
and the result follows.
We have now come to the demonstration for which this section has been developed.
The following result is essentially due to Morita (1951) (see also
.
I Nakayama and Shoda (1936), Clifford (1937), and Koshitani (1982~) 10.9. THEOREM.
Let F
p > 0, let N = 0 ,(GI P
be an algebraically closed field of characteristic and let { e
,... ,e
1 be a full set of representatives of
G-conjugacy classes of block idempotents of F N ,
e
where by convention
= 1~l-lc x
3SN
For each i E 11, el
= e?'
... , s } ,
let e? be the sum of all G-conjugates of
let Gi = G[e .I
and let mi = (G
:
Gi).
e
i (so that
Finally, write
for suitable positive integer n,.
for suitable cti E Z2(Gi/N,F*), 1 G i G s . number of conjugacy classes of (ii) If FGez group of
Gi
G
is a block of FG
Furthermore, s
contained in N
and FGe*
for some i E {l,
...,s } ,
coincides with the 2
F(G/IV).
then a Sylow p-sub-
is a defect group of FGe?. Z
(iii) If P E Syl (GI
P
i s normal in
G,
then each FGe?
Z
is a block of FG.
MORITA'S THEOREM
Proof. in
That s
241
1s equal to the number of conjugacy classes of
N is a consequence of Lemma 7.6(1).
G
contained
For the sake of clarity, we divide the
rest of the proof into two steps.
Step 1. Here we prove that if H FG
a central idempotent of
To begin with, put S S
and that S
Then e
e
=
=
2 FHe.
with
and note that e
Let
{ e ..I1 Q i , j
=
e
11
is
n} be a set of matrix units in F H e .
Q
z3
so by Proposition 1.5.1
n
11
e E FH
is the identity element of the ring
FGe = S a M (e S e Since e e
and if
Mn(Fl , then
FHe
FCe
+ e 22 + ... + enn'
11
G
is a normal subgroup of
, we have e 11 S e I 1
=
I1
e (FGe)e 11
) 11
11
e
=
I1
FGe
and thus
11
It therefore suffices to verify that
Owing to Noether-Skolem Theorem (see Reiner (1975)1 ,
for some a E Z 2 ( G / H , F * ) . given g E G , g e l l g Setting u
=
u'
+
is conjugate to e (1-el,
we
there exists a unit u E F H
see that u
unit ut E FH then y
e
such that ug
Let T denote a transversal for H
by an invertible matrix u ' E F H e .
11
-1
11
u = ge
11
g
-1
commutes with
in G ,
such that utt commutes with
e
with y t E F H
=
Hence, for each g E G ,
e 11
and for each 11
.
If y
can be written in the form
y = C y t ET
.
C (ytu-'t)u S!Z'
tt
and hence I1
. t E T choose a
is any element of
FG,
CHAPTER I11
242 -1 e l l (ytut ) e l l E ellFHell
Since
e it follows that e llFGell that e
11
u t # 0
FHe
11
and = e
11
11
(FHe)e
11
For each
FGe
t
E
11
,
(14)
.
Hence
Note
{elluttltE 7'). e
11
u
t have t
{ e u tl t E TI is an F11
t
.
T, set
=
e
11
h E H , t E T.
Then, by the foregoing, e l1 FGell
u t.
t
{sit
associative F-algebra with basis with
I1
is the F-linear span of the set
disjoint supports (lying in the cosets of H t ) 11
Fe
(since utt is a unit) and that the elements
t
basis for e
=
Y by (14), there exists a nonzero CX(z,y1 in F e
Let x , y E T and write
E TI.
(utt)-l(uxx) [u y )
Then
i s an
is a nonzero element in FH
=
ht
and so,
such that
-1
11
((uttl (uxx) (u y 1 ) e = aC;c,y)ell Y 11
But then
--IxyH
xH
= e
11
((utt1-l(uxx) (u y ) ) e Y
=
a(x,y)el,
11
proving (131 and hence the required assertion. S t e p 2.
Completion of the proof.
Observe that the second isomorphism of (i)
i s
a consequence of the first. Apply-
ing Lemma 10.8(11 and Step 1, we have
.
proving (i)
Assume that i E {l,
...,s }
is such that FGe?
Proposition 10.6(ii), there is a p-subgroup Di group of FGe;.
of
is a block of FG.
G ( e . ) which is a defect
Invoking Proposition 10.6(iv), we see that D.
Sylow p-subgroup of
G(eil
,
proving (ii)
Finally, assume that P E Syl ( G )
P
P is a defect group of any block
of
is in fact a
.
is normal in G.
FG.
Then, by
Then, by Corollary 9.13,
Hence, by Theorem 9.23Lii1, the
AN APPLICATION
number of blocks of FG defect group P.
C C_ N.
is equal to the number of p-regular classes of
But a p-regular
C
of
has defect group P
G
The desired conclusion is therefore a consequence of (i).
closed field of characteristic p .
D, then B
Y
If B
Let Di
Gi, 1
i C.S.
2
Since
is a block of
.
and the
i s p-nilpotent
Gi/N
Hence, by Theorem 10.9(i),(iv),
FGe;,
1
< i < s,
be a Sylow p-subgroup
Di and hence, by Theorem
Since FDi is indecomposable, FGe;
FG.
10.11. COROLLARY.
that G
G
FGe? 2 MnSm ( F D i l . z i
lo.S(i) and Lemma 4.9,
true.
.
if and only if
Mn (FD) for some n ? 1.
We keep the notation of Theorem 10.9.
FGe?
with
is a block of FG with defect group
Proof. Q
G
be a p-nilpotent group and let F be an algebraically
Let G
10.10. COROLLARY.
of
243
are all blocks of
Di is a defect group of FG.
So the corollary is
Further to the assumptions and notation of Theorem 10.9, assume
is p-nilpotent and let Di
be a Sylow p-subgroup of
Gi, 1 < i
s.
(i) For any block idempotent e
is the unique block idempotent of i of FN,e? 2
FG which covers ei. (ii) dimJ(FGe?l
F (iiil
’
=
n?m?(lD.I-l) z z
z
did(FG1 = n~m~(IDiI-ll F i=1
Proof. As we have seen in the proof of Corollary 10.10, all blocks of FG and FGe? z is a consequence of (ii).
2
Mnam ( F D i ) .
FGe; ,...,FG*es
This proves (i) and also that (iii)
z i
Since JCFDil
= TCDi),
we have
and hence
as required.
are
.
11. AN APPLICATION : CRITERIA FOR J(FG) = FG*J(Z(FG)) Throughout this section, G denotes a finite group and
F an algebraically
244
CHAPTER I11
closed field of characteristic p
3
0.
Tt is clear that
The problem that motivates this section is to discover necessary and sufficient conditions under which the equality FG.J(Z(FG))
=
J(FG)
holds.
Our main tool
will be a consequence of Morita‘s theorem (Corollary 10.10). Let H
11.1. LEMMA.
G, let R be an arbitrary ring and let
RG( C k ) and, in particular, L ( I ( G ) ) = R ( C g l . M H SEG Proof. Suppose that x = C x g E R ( I ( H ) ) , xg E R. Then x ( k - 1 ) = 0 or gEG = z k , for all k E H . Hence x = z for all g E G,k E H . Therefore
Then k ( l ( H ) )
3:
be a subgroup of
=
gk
9
is constant on the cosets g H
of H .
3:
This proves that
The reverse inclusion being obvious, the result follows. 11.2. LEMMA. If t
Let P be a p-group and let
is the nilpotency index of
R be a field of characteristic p.
JCRP), then J(RPIt-l = R ( E g
By Corollary 1.2, J(RP) = I(P) and so
Proof.
It follows, from Lemma 11.1, that J(RP)
t-1
C R ( C g).
R( C g) is one-dimensional subspace of RG.
&P --
SEP
But J(RPIt-l # 0 and
Hence
sfp
as asserted. 11.3. LEMMA.
Assume that there exists a primitive idempotent u
that all composition factors of FGu Then G
i s
p-nilpotent.
of FG
such
are isomorphic to the trivial FG-module.
9
245
AN RPPLICATION
Proof.
Denote by
lG the trtvtal FG-module.
(ii) and Theorem 3.6(ii), the multiplicity of
Then, by Proposition 1.10.16
lG
FG
as a composition factor of
is dimFGu. On the other hand, by hypothesis, dimFGu is equal to the multiF F plicity of lG as a composition factor of FGu. It follows that if
FG FG
is a decomposition of
i' 2
then none of the FGe
... @ FGe,
@
FGe
into principal indecomposable modules with u G
is linked to none of the e that FGe
=
i G n , has 2
z"
i
lG as a composition factor.
n,
=
e ,
Thus
e
which implies, by Proposition 1.10.14,
is the principal block of FG.
Since lG is the only irreducible
module belonging to FGel, we conclude, from Proposition E.lO(ii),
that G
is
p-nilpotent. 11.4. LEMMA.
then G
Let B
=
B(e1
be the principal block of
FG.
If
is p-nilpotent.
Proof.
Let u
be a primitive idempotent o f
the trivial FG-module. (g-1)u E J (FGI u
FG such that FGu/J(FG)u
Then, for a given g E G , p
and therefore I (GI u
5 J (FG)u.
is trivial, we conclude that I ( G ) u = J C F G I U .
u(mod J(FG)u),
Hence we have
I ( G ) J ( F G )=~ J(FG)~+S, ~ for all n
1
acts trivially on each factor of the series
FGu
3
JLFGIu
3
...
2
so
Since the opposite containment
Recurring this, we get
which shows that G
is
J(FG)'u = 0
CHAPTER I11
246
This implies t h a t each campasitton f a c t o r of
PGu
i s isworpNc
to the t r i v i a l
Now apply Lemma 11.3.
FG-module.
W e are now ready t o prove 11.5. THEOREM (Tshushima (1978b), KIilshammer (1979)1 .
The following c o n d i t i o n s
are equivalent. (i) J(FGl =
FG*J(Z(FC)) B of
(ii) J ( B 1 = B * J ( Z ( B ) ) f o r each block
FG
(iiil
J(B1
(ivl
G i s p-nilpotent with a b e l i a n Sylow p-subgroups. proof.
blocks of
=
B'J(Z(B11
(i)* ( i i ) : Write
FG.
f o r t h e p r i n c i p a l block
FG
= B
@
... @ Bn,
B of
where
El
FG
,...,E
Then
and
Hence
Thus
J ( B . ) = BiJ(Z(Bi)) f o r a l l
(iil * (iiil: (iii)=*
iE
{l,
...,n},
a s required,
Obvious
( i v ) : Write
B = B(el
so t h a t
J(B) = J(FG)e.
Since
are a l l
AN APPLICATION
proving t h a t Let
G
is p - n i l p o t e n t , by applying Lemma 11.4.
G.
be a Sylow p-subgroup of
P
247
B
Then, by Proposition 1 . 2 0 ,
FP
and
thus
To prove t h a t
P
is a b e l i a n , w e use induction on t h e o r d e r of
t h a t f o r any normal subgroup
Let
2
H of
be an element of o r d e r
c p g and
cT=
'I=
P/
of
Let
.
Given
x unless x
f 2.
x
P,
E
in
2
and p u t
c y
is a b e l i a n , by t h e induction hypothesis.
A s s u m e by way of c o n t r a d i c t i o n t h a t
P' = < z >
p
if=
SEP W e may assume t h a t
I t is c l e a r
P.
P, l e t z
be t h e c e n t r e of
P.
w e have
t be t h e nilpotency index
of
is not a b e l i a n .
IxPcP'I= p
J(Z@'P))
Eence
P
and
Then we have
zpr is the conjugacy c l a s s
is t h e F - l i n e a r span of t h e s e t
JVG).
By Lemma 1 1 . 2 , w e have
J(FZltcl
= FT
(3)
W e now claim t h a t it s u f f i c e s t o prove t h a t
Indeed, i f (4) is t r u e , then taking
y E P-2,
w e have
J ( F d = 0
(y-l)T
a contradiction. To prove ( 4 1 , we note t h a t
by v i r t u e of (1).
# 0 and, by ( 3 1 ,
'I
On t h e o t h e r hand,
t-1 E J(FZ) , so
CHAPTER I11
248
,...,Zt - 1 E
for all z
Z,
P-2.
t E
Finally, by definition of
6
(zi-l)
t,
= 0
(7)
i=l where z ,...,zt
are arbitrary elements of
z.
Taking into account (2) and
(5)-(7), we immediately deduce that C4I holds, as required.
(iv) * (i): By Corollary 10.10, for each such that Bi
p-group Di
J(Bi)
2
M
ni
M, (J(FDi)) and Z(Bil
i E {I,. . . , % I ,
(FD.) for some ni
1
z
FDi.
> 1.
there exists an abelian Hence
It follows that J ( B i )
i
GZ
M
( J ( 2 ( B i ) ) and
ni
hence J ( B . ) = BiJ(Z(Bi)), 1 1.
Q
i
n.
Thus
and the result follows. 12. GROUP ALGEBRAS WITH RADICALS OF SQUARE ZERO Throughout this section, G characteristic p > 0 .
denotes a finite group and F
an arbitrary field of
Our aim is to discover the structure of G
in case
J(FG)’ = 0. 12.1. LEMMA.
Assume that the Sylow 2-subgroups of
G
are cyclic.
Then G
is
2-nklpotent. Proof. Let P is a 2-group.
be a Sylow 2-subgroup of
G.
Since P
is cyclic, Aut(P)
Hence the required assertion follows by virtue of property (ii) in
the proof of Lemma 9.22. 12.2. LEMMA. Let B of G
be the principal block of FG.
are defect groups of B .
then J V G )
=
Then the Sylow p-subgroups
In particular, by Theorem 6.4(1),
if J ( B ) = 0
0.
Proof. Let e
be the block idempotent of FG
belonging to B.
We may
RADICALS OF SQUARE ZERO
write
e
2 49
in the form
e for some p-regular classes C 1
module.
Since
V
=
x 1 c+1 +
,...,Cr
belongs to B ,
of
... + xrC+r G.
we have
eu =
U
the trivial FG-
V
Denote by
for all u E V .
Hence for all u E V
which implies that p
ICI
for some p-regular class C
p-subgroup of
G is a defect group of C, as required.
12.3. LEMMA.
Let B
be a block of
Thus a Sylow
8
Then J ( B ) *
FG with J ( B ) # 0.
and only if all irreducible FG-modules in B indecomposable FGmodules in B
in Suppe.
=
0 if
are isomorphic and all projective
have composition length 2.
Proof. Assume that J ( B 1 2 = 0 and write
B = Be @ where
e
1
J(B1 # 0,
> 1.
,...,en
are mutually orthogonal primitive idempotents of
there exists
i
E 11,
...,nl
Since J ( B ) ( J ( B ) e i ) = 0 ,
irreducible and Bei/J(B)ei length 2
... @ Ben
2
such that Bei
... ,e n
As
has composition length
it follows from Theorem 3 . 8 that J ( B ) e i
J(B)ei.
This shows that Bei
and that the composition factors of Bei
idempotents e ,
B,
is
has composition
are isomorphic.
Since the
are linked (Proposition 1.10.14) and Be ./J(B)e .%Be ./JIB) e 2 z 3 j
implies Be. 2 Be (Proposition 1.9.41, the required assertion follows. Conversz j ly, assume that J(B)' # 0. Then J ( B ) * ei # 0 for some i E {l, ...,n] and so
Thus the composition length of Be
i is greater than
2,
as
required.
We are now ready to prove the main result of this section.
.
The equivalence
of (i) and (iii) in the theorem below was established by Wallace (1962a).
12.4. THEOREM. Let F be an arbitrary field of characteristic p > 0 and let G be a group of order p Um
with a
>1
and
( p , m ) = 1.
Then the following
CHAPTER I11
250
conditions a r e equivalent:
o
( i ) J(FGI* =
(iil
J ( B 1 2 = 0,
(iii) p'
p
(ivl
B
i s t h e p r i n c i p a l block of
FG
= 2 =
2
*
G
and
(il
Proof. (ii)
where
has a normal subgroup of odd order and of index
( i i ):
=*
Obvious
p
(iii): By hypothesis,
J ( B ) # 0.
Hence by Lemma 1 2 . 2 ,
d i v i d e s t h e o r d e r of
Since t h e t r i v i a l FG-module belongs t o
But
a p ldimV
(iiiI *
by Lemma 8 . 2 .
Thus
p
G
and so
J(FG) # 0 .
V be a p r o j e c t i v e indecomposible module i n
Let
B.
F (iv)
2
U
= 2
B,
Lemma 12.3 implies t h a t dimV = 2. F a s required.
By hypothesis, t h e Sylow 2-subgroup of
G
i s c y c l i c of o r d e r 2.
Now apply Lemma 12.1.
*
(iv)
(il : By hypothesis,
of order
G
is 2-solvable and t h e Sylow 2-subgroup of
is
G
Hence t h e required a s s e r t i o n follows by v i r t u e of Proposition 1.9..
2.
13. GROUP ALGEBRAS WITH CENTRAL RADICALS Throughout t h i s s e c t i o n ,
p > 0.
characteristic
G
denotes a f i n i t e group and
F
an a r b i t r a r y f i e l d of
Our aim i s t o p r e s e n t S p i e g e l ' s r i n g - t h e o r e t i c proof
(1970) of Wallace's theorem 11962bI which determines t h e s t r u c t u r e of
13.1. LEMMA.
subgroups. Proof.
J(FG)
If
5Z(FG),
In p a r t i c u l a r , Let
K
p
G
then
G
i n case
i s p - n i l p o t e n t with a b e l i a n Sylow p-
( c ' 1.
be t h e a l g e b r a i c c l o s u r e of
F.
Then, by Corollary 1.18,
J(KG) = K 8 J(FG) C - K 8 Z(FG) = Z(KG) F F and so, by Theorem 11.5, t h e l a t t e r implies t h a t
13.2. LEMMA.
Let
equivalent : (i)
XC Z(FG)
X
G
G'
.
is p - n i l p o t e n t with a b e l i a n Sylow p-subgroups.
5 O p t (G) ,
be an i d e a l of
t h e r e s u l t follows. FC.
Then t h e following c o n d i t i o n s a r e
Since
CENTRAL RADICALS
251
(ii) X * I ( G ' ) = 0
X
(iii)
5FGs,
Proof.
*
(i)
a,b E G.
and
z(ab~-~b-~-ll = 0.
Hence
~ba-'b-~,a,bE G ,
X
z E
(ii): L e t
x(ab-ba)= 0.
and
s =
where
Then
Since
G'
i s generated by t h e
t h e required a s s e r t i o n i s a consequence of P r o p o s i t i o n 2.1.10.
( i i l * (iii): D i r e c t consequence of Lemma 11.1 (iii)=) ( i ) : C l e a r l y
{pig E
G)
p
(iil
J(FG)
# FGs,
FGs
i s t h e F - l i n e a r span of t h e s e t
t E G, w e have
5 Z(FG),
then t h e following a r e e q u i v a l e n t :
s =
where
lG'l
Proof.
(i)
s2 = ( G ' ( s = 0 (ii)* (i):
Thus
Since
5 Z(FG).
FGs
If
]LEMMA.
(i) J(FG)
Z(FG).
and s i n c e f o r every
it follows t h a t
13.3.
E
*
(ii):
FGs
and
p
If
J(FG1 # FGS
1
J(FGl
By Lemma 1 3 . 3 ,
5 J(FG).
1,
IG'
Hence
then
e
=
5 FGs.
J ( F G ) = FGS
p
If
I
IG'I,
then
as required
I G r 1-l~i s a nonzero idempotent i n
FGs.
and t h e r e s u l t follows.
The following theorem w a s proved by Wallace (196213) f o r group a l g e b r a s over
I t s g e n e r a l i z a t i o n and t h e proof given h e r e a r e due
a l g e b r a i c a l l y closed f i e l d s . t o Spiegel (1970).
13.4.
THEOREM.
Let
be a f i n i t e group.
F
be an a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
Then
J(FG1
i n g t h r e e types: (i)
G
has o r d e r prime t o
p
5Z V G l
i f and only i f
G
p
and l e t
G
i s one of t h e follow-
CHAPTER I11
252
Ciil
G
abelain
its
P
(iii) I f
is a Sylow p-subgroup of P
complement Proof.
G
G'P
then
i s a Frobenius group with
G'.
and k e r n e l
Let
G,
be a nonabelian group whose o r d e r i s d i v i s i b l e by
p.
It
c l e a r l y s u f f i c e s t o prove t h a t J(FG)
5 Z(FG)
i f and only i f (iii)holds
By Theorem 8.11, t h e l a t t e r i s equivalent t o JCFG) C_ Z(FG) Assume t h a t
i f and only i f
5 Z(FG1
J(FG)
and put
S
dim[J(F(G'P)] = ( P I - 1 F
J(FG) index
c
FGs
(G:N1
and
p )' I G ' 1 .
Clearly
is not d i v i s i b l e by
p.
.
=
G :
and
p
(G'(
(1)
By Lemmas 13.1, 13.2 and 13.3,
rg
N = G'P
i s a normal subgroup of
G
whose
Hence, by Proposition 1 . 8 ( i i ) ,
didCFG) = (G:N)dimJUU) F F Since
JCFG)
C
FGs,
we have
Invoking (2) and (3) w e t h e r e f o r e d e r i v e
did(FN1 C F
on
t h e o t h e r hand, by Theorem 8 . 4 ( i ) ,
a s required.
Then
FNe a FP
Conversely assume t h a t
so
dimJ(FN1 a [PI-1. Thus dimJ(FN) = IPI-1, F F dimCFN) = (PI-1 and t h a t p IG'I. Let F
1
dfmJ(FNe) = [PI-1 = dimJ(FN) F F J(FN) = J(FNe) =
Since that
N
(PI
i s a normal subgroup of
JLFG) = FG*J(FN).
G
J(FN)e
of p'-index.
and thus
J(FN) = J(FNe).
5 FNs proposition 1 . 8 ( i i )
Invoking (41 w e t h e r e f o r e conclude t h a t
Hence
(4) ensures
COMMUTATIVITY OF THE RADICAL
Hence J(FG)
c- Z(FG)
13.5. REMARK.
by Lemma 13.2.
7.11, is in fact cyclic.
.
Thus completes the proof of the theorem.
Let G be a group such that G'P
plement P and kernel G'.
253
Then P
G'P/G'
is a Frobenius group with com-
.
is abelian and hence, by Remark
Furthermore, by Lemma 7.1,
G'
is nilpotent.
14. COMMUTATIVITY OF THE RADICAL OF THE PRINCIPAL BLOCK Throughout this section, G characteristic p > 0, lG
FG.
denotes a finite group, F an arbitrary field of
the trivial FG-module and B,
the principal block of
Our aim is to discover necessary and sufficient conditions for the commu-
.
tativity of J (Bo) Let A
We say that B
block of A .
@ 1
sum
If K
be a
... @ B e n
of principal indecomposable modules, each Be
same multiplicity.
and let B
is quasi-primary if in decomposition B = Be
as a direct
K
be a finite-dimensional algebra over a field
is a splitting field for B ,
i appears with the
then the following
conditions are equivalent: (il B
is quasi-primary
(ii) All irreducible B-modules have the same K-dimension (iii) B / J ( B )
i5 a direct product of isomorphic simple algebras
Indeed, the equivalence of (ii) and (iii) is trivial, while the equivalence of (i) and (ii) is a consequence of Proposition 1.10.16(ii). Recall that if
V
and
W are FG-modules, then the tensor product
v@ w
is
F
an FG-module via
14.1. LEMMA.
Let
cible FG-module.
V
beanFG-module of dimension
Then V @ W
1 and let W
is an irreducible FG-module
F Proof. By hypothesis,
be an irredu-
v = .8'V
for some nonzero V
in
v.
Let X
be a
CHAPTER I11
254
nonzero submodule of
v 8 W.
Then every
x E X can be written as
X = U
8W
F for a w E W.
It is a consequence of the deftnitton of the action of G on
V 8 W that the set F
x is a submodule of W
= { n ) E w ( u 8 w E X 1 # 0
and hence X
x
=
1
Iu
W.
=
Qpwlw E
But
x1
V
has dimension 1, so V 8
=
F
w ,
which proves the lemma. 14.2. LEMMA.
and W be FG-modules with
Let U,V
U
V.
Then as FG-modules
Proof. The canonical map
is surjective F-homomorphism whose kernel is U 8 W.
Since the map obviously
F preserves the action of G, the result follows. 14.3. LEMMA.
Let U be any indecomposable projective FG-module.
sition factors of P ( 1 1
G
tion factors of Proof,
U
If the compo-
have dimension 1, then the dimensions of all composi-
are equal.
Put V = U/J(FG)LI.
summand of P U G ) 8 V .
We
know, from Lemma 8 . 3 , that U
is a direct
Let
F
be a composition series of P ( l G ) .
Owing to Lemmas 12.1 and 12.2,
P ( 1 1 8 V = X 8 V 3 X 8 V 3 .. . > x 8 V = O G F 'F 2 F n F is a composition series of p(l I 8 V
and all composition factors have the same
'F dimension, namely dimV.
F
This proves the required assertion.
COMMUTATIVITY OF THE RADICAL
14.4. LEMMA.
255
The following conditions are equivalent:
(i) All irreducible FG-modules belonging to the same block have the same Fdimension Cii) All irreducible FG-modules belonging to the principal block have F-dimension 1. (iii) All composition factors of P ( 1
have F-dimension 1.
G)
Proof. (iil
(i)
=*
(ii): Obvious
(iii): Direct consequence of the fact that all composition factors of
=*
P(lG) belong to Bo. (iii)
=*
(i):
Apply Lemma 14.3 and the fact that the indecomposable projective
modules in a block are linked. The proof of the implications (il
=*
(ii) and
(ii) * (iii) below is extrac-
ted from a work of Hamernik (1975b). 14.5. PROPOSITION (Morita (1951)).
characteristic p > 0
and let P
Let F
be an algebraically closed field of
be a Sylow p-subgroup of
G.
Then the follow-
ing conditions are equivalent: (i) PO ,(G) 4 G
P (ii) Bo
G/PO ,(GI
is abelian
P
is quasi-primary
(iii) Each block Proof.
G
and
B of FG
(i) * (ii): Set
is quasi-primary
5
=
G/O ,(GI
P
and
=
PO ,(G)/O
P
P
,(G).
Since
is p-solvable (and hence p-constrained), it follows from Proposition 1.20 that
B
FE.
-
By hypothesis, G
1.
Since the same is true for Bo,
the assertion follows.
(ii) * (iii) : Follows from Lemma 14.4. (iii) * (i) : Write B
0
=
B0 (e) and note that, by Proposition 8.1O(i), 0 ,(GI = { g E Glge = el
P
Let
and
Hence FZ has only irreducible representations of dimen-
abelian factor group. sion
-
is a group with normal Sylow p-subgroup P
256
CHAPTER I11
G afforded by the FG-module FGe.
be the matrix representation of
g
if and only if ge
E Kerr
regard
r
as
e,
we have
Kerr = 0
a faithful representation of
G/Op, ( G I .
=
belong to Bo,
factors of FGe
P
that the set of all elements g E G Sylow p-subgroup. say !C
of
33
Moreover, Q
G/Op,CGI.
is abelian.
P
14.6. LEMMA.
Let F
such that J ( B )
Proof.
P
This shows that PO
P'
be a splitting field for FG and let B
(G) 4
G and
be a block of FG
is commutative.
sition factors of Be
If B
is normal in G/O , ( G I
So the proposition is true.
(i) If e is a primitive idempotent of B
(iil
Then it is easily verified
such that r . . ( g ) = 1 for all j forms a
and the corresponding factor group i s abelian. that G/Po '(GI
is quasi-primary implies
Thus we may assume that the entries
are zero if i < j.
(rij(gl)
of the matrix
We may therefore
Since the composition
the assumption that Bo
that all these factors are one-dimensional.
r..Cg)
(GI .
Since
with J ( B ) ' e # 0 ,
then all compo-
are one-dimensional
is the principal block, then B
is quasi-primary
(i) We claim that it suffices to show that dimSoc(Be1
1.
=
Indeed,
F assume that dimSoc(Be) = 1 and let V
be a composition factor of
Be.
Since
F the Cartan matrix of FG is symmetric (Theorem 3.6(vi)), Be/J(B)e tion factor of P ( V ) .
and so PCn
is a composi-
But, by Theorem 3 . 8 ,
If J(BI2P(V) = 0,
has a composition factor of F-dimension 1.
then the composition factors of P ( V l
have the same dimension equal to dimV
.
F Hence dimV = 1.
If J ( B 1 ' E " V )
# 0 then SocP(V)
V
is one-dimensional, by
F hypothesis.
This substantiates our claim.
Since J ( B ) ' e # 0, we have Soc(Be)
5 J ( B ) ' e 5 J ( B ) '.
show that all composition factors of J ( B I '
are one-dimensional.
will follow provided we prove that for all x , y E J ( B )
@a) ( ~ 1 . This is of G
so
since then G'
afforded by J ( B l 2 .
Thus we need only The latter
and U , b E G ,
(ab) (Ccy) =
will be in the kernel of the representation
COMMUTATIVITY OF THE RADICAL
Using the assumption that J ( B I
257
is comutative, we have
as required. (ii) By Lemma 14.4, it suffices to verify that all Composition factors of P(lG) are one-dimensional.
By (I), we may assume that J ( B ) 'P(lG)
composition factors of P(1 1
are isomorphic to
G
=
0.
.
But then all
So the lemma is true.
lG.
We have now come to the demonstration for which this section has been developed. Let F
14.7. THEOREM (Hamernik (1975)).
be an arbitrary field of characteristic
p > 0, let G be a finite group and let Bo be the principal block of FG. Then the following conditions are equivalent: (i) Bo
is commutative
(ii) J ( B o l C_ Z(Bo) (iii) J ( B (iv) G
is commutative
p-nilpotent with a b e Lan Sy )w p-subgroups
is
(i) * (ii): Obvious
Proof. (ii)
=)
(iii): Obvious
(iii) * (iv): Let K
be the algebraic closure of
F and write Bo
= Bo(e)
.
Then
e = e for some block idempotents e lFG,one of the
1 G i G s,
e
i'
say e
1
1
1
,...,e ,
+ e
2
+...+
of KG.
e Since e
acts as identity on
acts as identity on
lKG. Setting
B? = KGe,
it follows that
KQ B F o and that B*
1
= KGe = B*
is the principal block of KG.
... Q3 B: Invoking Corollary 1.18, we derive
CHAPTER I11
258
J ( K Q Bo’ = K Q JWOl = J(B;l
F
... CB J ( B S )
a3
F
which shows that J(B:)
is commutative.
Hence we may harmlessly assume that F
is algebraically closed. By Lemma 14.6, Bo
is quasi-primary and so, by Proposition 14.5,
PO ,(G) d G and G/PO ,(GI is abelian P P where P E Syl (GI.
P
Because G
e
=
is p-solvable, Proposition 1.20 implies that
1 0 P ,K)I-’
c
r
as0 , ( G I
P
Setting H = PO ,(G), it follows from ~emma7.3.
is the block idempotent in Bo.
P
that
on the other hand, by Proposition 1.8Cii.1,
J(FGI
=
FG.J(FH).
Hence
and therefore FG*I(P)e is commutative. Choose an arbitrary s E P
and u E NG(P).
suffices to check that U3u-l = we
d.
By Burnside’s criterion, it
To t h i s end, we put u8u-l
have
Since for any x E FP, ex = 0
implies x = 0, we deduce that
’
(1-s) cl-sl 1 = (1-s)
= 6
E P.
Then
CRITERIA FOR THE COMMUTATIVLTY
OF
259
J(FG1
and hence that s -ss 1
If p # 2
t h i s yields
which case either s
s = s 2 = 1.
=
1
=s-s2
s = s.
Ef p = 2 ,
s
=
or s
then we have s
But if s = s 2 ,
s2.
= s+s2,
in
then s = s 3 and so
and therefore in all cases s
Hence S = 1
+ ss
=
s.
This completes
1
the proof of the theorem. 15. CRITERIA FOR THE COMMUTATIVITY OF J(PG) Throuthout this section, G denotes a finite group and F an arbitrary field of characteristic p > 0.
J(FC1
is commutative.
Our aim is to determine the structure of
G in case
The following lemma is stated in a stronger form than
required. Let FI
15.1. LEMMA.
be a normal p'-subgroup of G.
Then the following condi-
tions are equivalent: (i) J (FG) J (F (G/N) 1 (iil dimJ(FG1
=
dimJ(F(G/N)l
F
F
(iii) Every block
B
V
of
FG with J ( B ) # 0 contains an irreducible FG-module
5 Kerv.
such that N
P E Syl G), PN 4 P
Furthermore, if for
G
then ti), (ii) and (iii) are equi-
valent to each of the following conditions: (iv) J(FG1 = FG'I(P)e, (v)
dimJ(FG1 =
that FGe
1
e
=
1
IN(- E g
1 GI ( /PI -1)
F Proof.
where
SEG
Ipl IN1
Owing to Proposition 2.1.9,
e
is a central idempotent of FG
such
F(G/N).
(i) * (ii1: Obvious (iil * (iii) : Since J ( F G 1 = J(FGe1 @ J(FG(1-e) l dimJ(FGe) = dimJ(FG),
F J(B1
we have
J ( F G ( 1 - e l ) = 0.
and, by hypothesis, Let B
be a block of FC
with
F # 0 and let V be an irreducible FG-module in B .
we have B CFGe and hence eU = U
for all U E V.
Since J(FG(1-el) = 0,
Thus, if
n E N, then since
CHAPTER I11
260
ne = e ,
we have
nu This proves that N (iii) =. Ci):
5 KerV,
Let B
as required.
V and so ev
=
v
FG with J ( B 1 # 0.
be a block of
contains an irreducible FG-module on
for all v E I/
(ne)v = ev = v
=
for all V E
V with N V.
5 KerV.
Thus B C F G e
By hypothesis, B
Then N
acts as identity
and therefore
as required. Now assume that PA7
G.
Q
Then PN/N
and hence, by Proposition l.l(iii)
is a normal Sylow p-subgroup of
G/N
and Corollary 1.4,
and
BY Proposition 2.1.9, of F-algebras.
the map F(G/N1
--+
F G e , g N k ge
Since FG*T(P)e is the image of
determines an isomorphism
F(G/N)T(PN/N) under this map,
we deduce from (1) that
.
Thus, by ( 3 1 , (ivl is equivalent to ti), while by (21, So the lemma is true.
(v) is equivalent to (ii).
In the result below, the equivalence of (il , Cii) and (v) was established by Wallace (1965). 15.2. THEOREM. P E Syl
P
(il
(GI.
JCFG)
Let F
be an arbitrary field of characteristic p # 2
Then the following conditions are equivalent:
5 Z(FG)
(ii) JCFG)
is commutative
and let
(iv) p
)
and J(FG) = F G * I ( P ) e , where
IG'I
(v) The group
G
e
=
261
J(FG1
CRITERIA FOR THE COMMUTATXYITY OF
IGrI-l
is one of the following three types
(a) G has order prime to p (b) G (c)
is abelian
G'P
is a Frobenius group with complement P
Proof.
(i)
and kernel G ' .
* (ii) : Obvious
(ii) =* (iii): Owing to Corollary 1.18, we may harmlessly assume that F is algebraically closed.
Since J(FG)
the principal block of FG. Thus G'
is abelian.
50
P'
is commutative, so is J ( B
and therefore p
(GI
1
Bo
is
is p-nilpotent and P
G
Hence, by Theorem 14.7,
where
)
IG'I.
We now claim that if J ( B ) # 0, then
J(B1'
# 0
(4)
Assume by way of contradiction that J ( B I # 0 but J ( B ) ' = 0 . irreducible FG-module in B ,
then by Lemma 12.3,
If
dimP(V) = ZdimV.
F hand, by Theorem 4.27,
dimP('V)
=
12'1 (dimV)
Hence
On the other
F
IPI
=
2(dimV)
contrary
F p'
F P"
F
V is an
to the assumption that p # 2. With J ( B ) # 0.
N
=
G'
we certainly have PN 4 G.
Then, by (41,
cible module
Let B
be a block of
J ( B ) ' # 0 and so, by Lemma 14.6, B
V of dimension
But then N
1.
5 KerV
FG with
has an irredu-
and so the desired con-
clusion follows by virtue of Lemma 15.1. (iii) * (iv): Direct consequence of Lemma 15.1 (iii) * (v): let M = G'P.
Let G
be a nonabelian group whose order is divisible by
p
By the property (1) in the proof of Theorem 13.4, it suffices to
show that dimJ(FM) = I P I - 1 .
F But M
and
is a normal subgroup of G
of p'-index and so, by Proposition 1.8,
CHAPTER 111
262
as required. (v) =* (i): Direct consequence of Theorem 13.4.
m
The next example illustrates that the hypothesis p # 2 in Theorem 15.2 cannot be ommited without sacrifising the conclusion. 15.3. EXAMPLE (Koshitani (1979)).
Assume that F
p = 2.
Let the group G be defined by
Then G
is 2-nilpotent with 0
P
idempotents of FO
P
(GI
(G) = < a >
and
P
is algebraically closed and
=
ESyl ( G ) .
P
The block
are given by
e = ( l + a + a 2 ) e, = ( ~ + E u + E ~ u )and , e Furthermore b-le b = e
and so G ( e
=
>
= (~+E~u+Eu)
is of index
2
in G.
Hence,
by Theorem 10.9,
and therefore
The first factor is commutative and the second is also commutative, since
1a2I2= 0. J(FG)
Thus J W G )
is commutative and, by looking at the second factor,
Z(FG).
In order to examine the case p = 2, we need the following preliminary observation. 15.4. LEMMA.
1’
Z( F C ) Proof.
Let I be an ideal of FG.
If I is commutative, then
. Let x,y E 1 and z E FG.
Then
263
CRITERIA FOR THE COMMUTATIVITY OF J ( F G )
so that
commutes with yx and hence 1 '
Z
'
C Z(FG).
Turning our attention to the case p = 2, we now prove Let F be an algebraically closed field of characteristic 2,
15.5. LEMMA. G
P
is odd and a > 1, and let
be a nonabelian group of order 2arn, where rn E
Syl ( G )
and N = G'P.
J(FG)
is commutative
(i1
(ii) G
Then the following conditions are equivalent:
is 2-nilpotent, G'
(iii) G
let
is 2-nilpotent, G'
is of odd order and for each s E G'-1,4
1
ICN(z)I FN
is of odd order and each nonprincipal block of
is of defect 1 or 0. Proof
(i) * (iil:
Since JCFG)
is the principal block of FG. P
Hence, by Theorem 14.7,
In particular, G'
is abelian.
is commutative, so is J(Bol
5 02'(G)
N 4 G and therefore, by Proposition 1.8(i1,
and so
J(FN1
It will next be shown that G' = N'
commutative.
G'
G
where Bo
is 2-nilpotent and
is of odd order.
5 J(FG).
Hence J V N )
Now is
and hence that FN has a
unique (up to isomorphism) irreducible module of dimension 1. Put x
=
C ,g, y SEN
=
Z g,
and z =
.
Since N '
is of odd order, Lemma
SEP
1.3 implies that
whence
' 5 FGZ.
It therefore follows
1 by Theorem 12.4.
Hence, by Lemma 11.2,
On the other hand, by Lemmas 15.4 and 13.2,
J(FG)
that
But I ( P 1 2 # 0 ,
y
E
since otherwise a
I(P)' and therefore yz E FGz.
for all g E G ' .
=
It follows, from Lemma 11.1, that yz(1-g) = O
Since 1 E Suppyx, g E Suppyxg
=
Suppyx C N'P
and thus
264
C W T E R 111
G' C - N'P.
is of odd order, we conclude that N' = G ' .
Because G'
G'-1
Assume by way of contradiction that there exists z has at least two 2-regular class of defect d
Then N 9.23,
FN has at least
principal block B B
2
>
blocks with defect d 2 2.
with defect group D
Mn(FD) for some n b 1.
If n
such that
1, then
=
2.
with
Hence, by Theorem
Thus FN
[Dl > 4.
has a non-
By Corollary10.10,
B has a one-dimensional irreThus n > 1
ducible module, which is impossible by the foregoing discussion. and J I B )
Mn (I(D)) .
4, I(D) # 0
ID1
But since
4 I IcNIZ)1.
and hence J ( B )
is non-
commutative, a contradiction. (ii) * (iii): By hypothesis, each nonidentity 2-regular class of 1 or
0.
G
is of defect
The desired conclusion is therefore a consequence of Theorem 9.23.
(iii) * (i): Let B nilpotent, we have
be a block of
B
1
Since N
Mn (FD1 for some n P 1 (Corollary 10.10).
nonprincipal, then by hypothesis Since N
FN with defect group D.
ID1
is a normal subgroup of G
Mn(I(D) 1
Hence JCB)
2.
of odd index, we have J ( F G I
is 2-
If B =
is
0.
=
FG*J(FN).
Put
Then we have J (FGI = FGJ (FN)e
Since FG*J(FN)el SFGe Lemma 13.2.
1
, FG*J(FNle
Now every block of FN
@
FGJ (FN)e
is a central ideal of contained in FNe
FG, by virtue of
is nonprincipal.
Hence,
2
by the above, S(FNe212 = J(FN12e2 = 0.
and so FG*J(FN)e
is commutative.
Thus
This proves that J ( F G )
is commutative, as
required. We have now accumulated all the information necessary to prove the following result.
J(FGI
CRITERIA FOR THE COMMUTATIVITY OF
G
characteristic 2, l e t
P E Syl (G) and (i) J ( F G 1
(ii1
N
=
a
x
<
1,
a
Either
IP n
odd o r d e r and a
Either
E G'-l,
P"l
4 2
1, o r
I
ICN(z)
]
= 1,
or
a
7
E
N-P
4
G'
or
a > 1,G
or
= 1,
x
for a l l
odd, and l e t
s
in
P,
Frobenius group with complement
P/<s>
.
Proof.
l,G
i s 2-nilpotent,
cN ( s l / < S >
a
J ( F G ) * = 0.
= 0,
then
Thus if
commutative.
a
1,
J(FG1
Since
a l s o assume t h a t
a
SVGl = 0
3
then
a
and i f
JVG)'
= 0
#
G'
F
is algebraically
then by Theorem 12.4,
= 1,
and t h e r e f o r e
Hence, by Lemma 15.5,
1.
G ' # 1 i s of
i s e i t h e r a 2-group o r a
i s obviously commutative i n c a s e
1 and
1 i s of odd
G ' # 1 i s of odd
is 2 - n l l p o t e n t ,
By Corollary 1.18, we may harmlessly assume t h a t If
G' #
is 2-nilptentr
a > 1,G
G' = 1, o r
o r d e r and, f o r every i n v o l u t i o n
closed.
rn i s
where
Then t h e following c o n d i t i o r s are e q u i v a l e n t :
G'
1, o r
Q
order and f o r each
(ivl
2 rn,
be an a r b i t r a r y f i e l d of
i s commutative
Either
(iii)
U
be a group of o r d e r
G'P.
F
Let
15.6. THEOREM. (Motose and Nfnomiya (1980)).
265
J(FG1
G'
is
= 1,
w e may are
ii)
( i ) and
equivalent. (i)
*
(iii): A s s u m e t h a t
G'
equivalent,
xyx
-1 -1
y
i s of odd order.
E P n G'
= 1
and so
and hence, by Lemma 15.5, (iii)* ( i v ) :
C (s)/<s>
N
x E
S(FG)
Let
s
Let
x
E G'-1
y E C (z). N
( Pn
P"1
Since ( i l and (ii
i s commutative.
and
Then
This shows t h a t
2.
be an a r b i t r a r y i n v o l u t i o n i n
i s a 2-group.
y E P n ?.
are
Now assume t h a t
P.
C Cs) # P .
N
If
c N ( s )= P ,
then
Then, f o r a l l
Cu(S)-P,
since
IP n
complement
P"l
6 2.
P/< s >.
Hence, by Lemma 7.2,
CN ( s )/<s
i s a Frobenius group with
266
CHAPTER I11
(iv)
*
xE
(ii): L e t
Is1
4 2.
S,
in
G'-1
W e may assume t h a t
S.
# 1, i n which case t h e r e i s an i n v o l u t i o n , say
S
CN ( s ) , and hence
u E N
C N ( s )/ < s
>,
Thus w e have
S C_
9 P" = < s > ,
15.7. COROLLARY.
(Motose and Ninomiya (19801).
G
JVGI
F
Let
S
5'P
.
as required.
be an a r b i t r a r y f i e l d of
be a nonabelian group of order
be commutative.
Now
is a Frobenius group
N
?/<e>.
c h a r a c t e r i s t i c 2, l e t
is not a 2-group. C (s)/<s>
and t h e r e f o r e , by hypothesis,
with complement
and l e t
I t s u f f i c e s t o prove t h a t
Then
which shows t h a t f o r some
S E Syl ICIv(x)).
and l e t
(m,2) = 1, u
2'm,
G
Then t h e Sylow 2-subgroups o f
1,
7
are either
z2 x z
c y c l i c o r isomorphic t o
2
Proof.
Let
of odd o r d e r ,
If
P n P"
x E N-P,
f o r all
P and k e r n e l G'.
Hence
On t h e o t h e r hand, i f
s i n P,
then
N = G'P.
mz-'s-'
P
then
P/<s>.
P n G'
Since
?L
P
either cyclic or
mrcm
= 1.
f o r some Hence
P/<s>
is
x E N-P
cc E C ( s ) N
and some i n v o l u t i o n
-P
C N ( s ) = P.
and so
C (sl/<s> i s a Frobenius group
N
a b e l i a n and i s a Sylow 2-subgroup of
P/<s>
is cyclic.
Thus
P
is
'
OF BLOCKS AND NORMAL SUBGROUPS
characteristic p
B
a-1'
2
Throughout t h i s s e c t i o n ,
those blocks
is
i s a Frobenius group with complement
c N ( s ) / < S > , it follows from Remark 7.11 t h a t
16. THE
G'
i s c y c l i c , by v i r t u e of Remark 13.5.
P n 9 = <s> E
N
Thus, by t h e condition (ivl of Theorem 15.6, with complement
Then, by Theorem 15.6,
i s a b e l i a n and
P
= 1
and l e t
P E Sy12(G)
3
of
normal subgroup of
0.
FG
G.
G
F
denotes a f i n i t e group and
an a r b i t r a r y f i e l d of
Our aim i s t o determine t h e Jacobson r a d i c a l f o r which t h e d e f e c t group of
B
J(B1
of
i s contained i n a given
As a p r e p a r a t i o n f o r t h e proof of t h e main r e s u l t ,
we
THE RADICAl, W BLOCKS
261
shall prove a number of important properties of independent interest. It will be fundamental for future investigations to provide a useful character -rization of defect groups of blocks.
The following preliminary results will
clear our path. Let W
16.1. LEMMA.
be any subgroup of
G
such that W
VH = where the Vi,l
i
is H-projective.
v
...
@
Write
vn
Vi are indecomposable FH-modules, and denote by Qi Then for each i E 11,.
n.
Proof.
Owing to Lemma 2.6.3cii1
16.2. LEMMA.
-1
yi,
Let
,
we
vi
..,nl,
a vertex of
there exists gi E G
i s
i
.
2
conclude that g .Qg-l c Q, as required. t
i
-
W be an indecomposable FGmodule with vertex Q
is a Sylow p-subgroup of
such that
~ x ~ n ~ xprojective ; ~ for some x E G. i there exists .y E H such that x.Qzil n H 2 yiQyil.
By Lemma 2.6.2, each
Setting gi = xi
Q and let H
be an indecomposable FG-module with vertex
G.
Then
(P;&l
divides dimW.
5 P,
where
P
In particular, if
F dimW is prime to p ,
then P
is a vertex of
W.
F Proof. We first consider the case where be an indecomposable FP-module with vertex
F
is algebraically closed.
X and source S.
Since X
Let
V
is a
subnormal subgroup of P, it follows from Theorem 4.23 that S p is indecomposable.
Hence
V
1
S p and so dimV
= (P:X) did
F
F
Now write
where the
Vi are indecomposable FP-modules, and denote by Qi
Then by (11,
1< i dimW.
F
< n.
a
vertex of
‘i (P:Qi) divides dimVi and, by Lemma 16.1, (P:Q) divides (P:Qi), F Thus, for all i, (P:Q) divides dimVi and hence ( P : Q ) divides F
This establishes the case where
F
i s algebraically closed.
CHAPTER 1 x 1
268
Tn general, let E of
W corresponding to the vertex Q. EQ9
W
Then
=
... CD Wm
W @
F where the
P, and let U be a source
be the algebraic closure of
W.z are indecomposable EG-modules.
is a component of E '8 W
and E
W
Since for all $ E 11,.
8=
is a component of E
F
F
F
.. ,m), Wi G
(E @ U) , we F
8 UIG for the EQ-module E @ U. This F F proves that each W. is &-projective and so has vertex Qi 5 Q. By the algeinfer that Wi
i s a component of
braically closed field case,
(P:&.)
and dimW = dimW F E l So the lemma is true. Given subsets X,Y of
(E
(P:Qi) divides dirnWi. E , idimWm, we infer that
..
i
E
In particular, CFG(G1 = Z(FG).
(P:&l
divides each
(P:Ql divides dimW. F
FX the F-linear span of X and by
FG, we denote by
CX ( Y ) the centralizer of Y in X I
Since
1.e.
Let S
5H
be subgroups of
G.
We shall refer
to the map
introduced below, as the PeZative trace map. Let S
16.3. LEMMA.
for S in
H.
5H
be subgroups of
For each z E CFGtS)I
T
G
and let X be a left transversal
put
(z) =
c zzz-1 ECEX
(i) C F G W f CFG(Sl, T H , S ( ~ E) CFG(H)
and T
His
is independent of the choice of
A.
(it1 TH,s
: CFG(S)
CFG(H)
TH,s(CFG(S))
particular I If
-+
D
is a homomorphism of
i s an ideal of
is a subgroup of S ,
CFG (H)-bimodules.
CFGLH).
then for all z E CFGw),
In
269
THE RADICAL OF BLOCKS
(ivl
D
Tf
I (GI
where
Ci1
s E S,
Because any element of
S,
centralize
we have
CFG(H)
FG
5 C(,S).
which c e n t r a l i z e s If
y =
H must a l s o
xs f o r some
3:
E
X and
then
yzy-I = and so
hX
then
is defined i n Lemma 2.2.5.
D
Proof.
G,
is a p-subgroup of
TH,S
(21
b S ) Z (xs1-l =
xszs -1s-1 = xzx-1
X.
is independent of t h e choice of S
i s another l e f t t r a n s v e r s a l f o r
in
H.
Fix
h E H
and note t h a t
Thus
a s asserted. (ii) The map
z E C,,CSI.
T
H,S
obviously preserves a d d l t i o n .
Then, f o r a l l
x
A s i m i l a r argument proves t h a t
X, we have ys
TH,s’zy)
=
= T (z)y
H
Y a l e f t transversal for D i n
(iii) Denote by
transversal for
E
D
in
H.
Fix
y E CFG(HI
and
xy and t h u s
a s required.
s.
Then
XY
is a conjugacy class of
G.
is a l e f t
Thus
as a s s e r t e d .
Uv)
Let
-+
z = C E ID(G) , where
C
Fix
g E C,
CHAPTER I11
2 70
put
Q
Q
L = CG(g) and choose
5 D.
Then
(L1
.
b)
E
cFG(o) and,
L,
G : Q ) is prime t o p ,
by (21,
9.
z = TG,L(gl.
T (L'I C,D
TG,D(L+l =
and so e i t h e r
0
or
E
1,IG).
Co(g)
w e may d e f i n e
K
form an F-basis f o r
C,,(D),
it
But
C.
Thus i n both
'
TG,DIL 1 E ID(G1, a s d e s i r e d . Let
L+
is a d e f e c t group of
+
16.4. LEMMA.
be a subgroup of
G
and l e t
D,H be subgroups of
K.
X a s e t of r e p r e s e n t a t i v e s f o r t h e double c o s e t s DxH i n K k E K put Hk
=
and
kHk-l n D. for all for a l l
D.
Apply-
proving t h a t
TG,D(W) = 2 ,
Since t h e elements
s u f f i c e s t o prove t h a t
f o r each
we must have
L t h e D-conjugacy c l a s s containing g E G and by C t h e G-conjugacy
c l a s s containing
Denote by
z E ID(Gl, w e may assume t h a t
Because
TG,Q'g) i n two d i f f e r e n t ways, we f i n d
T L , & ( g ) = ( L : & ) g and
Denote by
cases
P
Moreover, by t h e d e f i n i t i o n of
ing ( i i i l t o compute
Since
E Syl
a
E
CFG(H)
a E CFG(H) ,b E C,,(D)
Then t h e s e t
Y is obviously a l e f t t r a n s v e r s a l f o r
=
u T(X)X 6 X 8
in
K.
Thus, f o r a l l
a E C,,(H),
we
THE RADICAL OF BLOCKS
271
have
as asserted. (ii) We have (by Lemma 16.3(iii))
(by Lemma 16.3(iii)
)
as required.
.
(iii) Direct consequence of (i) (iv) Follows from (ii). 16.5. LEMMA.
Let A
be a finite-dimensional algebra over F and let e If 11,1 2f...,In are ideals of A
primitive idempotent of A. e E J
+
... + In,
Proof.
1
+ eI
e. 2
with a E Z
a 9 JU). Since A
1
,b
Thus we may assume that A E
I
2'
then not both
is local, a has inverse a',
such that
...,n).
is the identity element of the local ring
implies e E e 1 e 1 = atb
for some i E c1,
Without loss of generality we may assume that n = 2.
primitive, e
If
then e E Ti
be a
eAe
Because e
and e E
is
I1 + I 2
is local and that e = 1.
a and b
lie in J ( A ) ,
which gives
say
1 = a2' E T I ,
as
required. It is now easy to provide a useful characterization of defect groups of blocks. 16.6. PROPOSITION.
Let B = B ( e ) be a block of FG.
Then the following
conditions are equivalent: (i) D is a defect group of B
(iil
D is a minimal element in the set of all subgroups H of G such that
272
CHAPTER 111
B.
Suppose that D is a defect group of
Proof.
Because e
Then, by Lemma 16.3 Civ) , e E IH (GI
e.
assumption that D is a defect group of Assume that H is a subgroup of G
and that
contrary to the
Hence (i) implies (ii).
such that e E TG,H(CFG-(H))
be a minimal element in the set of all such subgroups H 16.4tiv) (with X
ID(G),
Now assume that H c D
Lemma 16.3(iv) implies that e E TG,D(CFG(D)). e E TG,H(CFG(H) 1 -
E
of G.
and let D
Owing to Lemma
G), we have
=
Applying Lemma 16.5, we infer that for some g E G
Then the minimality of
-1
D ensures that gHg
3 D.
Now suppose that H
(and
Then ~ H g - l>_ D and in view of the
hence 9Hg-l) is a defect group of B . implication (i)
* Cii),
16.7. LEMMA. (e.g.
e
we obtain
gHg-' = D, as required.
Let D be a p-subgroup of G
is a block idempotent of FG
(i) For any FG-module (iil
V
If
ti).
If V
satisfies (ii), then eV = V
defined by
O h ) = wx
for all x E eV.
FD and let
with defect
and so (ii) is a consequence of
such that e = TG,D(WI.
all elements g with g E D, 8 E End(VD). in G
B=B(eI
is D-projective.
we may choose w E CFG(Dl
eV
D
is D-projective
To prove (i), we employ the relative trace map.
0 : eV -+
be an idempotent in I (G)
with D as a defect group)
is an indecomposable FG-module in a block
group D, then V Proof.
V, eV
and let e
I E
eV.
Then
Let 7!'
Owing to Lemma 16.3(iv),
Consider the mapping Since w
commutes with
denote a transversal for D
THE RADICAL OF BLOCKS
= e x = x and t h e r e s u l t follows by v i r t u e of Theorem 2 . 5 . 4 .
16.8. LEMMA. If
e.
B
Let
e 9 7,
then
Proof.
Let
=
Ie y
B(el
FG
be a block of
273
. I be an i d e a l of
and l e t
ZCFG) .
i s a n i l p o t e n t i d e a l of
be the i r r e d u c i b l e r e p r e s e n t a t i o n of
Z(FG)
a s s o c i a t e d with
By P r o p o s i t i o n 2.10.8,
We c l a i m that
1 5 Kery;
as required.
Assume by way of c o n t r a d i c t i o n t h a t
is a f i e l d and
Hence
e-x
=
re
i f s u s t a i n e d it wtll follow t h a t
0 # y(1)
y ( e ) = y(xl
Thus
+
is an i d e a l of
f o r some
z(1-e)
x E I
f o r some
y(1l # 0.
y(Z(FG)),
Since
y(Z(FG))
it follows t h a t
and t h e r e f o r e
r
E JCZVG1)
and some
2
E ZCFG).
Conse-
quent 1y ,
e = xe
+
re
and
Now
ZVG).
J(Z(FG))
i s a n i l p o t e n t i d e a l of
a l g e b r a over t h e f i e l d
F
ZVGl
of c h a r a c t e r i s t i c
ZPG)
and
p
3
0.
is a commutative
Hence t h e r e e x i s t s a
CHAPTER I11
274
n such t h a t
positive integer
.
n e = (ze)P
Therefore
16.9.
e
Let
LEMMA.
I,
E
a contradiction.
be a block idempotent of
FG
D
with d e f e c t group
and l e t
A
h
ID(G)
6 ( C ) < D. G i s a n i l p o t e n t i d e a l of Z (FG)
C',
be t h e F-linear span of a l l h
i s an i d e a l of
Proof.
Z (FG)
rD(G) i s
By Lemma 2.2.5,
with
ID(G)
Then
.
T (G) e D
and
C E CRCG),
n
an i d e a l of
Z(FG).
ID,CG)
Since
i s the
h
SUm
of a l l
IH(G)
H c D, we see t h a t
with
T D G ) i s an i d e a l of
.
n
e , e 9 .TDLG),
t h e d e f i n i t i o n of t h e d e f e c t group of p o t e n t i d e a l of 16.10. LEMMA.
E
Z(FG),
e
Let
by v i r t u e of Lemma 16.8.
be a block idempotent of
be any f i e l d extension o f
F
FG
Z(FG1.
By
h
IDb(G)e i s a n i l -
Hence
with d e f e c t group
D,
let
and w r i t e
e = e a s a sum of block idempotents of
EG.
+
1
... + en iE
Then, f o r a l l
{l,
...,n},
ei h a s
D
as a d e f e c t group. A
Proof.
PD(G)
Denote by
respectively, i n
EG.
Fix
and
h
IE(G)
i E {l,
TD(G) and
t h e c o u n t e r p a r t s of
...,nl
ID(G),
and note t h a t , by t h e d e f i n i t i o n of A
t h e d e f e c t group of
e E TD(G)
ei,
it s u f f i c e s t o v e r i f y t h a t
PD(G) and s i n c e PDD(G)
i s an i d e a l of
ei E TECG)-.TZ,(G). 2 (EGI
,
Since
w e have
On t h e o t h e r hand,
A
and so, by Lemma 16.9, A
Let
Z(EG).
Hence i f
A
e E PDCGl,
i
T E ( C ) e i s a n i l p o t e n t i d e a l of
then
ei = eie E T;(G)e,
S be a subring of a r i n g
i f any e x a c t sequence
R.
a contradiction. An R-module
M
So t h e lemma i s t r u e .
i s s a i d t o be S-projective
THE RADICAL
Y
0-
-+
OF
BLOCKS
X-
M-
275
0
of R-modules and R-homomorphisms which splits as an S-sequence, also splits as an R-sequence.
If L C_ S S R
M is ;%projective.
are subrings and
M is L-projective, then obviously
We state without proof the following basic fact (see Higman
C1955, Theorem 4 ) and Hochschild (1956, Proposition 2) 16.11. LEMMA.
Let S
be a subring of a ring R
.
and let
M be an R-module.
Then the following conditions are equivalent: (i)
M is S-projective
(ii) The canonical map
$ :R @ S
M+
i.e. there exists an R-homomorphism (iii) There exists an S-module V
M
given by
q ( 8~m )
=
mn
is a retraction,
R €3 M with $4 = 1. S such that M i s a direct summand of R 8 V.
4
:
M+
S
(iv) Consider any diagram of R-modules and R-homomorphisms
with exact row.
If there is an S-homomorphism
commute, then there is an R-homomorphism
q
:
M+
$
:
M+
X making the diagram
X making the daigram commutep
In order to take advantage of Lemma 16.9, we make the following simple observation
.
16.12. LEMMA.
Assume that the following is a commutative diagram of R-modules
and R-homomorphisms
CHAPTER 111
276
0 If
then the following are equivalent:
= ly,
( i ) Im($XaI
5 Ima
(ii) There is a $ : Z-+ C such that $0 = B$
and
k$
= lZ
(in particular, k
is a retractionl. Proof. Assume that Im($ha)
which shows that %$p = 0 ,
5 Ima.
Because Ima = KerB, we have
by virtue of the fact that U
is an epimorphism.
Hence
and thus $ = B$0-’
i s a well-defined R-homomorphism from
Z to C with the
required properties. Now assume that (it1 holds.
Then
and therefore Im($Xa] SKerB = Ima. The next observation will allow us to take full advantage of the results so far obtained. 16.13. LEMMA. that AR =
Let
s
be a subring of a ring R ,
let A C_S be a subset such
RA and let U be an $-projective R-module.
Then M = U/AU
i s an
277
THE RADICAL OF BLOCKS
S-projective R-module (the fact that AU
U is a consequence
is a submodule of
of the equality RCAU) = ARU = A U ) . Proof.
Consider the diagram below where a l l maps are canonical.
)M i
x
K
Bearing in mind that
we obtain
The desired assertion is now a consequence of Lemma 16.12.
.
We have now accumulated all the information necessary to prove the following
fundamental result. 16.14. THEOREM. (Green (19591, Knorr (1976)). be an arbitrary field of prime characteristic p
Let G be a finite group, let and let B
be a block o f
F
FG.
278
CHAPTER I11
If P
G, then the following conditions are equivalent:
is a p-subgroup of
P is a defect group of B
ti)
All indecomposable FG-modules in B
(iil
ducible FG-module M Proof.
in B
Suppose that
(i)
*
(ii) is true and that (ii) holds.
and there is an irreducible FG-module V is P-projective, Q
M
hand, because P
and
is a vertex of M.
such that P
Then all indecomposable FG-modules in B
a defect group of B .
Since V
are P-projective and there is an irre-
in B
Denote by are &-projective
Q is a vertex of
such that
is conjugate to a subgroup of
P.
On the other
is Q-projective, P is conjugate to a subgroup of
Q are conjugate.
Owing to Lemma 16.7,
then all indecomposable FG-modules in B
Q.
Hence
if P is a defect group of
are P-projective.
v.
B,
Thus we need only
verify the following assertion:
Tf
P
i s a defect group of
B,
such that P i s a v e r t e x of
then there i s an irreducible FGmoduZe
M
in B
M.
(3)
For the sake of clarity, we divide the proof into three steps.
Step 1. Let d
The case where F i s aZgebraicalZy closed.
be the defect of B
By Theorem 2.2.7,
and let p"
be the order of Sylow p-subgroups of G.
there exists an irreducibleFGmodule M
pWd
is the highest power of p
in B
such that
dividing dimM
F
Q a vertex of M.
Denote by
and so Q C P .
is' indecomposable, M
Because M
On the other hand, by Lemma 16.2,
(S:Q)
is P-projective
divides dimV, where
F
G S is a Sylow p-subgroup of G.
d IPI = p , this means that Q = P
Because
and
G thus P is a vertex of
Step 2.
Let
E
M, proving
(3).
be the algebraic closure of F.
every irreducibZe FG-module
M
owl aim i s t o prove that
i s FG-projective.
Owing to Corollary 1.18, FG/J(FG)
is a separable F-algebra and
JBG)
=
E 8 S(FG)
(4)
F Denote by
U
a projective EG-module such that M
U/J(EG)U.
Then, by (41,
279
THE RADICAL OF BLOCKS
J(EG)U
=
and so M
J(FG)U
conclude that M
Invoking Lemma 16.13 with A = S(FG),
U/J(FG)U.
denote the block idempotent of
FG
central idempotent of EG, we may write block idempotents of EG. a defect group of B ' .
V
EG-module
=
Put B'
=
e
0.
Because e
contained in B . =
+
e
... + e n ,
F.
where the
is a
ei are
EGel and note that, by Lemma 16.10,
P is
Applying Step 1, we deduce that there is an irreducible such that P is a vertex of
in B'
by (4) JCFGlV
we
i s FG-projective.
Here we complete the proof by establishing (31 f o r an arbitrary
Step 3. Let e
1
Thus
VFG
V.
Now J(EG)V
=
0
and so,
is completely reducible and so S
VFG
= .@ Mi
z=1 for some irreducible FG-modules identity on
V,
Mi
all
16.11 ensures that V
Mi.
since multiplication by
are in B .
By Step 2,
V
i s
e
induces the
FG-projective, so Lemma
is a direct summand of S
EG 8 V
=
FG V
But ME
is irreducible, so V
{ M ,,..,,M 8 1 .
FG 8 M
and thus
@ LEG @ Mi) i=l FG
is a direct summand of EG 8 M
for some
FG Denote by
Q a vertex of M.
Then M
i s a direct summand of
V is a direct summand of
FQ EG 8 (FG 0 M ) FG FQ Applying Lemma 16.11, we infer that projective and thus
is FQ-projective.
P is conjugate to a subgroup of Q.
ducible FG-module in B .
.
jugate to a subgroup of wished to show.
V
EG 8 M FQ
Consequently M P.
A fortiori, V Now M
is P-projective and so
This proves that P
and
is EQ-
is an irre-
Q is con-
Q are conjugate, as we
As a final preliminary result, we prove
16.15. LEMMA.
Let N
be a normal subgroup of
G and let V be an FGmodule.
Then the following properties hold: (i) J(FN1V (ii)
If
V
is an FG-module is N-projective, then
V/JCFGIt: is N-projective if and only if
200
CHAPTER IIL
V
(iii) If V/SCFG) V
is projective and indecomposable, then N
contains a vertex of
if and only if
Proof.
(i) For a given g E G ,
and hence of JCFN).
g
conjugation by
is an automorphism of FW
Bearing in mind that
we infer that FGJCFN) = J(FN)FG.
Hence
as required. (ii) Put M J(FN1V
=
5 3(FG)V
Because 'Cv/J(F#)VIN
f :M
-+
-
J(FGlV/SIFN1V
--t
V/J(FN)V
--+
M
-
The inclusion
0
is completely reducible, the above sequence splits as an
But M
V / J ( F N )V
is N-projective.
induces an exact sequence
0
FN-sequence.
and assume that M
V/J(FG)V
is N-projective, so there exists an FG-homomorphism
such that
Bearing in mind that M
Hence J ( F G ) V / J ( F N ) V
=
0
and f ( M )
are completely reducible, we derive
and J V G ) V = J ( F N ) V.
Conversely, suppose that J V G ) V
=
J V N )V.
Since A = J ( F N )
condition of Lemma 16.13, it follows that M = V/JVG)V= V//s(FN)V
satisfies the is N-projective
THE RADICAL OF BLOCKS
(iiil
281
Direct consequence of ( i t ] .
We are at last in a position to achieve our main objective, which is to prove the following result. 16.16. THEOREM CKnorr (1976)).
be a normal subgroup of G
with defect group D.
be a block of FG
B = B(e)
Let N
and let
Then the following are
equivalent: (i) D
5N
(ii) [J(FG)lnV = [J(FNllnV for all indecomposable FG-modules V
in B
and all
positive integers n (iii) J ( B ) = S(FN)B for all projective indecomposable FG-modules V
JCFGIV = JcFN1V
(ivl
Proof.
(i) * (iil:
By Theorem 16.14, V
Assume that
V
is an indecomposable FG-module in B .
is D-projective and so
v
is N-projective.
conclusion now follows from Lemma 16.5Cii) by induction on (ii) * Ciiil: Write
e
=
+
e
... +
The desired
12.
n as a sum of primitive idempotents of FG.
e
is an indecomposable module in B
Then each FGei
in B .
and so
Bearing in mind that S(FGle
=
J(FGlel +
... + J ( F C l e n = J(FGIFGel + ... + J(FG1FGe,?
we derive
n (iii)
* (iv): Write B
=
@
Ui , where the U. are projective indecomposable
i=1
modules in
B.
Then
282
CHAPTER I11
Since
5S ( F G I U i
J(FN)Ui
i,
for a l l
w e have
J (FG) Ui
Hence (iv)
B
*
=
J (FN)Ui
J(FG)U = J(FN)U f o r a l l p r o j e c t i v e indecomposable modules ( i ) : Owing t o Theorem 7.6.7,
D is
such t h a t
U i n B.
composable FGmodule
'I
B
in
V
t h e r e e x i s t s an i r r e d u c i b l e FG-module
V.
a, v e r t e x of
in
U/J(FG)U f o r some p r o j e c t i v e inde-
V
Now
i
for a l l
The d e s i r e d conclusion i s now a consequence of
.
Lemma 16.15 (iii) 16.17. COROLLARY.
B
J(FD1
=
I(D),
16.18. COROLLARY. i s a block of
Proof.
with normal d e f e c t group
Define
FG
Put
0 (G1 PIP'
1= 0
(G)
and
D
P
=
N.
=
0 G I / O (G) PIP'
by
with d e f e c t group
N
D,
we have
D.
Then
J ( B ) = J(FD)B.
'
t h e r e s u l t follows.
contained i n
0 (G).
(G/Op(G))
0
=
If
B
0 (GI, then PIP'
i s a normal Sylow p-sub-
J(FN) = F N * I ( P ) .
Hence, by Proposition 1.1,
-
P'
P
Then
P
PIP'
group of
FG
be a block of
By Theorem 16.16, applied f o r t h e c a s e
Proof. Since
Let
Since
D
5N,
the
r e s u l t follows by v i r t u e of Theorem 16.16. Recall t h a t
G
is c a l l e d p-solvable i f each of i t s composition f a c t o r s i s
e i t h e r a p-group o r a P'-group.
G
Thus
i s s o l v a b l e i n t h e ordinary sense i f
and only if i t i s p-solvable f o r a l l primes
p.
G
Clearly
is p-solvable i f and
only i f it has a s e r i e s of normal subgroups
1 = V
C V
0 -
f o r which each f a c t o r group
vi-l/Vi
C . . . C V
-
1 -
n
= G
(4)
i s e i t h e r a p-group o r a p'-group.
For
each such group we can t h e r e f o r e d e f i n e t h e upper p-series
l = P Clv 0 - 0
C P 1
C P C . ..CP 2
C N
R-
9.
= G
(5)
283
THE RADICAL OF BLOCKS
i n d u c t i v e l y by t h e r u l e t h a t
(G/P,)
Nk/Pk = 0
Pk+l/Nk
and
=
P'
0 (G/Nk)
P
Thus, i n p a r t i c u l a r ,
No
(G)
= 0
, PI
=
0
P'
R
The number
and
N
=
and i s denoted by
(G)
0
P '*PIP'
which i s t h e l e a s t i n t e g e r such t h a t
G
l e n g t h of
(G)
P'rP
!L ( G I .
NR = G i s c a l l e d t h e p-
Observe t h a t
P
!L (G) may a l s o be defined
P
a s t h e s m a l l e s t number of p - f a c t o r s occuring i n any series such a s (4), t h e minimum being a t t a i n e d f o r t h e upper p-series ( 5 ) .
Thus
G
i s of p-length
1
i f and only i f
G/O
(GI
i s a p'-group
P'tP With t h i s information a t our d i s p o s a l , we now prove 16.19. COROLLARY.
PE
Syl ( G I .
G
Let
be a p-solvable group of p-length
Then, f o r any block
P
J(B)
B =
1 and l e t
FG,
of
G))B
S(F0
P'rP Proof.
N
Put
p-subgroups of
G.
=
0 GI. P'nP
Since
G
i s of p-length
1, N
contains a l l
Now apply Theorem 16.16.
To p r e s e n t our f i n a l a p p l i c a t i o n of Theorem 16.16, we quote t h e following standard group-theoretic f a c t . 16.20. PROPOSITION. Then
G
i s of p-length
Proof.
16.21.
be a p-solvable group with a b e l i a n Sylow p-subgroups.
.
1.
See Gorenstein (1968,p.228).
COROLLARY.
and l e t
G
Let
Let
P E Syl (GI. P
G
be a p-solvable group with a b e l i a n Sylow p-subgroups,
Then, f o r any block
J(B) = J(F0
B
of
FG,
(GI ) B
P'rP Proof.
Apply Proposition 16.20 and Corollary 16.19.
.
CHAPTER 111
284
17. GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRrNClPAL IDEaLS Let G
be a field of characteristic p > 0 .
be a finite group and let F
The
problem that motivates this section is to discover conditions under which J ( F G ) = ccFG
FGx
=
for some z E F G .
The main result, due to Morita (1951),
solves this problem for the case where F
is algebraically closed.
We have
chosen, for the sake of clarity and generality, to deal with the problem within the framework of a general theory, namely the theory of uniserial algebras, although a shorter proof can be obtained by means of ad hoc arguments. Throughout this section, A arbitrary field F.
denotes a finite-dimensional algebra over an
Unless explicitely stated otherwise, all A-modules are
assumed to be left and finitely generated. Let
V denote an A-module.
Then the dual module
V*
=
Hom(V,F) is a right
F A-module with the action of A
Tf
on
V*
given by
V is a right A-module, then a similar definition gives a left A-module
The following properties of left Lor right) A-modules V
and
W
P.
are easily
verified : (a)
V is indecomposable if and only if V*
(b)
V
(cl
(V**)*
(dl
(V @ U *
V*
is projective if and only if 2
is indecomposable
is injective
v 1
V* @ WY
Let V be an A-module.
of submodules of
V
Then the descending chain
is called the (lower1 Loewy series of
nilpotent, there is an integer k ,
V.
called the Loewy length of
Note that if J ( A ) ~ V= J ( A ) { + ~ V for
i < k , then
Because J ( A )
V
such that
is
GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRINCIPAL IDEALS
a contradiction.
k
The conclusion i s t h a t , i f
285
V,
i s t h e Loewy l e n g t h of
then
of
V.
i s s a i d t o be uniseriaz i f t h e above chain i s a composition s e r i e s
V
The module
V
Thus
i s u n i s e r i a l i f and only i f i t has a unique composition s e r i e s .
Because a u n i s e r i a l module has a unique i r r e d u c i b l e submodule, it must be
indecomposabte.
We s h a l l r e f e r t o t h e a l g e b r a
indecomposable A-module i s u n i s e r i a l .
i s a u n i s e r i a l r i g h t module.
A
u s being uniseriaz i f every
Note t h a t t h e dual of a u n i s e r i a l module
Hence, i f
A
i s u n i s e r i a l , then every indecom-
posable r i g h t A-module i s u n i s e r i a l . 17.1. LEMMA.
Let
V be a ( l e f t ] module over an a r b i t r a r y r i n g R.
Then the
following c o n d i t i o n s a r e equivalent.
(il
V
is i n j e c t i v e
(ii) Every homomorphism
from any module containing
(iiil
If
T
V
W--+
o f R-modules can be extended t o a homomorphism
W to
is a l e f t i d e a l of
V.
R
and
4
E Homu,V1,
then
4 can be extended
R Hom(R,V). R (i) * ( i i l : Assume t h a t
t o an element o f Proof.
homomorphism of R-modules.
If
X
V
@
I s i n j e c t i v e and t h a t
i s an R-module containing
W
:
v-
and
V c1 :
i s t h e n a t u r a l i n j e c t i o n , then by d e f i n i t i o n t h e r e e x i s t s an R-homomorphism
$*
:
X+
V which renders commutative t h e following diagram:
\
is a
W+
X
286
CHAPTER I11
Hence $*
(ii) * (iii): (iii)
4.
extends
Obvious
* (i) : Assume that
v
satisfies (iiil and let X,Y,a,f3
be given as in the
following diagram
B
Put CC(X1
yo
:
Xo
=
xo-
Yo
V by
=
6a-l.
Now let 2
such that U
is a submodule of
Cu1 ,Y1 1 , (U2,Y
1
and y
extends yl. 2
=
an R-homomorphism @
{r E RlPy :
T-+
$
extends to an R-homomorphism
Y*
:
U+@
by
y E U,
Y* (tr+pY)
and because y
17.2. LEMMA.
Let
Set
M
E
(U,yl.
y
extends y
0'
given by
@* : R-+ V . + @*(r1
We show that
V be an A-module. V
u
=
If
Y.
Then 1 is a left ideal of R
Uj.
was arbitrary we have
= HomU,Vl
and
$(r1
=
and
y ( ~ y ) . By hypothesis,
With this map we can now define
.
This is easily seen to be well
Hence by the maximality of
injective A-module that contains Proof.
k'
= Y (tr1
CU,Y1 G (U+Ry,Y*1.
defined and
v,
(U1, y l l G ( U 2 , Y 2 ) if and only if U 1 c -U
contains a maximal element, say,
r
U-+
:
(u,Y)
(~o,yol E 2, it follows easily from Zorn's lemma
Since
Let y E y and let we have
be the set of all ordered pairs
Y,u 3 xo, y
are in 2, we write
2
that
a is injective we can define
and observe that because
U = Y.
(U,Y) we have
so the lemma i s true.
Then there exists a (finitely generated) as a submodule.
and note that
M is an A-module via
F
We claim that $ : I--t
M
is injective.
M be an A-homomorphism.
Indeed, let I Define
:
be a left ideal of A I+
V by
and let
207
GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRINCIPAL IDEALS
p
Then
is c l e a r l y a n F-homomorphism and hence we may extend
h
morphlsm
0
Then
:
V.
A-
Now d e f i n e
M by
$
and so
i s an A-homomorphism extending
V
Because t h e map
17.1.
0 : A-
--+
v
M given by
M
If
A
is a Frobenius a l g e b r a , then
f,(a) = av,a E A,v E V ,
with
i s an i n j e c t i v e A-homomorphism, t h e r e s u l t follows. 17.3. LEMMA.
is i n j e c t i v e , by Lemma
.
fv
A
t o an F-homo-
?A
i s an i n j e c t i v e l e f t A-
module. Proof.
Let
4
1 be a l e f t i d e a l of
and l e t
f
E Hom(1,A).
By Lemma 17.1,
A it s u f f i c e s to show t h a t
thesis, there exists
a E A,
define
clearly
ha
ha:
Q
f
E A*
r-A
can be extended t o an element of such t h a t
1a ( x )
by
(A,$)
= xa.
M
=
To t h i s end, note t h a t t h e mapping with k e r n e l
1’
{hala
dim(Hom(I,AI 1 F
Then each
Given
Horn(1,A)
and
of
End(A) A
by p u t t i n g
=
x;(Z)
ax2
E A)
A
a
ha E
By hypo-
A
can be extended t o an element
Thus we need only v e r i f y t h a t f o r
Endul. A is a Probenius algebra.
h a
i d F
i s an F-homomorphism of
A
onto
M
~ ( 1 ) .Therefore, by Theorem 3 . 6 ( i i i 1 , I = ] d i d = dim (Hom (T,Fll. d i & = d i d - dimr ( F F F F F F
and so prove (1) it s u f f i c e s to show t h a t t h e F-linear map
is i n j e c t i v e . of
9. A
Thus submodule
But i f
f(n = 0
C$fl
UI
= 0,
then t h e l e f t i d e a l
and t h e r e f o r e
W of an A-module
f = 0,
c a l l e d an i n j e c t i v e
huZZ
of
V
if
as required.
V is c a l l e d essential i f
i n t e r s e c t i o n with any nonzero submodule of
M
V.
.
f(I) i s i n the k e r n e l
An A-module
W
has nonzero
M containing
i s i n j e c t i v e and, f o r any i n j e c t i v e
V
is
CHAPTER 111
288
A-module M
V C M , C_M,
with 1
M
we have
=
M.
By Lemma 17.2,
M
such
always exists. LEMMA.
17.4.
V
Let
a submodule of M Then M
=
Let X
if X' C_ M
and X
-
Then W
be an A-submodule of M
-W - +
V c Wc K.
Let Y
IJJ
:
= W =
$(GI.
(i) M
so
V
Then
M/X.
Let
w,t) E W
I-+
w.
,
=
0 and X' n W # 0
W
be the image of
Since kf is injective, can be extended to an
say
Y
=
$CM/X), we have
$(z),
=
where
is
n @ # 0 and hence 0 # $ (Z)f- $(k = Y n W. since V
0,
5 = M/X
W.
is an essential submodule o f
is an essential submodule of
Et follows that
LEMMA.
17.5.
Let
be a nonzero submodule of K ,
(Y n w1 n V #
Y n V # 0,
+X
-W
M/X--t M (Lemma 17.1). Setting K
a nonzero submodule of M/X. It follows that
M,W
W
such that X n
is an essential submodule of
injective homomorphism
K
be
being an essential submodule of
X'.
is a proper submodule o f
the injective homomorphism
Hence
V
maximal with respect to
and let W
X @ W for a suitable A-submodule X of M.
Proof.
in M/X.
M
be a submodule of an injective A-module
K , therefore
and hence that M = X @ W , as required. a
V be a submodule of an injective A-module M.
is an injective hull of
V
V
if and only if
is an essential submodule
of M. is indecomposable, then M
If M
(iil
verse is true if SocV
(il
and so V
Suppose that Then
in Lemma 17.4.
V
c- W c- M.
morphism
By Lemma
M-+ W .
(iil
If M
Lemma 17.4.
M is an injective hull
of
V
W is an injective A-module with V
and let W
W CM.
be as W=M
Hence
Conversely, assume that V
is an
M and denote by W an injective A-module with 17.1,
Since V
homomorphism is zero.
The con-
V are isomorphic.
is an essential submodule of M.
essential submodule of
V.
is irreducible.
(iii) Any two injective hulls of Proof.
is an injective hull of
the inclusion map
W extends to an A-homo-
is an essential submodule of
Thus M = W
M, the kernel of this
is an injective hull of
V.
is an essential submodule of M,
by
and hence
is indecomposable, then V
Thus, by ti), M
V-+
M
must be an injective hull of
V.
Conversely,
GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRINCIPAL IDEALS
assume that M by Li),
is an injective hull of
V
V is an essential submodule of
and that SocV i s irreducible. Then,
M
V.
contained in any nonzero submodule of
and, by hypothesis Hence, if M
=
X 8Y
then S o c V C X n Y, a contradiction.
submodules X,Y of M ,
289
SocV
is
for some nonzero Thus M
is
indecomposable. Let M
(iiil
and M
be two injective hulls of
V.
Because
V is an
1
essential submodule of M A-homomorphism M2-
MI.
A-homornorphism M 1 +
M2
17.6. LEMMA.
Let A
1
,
it follows from Lemma 17.1 that there is an injective A
similar argument shows the existence of an injective
and thus
M
2
.
ks
A-module is injective.
If
V
Hence
is
be an A-module.
Then
projective.
By Lemma 17.3 the regular module
free A-module.
V
be a Frobenius algebra and let
V is injective if and only if V Proof.
M1
AA is injective.
Hence every free
V is projective, then V is a direct summand of a a direct summand o f an injective module and so is
injective. Conversely, assume that
V is injective.
Since V
is a direct
V
be an irreducible submodule of 1
V1.
injective hull of of
A
such that
V
Soc(Ae1.
Hence
An algebra
V
Owing to Lemma 17.5(ii),
But, by the foregoing, Ae
Thus, by Lemma 17.5(ii), 1
V is an
By Theorem 3.8, there exists a primitive idempotent e Soc(Ae).
injective A-module.
V.
of
V is indecomposable.
finitely many indecomposable modules, we may assume that Let
sum
Ae
Ae
is an indecomposable
is an injective hull of
by Lemma 17.5Ciii1, as required.
A is said to be of finite representation type if it has only
finitely many isomorphism classes of indecomposable modules. 17.7. LEMMA.
Let
V be a uniserial A-module.
Then
factor module of a projective indecomposable A-module. uniserial, then A
f
:
V is isomorphic to
a
In particular, if A
is
is of finite representation type.
Proof. There exists a projective A-module X and a surjective A-homomorphism n X* V. Write X = 8 Xi, where each Xi is indecomposable. If f(Xi) C V i=1
for all I, then f(Xi)
J(A)V
for all i.
Thus
f(x) 5 J ( A ) V ,
a contra-
290
CHAPTER I11
Consequently, f(X$
diction.
=
V for some z',
as required.
We are now ready to provide an important characterization of uniserial algebras. 17.8. THEOREM (Nakayama (1939)). (i) The algebra A
The following conditions are equivalent:
is uniserial
(ii) All indecomposable projective and injective A-modules are uniserial (iii) For any primitive idempotent e
of A ,
the modules Ae
and eA
are
uniserial. Proof. Properties (a)
- (d) of dual modules ensure that
(ii) is equivalent to
Since (i) obviously implies (ii), it suffices to verify that (ii) implies
(iii).
.
(i)
Suppose that all indecomposable projective and injective A-modules are uniserial. that
v
V denote an A-module.
It will be shown, by induction on dimv, F is a direct sum of uniserial modules. This will clearly prove the
result.
Let
Assume that dimV > 1.
If dimV = 1, then the assertion is obvious.
F If
V
F
is irreducible, then V
is obviously uniserial.
V has a proper irreducible submodule uniserial.
Denote by
W
By induction hypothesis, V'
8'.
W # 0.
respect to the property that
PI n X
must be
V with dimW as large as
a uniserial submodule of
Then we must have
possible.
Thus we may assume that
F Let X be a submodule of V, maximal with =
0,
Assume that X /X and X /X are irre1
ducible submodules of
XI n X2 n
w
# 0
since W
proving that Soc(V/X)
Then X n
V/X.
is uniserial.
W#
0
W # 0
and X
Hence X 1
and hence
n X # X and so X 2
is irreducible. By Lemma 17.5(ii),
V/X
of an indecomposable injective A-module, and so is uniserial.
exists a homomorphism g : p-+
V
X , 2
is a submodule
Invoking Lemma
17.7, we deduce that there is a surjective homomorphism f : P-+ V/X suitable indecomposable projective A-module P.
= 1
for a
Since P is projective, there
which renders commutative the following diagram:
291
GROUP ALGEBRAS WITH RADICAL EXPRESSIBLE AS PRINCIPAL IDEALS
Because P
is uniserial, so is g(P).
isomorphic to a submodule of
V/X.
w
Now
n
x=
0 implies that kr is
Therefore, by the maximality of
dimW ,
F
dimg(P)
F
dimW F
dim(lf/XI F
On the other hand, the commutativity o f the diagram ensures that dimg CP) 2 dim tV/XI F F
W.
Y = gCP)
+X
and g ( P ) n X = 0.
Hence g ( P )
V/X
V
and the desired assertion follows by induction hypothesis.
=
g(P) @ X
17.9. COROLLARY.
Let
It follows that
A be a Frobenius algebra.
only if for each primitive idempotent e
of
Then
Thus
A is uniserial if and
A , the A-module
Ae
is uniserial.
Proof. Apply Theorem 17.8 and Lemma 17.6. 17.10. LEMMA.
in
A.
Let I
be an ideal of A
such that I = Ax = yA
for some x,y
Then I = 4 = Ay.
Proof. Given an A-module The map A
d I, a
I-+
ax
Hence c ( I ) = c(A/k(dAl I .
M, let c W I
denote the composition length of M.
is a surjective A-homomorphism with kernel k(x) = k ( d ) . However, 4
c - I and k ( d ) 2 %(I]. Thus
CHAPTER I11
292
By combining these two i n e q u a l i t i e s , we deduce t h a t
I = Ay
I
and, by a similar argument,
C(T) =
c(Ay).
Consequently,
So t h e lemma i s v e r i f i e d .
= xA.
We have now a t our d i s p o s a l a l l t h e information necessary t o prove t h e following important r e s u l t . 17.11. THEOREM. (Morita (1951)., Azumaya and Nakayama (1948)1.
F.
algebra over an a r b i t r a r y f i e l d (i) There e x i s t
(iil
x,g
x EA
There e x i s t s
(iii) A
A
in
be an
Then t h e following conditions a r e equivalent:
J(A1 = xA
such t h a t
Ay
J ( A ) = xA =
such t h a t
A
i s u n i s e r i a l and each block of
Proof.
A
Let
Clearly ( i i ) implies (il.
=
Ax.
i s quasi-primary.
By Lemma 17.10,
( i ) implies (ii)and
hence we need only v e r i f y t h a t l i i ) is equivalent t o ( i i i ) . For t h e sake of c l a r i t y , we d i v i d e t h e r e s t of t h e proof i n t o t h r e e s t e p s .
Here we prove t h a t (iil impZies t h a t
Step 1.
J ( A ) = XA = AX
Assume t h a t idempotent of
A.
Let
of
a 9 J(A)l’+le.
a. 4 J ( A ) e ,
Aa
=
Ae,
such t h a t
a
SL4)e
since
=
J(Alpe Because
x’a0.
submodules of of
eA.
I = J(A)”e.
Ae.
Hence
S t e p 2.
A
and
e
be any p r i m i t i v e
JIA)’e = x’Ae,
Then
Ae.
17 - S(Al’+’e. Aao
5 Ae
=
Ax’a
=
Let
This shows t h a t
Similarly
Ae.
Hence t h e r e
a
there exists and
Aao Ae.
Then
be an element
a E Ae,
J ( A ) e , so t h a t Hence
S(A)’ao = J(A)l’e A e , J ( A ) e , J U ) ’ e , . . . a r e t h e only
eA,eJ(A) , e J ( A ) ’ , . .
.
exhaust a l l r i g h t submodules
i s u n i s e r i a l , by appealing t o Theorem 17.8.
We now prove t h a t liil implies (iiil.
By Step 1, we need only show t h a t each block of end, w r i t e
and l e t
A,
i s a unique maximal submodule of
I 2. Aa proving t h a t
in
i s a unique maximal submodule of
FI 2 1 such t h a t T
I with
x
I’ be any proper nonzero submodule of
I c J ( A l e s i n c e J(A1e exists
f o r some
i s uniserial.
A
A
i s quasi-primary.
To t h i s
293
GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRINCIPAL IDEALS
where t h e
i
e..
23
a r e mutually orthogonal p r i m i t f v e idempotents,
E {l, ...,f ( k ) } and
JUI’(= Ax’ =x”A) i s
Aeki ir Ae
Aeki
Aek,
if
k # t.
for
Ax’eki:
Ae . .x’ 13
i E [1,2,.
for a l l
Aek,
Then i t i s e a s i l y seen t h a t
e x p r e s s i b l e as a d i r e c t sum of
and s i m i l a r l y a s a d i r e c t sum of
Because
ti
Aeki
. ., f ( k ) 1,
we have
and
Ae tix’ where
Ax’e
ki
Ae .xu tJ
and
with
Ae tlx’
Ableki
= 0
and
(1
Ae x’ tj
=
j
Q
f(t))
0 a r e excluded.
Next, we prove t h a t
Ax’eki
Ax’etj
Ae x’ 2( Aekza’ tj Ax’eki where
Ax’eki
Ae x’ ti
k # t
if
t # k
a r e indecomposable l e f t i d e a l s
Ae .xu with Ax’eki = 0 tJ Then, by Ax eki zz Ax’etj.
and
Suppose t h a t
and
if
and
’
J(,4)’eki/J(A)’+leki
Ae x’ tj
=
0 a r e excluded.
S t e p 1, t h e r e e x i s t s
esm
such t h a t
Aesm/J(A)esm
and
J(A)’etj/J(A)’+’etj Choose
a E esmJ(A)’eki,b E esmJ(A)’e
tj
1
with
Aesm/J(A)esm a 9 J(A)’+l,b 9 J(A)’+l*
have
e JW’= sm
a~ = b~
Then w e
294
CHAPTER I11
and therefore e
J(A1’
is a homomorphic image of both
srn
e A kz
e A tJ
and
But
then
ekiA/eki J U ) and hence
k = t,
d
e t jA/e t j J ( A )
proving ( 5 ) .
Property (6) is a direct consequence of the fact that Ae
To prove (71, note that Axveki
morphic image of A e t j . of some Ae and Ae
ti
sm
x”
and Ae x’
tJ
x” is a homotj is a homomorphic image
”
is a homomorphic image of A e t j .
Thus both A X eki
are uniserial and therefore indecomposable, proving ( 7 ) .
Invoking ( 3 )
-
( 7 ) and the Krull-Schmidt theorem, we deduce that there is a
one-to-one correspondence
$
between the subsets of
{1,2,.
. .,n}
such that
and
and Ae xu which reduce to zero. tJ From (8) it follows further that
where we exclude those Ax’ekz.
if h u e k l # 0 ,
then JL41’ekl/J(d)”’1ekl
Hence the number f ( k ) A.
is the same f o r any Aeki
This proves that each block of A
Step 3.
A e @ ( k ,)l / J ( A ) e $ ( k ),1
( 10)
contained in a fixed block o f
is quasi-primary.
CompZetion of the p r o o f .
By Step 2, it suffices to show that Ciiil implies (ii1.
Then the proof of (5) shows that k # t
implies
To this end, we put
$(k) # $(t).
Since Aekl
and
are contained in the same block, we have f ( k ) = f ( $ ( k ) ) , since each Ae $ ( k l ,1 block of A is quasi-primary. For an element C k i such that
we have
GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRINCIPAL IDEALS
and ck.;A = e
Acki = J(Aleki
n Setting x
f(kl
Z
C cki k=1 i = 1
=
295
(11)
$ ( k ) ,i J ( A 1
, we obtain a direct decomposition n Ax =
f(k) Z Acki
C
k=1 i=1 Applying (ll), ekiA = e
$ ( k ) ,%
we infer that J ( A ) = A X .
.JtA)
and J ( A )
=
xA.
Similarly xA
Hence J ( A )
=
Ax
is a direct sum of
X A and the result
=
follows.
For the r e s t of t h i s section,
F
denotes a fieZd characteristic p > 0 . FG
next aim is to discover necessary and sufficient conditions under which finite representation type.
sum of elements in X
Given a subset X
and denote by
FX'
of
we write 'X
G,
the F-linear span of .'X
our is of
for the The next
three lemmas and Theorem 17.15 (below) will not be required in their full generality. Let G be a p-group.
17.12. LEMMA.
(il
If
I is a proper left ideal of F G ,
then F G / I
is an indecomposable FG-
module, (iil
FG has a unique minimal left ideal, namely FG'
(iiil Each nonzero submodule of FG Proof.
(il
is indecomposable.
Owing to Corollary 1.2, J V G )
=
Hence J(FG)/I is a unique maximal submodule of
I(G)
and so F G / J V G )
FG/I
and thus F G / I
F. is inde-
composable. (iil
It is obvious that FG'
is a minimal left ideal of FG.
is an arbitrary minimal left ideal of FG. afforded by
Since the representation of
X is irreducible the isomorphism FG/J(FGI
acts trivially on X for all g E G.
and d i d = 1.
F + But then x = XG
Hence X
=
Fx
for some h # 0
P
F
in F.
(iii) Direct consequence of (iil. Suppose that G = < g >
is cyclic o f order pn.
X
G
ensures that G
for some x E FG
asserted.
17.13. LEMMA.
Assume that
and gz=z
Thus X =' % 2
as
296
Iil
CHAPTER I11
Each indecomposable FG-module is a homomorphic image of FG
i E {l,2,..,,pn},
(ii) For any
has exactly one submodule of codimension i,
FG
namely J(FG)'
n
(iii) For any i E {1,2 ,...,p },
put
Vi
= FG/cJ(FG)~.
Then
(a) V .
is an indecomposable FG-module of dimension i, and
F-basis
Iv 1 ,...,v 2. } such that
(b) Each indecomposable FG-module is isomorphic to
n Proof.
ti1
We first note that FG
principal ideal domain, each FIX] cyclic FIX]-submodules. is cyclic.
Hence, if
4
F[Xl/@
Vi possesses an
,...,p n } .
V;
for some i ~ i 1 , 2
- 1).
Because FIX1
is a
(and hence FGI-module is the direct sum of
V
is an indecomposable FG-module, then
It follows that there is a surjective homomorphism
FG-+
V
1/
PG-
of
.
modules, proving (i)
correspond to the ideals 1 of FIX] for which
(ti) The submodules of FG
n
13 urp
-11
n Since FIX] is a principal ideal domain, I = cfl for some f dividing X p n n But f - 1 = (X-11' , so T = I I X - 1)' for some i, 0 Q i 4 p n , and the image
I* of T in FG is .T* = F G l g - 11 i
Because JVG) = T ( G )
and
Hence the submodules of FG
and
G is generated by 9 ,
are the
we
also have
-1
297
GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRINCIPAL IDEALS
as asserted. each V . z
(iii) Owing to Leunna 17.12(iI,
..,V
we obtain an F-basis
proving (a).
{Ul,.
Assume that
By ( i l ,
is an indecomposable FG-module.
V
FG/I
i for some i and hence
Let G = < a > x ,
17.14. LEMMA.
where < a >
Let
and < b > are cyclic groups of there exists an indecomposable FG-
Then, for any positive integer n,
module of dimension Proof.
Vi such that
.
(b).
order p .
of
Setting
By lii) , .T = FGLg - 1)
for some FG-submodule I .
proving
V
.I
is indecomposable.
2n+l.
V be a vector space over F of dimension & + l
and let
... ,xn y ,... ,Y, 1
‘xo ,t , be a basis of
V.
I
~
It is inmediate to verify that V becomes an FG-module by
setting
Let X and
Y denote, respectively, the F-linear span of { x ~ , ...,Xn 1 and
{Y~,.-.,Y~], and let
?T
:V-
Y be the natural projection.
left multiplications by (a- 1) and
Owing to U2),
db - 1) provide injections Y-+ X and map
to 0 Assume by way of contradiction that
V
=
V @ V 1
2
for some nonzero FG-sub-
x
CHAPTER I11
298
modules
v1
and
v2
of
v,
Put r
(a-1)V C V 1 -
=
and 1
dim.rr(V1)
SO
F
(a-1)V
that 1 g r
< n.
Then
= (a-l)7T(V1)
and therefore dim(kz-1)V
= P
F Bearing in mind that
(a- 1)V1, we derive
(b - 1)V1
(0 Q P Q n)
dimV 2 2 r + 1
F’ A
similar inequality is also true for
.
Setting
3 =
dimv(V 1 , F 2
we have
r + s > n and dlmV2 2 2s + 1
F Hence dimV 2 2 ( r + s l
+
2
? 2n
+2
F
.
contrary to the assumption that dimV = lemma.
2y1+
This completes the proof of the
1.
F
It is now an easy matter to prove the following general result. 17.15. THEOREM, (D G Higman (1954), Kasch, Kneser and Kupisch (1957)). be an arbitrary field of characteristic p of
3
0
Let F
and let P be a Sylow p-subgroup
G.
(i) If
P is not cyclic, then there exist indecomposable FG-modules of arbitrary
large F-dimension (ii) If P
is cyclic, then there exist at most
posable FC-modules. Proof.
IGl
nonisomorphic indecom-
Any indecomposable FG-module is a homomorphic image of
(i) Assume that
P is not cyclic.
abelian factor group of order p2.
(flP.
P has an elementary
Hence, by Lemma 17.14, for each n 3 1 there
exists an indecomposable FP-module Vn rnorphic to a direct summand of
Then
FG.
with dimV = 2 n + l . But Vn is isoF n Hence, by the Krull-Schmidt theorem,
Vn
GROUP ALGEBRAS WITH RADICALS EXPRESSIBLE AS PRINCIPAL. IDEALS
is isomorphic to a direct summand of
Ifil.
Y of
299
Wp for some indecomposable direct summand
It follows that dimW 2 dimVn = 2n + 1,
F
F
as required. (ii) Since P
is cyclic, it follows from Lemma 17.13 that there are
,
composable FP-modules V 1 , V2,...
where
dimV.
=
i
\PI inde-
and every indecomposable
F" FP-module is isomorphic to some module.
(V,)
G
.
Write V
P
= @
83
is an indecomposable FG-
is isomorphic to a direct summand of
F
with j
V
X i for some indecomposable FP-modules Xi.
Consequently, V
dimV 2 didi.
F
Assume that
V is P-projective, so V is isomorphic to a direct summand of
Then
mull-Schmidt theorem, V and
Vi.
Then, by the
4
for some i
is isomorphic to a direct summand of some
fl
< dimV.
Since d i m e 3
F
composable direct sununands o f
fi3
CG:P)J', the number of nonisomorphic inde-
=
of dimension at least j
(G:PI.
is at most
Hence the number of nonisomorphic indecomposable FG-modules is at most By Lemma 17.3(i), morphic image of Thus V
Vi is a homomorphic image of FP.
(FPIG
FG.
But
V
Hence
vGi
\GI.
is a homo-
is isomorphic to a direct summand of
is a homomorphic image of F G ,
83
as asserted.
We have now come to the demonstration for which this section has been developed. 17.16. THEOREM (Morita (195111. characteristic p group of
G.
3
Let F be an algebraically closed field of
0, let G be a finite group and let P be a Sylow p-sub-
Then the following conditions are equivalent:
(i) There exist x,y E FG such that JCFG) = zFG = FGy (ii) There exists x E FG (iii) G Civ) (v)
such that JlFG) = zFG = FGz
is p-solvable and P
is cyclic
PO , ( G ) a G and P is cyclic
P
PO ,(G) d G , P is cyclic and G/PO ,(G) P P Proof.
is cyclic of order dividing p
(i) * (ii): Direct consequence of Lemma 17.10.
(ii) * (iii): By Theorem 17.11, FG
is uniserial and each block of FG
is
- 1.
CHAPTER I11
300
quasi-primary.
Hence, by Lemma 17.7 and Theorem 17.15, P
Proposition 14.5, PO f ( G ) d G.
is cyclic and, by
Since the latter implies that
P
G
is p-solvable
(iii) is established. tiii) * (iv): By Proposition 16.20, G
PO , ( G I P (iv) * (v): Set C(P) =
5
=
since CLPI
G
G
subgroup of
Aut (PI.
G/O
P
(GI
a cyclic group of order
12.1.
i;
and
P P and
But
p-1.
p
n-1
o
(GI a G
=
PO
=
1.
P
Hence, if x E FN
P
,(GI.
:/F
Hence
z/p
=
G/PO ,CG).
pn
and p > 2,
(p-11.
Therefore G/PO , ( G I
=
P
we have
G
=
is such that J V N )
B be an arbitrary block of FG. Since P
G/pO
P
(G)
is
then Aut(P)
is
is a cyclic group of
PO f ( G ) , by virtue of Lemma
=
FNx
=
P
= 2.
xFN, then
=
FGtFNx) = (xFNIFG
=
FGx = xFG in which case G
is p-nilpotent.
Let
It suffices to verify that J(BI = xB = Bx
for
is cyclic, it follows from Corollary 10.10 that
M (FD) for some cyclic p-subgroup D of G.
Since J(B)
since
where g
and
and observe that, by Proposition 1.8(ii),
Thus we may harmlessly assume that G = N
some x E B.
Thus
P
IP(
45
is isomorphic to a
If
In case p = 2
PO ,(GI P
Then
Aut(P).
J(FG1
B
,(G)/O
Thus the required assertion is also true for p
(v) * (ii): Put N
1 and so
P'#P
a 5 and 0 ,(:I
isomorphic to a subgroup of
order dividing
=
is of p-length
is a generator of D, the result follows.
9
Mn(J(FD))
and
301
4 Group algebras of p-groups over fields of characteristic p In this chapter we confine our attention to the case where G a field of characteristic p .
and F
After examining dimension subgroups, we
provide Jennings' description of bases for S ( F G I n tency index
t(G1
of S ( F G ) .
and a formula for the nilpo-
As an application, we determine the right and left
annihilators of powers of I l G )
and exhibit a remarkable symmetry of the numbers for some individual p-groups P,
= did(Gln/l(G)n+l. We then compute t ( P ) n F such as metacyclic p-groups, p-groups of order
c
p4, etc.
I n the final section, all p-
second highest, lowest and second lowest values.
t(P)
7
are characterized.
p
I. DIMENSION SUBGROUPS IN CHARACTERISTIC Throughout this section, F G
and
for the subgroup of
G
3
0 ?&ID RELATED RESULTS.
and
G2 of
G,
The lower c e n t r a l s e r i e s
{y,(G)}
-1 -1
= z
of
G
y
with
xy
LC
E G
is central in
nilpotent if Yc+l(G) est such integer G.
If
= 1
G/Y,~(G).
for some integer
c with Y,+l(G)
= 1
1
,y
E G2
G
and
is defined inductively by
It is clear that these groups are, in fact, characteristic subgroups of Y,(G)/Y,~(G)
IG1,G21
we write
generated by all commutators
IXPYI
that
p > 0
denotes an arbitrary field of characteristic
Given subgroups G I
a finite group.
The results obtained
P in case t ( P ) attains its highest,
are applied to investigate the structure of
groups P with
is a finite p-group
Furthermore, c 2 0.
G
is called
When this occurs, the small-
is, by definition, the nilpotency c h s s of
n 2 1 is an integer, we put Gn = < g n l g E G>
CHAPTER 4
302
It is clear that all such subgroups Gn
are characteristic in G.
The Erauer-Jennings-Zassenhaus series for G
M
1 ,P
(GI
is defined inductively by
= G
and for n 2 2
M
nI
(G) = lMn-l
(GI , G I M i , p ( G ) P IP
P
is the smallest integer satisfying i p 2 n ,
where
One immediately verifies,
by induction, that this is a decreasing sequence of characteristic subgroups of
G. The Lazard series for G
is defined by
The following observations are easy consequences of the definition: (a) L
1I
(G)
G
=
P
(b) L q l p ( G )
is a product of finitely many of the characteristic subgroups
Y i (GI p3
is a decreasing sequence of characteristic subgroups of
(c) { L n , p ( G l }
G.
(GI contains the factor Y1 (G)" = G and so L (G) = G. The #P 1 tP reason for (b) is that, since L I G ) >_ Y n ( G I P 0 = Y n [ G ) , all additional factors n,P with i n are redundant. A l s o , for i n only the minimal j with ip' n Indeed,
L1
is really required,
Finally, (c) is a consequence o f the fact that increasing n
decreases the number of defining factors for L The dimension subgroups
Let G = G G.
Then
1
>_
G
3
2 -
Dn(FG)
of
... 3 Gn -3 ...
G
(GI.
are defined by
be a decreasing sequence of normal subgroups of
is said to be an N -sequence if the following two conditions hold:
{Gi}
P
IGitGjl
g E G~ Of course, if
n,P
G
Gi+j
implies g p E G
iP
is a p-group, then the above implies that Gi/GPi
(and hence
DIMENSION SUBGROUPS
G
303
is elementary a b e l f a n .
1.1. LEMMA.
FG
i d e a l s of for a l l
I(G1 = I
Let
i.
3
1 -
satisfying
I
3
2 -
... 3- In -3 _.. be
IiIj c I i + j f o r
a decreasing sequence of
i,j,
all
Gi
and l e t
=
G n (l+l.)
Then G = G
i s an N -sequence.
2 G
1 -
... 3- Gn -2 ...
3
2 -
I n p a r t i c u l a r , t h e dimension subgroups
P
D (FG)
form an N
P
-
sequence. Proof. Since
I(G)
Since
=
I
and
li 2 li+l,w e have
Ii i s t h e kernel of t h e homomorphism G G.
normal subgroup of
x
E G
i'
y
E
G
ri*j
=
Finally, f i x
IjIi
and
a r e contained i n
[x,y
E Gi+j
8 E Gip
1 . 2 . LEMMA.
Let
G
=
G.
g+Ii,Ii i s a
is a decreasing chain
g e Gi.
=
-1 -1
x y
Then
ri + j Furthermore,
(xy-yx) E [Gi,Gj] 5 Gi+j
and hence t h a t
.
g - 1 E Ti so t h a t
gp-1 Hence
... 3- Gn -3 ...
3
-
(x-1) (y-1) - (y-1) (x-1) E
x-ly-lxy - 1 which shows t h a t
tG+Ti)/Ii,g*
and
and note t h a t
j'
w-yr s i n c e both
G2
3
1 -
3 Gi+l
G.
of normal subgroups of Now l e t
G = G
Thus
-
Gi
=
(g-1)P € I? c I. z - zp
and t h e r e s u l t follows.
G
be a group of order
G = HI 3 H2 3 be a composition s e r i e s of
G.
pn
and l e t
... 2 Hn
For each
i>
3 Hn+l
= 1
1 choose
gi E H.-Hi+l.
Then t h e
elements
x (al,a2,... ,an) =
n a n (gi-u i i=l
with
0 6 ai < p
h(0,0,
...,01 = 1
Proof.
form an F-basis f o r
FG.
Moreover, t h e s e elements with
d e l e t e d form an F-basis f o r
I(G).
Since t h e number of t h e given elements
h(al,...,an
i s precisely
CHAPTER 4
304
G,
the order of
it suffices to prove that they span FG. \GI.
being trivial, we argue by induction on
The case G = 1
IH21 < I G l ,
Since
it follows
from the induction hypothesis that the elements
span FH
.
Now
is cyclic of order p
G/H2
,...,
and is generated by the image of
is a transversal for H 2
<-I} Hence { l , g l gl. of FG can be written as
G,
in
and each element
p-1 i x = cgx.
i=o
with x E FH2.
’
1
Finally, bearing in mind that
the result follows. G
Let
be a finite p-group and let G = G
be an N -sequence of
P
written A
=
Ala
3 G
1 -
I . . . > C I G = 1 - d - d+l
2 -
For each g # 1 in G ,
G.
for which g E Gm, and put
to be the largest m
W(g1,
,...,a
)
define the height of g,
is as in Lemma 1.2, we define the weight
W(A)
V ( 1 ) = m.
of
X
If
by
n k Let Ti be the F-linear span of all products of the f o n ( ~ ~ - 1for ) some k k j=1 with 5 E G and c V(x.1 2 Clearly, Ii is an ideal of FG because it is
n
i
j=1
<.
J
closed under multiplication by 1 and 9-1, for any g E G. of
V,
From the definition
it follows that
rw
r1
=
I
3 -
3 2 -
is a decreasing sequence of ideals of FG this sequence as the f i l t r a t i o n of 1.3. LEMMA.
I(GI
... 2 I z. -3 ... with I i I j C I i + j .
determined by t h e N -sequence
P
Let G be a finite p-group, let G = H
3 1
H 2
3
We shall refer to
...
3
Hn+l
=
1
{Gi).
DIMENSION SUBGROUPS
G
be a composition s e r i e s of
G = G
and l e t
{I.}
Cil
A
span
,...,a,)
It modulo
It+l
=
(ii) The elements
(il
Proof. then
which r e f i n e s a given > G
1 -
=
,...,a
P
-sequence
- d -3 Gd + l = l
I(G)
determined by t h i s sequence.
i s as i n Lemma 1 . 2 ,
X(a
N
3 . . . > G
2 -
be t h e f i l t r a t i o n of (a
If
305
then t h e elements
with
w ( A ) 2 t form an F-basis f o r
with
Note f i r s t t h a t , by d e f i n i t i o n of
W(x)
It, i f
and
W ( x ) =
It
=
W(x1
t
t,
W e a r e t h e r e f o r e l e f t t o v e r i f y t h a t any product
E Tt.
(x.E G) 3
k
c
with
i
V ( z .)
j=1
can be w r i t t e n a s an F - l i n e a r combination of
3 plus a term i n
We argue by induction on
i E 11,...,k l
To begin w i t h , we f i x
a,B E FG.
Put
refinement of
V(x.1
z
{Gs},
f o r some i n t e g e r
r.
=
m
so t h a t
t
k.
and w r i t e
g E G,.
'S with W ( x ) =
u = C&(z.-l)@ f o r some
Since t h e sequence
{Hsj
is a
w e have
Invoking Lemma 1 . 2 , w e i n f e r t h a t
x.-1
z
i s an F-linear
combination of terms of t h e form
with t h e
as
n o t a l l zero.
However, each
gs
above s a t i s f i e s
VCg
)
2 m
= v(x .)
and so
f o r some
bs E F,y E T,+l.
Hence
and t h e r e f o r e it s u f f i c e s t o v e r i f y t h a t each product combination of t h e each
zi-1
height.
If
A's
modulo
It+l.
Moreover, by applying t h i s argument t o
i n t u r n , we may assume t h a t each
k = 1, then
u = g
j
-1
a ( g -1)8 i s an F-linear
xi i s
f o r some
j
some
with
gi,
with t h e same
V ( g . ) 3 t. 3
Hence
CHAPTER 4
306
u
is a basis element
h
> t.
of weight
W ( 1 ) > t implies h E It+l
But:
and thus the result is true for k = 1. Assume that the result is true for all such products with less than
k
terms
and write u = c1 (g .-1)( g .-I)B % 3
where
(gi-l)
Put r
( g .-1) are adjacent factors. 3
and
v(gi)
=
,S
=
v ( g .) 3
and
consider the identity -1 -1 (gi-l) ( g .-1) = (g .-1) (g.-1) + ( g . g . g.g .-1) 3 3 % 3 Z 3 -1 -1 + (9*g .-1) (9; g j gigj-l) 31-
-1 -1 Setting h = gi g j gigj
it follows that
h E [GrrGsl SO
v(h)
> r+s
and thus
5 Gr+s
t
(g.g.-l) (h-1) E 7r+s+l. 31-
Therefore
u = a ( g .-1)( g .-1)f3 2 3 -
= c1 ( g .-1)(gi-l)
3
Observe that a ( h - 1 ) B at least t.
i s a product of at most
+ a ( h - 1 )B
k-1
(mod It+l)
factors but with total height
Consequently, by induction, ~ . ( h - l ) B is an ?-linear combination
It+lof the h ' s
modulo
B
a ( g .-11 ( g .-1)f3 3
of weight
t.
Thus we are left to verify that
is such an F-linear combination of the
A's,
By the foregoing, in proving the result we may harmlessly interchange adjacent factors.
This implies that we can put all the factors in their natural order.
Thus it suffices to consider u
n
n with
bi 2 0,
u = h(b
1,
...,b
that bi 2 p
Z b. = k
and
and v ( h ) 2 t
for some i.
is a product o f fewer than k so
that
8 E Gsp
C
biv(gi) 2 t.
If b . < p
i=1
2
)
of the form
2
for all
so that the result follows.
Then
(gi-lIp = (9:-1)
terms.
and V ( G 1 2 s p .
Finally, assume
occurs in u
Moreover, if V ( g
.)
z
=
i, then
and
s , then gi E Gs
The conclusion is that in this new
DIMENSION SUBGROUPS
u,
representation of
307
the sum of the heights i s still 2 t .
Hence the required
assertion follows by induction. (iil E
It.
are linearly independent and that w ( A 1 2 t implies
x's
We know that the
repeated application of (il , we see that the set span It modulo
Is+l
for any s 2 t.
for all sufficiently large Since Gd+l
=
of
span I t .
x's
It therefore suffices to verify that these
with s
X'S
Invoking a
> W(x1
Thus we need only show that
2 t
Is+l = O
S.
1, it follows that V ( g )
Q
d
for all g E G,g # 1.
There-
fore in the product
.. (+,-1)
u = (+ -1) (xg-l).
... +
we have V ( x 1 + \)(r21+
dk.
V(Zkl
Hence I d ,
5 I (G)
and since I (G)
is nilpotent by Corollary 3.1.2, the result follows. We are now ready to prove the following result. Let G
1.4. THEOREM. (Jennings(l94111.
G = G > G 1 -
be a finite N -sequence with
P
filtration of I ( G l
be a finite p-group, let
>...>G >G - 1 - d - d+l-
2 -
IGi/Gi+ll
=
i Q d , and let 11.1 z
pei, 1
Put To = FG,
determined by this sequence.
rn
be the
d C (p-l)nen n=l
=
and define the integers cn by
d X(P-lli)ei n ( 1 + ~ i++... ~2i $=1 +
(i) I ,# 0 (iil
but
=
en = diyn/rn+l
(iii) G n (l+Iii) = G.
- m - c en? n=o
0
for all
n E {O,l,
...,ml
for all i E {1,2 ,...,d + l }
Proof. Note that the definition of N -sequence allows some of the e ' s
P
zero, but this does not effect our calculations. any refinement of chosen, then e;
{Gs} i s
{Hsl
observe also that if
into a composition series and if gi E H . - H i + l z
precisely the number of gjs
to be is
is
of height j .
d
Keep the notation of Lemma 1.2 so that the elements (i)
h (al,.
n
.., a , l = n
i=1
a
(gi-l) i
3 08
CHAPTER 4
form an F-basis for FG.
h occurs when all
Then the largest possible wieght of
ai are equal to p-1.
Since there are precisely
e
of the g ! s
j
with height
we have in this case that
j,
d
n
1 # 0 but
Hence, by Lemma 1.3(i),
=
m
0.
(ii) Owing to Lemma 1.3(i), the number f n = d i dn/In+l 1's
number of
of weight n.
coincides with the
On the other hand, the weight of
=
A(a
,...,an )
is
w(X)
a v ( g l ) + a 2 v ( g 2 )+
... + anv(gn)
is equal to the number of ways of choosing 0 4 a .
which shows that f such that
=
But the latter is the coefficient of
V ( h ) = II.
A ( 1 + Xv(g$
2V(P.)
+x
... + x
+
x"
< p-1
in the product
(p-1)v(gil )
i=1
Finally, since there are precisely
m
f,x"
C
=
(iiil Fix i E
n {1,2
S
=
g.
W(g-1) =
follows.
j
we get
If
...,m}.
E {o,l,
g E Gi, then v(g) 3 i so that g-1 E Ti
Conversely, let g E G-G {Gs}
of
i so that V ( g ) < i, and let
into a composition series.
for some j 2 1 and in choosing our generator for H j / H j + l
gj
=
X(p-l)ilei
,...,d+l}.
be a refinement
2
n ( l + x i + ~...* i+ +
fn for all n
=
by definition of Ti. {H }
i
i= 1
m=o
This proves that c
with v(gi)
of the g !S
e
d
Since g f l , g E H j - H j + l we can clearly take
With this choice of generator, g-1 = 9.-1 i s a basis element of weight 3 v(gl C i. Hence, by Lemmas 1 . 2 and 1.3(ii), 9-14 Ti and the result
'
In order to apply the previous result to the powers of the augmentation ideal
of F G ,
we characterize the dimension subgroups in a purely group-theoretic
manner.
Assume that
{Gi] 2
{Gi}
{Hi} if and only if
and G. 2 2 -
1.5. THEOREM. (Jennings (1941)). the filtration of T(G1
{Hi} are two N -sequences in
P H. for all i. 2 Let
determined by
G
G.
Then we write
be a finite p-group and let
IDi CFGI 1.
{Ii) be
DIMENSION SUBGROUPS
ti1
ID ( F G ) } is the unique smallest N -sequence z' P
the sequence
(ii) 1;
=
3 09
i for all .i
I(G1
ti) By Lemma 1.1, the dimension subgroups D . ( F G )
Proof.
Since I ( G )
sequence.
{ G .}
Assume that
P
is an N -sequence
{Di(FG)}.
P
P
P
filtration of
I(G)
-
Then obviously
{Gi n Di ( F G ) 1
It therefore suffices to verify that there is no
{Di ( F G ) 1.
N -sequence smaller than Fix an N -sequence
P
for some integer d .
is nilpotent, Dd+l(PG) = 1
is an arbitrary N -sequence.
form an N
{Gi)
IDi ( F G ) 1,
{Gi}
with
{Gi}.
determined by
and let {Jil
Since I ( G )
=
Jl
be the
we have
On the other hand, let u = ( 3 -1) (cc -1). 1
2
..(zk-l)
be any generating element of S. so that V("
Then
"i
GV(CC.1
C
-
DV(3c.l ( F G )
)
+
V(Cc2)
-I.
... +
V(Z
k1 P i
so that xi - 1 E I ( G 1
VC.qi)
and
z
as required.
i
(ii) By (i1, { I ( G ) } {Gi}
=
{Di(FGl}.
=
{Ji} is the filtration of I ( G 1 i
Thus J .
z
=
1, = T(G1
z
determined by
and the result follows.
The following result provides a more explicit characterization of dimension subgroups.
Although it is valid for an arbitrary group G ,
the special case
below will suffice for our subsequent investigations. 1.6. THEOREM.
Let G
(i) (Jennings (19411)
be a finite p-group.
IG)
Dn(FG) = Mn IP
Then, for all n 2 1,
CHAPTER 4
310
(Lazard (19531 1
(iil
Proof.
Dn(FG1
Lnlp(G1
=
D (FG)
We f i r s t show t h a t
n
n.
t r i v i a l , w e argue by induction on
< n.
is t r u e f o r a l l subscripts
N -sequence of P
G.
2 M,
(GI.
The case
n
= 1 being
IP
n
Assume t h a t
>
and t h a t t h e a s s e r t i o n
2
{D.VG)}
We know, from Lemma 1.1, t h a t
i s an
Since
t h i s yields
Let
i
be minimal with
we have
Dn(FG) 3 DiVGlp Bearing i n mind t h a t derive
Dn(FG)
2 M,
Mn, p
(G)
3.Mi,p(G)P
i s generated by
,GI
[Mn-l,p(G)
and
Mi
(GIp,
we
IP
(GI. I P
We next show t h a t
M (GI n tP
3L
Mi ,P 'Z
which i m p l i e s , by induction on
Taking i n t o account t h a t
Mn, p
(GI
contains a l l
rP
Since
>_ yi ( G ) .
GI
for a l l
Furthermore,
M
iP ,P
CG)
i
3M i , p ( G ) P
j, t h a t j
j
. (GI ip3 , p
2 Mi,p ( G I p 2 yC(GIP
{MnSp(G)1
YC(GIPJ
CG).
1. [Mi,l,p(Gl,
GI
M, -(Gl
it follows by induction t h a t
M
nrP
i s a decreasing sequence, we conclude t h a t
with
ipj 2 n
and hence
M (GI ntp
2 LnIP(G).
We
have t h e r e f o r e shown t h a t
Dn(FG) Thus it s u f f i c e s t o prove t h a t Passman (1977)) t h a t
(GI}
{L
L
n,P
(GI
2 Dn(FG). P
is t h e unique minimal N -sequence of
t h e r e s u l t follows.
I t can be shown (e.g. s e e
i s an N -sequence of
I P
{Dn(,?GIJ
1. M n , p ( G ) 3 L n l P ( G )
P
G.
G. Hence
But, by Theorem 1.5,
(GI
L I P
3 DnIFG)
and
DIMENSION SUBGROUPS
311
We now turn our attention to the powers of the augmentation ideal o f
FG.
Since the corresponding result entails a fair amount of notation, it will be especially useful to assemble most of it in one place.
G is a finite p-group G. =
z
M
G = G
G = H
i,p
(GI
3 G
1 -
3
1
H
IGi/Gi+ll
and
=
pei
3
... 3 Gd 2 Gd+l
=
3
... 3 Hn
= 1
2 2
3
Hn+l
1 G
is a composition series o f
which refines
IGi). gi E H.-H z
V(gil t(G)
i+l’
1 G
i
G n
is the largest
m for which
9;
is the nilpotency index of J(FG1
1.7. THEOREM. (Jennings (1941)).
Grn =
I(G1
With the notation above, the following proper-
ties hold:
z
(i) For all k
the elements
1,
n
n n Qi-ll
U
i with 0
a. c p 2
2=1
uiv(gi) 2 k form an F-basis for TCG)k n i=1 U i with 0 G (ii) For all k 1, the elements Cgi-ll n z=1 k+l C uiv(gil = k form an F-basis for I ( G ) modulo T ( G ) i=1 d (iii) t C G l = 1 + (p-1) C i e d i
n
(iv) Let m
=
C (p-l)z’ei and define the integers C k
i=1
d
n( 1 + ~ i + ~ 2 +i ...
+
X(P-lli)ei
k
Then ck = dimICG) / I ( G )
k+i
for all
k=0 1
c p
and
by
=
i=1
Ui
and
k
p .
k Gm.
F Proof. Let Theorem 1 . 6 C i ) ,
(1.1 be the filtration of I C G ) Z G . = D.(FG) z z
determined by
and so, by Theorem 1.5(ii),
{GiI.
Ii = ICGIi
By
The required assertions now follow by appealing to Lemma 1.3 and Theorem 1.4. As
an application of Theorem 1.7(iv), we prove the following result.
1.8. THEOREM. Let G
and
be a finite p-group and let rn
=
tCG1
- 1.
.
for all i.
Then
CHAPTER 4
312
d Proof.
We keep t h e n o t a t i o n of Theorem 1.7 so t h a t
=
xmnd
...
+
(l+~-i
+
m
c
=
C ~ - l l & ? ~ . and
x-(p-l)i ei 1
i=1
m
=Pc~x” n=o
m
c c m-n x” n=o
=
Comparing c o e f f i c i e n t s , we deduce t h a t c
m
Hence, i f
1
< m-1,
n
= c
m-n
then
On t h e o t h e r hand, by Lemma3.11.2, d i d ( G l m = 1.
Since
I(Glm+l = 0,
we
F conclude t h a t d i d ( G ) O / I ( G ) = dimFG/I(G) = 1 = dimICGlm/l(G)m+l
F
F proving t h e case n = 0. If
1
n
=Gm ,
,
F
then by t h e foregoing
n-1 dimFG/I(Gln = dimFG/ICG1
F
F
+ C did(G1 k/ I ( G)k+l k=1 F n- 1
= dimT(Glm
+ C
k=l
F
dimZ.IGlm-k/l(Glm-k+l
F
= did(G)wn+l
F Therefore
a s asserted.
.
I GI
= dimFG = diml.IGln+ dimICGlm-n+l
F
F
F
Next w e determine t h e l e f t and r i g h t a n n i h i l a t o r s of powers of
1.9. COROLLARY. (Hill (19701). Then
Let
I(G).
G be a f i n i t e p-group and l e t m = t(G) -1.
DrMENSION SUBGROUPS
313
On t h e o t h e r hand, by Theorem 3.3.6,
and dimr(T(G)nl = [ G I
-
F
diraI.(.Gjn
F
Applying Theorem 1.8, we t h e r e f o r e deduce t h a t dimR(TCGlnl = dimrU(Glnl = dimI(G)m-i-l+l
F
F
F
Thus
as a s s e r t e d . W e c l o s e this s e c t i o n by providing some observations on t h e nilpotency index
t(G)
of
JCFG).
1.10.PROPOSITION.
(Wallace (196811.
Let
G be a p-group of order
pn.
Then
n ( p - l ) + l C t CGl G pn By P r o p o s j t i o n 3 . i . 8 ( i ) ,
t(G1
[GI
I n p a r t i c u l a r , i n our c a s e we have
t(Gl
pn.
proof.
f o r an a r b i t r a r y f i n i t e group
G.
I n t h e n o t a t i o n o f Theorem 1.
we have
d d Since
C e . = n,
we have
d
a s required.
1.11. THEOREM. (Wallace ( 1 9 6 8 ) ) . G.
Then
Let
P be a normal p-subgroup of a f i n i t e group
CHAPTER 4
314
Proof. By Proposition l.l(ii1, we have
Set n = t Ip) and m = t G / P )
proving that t (G)
.
Since J V P )
=
I(.P)
we have
< t (P1t (G/P1.
We now prove the second inequality.
Since
[ J CF (C/P) 1
Imp'
# 0
it follows from C21 that there exists
be a transversal for P
Let T
s E FP and ti i
E
T.
in
G
and write
# =
t s + 1 1
Then there exists j E {l,21...ln)
... + tnsn
such that s j
with
9 I(P1,
i.e.
Consider now I ( P I n - l
But
(
c
which, by Lemma3.11.2,is spanned by
Z' y . LFp
Then
y)W # 0 , since
!@
and in this expression of
(
C y)rJ
as a sum of elements whose support lie in
.#P distinct cosets of P
in G1 we have
upon noting that the coefficient of It follows that
1 in
( C a#)( X y ) 9Y=p
is
X a # 0. SEP
315
Remark. p = 2
The bounds given in the above theorem cannot be improved. and
IGI = 4.
If
Indeed, let
P is a subgroup of order 2 , then t(P) =
t(G/P)
2
=
and so 4 = 2.2
If G
3 t(G) 2 2 + 2 - 1
is cyclic then t(G1 = 4 and if G
2. COMPUTATION OF
t(P1
L
z2
x
= 3
z2
FOR SOME INDIVI'DUAL p-GROUPS
then t(G)
J ( F P 1 = I(P1
3.
P
Throughout this section, P denotes a finite p-group and F of characteristic p .
=
an arbitrary field
Our aim is to compute the nilpotency index t ( P )
P.
for some individual p-groups
Mn
=
M
n,P
of
For convenience, we put
(PI
for all n 2 1
so that if
P = M1 -> M 2 -2 . . . z M d > M d + l = 1 and
then by Jennings' formula (Theorem 1.7 (iii)I
Let
A be an abelian subgroup of P, let
that in 2.1. 2 E
P.
is a normal subgroup of
P.
z E
Z(P) be of order p and assume
Then both
Ap and < A p , , >
are normal
With this observation, we now prove
THEOREM. (Motose (197811.
Z(P1 be of order p
Let
A be an abelian subgroup of P, let
and assume that d P
and that P/
is
CHAPTER 4
316
elementary abelian.
p (i) If
b
= (P:A),
P/Ap
Put
pCt
t
t+1
= (Ap : A p
S
= the exponent of
A
( t 2 0)
is elementary abelian, then
t ( P ) = Cb (ii) If
p
and
P/Ap
+
s-1 t C ctp ) ( p - 1 ) t=o
+
1
p,
is not elementary abelian but of exponent
t(P1
=
P/Ap
(iii) If the exponent of
s-1 (b + 1 + C C $ t ) (p-1) t=o
p,
is larger than
1
(3)
then
s-1 C ctp ) (p-1)
+
t ( P 1 = (b+p-l
+
then
+
1
(4)
t=o
Proof.
By hypothesis, P' C < A p I z >
P'
(i) O u r hypotheses ensure that
and so
5A p
= A'.
and
Invoking (5), we there-
fore deduce that
M2
=
M3
Mp+1 = Mp + 2
=
=
... ...
=
M
=
M
= Ap
P
= AP2
P2
- - - - - - - - _ _ _ _ _ _ _
M PS-l+l
=
Hence
9-2 = Cs-2'eps-2+1
=
P
P
s-1
... = eps-Ll = o
= c6-1
and therefore, by (11, we have t(P) = 1
+
@-I)
PS-l C ie
2'= 1
i
= 1
+
5-1
t
( p - 1 ) ( b + C ctp )
t=o
t(P)
COMPUTATION OF
317
as required. (iil
Since
€'/<Ap,.>
Ap C- C- <Ap,,>.
is elementary abelian, we have
i s not elementary abelian but of exponent
By hypothesis, p/Ap
p.
easy to see that = < A p , , >
p # 2,
Applying (51, we infer that if
M
M
P+l
and
'A
=
9
then
=
...
=
... = M
=
M =Ap
P
=
AP2
P
M
= I
ps-1+1
while if
p = 2,
then
M p 1fl
=
1
b+c -1 Now
IP/<AP,z>[
=
p
Hence, in both cases p # 2 (iii) Since
P'
since
p
and
5 <Ap,,>= 9 , M
= 2
M
an application of (51 shows that
.,, = MP =
!?+I
(3) follows from (1).
= 2,
...
=
=
<Ap z > =
M P2
APZ
Hence It is
CHAPTER 4
318
'
Thus (41 follows from 111. 2.2. 2 E
COROLLARY.
[PI
Let
=
pr
and assume that there exists a central element
and an element h E P
P of order p
of order ps
< h , z > 4 P and P / < h p , z > ( i ) If
P/
such that
is elementary abelian
is elementary abelian, then
t ( ~= )ps
t
( r - s )(p-1)
(ii) P / < h p > is not elementary abelian but of exponent p ,
(iii) If the exponent of P/
t (PI Proof.
=
Setting A = < h > ,
is elementary abelian.
is larger than p ,
then
then
p s + (r-atp-1) (p-1)
we have Ap
=
,
d P and P/<Ap,z>
Hence the hypotheses o f Theorem 2.1 are satisfied.
Furthermore, in the notation of Theorem 2.1,
we have
ct
= 1
for all t.
Hence
s-1
(
c
c t p l(p-11
=
pS-1
t=o
and the result follows by appealing to Theorem 2.1. Recall that a group G
i s called rnetacyczic if
group whose factor group i s also cyclic. elements a
and b
where the positive integers m,r,s, Set d = h ,t ) and k = t / d . also assume that t Im THEOREM.
Therefore, G
can be generated by two
with defining relations
am = l , b s
2.3.
G has a cyclic normal sub-
=
at
and b-lab
and t
=
ar
satisfy rs
Then, replacing a
by
l(mod
m ) and
bfItk-1).
ak , i f necessary, we may
.
(Koshttani (1977a1, Motose (1978)).
Let P
be a metacyclic p -
t(P)
COMPUTATION OF
=
Proof.
P
If
if
s < k
if
s > k trivial.
i s c y c l i c , then t h e a s s e r t i o n i s
s 3 1 and
assume t h a t
I
Pn+pS-l n+s-k kP +P
3 19
k
1.
t
[b,al c < a p > , t t+l > p =
Since
,g
one e a s i l y v e r i f i e s t h a t
,gt+'>
and
Hence i f
s
f
k.
then
M pn-l+l while i n case
s
3
k,
= 1
we have =
...
M pk-2*l
=
... = MPk-l
M pk-51
=
... = M
M2
M
=
Mz7
=
P+l
P
k-1 =
=
k-1
,ap
>
Hence we may
>
CWTER 4
3 20
P
n+s-k-l
+1
=
The d e s i r e d conclusion now follows by applying (1). W e next record some group-theoretic information.
Dn given by t h e
A group
defining r e l a t i o n s :
2n-l
D, i s obviously of order
=
1,b-lab
= l,b2 =
(n
=
and is c a l l e d t h e dihedraZ group of order
2n
2)
The
2".
Qn given by
group
i s again of o r d e r
and is c a l l e d t h e generalized quaternion group of o r d e r
2'
p # 2,
For every prime
M(p)
t h e group
2n.
i s given by t h e following d e f i n i n g
relations:
M(p1 =
P
=
M (GI P
=
bP =
P1.
P
(i.e.
P satisfying
of a l l maximal sub-
the gntersection
s ( n ) =
Mn(p)
Z r 1 = %
=
(61 i s c a l l e d
The semidihedral group S(n) f o r
Z P
(6)
extra-special.
n
3
=
l , b 2 = 1 , b -1ab = a-
3
i s defined f o r
2n
i s a group of o r d e r
2n-l
n >4
if
p
1+zn-2
= 2
n-1 Mn(p) =
=
l,bp
=
l , b - l a b = a'+'
and f o r
As an a p p l i c a t i o n of Corollary 2 . 2 and Theorem 2.3,
2.4.
COROLLARY.
The following p r o p e r t i e s hold:
ti)
*con,
+
1
n-2
we now prove
defined by
>
p # 2.
= 2n-1
Denote by
Then
P' = O ( p ) Any p-group
p3 and has exponent p .
is obviously of order
@(PI t h e F r a t t i n i subgroup of groups of
1 =1, a- ba = bc,a-lca = c,b-lcb = c>
cp
>
n
>3
if
COMPUTATION OF
(ii)
tcg, = 2n-1
+
321
t(P1
1
(iii) t ( M ( p ) ) = 4P-3 (iv)
+
t ( S ( n ) )=
1
+ p-1
(v) t (Mn(p) ) = p n - l
Proof. Properties (i) ,(ii),(iv) and (v) follow from Theorem 2 . 3 , while
.
property (iii) is a consequence of Corollary 2.2 (ii) We now determine formula for t ( P )
t(P)
\PI Q p ' .
in the case
can be obtained from Corollary 3.2.5.
corresponding formula follows from Corollary 2.4. is nonabelian of order p 4 .
that P
well known (e.g. see Huppert (1967)).
In case
IPI
The defining relations for such
=
p3
the
P are
The results below can be easily verified Corollary 2 . 4 , Corollary
and Jennings formula (11 (for direct calculations refer to Koshitani (1977a). TABLE (Nonabeliangroups of order p'
2.5.
is abelian then the
Thus, we may harmlessly assume
Corollary 2 . 2 , Theorem 2.3,
by applying Theorem 2.1, 3.2.4
If P
where p 2 3 )
Defining relations of P
2p2
-1
I
p2+3p-3
I
P=
< a,b ,c12= bp = 2' = 1 ,a-'ba
= bc?
,a- 1ca = bc ,b-'cb
=
c>
p2
+ 3p - 3
P = < a , b , c I z = b P = c ? 2 = 1 , a - 1 b a = bcnP,a-'ca=bc,b- 1c b = c > where n
P
=
is a quadratic nonresidue modulo p .
b-ldb
=
d,a-'ba
=
1
2
=
d P = 1 , b - l c b = c,c-'dc
b,a- ca = bc,a-ldu = cd>
p2+3p-3 =
d,
-1
322
CWTER 4
2 . 6 . TABLE monabelian groups of order
OF ORDER pa
3. CHARACPERIZATION OF GROUPS P t ( P ) =pa-’+p
- l,t(PI = pa-1
Throughout this section, P
AND t[P) =
of J(FP1
WITH
t(P)
=
U(p-1)
U
+ l,t(PI = p ,
(a+l)( p - 1 ) + 1 a
denotes a group of order p ,a 2 1, and F
arbitrary field of characteristic p index t ( P )
24)
3
0.
an
By Proposition 1.8, the nilpotency
satisfies a (p-1)
+1
t (PI
a
p
It i s therefore natural to investigate those groups P t ( P ) = a@-1)
+
1.
for which t ( P ) = pa a If P is an abelian group of order p ,a 2 , then by
Corollary 3 . 2 . 5 , the second highest value of t ( P )
is
pa-’ + p
case P i s not cyclic but has a cyclic subgroup of index p . generalize this fact by characterizing arbitrary groups P for which t ( P 1 = pa-’ groups P
for which
+ p - 1. t(P1 = p
-1
and in this
In what follows we a
of order p ,a 2 2 ,
In the final part of the section we examine those or
t ( P ) = (a+l)(p-1)
+ 1,
the latter being
the second lowest value of t P ) , by virtue of Jennings’ formula (Theorem
1.7 l i i i l
.
or
CHARACTERXZATION OF GROUPS ? .
323
We start by r e c a l l i n g t h e following p i e c e of information.
@(GI
G = <X,g>
g
E
MI
so
@(GI
G/@ GI
ticular, i f
Thus
5M
G'
2
G
Thus
G
G/CJ(GI
x
E G.
p.
has order
CG:@(Gl) = p , then
P
X
M
E
for a l l
.
2 E
M of
G/@(G)
d
G,
.
(G:M) = p.
G as
(G:M)
@(G)
then
p.
5
@(GI
G/@(G).
is a
G.
i s a generating s e t of
may be omitted from any generating s e t of
(G:O(G))
and so
=
p
d
,
G;
Howhence
as asserted..
W e are now ready t o prove
3 . 2 . THEOREM.
blotose and Ninomiya 11975a11.
Let
P be a p-group of o r d e r
pa,a 2 1. (il (iil
t ( P l = a@-1) + 1 i f and only i f
t(P) Proof.
=
pa
(il
i f and only i f For a l l
n
a
1,
P=M 3 M 1 -
e If
IMi/Mi+ll
= p
i,
P i s elementary a b e l i a n
P is c y c l i c Mn = M
(PI and w r i t e n*P
put
3 . . . 3 M
2 -
3 M
- d-
d+l
But
On t h e o t h e r
..
.
=
i s elementary a b e l i a n .
{ g l @ ( G l , , ,grOtG) 1
IX,gl ,,, ,grj
d
G'
w e have
d elements, so d 3 r . and
I t follows t h a t
1
p
r is t h e minimal number of g e n e r a t o r s of
G/@(Gl,
ever t h e elements of =
I n par-
i s a b e l i a n and t h e image of each
This proves t h a t
is a g e n e r a t i n g s e t of
g e n e r a t i n g set of
G
G.
G, then M 4 G and
G/@(Gl
Hence
G has a generating set c o n s i s t i n g of hand, i f
But
which means t h a t
i s elementary a b e l i a n of o r d e r
is a b e l i a n , and
r
If
<X>,
=
G.
of
is cyclic.
i s a maximal subgroup of
for a l l
x i n G -@(GI
M
is a
H
If
may be o m i t t e d from any g e n e r a t i n g s e t of
G/M
as
E @(GI
.
<X>
=
f o r some maximal subgroup
Because t h e above holds f o r each maximal subgroup and
H
and l e t
a contradiction.
Then
M
rf
CM
i s c y c l i c , then
G.
g e n e r a t o r s of
E @(G)
G.
G be a f i n i t e p-group and l e t d be t h e minimal number of
3 . 1 . LEMMA. L e t
Proof.
CM,
G
a l l elements of
g
with
G, then H
proper subgroup of then
G i s c a l l e d t h e F r a t t i n i subgroup of
of a l l maximal subgroups of
Suppose t h a t
The i n t e r s e c t i o n
= 1
then by Jennings' formula (Theorem 1 . 7 ( i i i ) )
CHAPTER 4
324
t(P1 = 1 Since
{Mil
+
d ~p-11 2 iei i=l
is an N -sequence, each M./M
P
assume that t ( P 1 = a(p-1)
+ 1.
is elementary abelian.
d
i+l
2
d
c iei
Then, by (1)
i=1
= a.
But
Now
1e . = a
'
and
thus
= a and e
e
Hence P
i
= O
for i # 1
is elementary abelian.
Conversely, assume that P
a
c
t(P1 =
is elementary abelian.
-
p
(a-1) = p a - (a-1)
=
Then, by Corollary 3.2.5,
a(p-11
+
1,
i=1
as requifed. (ii) If P
is cyclic, then t ( P 1 = p a , U
sely, assume that t(P1 = p ,
by virtue of Corollary 3 . 2 . 5 .
Conver-
Then, by Theorem 1.9,
whence
tLP/QPlI = I P / W BY Lemma 3.1,
I
=
pb
(b Q a )
P / @ ( P ) is elementary abelian and so by (il
tCP/Q(Pll = bIp-11 + 1 It follows that
which is possible only in the case b = 1. cyclic.
But then P/Q(P1
(and hence P )
is
So the theorem is true.
The following simple observation will be required for the characterizations of those
P for which t C P ) Let q 2 2
3 . 3 . LEMMA.
where the e
i
= pa-'
and n
+ >
2
p
- 1. be positive integers and l e t
are positive integers with
ing conditions are equivalent:
e
1
e
2
... 2 e n'
F =
n C ei i=1
Then the follow-
P
CHARACTERIZATION OF GROWS
n
c
(il
e
-
q
325
r- 1
(n-11 2 q
i=1 (ii) n = 3 , r = 3 , q = 2
or n=2,e
= 1 2
Proof. The implication (ii) Obviously, if x
If n ? 3
>
2
and y
+
>t >
(il 2,
is obvious.
So assume that (i) holds.
then
then by ( 2 ) we have i=1
e +e -1 2 4
1
2
e +.,.+e + q 3 n+qn-2-
2-
n e ~ . q ' + n - l
1=1 e +e -1 e e > q l - q ' - q 2 + ( n - 2 ) q + n - 3 =
Thus n = 3 , e
e
= 2
r
= 3.
(qP- 2
and
= 1
e e -1 (q ' - 4 ) ( q
3
Finally, if n
=
2
-
2 ) (U-1)
-
1 1 + (72-3) ( q i 1 1 2
= 0
o
which implies that q
=
2
and
then e e -1 q - 1 2 ( q I - q ) ( q * -1)
and so e 3.4.
=
1, as required.
THEOREM. (Koshitani (1977a), Motose (19781).
a of order p . (i) t ( P )
=
(ii) t ( P ) (iii) P
+ p -1
pa-1
contains an element of order p
(iii) =* (i): Let h E P
Cii)
.
be of order p
a- 1
.
Then
is normal.
is not cyclic, for otherwise there exists x E P such that
P is cyclic. t ( P ) _pa-1
a-1
(il * (ii): Trivial
Proof.
P/
is noncyclic
Then the following conditions are equivalent:
pa-'
7
Assume that P
+
Hence P / < h p >
9=
h
and
is elementary abelian and, by Corollary 2.2(i),
p - 1.
* (iii): First assume that
NOW
P
is abelian, say P
z Pe
Then, by Corollary 3 . 2 . 5 ,
k=l
1
32h
CHAPTER 4
n C ek and n 2 2. O e n g to Lemma 3.3, (31 holds only for k=1 n = 2,e = a - l , e = 1. n u s P z Z x Z and so P contains an element of 2 a- 1 Pa-l P order p
where a =
.
By the foregoing, we may assume that P is nonabelian so that a
We
If a = 3, then either P contains an element of
argue by induction on a. order p2 or P
3.
M(p).
However, t(Mlp))
=
4p-3
by Corollary 2.4, and
so the case a = 3 is established.
From now on, we assume that a p, and put
P/.
=
Then
> 4.
-P
Let z
be a central element of order
is not cyclic and, by Theorem 1.9, t ( F )
x
Hence, by the induction hypothesis, there exists an element
F.
Because
is not cyclic,
implication (iii)
* ti).
normal and P/
is elementary abelian, by virtue of
Thus there exists an element h is elementary abelian.
such that a- 2 k
hp
Now if
On the other hand, if h
is of order pa-’.
then h
p/<xp>
of order pa-2 in
=
z (0 C
is
k < p),
is of order pa-2
then,
by Corollary 2.2, t(P) =pa-2+2(p-1)
or,
t t ~ =pa-2+3(p-l) )
Since in all of these cases *(PI d pa-’,
t ( P ) = pa-2 + p 2 - 1
or
we conclude that h
is of order pa-’,
as required. a be a group of order pa,a 2 1.
Let P
3.5. THEOREM. CMotose (1978)).
Then the
following conditions are equivalent: (1) t ( P ) = pa-’
(ii) P
M(3)
Z x Z x Z
or P
2
2
2
Proof. The implication (iil and Corollary 3.2.5.
P Iz
x
... x
So assume that t(P1
,
Z
Pe1
* (il is a consequence of Corollary 2.4(iii1
then n 2 2
.
pa-
If P is abelian, say
and
7 n
n
pa-’
=
= t(~1 =
e
1p
i
-
(n-1)
i=1
by Corollary 3.2.5. Assume that P exponent p
Hence, by Lemma 3.3, P 2 z is nonabelian of order p3.
and so P
M(p1.
x 2
z
x 2
z
2
.
By Theorem 3.4,
P is of
However, p 2 = t (PI = 4p-3, by Corollary 2.4(iiiI ,
P
CHARACTERIZATION OF GROUPS
and thus p
=
3.
Finally, assume that P be of minimal order with
p
and put
321
=
P/<x>
then, by Theorem 3 . 4 ,
. -P
exists an element h E P
i s
nunaheliad an6 that a 2 4 .
t(P) = pa-1.
Let
t(F)
Then
pa-2
x E P be a central element of order by Theorem 1.9. of order p
contains an element
of order p
a- 1
pa-'
=
t (PI= p
or pa-'
=
+ p -1
a-1
+
pa-2
or pa-l
3 ~ p - 1 1 or
then a-1 2 4
p a-1
and so our choice of P
pa-2
= =
pa-2
and hence
P
2
z2 x z2
F
M(3J
x
z2 ,
or
=
.
=
23.
__-- -
Hence there
X
z2 '
+
is
p 2 - 1
t(F)
c3
=
pa-2.
is abelian.
which contradicts a
MC3)
and we can write
--
- = c , b'-1-c -b
--1--Il,a b a = b c ,a c a
M = M = M =Mt,M 4
5
=
= 1
6
which implies that 2 < e l < 3 , e
2
< l , e 4 = e
< 2 , e
=o
and e
But then, by Jennings' formula, 3 3 = t(P) = (e +2e +3e +6e ) (3-1) 1
2
> 5.
3
6
+
1 < 33
This completes the proof of the theorem.
But Thus
If
This contradicts
5 <x,c> , M3 5 < x >
1 # M2
If
by the foregoing discussion.
Thus L
=
=
Keeping the notation o f Jennings' formula (1) in Sec. 2 , we see that
a contradiction.
> pa-2
2 (p-1)
then t @ )= 5 or 6, by Corollary 2.2.
the assumption that t(P1
F
- 5 X z2
+
ensures that
= 23
then by the abelian case proved above,
-
a- 2
t(p)
a- 2 such that P/<x,hp> or p
Since the above equalities are impossible, we infer that 5,
If
Therefore, by Corollary 2 . 2 ,
elementary abelian.
a 2
We may choose P to
c'>
CHAPTER 4
328
Assume that a group' P of order pa
3.6. THEOREM. (Motose (19781). elementary abelian. (i) t ( P 1
Then the following conditions are equivalent:
( a + l )(p-11
=
(ii) t ( P ) < (a+2)@-I)
+
1
+
1
(iiil There exists a central element z
(iii)
(i)
of order p
in P
When this is the case, p = 2
is elementary abelian. Proof.
is not
such that P / < Z > is of exponent p .
or P
(ii) Trivial
=)
.
(i): Apply Corollary 2.2 (ii), (iii)
=)
(ii) * (iii): Put
M,
=
Mn ' P (P) and let d 2 1 and e i >
P = M
3 M
0
be such that
> . . . >-M d -> M d + i = l
2 -
and
Then, by Jennings' formula (Theorem 1.7(iii))
; condition
(ii) implies that
d
Since a
,
Cei
=
we conclude that
i=1
d 2
Hence
e
= 3
e
= 4
... = ed
=
P is elementary abelian.
3
C (i-lle i i=1
0 and e E Thus e
= 1
en M and hence
e
1
= a-1,
=
an
.
so
proving that
(p-1) + 1 Furthermore, M 2
by virtue of (41. Let z
be a generator of M 2 .
assume that the exponent of P
(5)
is a central subgroup of
Then P / < z >
P of order p .
is elementary abelian.
is greater than p .
Finally,
Invoking Corollary 2.2Ciii)
and (51, we derive
(a+p-ll (p-1) + 1 and so p = 2.
= t ( P ) = (a+l)(p-1)
+
1
This completes the proof of the theorem.
4. CHARACTERIZATIONS OF p-GROUPS Throughout this section, F
P WITH t ( P )
7.
is a field of characteristic p > 0
and
t(PI
is
the nilpotency index of J ( F P 1 ,
where P
aim is to characterize those P
for which t ( P )
is a p-group of order
7.
a p ,a 2 1.
By hypothesis, t(P)
Our 3
1.
The case t(P) = 2 being a consequence of a general result, namely Theorem 3.12.4, we may assume that t(P)
2.
All the results presented below are extracted from
Koshitani (1977a1, Motose (19781 and Motose and Ninomiya (1975a) 4.1. PROPOSITTON.
The following conditions are equivalent:
t(P) = 3
(il
(ii)PEz!
PzZ
or
2
3
x Z
2
Proof. That (ii) implies (il is a consequence of Corollary 3.2.5. that t ( P ) = 3.
a = 2 and p
=
must have P follows.
By Proposition 1.8, 3 2, or a = 1 , p = 2.
z!
IPI
or 3
=
1 and hence a
In the latter case t ( P )
Since
4.
> a(p-11 + P
Iz!
=
=
Assume
l,p
=
3, or
2, and so we
implies t ( P ) = 4, the result
'
4.2. PROPOSITION. The following conditions are equivalent: (i) t(P) (ii) P
=
4
is one of the following types:
(a] p = 2 and P (bl p = 2
is cyclic of order 4
and P z z !
2
x z
2
xz!
2
Proof. The implication (ii) * (il follows by virtue of Corollary 3.2.5. Assume that t ( P ) = 4.
or a
=
have
t(P) < 3
By Proposition 1.8, 4
or a = 2,p
1,p = 3 ,
=
2, or a = 3,p = 2.
[PI = 4 or
and thus
a(p-1)
IPI
= 8.
+
1 and hence U=l,p=2,
In the first two cases we
The desired conclusion now
follows by appealing to Corollary 3.2.5 and Corollary 2.4Cil,(ii).
4.3. PROPOSITION. (il (iil
The following conditions are equivalent:
t(P) = 5
P is one of the following types:
(al p = 2
and P z z !
(bl p = 2
and P Z D
Ccl
p = 2
and P E Q
(d) p = 2
and P z z !
4
xz!
2
3 3
2
xz!
2
x z xz! 2
2
'
CWAeTER 4
3 30
p = 3
(el
(f) p = 5
and P E z
3
x z
3
z
and P
Proof. The implication (ii)
* Ci) is a consequence of Corollary
Assume that t ( P ) = 5.
Corollary 2 . 4 ( i ) ,(ii).
and, by Jennings‘ formula p - 1 divides 4. If p = 5 then a = 1 and so P z
a
= 1 implies
t(P1 = 3,
we
z
.
By Proposition 1.8, 5 S a C p - 1 ) + 1 Thus aCp-1)
If p = 3 ,
have p = 3,a = 2
<4
then a
and p € { 2 , 3 , 5 ) . Since p = 3 ,
2.
z3
in which case P
Finally, assume that p ’ = 2.
virtue of Corollary 3.2.5.
3.2.5 and
Then a
x 4
z3 ,
by
and the
result follows by Corollary 2 . 4 , Corollary 3.2.5 and Table 2.6. 4.4.
PROPOSITION. The following conditions are equivalent:
ti)
t(Pl = 6
(ii) P
is one of the following types:
(a) p = 2
and P ” z
(b) p = 2
and P z D x a!
(cl p = 2
and P z Q
4
xa!
3
3
x
xz2
2
2
z2
(dl
p = 2
and P ” < a , b , e l a 2 = b 2 = c 4 = l ,-1 a ba=bc2,a-lca=c,b-1c b = c >
(el
p = 2
and P
Proof.
is elementary abelian of order Z 5
The implication (ii)
*
(i) follows by virtue of Corollary 3.2.5
By Proposition 1.8,
6
> a@-1) +
Table 2.6.
Hence p = 2
divides 5.
and
5.
0
virtue of Table 2.6, Corollary 3.2.5 that p = 2
and a = 5.
a normal subgroup Q
of P
If
If a
4,
then the result is true by
and Corollary 2.4Ci) ,(ii).
of index 2
By Theorem 1 . 9 , t (PI
t(&l< 5.
But
= Z4
1 and, by Jennings’ formula p-1
such that Q
a t (Q) + t (&/PI
-
1
contains an element of =
t (Q) + 1 and hence
which contradicts Propositions 4 . 1 - 4.4.
elementary abelian and the result follows. 4.5.
PROPOSITION. The following conditions are equivalent:
(i) t[P1 = 7
P is one of the following types:
(iil (a)
P
1
Z
Finally, assume
P contains an element of order 4, then there is
order 4.
141
and
Thus
P is
,?
CHARACITRIZATION OF p-GROUPS
(b) P
is elementary abelfan of order
(c) P = z (d)
4
x z
<7
331
26
P is not elementary abelian and P contains a central
2',
such that P / < z >
is elementary abelian
(el P (f) P
or
tcP)
4
P is of order
involution z
33
WITH
=
Proof. divides 6 .
=
c4
=
1
1.a-lba = b,u-lca = bc,b- cb
By Proposition 1.8, 6 2 a@-1) Hence the order of P
is
7
=
c>
and, by Jennings' formula p-1
or
3'
(aG3) or
2'
(aQ6).
desired conclusion is now an easy consequence of Corollary 2 . 2 , Theorem 3 . 4 , Theorem 3 . 6 and Table 2 . 6 .
The
This Page Intentionally Left Blank
333
5 The Jacobson radical and induced modules The aim of this chapter is to exhibit some close connections between the Jacobson radical and induced modules. terms of
ann(V).
We start by describing
ann(F)
exclusively in
We then examine simple induction and restriction pairs and
prove a number of their important properties contained in a work of Kn&r
(1977).
As one of the applications, we provide necessary and sufficient conditions for J(FG) = FG*J(FG),
H is a subgroup of G.
where
Further applications are
exhibited in the last section, where a detailed analysis of p-radical groups is As a consequence of one of our results, we establish a purely group
presented.
theoretic fact:
the number of p-regular classes of
of double (P,P)-cosets in
G does not exceed the number
G, where P E Syl (GI. P
1. ANNIHILATORS OF INDUCED MODULES
Throughout this section,
G.
finite group
F denotes an arbitrary field and H a subgroup of a
Unless explicitely stated otherwise, all modules are assumed to
be left and finitely generated. the annihilator of
V
Given an FG-module
V, we write ann(V)
defined by
The annihilator of a right FGmodule is defined similarly.
is an ideal of FG.
cases ann(V) ann(fi)
where
Let M
V
for
Note that in both
Our main result in this section describes
is an FH-module.
be a right FG-module.
underlying vector space
Hom(M,F)
Then M*
denotes the (left) FG-module with
and group action
F (g$l (rn)
= Qhg)
( g E G,m E M,@ E Hom(M,F))
F If
M is a (left) module, then a similar definition gives the right module M*.
1.1. LEMMA.
Let M
be a right FG-module and
V
a right FH-module.
Then
CHAPTER 5
334
(1) ann(M) = ann(M*l
cI6*
(ii)
(V*)
Proof.
M*.
(il
G
Then, f o r a l l
x
€
E ann(MI
(4) On) =
m E M,
v e r s e l y , suppose t h a t
ann(M*).
@
and l e t
be an a r b i t r a r y e !merit of
Then, f o r a l l
E annm)
,
@
..
(V*IG
element of
e M*
and a l l
Con-
m e M,
as required.
{ g l,g2,, ,g 1 be a l e f t t r a n s v e r s a l f o r n
Let
x E ann(M*).
@ h z )= @(O) = 0, so
m = 0 f o r a l l m E M so t h a t x
Hence (ii)
x
Assume I a t
H
G.
in
Then a t y p i c a l
can be uniquely w r i t t e n i n t h e form
n with
-1
-1
Because
{ g l 19,
,...,g- '1 n
is a r i g h t transversal for
in
H
G,
Qi
v*
each element
n
v"
of
can be uniquely w r i t t e n i n t h e form
n
c vi
8 gil
with
vi
E V
i=1
0 : (V*)G-+
I t follows t h a t t h e mapping
Cfi)*
d e f i n e d by
n
[e(.c
gio$i)1(:vi8g;1)
i s an i n j e c t i v e F-homomorphism.
write
ggi
=
g .h 3
x
f o r some
'
FJ1.w.) '
=
(V*IG and
Since t h e F-dimensions of
0
a r e t h e same, we deduce t h a t l e f t to verify that for
=
i=1
2=1
i s i n f a c t a n F-isomorphism.
gi 8 $i
and
g e G,@(gx) = @(XI.
.
j E { 1 , 2 , . . ,nj and h E H .
Then
(PI*
We a r e t h e r e f o r e To t h i s end,
gz = g . 8 hqi 3
and so
For any a d d i t i v e subgroup
x
of
FG,
let
Id(X)
denote t h e sum of a l l two-
ANNIHILATORS OF INDUCED MODULES
FG
s i d e d i d e a l s of
X.
contained
V
(Knorr (197711.
A = ann(V),
If
A
H
Let
arm(?)
then
i s an i d e a l of
Proof.
2.4.3
FG
i s t h e l a r g e s t i d e a l of
G
be a subgroup of a f i n i t e group
and
be an FH-module
(i) If
(ii)
Td(X)
Then
We a r e now ready t o prove
1.2. THEOREM. let
X.
contained i n
335
(i) L e t
K
FH,
5 FG-A
1dCFG.A) = Id(A*FG)
then
FG.
be an i d e a l of
K(@I
(ii) t e l l s us t h a t
= Id(FG*A)
0
=
.
Since
K C_ a n n ( p )
Hence
To prove t h e opposite containment, f i x an F - b a s i s , say
AV
= 0,
Proposition
and t h e r e f o r e
U1,...,Vn
of
V.
Then
t h e map
f = v x v x ... x v (33 ,xu ,...,33 1 n
FH-+ 2-
1
2
i s an FH-homomorphism whose k e r n e l i s
0 Tensoring t h i s sequence with account t h a t
FG,
A.
FH/A
v"
using Proposition 2 . 4 . 3 ( i ) , and t a k i n g i n t o
i s a f l a t FH-module, one o b t a i n s an e x a c t sequence
FG
This shows t h a t
5 FGoA,
am(?)
V = FH/A.
by Lemma l.l(ij
Hence t h e r e i s an e x a c t sequence
- -
O+FG/FG-A-+
(ii) Put
(n t i m e s )
Then
and ( i ).
V
G
@
G V
@
... @ fl
(n times)
proving (i).
A = ann(V,l
and
Again using Lemma l . l ( i ) and t h e r i g h t analog of
( i ),
one has ann[(V*l
G
1
= Id[(annV*)FG] = Id(A°FG)
The d e s i r e d a s s e r t i o n now follows by appealing t o Lemma l.l(ii1.
{I I y E
1 . 3 . COROLLARY.
Let
r,
nI YET y '
and l e t
I =
Y
Then
r]
be a set of i d e a l s of
FH
.
f o r some index set
CHAPTER 5
3 36
Proof. Put V
=
Because 8
FH/T
Y'
Y
fl "-
Y Y
(@
Vy)
G , the required assertion
Y
follows from Theorem l.Z(i.1.
2. SIMPLE INDUIXION AND,RESTRICTIONPAIRS Throughout this section, F denotes an arbitrary field of characteristic p > 0 ,
H a subgroup of a finite group G and e and f central idempotents of FG
.
and FH, respectively reducible module.
We recall that by definition 0 is a completely
Our aim is to present some general results to be used later.
These results are extracted from a work of Knorr (19771. 2.1. LEMMA. Let
eW/eV
2
VCW
Then eV
be FG-modules.
5W e
are FG-modules such that
e(W/V).
Proof. The fact that eV tion that e
5 eW
are FG-modules is a consequence of the assump
is a central idempotent,
Let f : W
d W/V be the natural homo-
Then f restricts to a surjective homomorphism
morphism.
and Kerf* = eW n
v.
But eW II
v=
ev,
hence the result.
.
Following Knorr (19771, we define simple induction and restriction pairs as follows.
We say that
( e , f } is a szrnple induction pair if eVG is a completely
reducible FG-module for all irreducible FH-modules called a simple r e s t r i c t i o n p a i r , if all irreducible FG-modules
M
rephrased in terms of JVG) 2.2. PROPOSITION. (i1 (a) (e,fl
5 FG*S(FHI
(c1 fJ(FGle
5J(FH1FG
(ii)
(e,fl
eM.
fV.
The pair
(e,f)
is
is a completely reducible FH-module for
Not surprisinq, the above definitions may be
and JVH), as shown below
.
The following conditions are equivalent:
is a simple induction pair
eJCFG1f
(b1
=
fMH
V =
is a simple restriction pair if and only if
337
SIMPLE INDUCTION AND RESTRICTION PAIRS
(iii) Let Z? C F Then
FH
(e,f)
be the smallest field containing the coefflcients of e
is a simple induction (respectively, restriction) pair for FG
if and only if it is for EG
f.
and
and EH.
li) Put V = FHf/J(FH)f.
Proof.
Because V = f V
and contains a copy of every irreducible FH-module S
(e,f)
and
efl
is a simple induction pair if and only if
=
is completely reducible fS,
it follows that
is completely reducible.
w i n g to Proposition 2.4.3(i),
f
FGf/FG*J(FH)f
and so, by Lemma 2.1,
e 8 Thus efl
eFGf/eFG*JCFH)f
is completely reducible if and only if d ( F G ) f c - eFGJ(FH)f.
Since
- FG*J(FH), the equivalence of the latter containment is equivalent to eJ(FG)f c (a) and (bl follows. Assume that eJ(FG1f
An
5. FG'JCFH) .
Then
application of Theorem 1.2Cii) shows that
Thus fJ[FG)e (iil
5J(FH)FG.
Put M = FGe/J(FG)e.
The converse follows by symmetry. Since M = eM
is completely reducible and contains
a copy of every irreducible FG-module W = eW, simple restriction pair if and only if fMH
it follows that
(e,f)
is completely reducible.
is a Taking
into account that, by Lemma 2.1,
fu 2 fFGe/fJlFG) e w e see that
fMH
is completely reducible if and only i f
as FH-modules
CHAPTER 5
338
Since the latter containment is equivalent to e J ( F H ) f C_ J t F G I , (ii) follows (iii) Direct consequence of (il, (iil and Corollary. 3.1.18. 2.3. COROLLARY. (il
The following conditions are equivalent:
V,
For every irreducible FH-module
8
is completely reducible
(iil JVGl C_ FG*J(FHl (iiil
.
J(PG1 C_ JCFHlFG
Proof. Apply Proposition 2.2(il for the case e = f = 1. 2.4. COROLLARY. The following conditions are equivalent: (i) For every irreducible FG-module M,MH
JWHl
Ciil
is completely reducible
5J(FGl
Proof. Apply Proposition 2.2Ciii for the case e It is obvious that
f
= 1.
( e , f ) is a simple induction (restriction) pair if e f = 0 .
To investigate the case ef # 0 , 2.5. LEMMA.
=
we need the following preliminary observation.
The following conditions are equivalent:
ef # 0
(il
(ii) There exists a projective indecomposable FH-module
V =
fv
such that
evC # 0. (iiil
There is an FH-module
v = fv
such that e 8 # 0
(ivl There is a projective indecomposable FG-module P = eP
(v) There is an FG-module M Proof.
ti)
=)
(iil :
= eM
such that
such that ?j
# 0.
fM # 0.
Owing to Proposition 2.4.3 (il , we have
e(FHflG
1
eFGf = FGef # 0
Therefore there exists at least one indecomposable component V of
FHf
such
that e f i # 0. (iil
* (iiil: Trivial
(iii) * Civ) : Let M be an irreducible FG-module in Soc(e8l be the projective cover of M. that
Since V = fV,
and let P = P ( M
it follows from Theorem 2.4.9ci.i)
SIMPLE INDUCTION AND RESTRICTION PAIRS
0 # Hom(P,fll
1
Hom(PH,Vl
FH
FG
339
1 Hom(fspH,~l
FH
and thus J@E' # 0 . (ivl
Cvl:
=*
Trivial
(v) =* (i): By hypothesis, 0 # fM = feM = efM and f
If e
are block idempotents, then
groups of corresponding blocks;
and so
ef # 0 .
6 ( e ) and 6Cf)
the vertex of a module M
denote the defect
vx(M).
is denoted by
We are now ready to prove the main result of this section. 2.6. THEOREM.
Let e
and f be block idempotents of FG
and FH, respectively,
and assume that ef # 0 .
( e , g ) is a simple induction pair, then fMH # 0 for all FG-modules
(i) If
0# M
=
eM; moreover, 6 ( f ) 2G & ( e l . n
(ii) If 0
# V
=
( e , f ) is a simple restriction pair, then ev" # 0
fv;
proof.
moreover, d tfl gG 6 (el (i) Let V = eV
wish to show that fpH = 0 contrary.
P of V.
it follows from Theorem 2.4.9tii)
0 # Horn(P
,W1 FH H
ekF
fV H
be an irreducible FG-module such that
=
0.
We
Assume the that there exists
W = fw such that
an irreductble FH-module
But
.
for the projective cover
Since eP = P,
for all FH-modules
HomCP,ehFI
1 Hom(p,#)
.FG
FG
is completely reducible by hypothesis, so 0 # Hom(V,hFl
FG
HomCVH,W) = HomCfVH,W) = 0 , FH FH
a contradiction.
Now assume that eM
fVH
=
=
M # 0 and fMH
=
0
0 for any composition factor V of M.
for some FG-module M . Since e
Then
is a block idempotent,
all indecomposable projective modules in the corresponding block are linked (Proposition 1.10.14).
Hence by repeated use of the above argument, fpH = 0
for all projective modules P = e P ,
contradicting Lemma 2.5(iv).
The first
assertion is therefore established. By Theorem 3.16.14, there exists an irreducible FGmodule
V
=
eV
such that
340
CWTER 5
Then, by t h e foregoing, Soc(fvHJ.
Since
fVH # 0
eV = V,
and so t h e r e
FH
Homt8,Y) FG
Hom(e#,V) FG
e 8 i s completely reducible by hypothesis, and
8
so
V
in
i t follows from Theorem 2 . 4 . 9 ( i ) t h a t
0 # Hom(W,VH) But
W
an I r r e d u c i b l e FH-module
its
#.
is a d i r e c t summand of
6 3 8
i s a d i r e c t summand of
Therefore, by Theorem 3.16.14,
gS(f)
61e) = v.z(V) GGuv3:(W) G
(iil
The f i r s t statement i s an easy adaptation of t h e i d e a s i n t h e proof o f ( i l .
To prove t h e a s s e r t i o n on t h e d e f e c t groupsr choose an i r r e d u c i b l e FH-module
W
fw
=
such t h a t =
V.z(kr1
V
Let
Since
=
fVH
W
fVH
i s completely reducible and
W
VH.
i s a d i r e c t summand of
i s a d i r e c t summand of
hence
aCf1
eY be an i r r e d u c i b l e FG-module i n
deduce t h a t
V
H
that
Then
is a d i r e c t summand of
A
Let
G
[(VA’A1 .1,
W is
VH,
be t h e v e r t e x of
tV,l G , so V is a d i r e c t summand of tl
i s a d i r e c t sumnand of
s i t i o n (Theorem 2.6.11
Soc(ebf).
[(V,)
we
V. G
1,
Then and
I t follows from Mackey‘s Decompo-
a d i r e c t summand of
f o r some
g E G
and thus
a s required. The following a s s e r t i o n s a r e d i r e c t consequences of t h e above proof.
2.7.
REMARK.
(il
Assume t h e c o n d i t i o n s of Theorem 2.6(11.
Then
Soc ( f V H ) a f v H / J (FH) fVH f o r a l l i r r e d u c i b l e FG-modules
(iil
V = eV.
Assume t h e conditions of Theorem 2.61111.
Then
SXMPLE INDUCTION AND RESTRICTION PAIRS
f o r a l l i r r e d u c i b l e FH-modules
2.8.
REMARK.
and
FH,
e
Write
=
Ce
respectively.
f
and
i
'
fW.
=
=
zfj
a s a sum of block idempotents of
(e,fl
Then, by t h e d e f i n i t i o n ,
(induction1 p a i r i f and only i f
341
FG
i s a simple r e s t r i c t i o n
a r e simple r e s t r i c t i o n (induction) p a i r s
Cei,fj)
i and j .
for all
W e c l o s e by providing some a p p l i c a t i o n s of Theorem 2.6.
The following pre-
liminary observation w i l l be very u s e f u l .
2.9.
LEMMA.
H C S be subgroups of
Let
G
and l e t
W
be an i r r e d u c i b l e
FS-
module.
( e , f ] i s a simple induction p a i r and
(i) I f
flH
# 0,
then
i s completely
elf
reducible.
(iil
(e,f)
If
i s a simple r e s t r i c t i o n p a i r and
e8
# 0, then
fwH
is
completely r e d u c i b l e . Proof.
'L
module
fWH #
( i ) By hypothesis,
in
Soc(fwHl.
Then
fV
0, so w e may choose an i r r e d u c i b l e FHand
=
0 # Hom(V,WH)
Hom(#,U) FS
FH Because
W
i s i r r e d u c i b l e , t h e r e i s an e x a c t sequence
8s-w-0 Therefore t h e r e e x i s t s an e x a c t sequence I
Since
e#
(e,f)
I
i s a simple induction p a i r ,
e P
i s completely r e d u c i b l e .
is a l s o completely reducible.
(ii) The proof i s analogous and t h e r e f o r e w i l l be omitted. We a r e now ready
2.10.
COROLLARY.
idempotents of p o t e n t s of
.
to provide a number of consequences of Theorem 2.6.
Let
FG
Then
H
and
FS such that
5s
FH, Eif
be subgroups of respectively.
# 0.
Put
G
{&.I
Let E
= ZE
and l e t
i
e and f
be c e n t r a l
denote a l l block idem-
and assume t h a t
(E,f)
is a
CHAPTER 5
342
simple induction pair.
(e,f)
(i)
Then the following conditions are equAvalent:
is a simple induction pair
l e , ~ ] is a simple induction pair.
(iil
Proof.
(il
* (iil : Let W
fWH #
and Theorem 2.6(i), e#
by Lemma 2.9(i),
0
= EW
since
be an irreducible FS-module.
By Remark 2.8
is a simple induction pair.
(E,f)
is completely reducible which means that
Hence, is a
(e,E)
simple induction pair. (iil
* (i) : Let V
=
fV be an irreducible FH-module.
are completely reducible, since Invoking Lemma 2.5,
e8
Hence
and
(E,f)
5s
Let H
be subgroups of
potents of FS such that Eie # 0. simple restriction pair.
(iil
Let N
FN and let B
Let V ,V , - . . , V 2
t
denote all block idem-
C E ~ and assume that
Ce,El
is a
.
i
=
be a normal subgroup of
let b = b(f1
G,
5 G m , be the blocks o f
B.(ei), 1
z
.
Vi, 1
be a
FG covering b . and let H
i
i 4 t.
..
m 4 mini (H :Nl , (H :Nl ,. ,( H t : N ) 1 If m = [ G ( f ) :NI
(iil
.
where
G(fl
is the inertia group of f ,
J(FG)f C FG*J(FN)
Proof. Since
N4
tion pair. 1
E =
Z
be all nonisomorphic irreducible FN-modules in b ,
be the inertia group of ti1
Let f E . 1
The proof is analogous and therefore will be ommitted.
2.12, PROPOSITION.
E?.
G and let e and f be central
is a simple restriction pair.
Proof.
block of
fl =
that
Then the following conditions are equivalent:
is a simple restriction pair
(E,fl
1
Put
E
k , f ) is a simple induction pair.
idempotents of FG and FH, respectively.
(i) (e,fl
are simple induction pairs.
it follows from the construction of
is completely reducible and so
2.11. COROLLARY.
(e,E)
and e (E?IG
Then E f l
< i < m.
(i) Let G,
and
then
6 ( e .) = 6 ( f l
z G
V be an irreducible module in b
and
H its inertia group.
it follows from Clifford's theorem that (1,l) is a simple restricTherefore, by Remark 2.8, Also,
by Lemma 3.10.2(iii),
(ei,fl
is a simple restriction pair,
eif # 0
for all i E
11,...,m}.
APPLICATIONS
343
# 0, 1
Invoking Theorem 2.6(fi), we conclude that tV
i =z e '
R.2 > 0 be the composition length of Vi and M
Let
ny.
its composition factors.
ij
Because.
(v5N v.= SET where T is a transversal for N
in G, it follows from Clifford's theorem that
did4 = Pij(G:H(dimV F ij F for some integer P..
23
3
0.
Therefore
I G:NI dimY = dim$ F
m
=
F
rn =
c dimMi i=1 F
'i
C C r..IG:HldimV i=1 j=1 ' 3 F
and
as required-.
ZH,
(ii) Since G ( f )
rn and thus
=
it follows that
IGC~):N(
I H : N I = rn.
3 JH:N
By C*) this implies k .
z=
is completely reducible. by Proposition 2 . 2 ,
Hence
(1,f)
1 for all i, so
8
= @
ei?
is a simple induction pair and therefore,
J(FGlfZFG*JL$"I.
The assertion on the defect groups
follows from Theorem 2 . 6 and Remark 2 . 8 .
3 . APPLICATIONS
Throughout this section, F
denotes an arbitrary field of characteristic p > 0 ,
H a subgroup of a finite group G and e and F H ,
respectively.
(1977).
central idempotents of
All the results presented are contained
Some particular cases of these results may be found
in Motose and Ninomiya (1975b) and Khatri (1973).
3.1. PROPOSITION.
FG
Our aim is to tie together the concepts of induction,
restriction and the Jacobson radical. in a work of Kn&r
and f
The following conditions are equivalent:
CHAPTER 5
344
(efllH is completely reducible for all irreducible FH-modules V
(il
(ii) eJ[FH)FG (iiil
5 FG*J(FH)
eJ(FH)FG = FG*J(FHle
Moreover, if these conditions are satisfied, then
e~ WHI n~~ Proof.
(il
* (ii): Put W
(@@IH
so by assumption
=
for all n
= (eJ(PHI*FC)
FH/J(FH).
Then K
is completely reducible.
1
is completely reducible,
Therefore eJ(FH)
5 ann(8).
But
G
ann (W 1 = 1 d V G - J @HI 1 by Theorem 1.2 (i), so eJ IFH)FG
5 FG *J (FH].
(ii) * (iiil: It is clear from (iil that
and therefore, by Theorem 1.2(iilI
- eJ(FH)FG. It follows that eFG*J(FH) c
(iil
The opposite containment is obvious from
.
(iii) =$ (il: it is obvious from (iii) that FG-JCFHIe is an ideal o f
V be an irreducible FH-module and let A = ann(V1. FG-J(FHle
5 FG*A.
(efl]
= 0
and therefore
assertion follows by induction on n , 3 . 2 . PROPOSITION.
(il (ii)
5A
5 Id(FG-A) =
and
(efl)H
am(@)
is completely reducible.
The final
using (iii).
The following conditions are equivalent:
WBlGis completely reducible for all irreducible FG-modules
f
Let
Invoking Theorem 1.2Li1, we deduce that J(FH)e C FG*J(FH)e
Thus J(FH1
Then J ( F H )
FG.
M
is completely reducible for all irreducible FH-modules V =
fMH is completely reducible for all irreducible FG-modules M.
fv,
and
APPLICATIONS
Wil
345
JVGIf = F C - d ( F H I f
(iv) f J r P G ) = f J ( F H ) F G (1,f) is a simple induction
Proof. Condition (ii] may be reformulated as:
The equivalence of (ii), (iii), and (iv) is
pair and a simple restriction pair.
therefore a consequence of Proposition 2 . 2 .
Since (ii) obviously implies (i), we
are left to verify that (i) implies Cii). Let M be an irreducible FG-module.
Then, by hypothesis,
is
f l H is completely reducible.
completely reducible and so, by Corollary 2.4.5, Let V =
CfMH)G
fV be an irreducible FH-module and let M be an irreducible FG-
module in Soc
(fl). Then
Hence there is an exact sequence fMH
-V
0
which gives rise to an exact sequence
(pH) 4 vG Because
(fMHIG
is completely reducible, VG
+
0
is also completely reducible.
So
the proposition is verified. To prove our next result, we need two preliminary observations. 3 . 3 . LEMMA.
(il (iil
The following conditions are equivalent:
M
For any FC-module
FGJ(FH1FG Proof.
(i)
=
eM,
if MH
M.
is completely reducible, then so is
2 J(FG)e * (iil
completely reducible.
:
Hence, by assumption, M
J CFG) e C - FGJ CFHl FGe
5 FGJ (FH) FG.
(ii) * (1): Let
=
M
Then M = eM
Put M = F G e / F G J ( F H ) F G e .
2 JCFGIe,
MH is
is completely reducible.
eM be an FG-module such that' MH
Then JCFII1M = 0 and if FGJ(FH)FG
and
then
Thus
is completely reducible.
CHAPTER 5
346
Thus M
is completely reducible. Let M
3.4. LEMMA.
Proof.
8
be an FG-module such that M/J(FG)M
Then Y = M/(X+J(FG)M)
Put X = FGJ(FH)M.
is H-projective.
is a direct summand of
the completely reducible and H-projective module M/J(FG)M, reducible and H-projective.
sequence.
But
is completely
( I Y / X ) ~ is completely reducible.
Y is H-projective, so
(*)
Thus
(*)
splits as an FG-
Hence
and multiplying both sides by J V G )
since Y
so Y
Consider the exact sequence
- X, it follows that Since JIFH)M c splits as an FH-sequence.
Then
is completely reducible.
and therefore J(FG)M
5 X,
as
gives
Thus
asserted.
.
We are now ready to prove (i) If J CFGl e
3.5. PROPOSITION.
M
=
eM
5 FG * J CFH),
are H-projective.
(if) If all irreducible FG-modules ely reducible implies Proof.
(il
Write e = Ce
i
(e 11
then all irreducible FG-modules
M
=
(e,U
FG.
= eM
be an irreducible FGmodule.
Theorem 3.16.14,
Then
M is dIei)-projective.
virtue o f Lemma 2.5.8 (iil
.
M
eV.
Then, by Remark 2.8, each
=
eiM
6(ei)
H. Let G for some i and hence, by
Hence, by Theorem 2.6(i),
i’
M
=
is a simple induction pair.
a sum of block idempotents of
is a simple induction pair.
VH complet-
are H-projective, then
V completely reducible for any FG-module V
By Proposition 2.2(i), as
eM
Therefore, M
is H-projective, by
APPLICATIONS
(ii] Assume that all irreductble FG-modules 3.3(ii), we need only verify that FGJ(FH)FG Lemma 2.5.8(il, FGe/J(FG)e
M
347
= eM
are H-projective.
2 J(FG)e.
is &projective.
By Lemma
By hypothesis and by
Therefore, by Lemma 3.4,
as required. 3.6. LEMMA. V,
If
(eFIH
then J ( F H l e
is completely reducible for all irreducible FH-modules
J(FG). n
Proof. Assume that for every irreducible FH-module 8, ( c f ) H is completely Then, by Proposition 3.1, eJ(FH1FG
reducible.
is a nilpotent ideal of FG.
Therefore
as required. 3.7. THEOREM. (il
J(FG)e
(ii) J(FGle
=
The following conditions are equivalent: FG*J(FHle
=
J(FH1FGe
(iiil efl is completely reducible for any irreducible FH-module V completely reducible for any irreducible W-module (ivl efl and (v) (e#lH
(e'/;IH
eM
is completely reducible for any irreducible FH-module V
M
=
and all
eM are H-projective
is completely reducible for any irreducible FH-module V
completely reducible inplies M Proof.
=
is
(efllH are completely reducible for all irreducible FH-modules V
irreducible FG-modules (vi)
M
and MH
and
MH
completely reducible for any FG-module M = eM.
Condition (iii) may be reformulated as follows: (e,l) is a simple
induction and a simple restriction pair.
Hence, by Proposition 2.2, (iii) iS
equivalent to either of the following conditions:
But (11 is equivalent to (i) and (2) is equivalent to (ii), hence (i), (ii), and (iii) are equivalent.
CHAPTER 5
348
The implications (ivl the implication (ivl
=*
=*
(v) * (vil follow by Propositions 3.5 and 2.2,
while
(iiil is a consequence of Lemma 3.6 and Proposition 2.2.
We are therefore left to verify that (vi) implies (iv) and (ii) implies (ivl. (vil
FGJ(.FH)e = J"f)FGe
(ivl: By Proposition 3.1,
FGJ FH)FG
and, by Lemma 3.3,
2 J (FGIe
It follows that
e?
and hence, by Proposition 2.2,
is completely reducible for any irreducible
FH-module V. Direct consequence of implications (ii) * (iii], (ii) * (i) and
(ii) * (iv):
Proposition 3.1.
9
For any subgroup H
of G,
the nomai? closure of H
-
the intersection of all normal subgroups of 3 . 8 . LEMMA.
Let
FG-I(QI*FG
On
G containing H.
Then
By Sylow's theorem, we may assume that Q
Then I($) Z I C P )
is defined as
P be a p-subgroup of G, let N be its normal closure in G
and let Q be a Sylow p-subgroup of N .
proof.
in G
=
3P
since
FG-I(8) FG
for all g E G
and one inclusion follows.
the other hand, put
V Given z E P,g E G we have
=
FG)FG*I(P) *FG
b-llg E IIPlFG ccg
5 FG*I(Pl*FG, so
g(mod FG*T(P)*FG)
5 KerV
5 KerV.
and hence
P
FG.ICQ1 V
0 and therefore FG.I(Q) *FG 5 FG-I(P)FG , as required.
=
3.9. LEMMA. Then XN
KerV.
Let
Thus N
and Q
It follows that 9
X be an indecomposable N-projective FG-module with N
is a direct sum o f G-conjugate indecomposable FN-modules.
4 G.
APPLICATIONS
Because X
Proof.
U
such that X
is N-projective, there exists an indecomposable FN-module
is a direct summand of
where T is a transversal for
UG.
N in G.
As
The
modules which are conjugate under the action of summand of
&IN,
in G.
t 8 U are indecomposable FN-
.-
Since XN is a direct
G.
be a p-subgroup of
G
and let N be its normal closure
Then the following conditions are equivalent:
(i) For any FG-module (ii) If
N 4 G, we have
the result follows.
Let P
3.10. THEOREM.
349
V = eV
V = eV,
V p is completely reducible, then so is
is an N-projective FG-module such that
posable and
FG-I(P1V c V ,
(iiil If M
=
eM
if
Y/J ( X I V
then
V/J(FN)V
V.
is indecom-
is N-projective
is an irreducible FG-module such that P
5 KerM,
then M
is
N-projective. Proof.
(il * (iil: Assume that
v
is as in (ii).
By Lemma 3.16.15, it
suffices to show that
JUG1 V The inclusion JCFNI V C_ J(FG1 V
=
JCFNI V
being obvious, put
W
Then
WN is
W is N-projective.
Since
= V/J(FN) V .
completely reducible, say
WN where the Si are irreducible.
=
s1 @ s2 CB
... @ sr
By Lemma 3.16.13,
W is indecomposable by assumption, it follows from Lemma
si = gi
3.9 that
8s
(1
for some gi E G and some irreducible FN-module S .
Now let Q be a Sylow p-subgroup of N Then
and choose n i E N
such that
i
r)
CHAPTER 5
350
0.
It will next be shown that I ( & ) S =
for all i
Then I(&)si# 0
Assume the contrary.
and therefore FNI(&)Si = Si, 1
i
F.
It follows that
W ZI FG*T(Q)*FGGW FG*I(&)W
=
F
and so W = FG-I(&)FGW.
by definition of
W.
By Lemma 3.8 then, FG-I(P)FGW =
Since
W and therefore
s(FN)v 5 J(FG)V, it follows from Proposition 1.6.9
that
FG-ILF') V
=
FG*Z(P) *FGV = V
contradicting the hiypothesis. Thus I ( & ) S= 0 I(Q)W
=
0
for all i E {l ,...,PI.
and so I(&)si= 0
Therefore W
and hence I(P)W = 0 .
P
is completely reducible, so W
Hence J ( F G )W = 0
is completely reducible by hypothesis.
This implies
and J V G )V
5J ( F N V ,
as required. (ii) * (iii): Let M be as in (iii) and L eL = L
is N-projective and L/J(FFJIL
since P
5 KerM.
reducible.
Assume
If
i s indecomposable.
Then
Furthermore, I(P)M=O
Hence I(P1.L C - J(FG)L and therefore
By assumption then, M I L/J(FG)L (iii) * (i) :
a projective cover of M.
W
V = eV
is &projective.
is an FG-module such that V
P
= V/J(FG) V ,
is completely
then
W = M
@M 1
@...@Mn 2
for some n 3 1 and irreducible FG-modules
Mi'
Since P 5 KerV, we have
APPLICATIONS
P 5 KerMi for all
351
Hence, by hypothesis, each M
E {l,.
projective and therefore Q-projective, if again Q
N.
Thus W
denotes a Sylow p-subgroup of
is Q-projective and
But Q
by Lemma 3.4.
Thus V
i is N-
5 N 5 KerV,
so I ( Q V) = 0
is completely reducible, as required.
3.11. COROLLARY.
and, by the above, J(FG)V
=
0.
'
P be a p-subgroup of G and let N be its normal
Let
closure in G. (i) If all irreducible FG-modules M
=
eM
are N-projective, then
(ii) If FG*I(P)*FG 2 J(FG)e
=
eM
is an irreducible FG-module such that
P LKerM, then v z ( M
=
Q
M
(iii) If I ( P ) e c - J(FG) modules M = eM Proof.
M
and Syl(N)
E
and M
has trivial source.
P
and FG*I(p)*FG 2 J(FG) e ,
then all irreducible FG-
have the same vertices and trivial source.
(i) BY Lemma 3.3,
tion (i) of Theorem 3.10.
FG*I(P).FG 3 - I(FG)e is equivalent to the condi-
Since our assumption implies condition (iii) of
Theorem 3.10, the assertlon follows.
M is N-projective, hence &-projective, and clearly
(ii) By Theorem 3.1o(iii),
MQ is completely reducible. X
5 Q.
Now M
hence M
Setting X = V z ( M ) ,
(Mx)G and Mx
is a direct summand of
is a direct summand of
G
(lX)
,
where G [(lX) IQ.
Therefore M
is a direct summand of
and since
is a direct summand of MQ,
Q
lQ
1
since each
(1
)'
=
we may therefore assume that
(1
lX
is completely reducible,
is the trivial FX-module.
By Mackey's decomposition,
we conclude that
flw
for some g E G
IQ
is indecomposable by Theorem 3.4.24.
Thus
.@w= Q
and
Xg% X
=
Q, proving both assertions.
(iii) If I ( P ) e 5 J(FG),
then by Proposition 2.2 M p
is completely reducible
CHAPTER 5
352
M
for all irreducible FG-modules 3.12. LEMMA. module
V.
Proof.
=
eM and therefore
Then H
is of p'-index in G,
Let B
B(el
group P of G
be the principal block o f
is a defect group of B .
irreducible FG-module
V in B
VH.
irreducible submodule of
Hence
8.
Let W
be an
Hom(hF,Vl
Hom(ewG ,Vl
FG
FG
8
and
8
is completely reducible by assumption.
8.
V is a direct summand of
.
V is &projective and so H contains a conjugate of P.
is of p'-index in G.
3.13. LEMMA.
Then a Sylow p-sub-
is a vertex of
e 8 is completely reducible and therefore
It follows that H
of
FG.
Since eV = V, we have
FH e 8 is a submodule
Now applycii)..
By Theorem 3.16.14, we may choose an
such that P
0 # Hom(W,VH) But
5 KerM.
is completely reducible for any irreducible FH-
Assume that
=
P
Thus
The following conditions are equivalent:
(i) All irreducible FG-modules are H-projective
(ii)
H is of p'-index in G The implication (ii) * (il is a consequence of Corollary 2.5.5.
Proof.
Conversely, assume that (il holds.
Let B = B ( e ) be the principal block of FG.
By Theorem 3.16.14, we may choose an irreducible FG-module V P E Syl(GI
is a vertex of
P
contains a conjugate of
P.
V.
By hypothesis, Thus
V
in B
such that
is H-projective and so
H is of p'-index in G,
as required.
. H
We close this section by providing the following application of Theorem 3.7. 3.14. THEOREM.
Let H
be a subgroup of a finite group G
of characteristic p > 0.
and let F be a field
Then the following conditions are equivalent:
(i) J(FG) = FG*J(FH) (ii) J(FG1 = J(FHl*FG (iiil There exists a normal subgroup N
of
G such that N C_ H and N
has p'-
index in G (ivl
fi
is completely reducible for any irreducible FH-module V
completely reducible for any irreducible FG-module
M.
and MH
is
APPLICATIONS
353
fl and (f),are completely reducible for all irreducible FH-modules V (vil (fll, is completely reducible for a l l irreducible FH-modules V and all
(v)
irreducible FG-modules are H-projective (viil
G
(V
IH
is completely reducible for any irreducible FH-module V
M completely reducible for any FG-module M.
completely reducible implies
(6lBis completely reducible for all irreducible FH-modules
(viii)
and MH
V and H
G.
is of p'-index in
(ix) (MHIC is completely reducible €or any irreducible FG-module M. Proof.
Applying Proposition 3.2 for f = 1 and Theorem 3.7 for e = 1, we
see that the conditions (il ,(it),(iv), (vl ,lvi) , (viil , and (ix) are equivalent. Furthermore, by Lemma 3.13, (vil and lviiil are equivalent.
We are therefore
.
left to verify,that (iii) is equivalent to one of the conditions (il , (iv), (viii) For the sake of clarity, we divide the rest of the proof into two steps.
reducible.
Our aim is to prove that f o r all g J V H ) = FamJ(FL1
Put V (F),,
V, iflIH is completely
Assume that fpr m y irreducible FH-module
Step 1.
=
FH/tJ(FH1.
Then
by hypothesis.
V
where
and
S =
Setting k'
FL/A.
#
Proposition 2.4.3(1),
=
H' n gHil
to be the restriction of
8
5 FH-A
J(FH1
is completely reducible.
5 A.
But
FH n gJ(FH)g-I
is a nilpotent ideal of FL.
A and therefore
=
FH n gJ(FH1g-1
=
Put
Hence by Corollary 2.4.5,
C_ JVL) fl gJ(Fmg-'
to FL, it
and so by
FL n g(ann~1g-l= FH n ~ ( g ~ g -n~gJ(FH)g-' i =
since FH
v'
is completely reducible.
By Theorem l . Z ( i 1 ,
S is completely reducible and thus J(FL1 A
=
G,
i s a completely reducible FH-module, hence so is
fallows from Mackey's decomposition that A = ann(fl
L
E
Hence
J(FL)
354
CHAPTER 5
Replacing
g by
g-l,
it follows by t h e same argument t h a t
FH n
g - l J (FH)g = J (F (H n g-lHg) )
Since
J(FH)
Thus
1 J(FL)
S t e p 2. CompZetion
Of
argue by induction on
L = H (1 gHg-l
that
J(FH) = FH.J(FL), a s a s s e r t e d .
and t h e r e f o r e
t h e proof.
IHI
.
To prove (iii), we
Assume t h a t ( v i i i ) holds.
If
H 4G
then we a r e done with
i s a proper subgroup of
tl
N
=
g E G.
f o r some
H.
Suppose
Since (i)and
( v i i i ) are e q u i v a l e n t , it follows from Step 1 t h a t
Hence, by induction, t h e r e i s a normal subgroup
N
hasp’-index in
G.
Since
L
5H ,
F i n a l l y , assume t h a t (iii)holds. Since
(PIN i s a
N
G such t h a t N
of
Because
N
and
(iii) follows. Denote by
an i r r e d u c i b l e FH-module.
d i r e c t sum of FN-modules of t h e form
completely reducible.
5L
hasp’-index i n both
’Y,g E G,
(FIN i s
H
and
F
a f i e l d of character-
G,
it follows
from t h e equivalence of (i)and ( v i i i ) t h a t
Thus
proving ( i l and hence t h e r e s u l t .
4. p-RADICAL GROUPS
Throughout t h i s s e c t i o n ,
istic
p > 0.
If
G
denotes a f i n i t e group and
X i s a s u b s e t of
FG,
we write
r(X)
=
rG(X) and
p-RADICAL GROUPS
R(X1 = RG(X) As
for the right and left annihilators of
usual, if X is a subset of G, then 'X Let H
4.1. LEMMA.
of FH.
7
(i)
(ivl
Let
be a subgroup of G
fl = g - l I g
and let
?
=
FH,
X in F G ,
respectively.
denotes the sum of elements in X.
and let I
be a nilpotent left ideal
n FGsfl.
9EG
is a nilpotent ideal of FG
If I is"an ideal of
355
then
T=
nPgFG SEG
Proof.
and thus
Fix x
in G.
Then
is an ideal of FG.
It follows that
for any integer m 2 1.
Hence, by induction,
for any integer rn 2 0.
Since
potent ideal of
I is nilpotent, we deduce that
FG.
(ii) By Theorem 3.3.6(iii),
I = kH (21H (I11 which implies that FG.1 = LG(rH(I)FG)
since FG
is a free FH-module.
Hence, if q E G ,
then
i
is a nil-
CHAPTER 5
356
a s asserted.
tiii)
Applying (ii) and Theorem 3.3.6
( i i i ) ,we have
a s desired. (iv)
FG
Owing t o Theorem 3.3.6(vI, ( o r FHI
coincide.
Let
Proof.
(Cx
99
+ )H
=
(il
= LH(H
H
be a subgroup o f
G.
Then
The second e q u a l i t y i s a consequence of t h e f a c t t h a t
Z I = 0 f o r a l l g E G. Since h - 1 E RH(H') h E H gh + I U f l 5 R H ( H 1. But I(H) i s of codimension 1, hence
0 i f and only i f
it follows t h a t
I(H1
Invoking (ii), w e t h e r e f o r e d e r i v e
.
a s required.
4.2. LEMMA.
t h e l e f t and r i g h t a n n i h i l a t o r s of an ideal i n
+1 .
Since
is a f r e e FEi-module, w e conclude t h a t
FG
R (H+) = FGR (H+) = F G * I ( H ) , G H as required. (ii) The proof i s s i m i l a r and t h e r e f o r e w i l l be omitted.
4.3.
(il
LEMMA.
Let
n FG*I(#) SEG
P =
be a Sylow p-subgroup of
n I($)FG SfG
G.
i s a n i l p o t e n t i d e a l of
FG
.
p-RADICAL GROUPS
=
Proof.
&G
uzzl
C
a
fiSy"
= 0 for all
357
z E G
and all S E Syl ( G I )
P
(i) Since I(P) is a nilpotent ideal of
FP, the required assertion
is a consequence of Lemma 4.1. (ii) put X
=
fl
FG.I(fl)
and Y =
SEG
X
n I(@)FG. SEG
Then, by (i),
nFG*I(S)= niT(S)FG
= Y =
S s y l (GI
P
E S y l (GI
P
and so the desired conclusion follows by virtue of Lemma 4.2. Following Motose and Ninomiya (1975b), we say that G subgroup H
of p'-index in G I J ( F G 1 c J(FHIFG.
J(FG) C - J(FH)FG
Note that, by Corollary 2 . 3 ,
is equivalent to J ( F G ) C_ F G . J ( F H 1 ,
Corollary 2.3, if J ( F G ) C J I F H ) F G ,
then H
is p-radicai! if for any
and that, by Lemma 3.12, and
is of pf-index.
4.4. THEOREM. (Motose and Ninomiya (1975b)).
The following conditions are
equivalent :
G is p-radical
(il
(ii) S ( F G )
FG-I(P1
for some (and hence all) P E Syl ( G )
P
r (viii) J ( F G ) = { C z @G
Proof.
I
C z
6 s sg
=
11
nd all S E Syl ( G ) }
g G G
0 for all g
E
P
G
and all S E Syl ( 0 1
P
The equivalence of (iii), (iv), (vii) and (viii) follows from Lemma
4.3. (i) * (ii): J(FG)
If P E Syl (GI then P is of p'-index, so by hypothesis
P
5F G - I ( P ) .
(ii) * (iii): If J ( F G )
5 FG*I(P)
J ( F G ) = J(FG)'
for some P E Syl ( G ) I
P
5FG*I(P)g =
FG*I(Pg1
then for all g E G
CHAPTER 5
358
and hence J(FG1
5
n F G * I (S) E S y l (GI
P
Thus,
=
n F G * I ( S ) , by virtue of Lemma 4.3(1). F s y l (GI
P
H be a subgroup of G of p'-index.
(iii) =* [i): Let
S E syl (G) such that S C_ Ii.
P
By hypothesis, J l F G )
Then there exists
5F G * I ( S ) ,
completely reducible for every irreducible FS-module V ,
fl S($lG,
Since
it follows from Corollary 2.4.5 that
reducible, for every irreducible FS-module J(FH)
5 FH*I(S)
V.
so
rf;
is
by Corollary 2.3.
f
is completely
Hence, by Corollary 2.3,
and therefore, by the implication (ti)
=*
liii)
,
Thus
as required. (iv) * (v1 and (iff)
* (vil : Each of the statements follows from the other by
taking annihilators. As
an application of Theorem 4.4, we derive the following two results.
4.5. COROLLARY.
Let H
be a subgroup of
G
of p'-index.
If G
is p-radical,
then so is H. Proof.
Let S be a Sylow p-subgroup of G with
of the implication (iii)
=*
Hence, by Theorem 4.4, if 4.6. COROLLARY.
Let P
trivial FP-module. reducible.
s 5 H.
Then, by the proof
(1) of Theorem 4.4,
G
is p-radical, then so is H.
be a Sylow p-subgroup of
Then G
G
and let lp
is p-radical if and only if
(I
(Ip)
be the
is completely
p-RADICAL GROUPS
Proof.
If
V
=
FP/I(P),
V
then
l p and, by Proposition 2 . 4 . 3 ( i ) ,
fi 8
Hence
359
FG/FG*I(P) J(FG) c FG'I(P).
i s completely r e d u c i b l e i f and only i f
l a t t e r , by Theorem 4.4, i s equivalent t o
G
being p - r a d i c a l ,
Since t h e
the r e s u l t
a
follows.
The next r e s u l t , with t h e exception of t h e f i r s t property of ( i ) ,i s due t o Khatri (1973). 4.7.
THEOREM.
(il
If
G
N
then so a r e
G/N
i s a p-group and
G/N
(iii) I f
G.
be a normal subgroup of
i s p-radical,
N
(ii) I f
N
Let
Is a p'-group,
G/N,
and
is p-radical,
then
then
G
i s p-radical
G i s p-radical i f and only i f
N is p -
radical. (i) Assume t h a t
Proof.
choose a Sylow p - s u k ~ r o u p Q
5 J(FG1
J(FN1
But
i s p - r a d i c a l and l e t
N
of
with
Q
5P.
P E Syl (G). P
W e may
By P r o p o s i t i o n 3 . 1 . 8 ( i ) ,
and so, by Theorem 4.4,
xQ+ and x ( P - Q ) +
Lemma 4 . 2 ( i ) ,
G
have d i s j o i n t s u p p o r t s , so
z E FN-TIQ).
Thus
S ( F N ) C_ F N ' I C Q )
xQ+ = 0
and t h e r e f o r e , by
and so, by Theorem 4.4,
N
i s p-radical. If
H
=
PN, then H
i s of p'-index J(FG)
Thus, by Corollary 2.3, modules
V.
fl
G,
so by hypothesis
5 FG*J(FHI
i s completely r e d u c i b l e f o r a l l i r r e d u c i b l e FH-
In particular,
i s completely reducible.
in
But
(lH1
H/N
i s completely reducible. i s a Sylow p-subgroup of
G/N
Hence
G/N,
(~H/N) so
G/N
is
p - r a d i c a l by Corollary 4.6.
(ii) Assume t h a t
N
i s a normal p-subgroup of
G such t h a t
G/N
i s p-radical.
360
CHAPTER 5
Since N f P, we have H/N = P/N.
By Corollary 4.6, (lp,NIG'N
is completely
n
reducible.
Hence
(lplu
is completely reducible and therefore, by Corollary
G is p-radical.
4.6,
(iiil Assume that G/N Corollary 4.5.
If G
is a p'-group.
Conversely, suppose that
N is p-radical.
group, there is a P E Syl (G) with P C_ N.
5 FN'ItP).
so
is PI, by
Since G/N
is a p ' -
Owing to Theorem 4.4,
P
J(FN1
is p-radical, then
On the other hand, by Proposition 3.1.8, J ( F G ) = FG'JCFN).
Thus
and so
.
G is p-radical, by virtue of Theorem 4.4.
Turning our attention to Frobenius groups, we now prove the following result. 4.8. THEOREM. UChatri (19731,Motose and Ninomiya (1975bI). Frobenius group with kernel N (il If p (ii) If
divides the order o f
Let G
be a
H.
and complement
N, then G is p-radical
p divides the order of H I then G is p-radical if and only if
so
is
H. Proof.
(i) By Lemma 3.7.1 iv), N
Theozem 4 . 7 ( i ) ,
i s
nilpotent and so
But, by Lemma 3.7.1(iiiI, G,"
2
N is p-radical, by
H is apLgroup.
Hence G
is
.
p-radical, by Theorem 4.7 (ii)
(id.) By Theorem 4.7(il, we need only verify that if 8
G.
If P
is a Sylow p-subgroup of H I
is p-radical, then so is
then J(FH) f F H ' I ( P )
by Theorem 4.4.
By Theorem 3.7.7,
and therefore
JIFG) Sinae
P E Syl ( G )
P
C_ FG*I(P)
we conclude, from Theorem 4.4, that G
4.9. THEOREM. Uhatri (1973)I . if so are
5 FH-I(P1 (N'I
G I and G
If G = G
X 1
G
then G 2
is p-radical. is p-radical if and only
p-RADICAL GROUPS
Proof.
GI
Assume t h a t
G
and
Pi E Syl ( G . 1 , i = 1,2.
P
Then, by Theorem 4.4,
a r e p-radical.
5FGi*I(Pi)
J(FGi) where
361
Since, by Theorem 3.2.3,
2
we have
!thus
G
The converse being a consequence of
is p - r a d i c a l , by Theorem 4.4,
Theorem 4.7(il, the r e s u l t follows. 4.10.
LEMMA.
E
r e s p e c t to Proof.
E
Let
If
J(EG1 C - EG*I(PI
P
(,NI, then by Theorem
E Syl
P
JCEGJ
5E G * I ( P I .
by Corollary 3.1.18.
J(FG) C - E G * I ( P 1 n FG
=
FG*I(P)
N.
p-complement
Then
G
5 F G g I ( P ) , then that J ( E G 1 5 EG.I(P).
So t h e Lemma i s true.
by Lemma 4.2(11.
Let
P E
G be a p - n i l p o t e n t group with a b e l i a n
Let
i s p-radical.
Syl ( G )
Since
F
e2JG
(eiFNIG
F
is algebra-
and Let
P
l = e be a decomposition of
.
groups.
Owing t o Lemma 4 . 1 0 , we may harmlessly assume t h a t
i c a l l y closed,
F.
J(FG)
If
Conversely, assume
'l"heorent. ( T s h u s b a (1986)).
Proof.
i s p - r a d i c a l with
it s u f f i c e s t o v e r i f y t h a t
4.4,
The next r e s u l t provides another c l a s s of p-radt-1
4.11.
G
Then
i f and only i f it is p - r a d i c a l w i t h r e s p e c t t o
JVGI C - F G - I ( P I i f and only i f
Then
F.
be a f i e l d extension of
1
+ e
2
+...+
e n
1 a s a sum of orthogonal p r i m i t i v e idempotents of
is a s p l i t t i n g f i e l d foc i s of dimension
IPI
E",
.
each Now
e.FN 2
G = NP,
theref ore
e.PG = (eiFFJ1 (FP) = eiFP 2-
i s one-dimensional!. so
FG = (FN)( F P )
FN. Hence and
CHAPTER 5
362
It follows that the map
I
FP-
eiFG
x I+ eix
is a surjective homomorphism of right FP-modules of the same F-dimension.
FP 2: eiFG
as FP-modules
and so ezZG*T(P) i s a unique maximal FP-submodule of ePG. s
proper FP-submodule of eiFG,
Thus
(1 G
i
G ?z)
But eiJ(FG) is
hence
eiFG*r(P) 3 J(FG1. and therefore FG*I(P) -
2 eiJ (FG)
i G n)
The desired conclusion now follows by virtue of
Theorem 4.4. The result above need not be true f o r an arbitrary p-nilpotent group
c.
The
following example is due to Saksonov (1971) (see also Motose and Ninomiya (1975b)). Let p = 3
and let F be an arbitrary field of characteristic 3 .
Let G
be
the group of order 24 defined by
G =
= 1,
z z ~ - ~ = y , z y =z -q~>
be the class sum of the conjugacy class containing g E G. c
=
X
Then we have
x+x3+ y +q+x2y +x3y
+ zzZy + .ZX~+ Z&
=
c 3 = l - c
X
c 2
x
= c
x
and therefore
Hence c
x
+ cz - 1
=
ac+x3+y+xy+x2y+x~y+z+zx2y+~x3+zx3y - 1
i s a central nilpotent element of
cx
+ cz
-
FG and so cx + cz - 1 E J(FG1.
However,
1 does not satisfy condition (viii) of Theorem 4.4, since the sum of
the coefficients of c
X
+
c
Z
-
1 over the coset {x,zx,Z2x} of
is
1 # 0.
P-RADICAL GROUPS
363
G is not p-radical.
Thus
Remark.
According to Tshushima 119861, a p-nilpotent group
G is p-radical if
and only if
10pr(G1 ,Dl n Co
10) = 1
P for any p-subgroup D of G. Let
U and V be any FG-modules.
V, written i ( U , V ) ,
and
Recall the intertwinning nwnber for U
is defined to be
i(U,V) = dim Hom(u,v)
F FG Since
we have
for any FG-modules U,V and
W.
Similarly
and therefore
r
i( @ un, n=l
k
r
@ Vm) =
c
k
c
i(Un,Vm’
n = l m=l
m=l
For the rest of this section, E denotes the algebraic closure of F and
.
{ V l ;. .,Vr 1 a full set of nonisomorphic irreducible EG-modules. module
Given an EG-
M, we write 21
M = C akVk k= 1 if
appears ak
Vk
times as a composition factor of M.
irreducible EG-module, then i(V,V) = 1.
Hence i f
Note that if
V is an
M is completely reducible,
r
c a;. With this information at our disposal, we now prove k=1 r r 4.12. LEMMA. Let M % C akVk and N = C bkVk be two EG-modules. k=1 k=1
then by (11, iCM,M) =
r
(il (ii)
iCM,NI
M
Z akbk k-1
is completely reducible if and only if &(M,M) =
r Z a; k= 1
CHAPTER 5
364
r Proof.
M is irreducible and
so M
3
j
Tf
C ak = 1, then
k=l
...,r}.
for some j E {l,
V
11
C ak. k=1
(il We argue by induction on the value
Hence
Hom(M,N) = HomLM,SocN)
EG
EG
and ttierefore, by (11, P
i ( M , N ) = i(M,BocN) G bj = k=1 “kbk Now assume that
Mo is a submodule of M. 0
-+
By looking at the exact sequence
HomIM/Mo,N) +Hom(M,N)
EG’
--f
Hom(M , N )
EG
EG
we deduce that i(M/Mo,N)
+ C(Mo,N)
(2)
3 C(M,N)
r If Mo is a nontrivial submodule of M with M/M
%
C a{Vk k=1
and
M O
r Z aiv, k=l
then by the induction hypothesis
and
r
But M
r GZZ
k=1
(ai+a{lVk, so it follows from (21, (31 and (4) that r r r
as required. Ciii)
I’f M
is completely reducible, then as has been observed earlisr
r
r .L’IM,M) =
C a;. k=1
Conversely, assume that -f(M,M) = C a; k=1
M/SQCM
=
r
r C
aiVk ,, s o d
k=1
Then, by (il, we have
k C= l afVk
and write
P-RADICAL GROUPS
365
and so applying ( 2 ) , we d e r i v e
r
P
P
Thus
and P
i(socM,M) =
C a% k k k=l
(61
Since P
i(SocM,M) =
i (SocM,SocM)
=
C (a")
k
k=1
a{
it follows from (6) t h a t
ak # 0.
Then
L
a submodule of
M.
and Thus
if
a{ # 0.
a; # 0 and so by (51,
with
SocMf:L c M
i(M/socM,M) # 0.
such t h a t
M/L
..,PI
and hence
3
= af
Syl ( G )
3' k
for a l l Let
{l,.
E
uk
= dimUk,
P
Proof.
..,PI
a = j
a'!
-3
and t h e r e f o r e
0 and
Hence t h e r e e x i s t s
a! 3
such t h a t =
M = SocM,
be a p r o j e c t i v e cover of
Uk = P ( V 1 k
and p u t
then
=
i s isomorphic t o a submodule
a" # 0,
But i f
a{
Now assume t h a t
j e (1,.
4.13. LEMMA.
PE
=
ak
The l a t t e r implies t h a t t h e r e e x i s t s
a'! # 0.
ak
ak
=
Vk, 1
0,
a! # 0 3
a contradiction. a s required.
k G P,
let
Then
E
(IP)'
If
%
c
akVk, then by Lemma 1 . 1 0 . 1 2 ,
k=1
On t h e o t h e r hand, by Theorem 2.4.9,
as r e q u i r e d .
.
W e a r e now ready t o provide a n o t h e r c h a r a c t e r i z a t i o n of p - r a d i c a l groups. 4.14. THEOREM. (Saksonov (19711).
Let
E
be t h e a l g e b r a i c c l o s u r e of
F,
let
CHAPTER 5
366
{ V L ,...,V p } be a full set of nonisomorphic irreducible EG-modules and, for each k E {l,..
.p ) ,
Let P E Syl (GI
let uk = dimP(Vk).
E of double (P,P)-cosets in G.
and let
P
dp be the number
Then
r
with equality if and only if G Proof. (Tsushima (1986) E
=
F.
is p-radical.
.
By Lemma 4.10, we may harmlessly assume that
Then, by Lemmas 4. 2(iil and 4.13,
(lpIG is completely reducible if
and only if
Let T be a set of double coset representatives for
(P,P) in G.
In view of
the above equality, Lemma 4.12(1) and Corollary 4.6, we are left to verify that
BY Theorem 2.4.9 and Mackey decomposition (Theorem 2.6.11, we have Hom(lg, $1
Hom(lpIp, G lp)
FP
FG
P
Horn( @ (1 1 , I ) FPET tPt-lW P @
Hom((1
t€T
FP
P @
P
1 , lP) tPt-hP
Hom
(1
,1
)
t f ~ F (tpt-lw) tpt-lnp tpe-lnp
G G and thus i(lp,lp) = [TI,
4.15. LEMMA.
Let A
proving ( 7 ) .
This completes the proof of the theorem..
be a finite dimensional algebra over a field and let V
a (finitely generated) A-module
I
=
.
be
Then
I$ E End(V) I $ ( V )
5J ( A l V }
A
.
is a nilpotent ideal of End(.V)
A Proof. End(V1. A
It is clear that I is a subspace of the finite-dimensional algebra
If $ E I and I) E End(V),
(4'4) (Vl and
then
A =
$ ( J l ( V l ) C_ $CV) C_ J(A)V
p-RADICAL GROUPS 367
so that $$,$$ E I.
Let n 2 1 be such that
Thus I is an ideal of End(V1. A
Then, for all $ E I
J ( A I n = 0.
and so $n = 0 .
Thus
I is nilpotent, by virtue of Proposition 1.10.5.
The next result is extracted from Tsushima (19861. Let P E Syl (G1
14.16. THEOREM.
End[(l
and let
dP be the number of double (P,P)-
G Then the algebra End[(l ) I P
cosets in G.
,.
P
FG is semisimple if and only if G
is of F-dimension
dp and
is p-radical.
FG Proof.
Put Y
=
and A = End(V1.
Since in the proof of (7) we did
FG not use the assumption that F
is algebraically closed, we have did
F
=
i(l
G G
P’ 1P1 =
dp
Owing to Corollary 4.6, we are left to verify that and only if A If
V
V
is completely reducible if
is semisimple.
is completely reducible, then A
Proposition 1.5.2.
is semisimple by Corollary 1.3.8 and
Conversely, assume that A
is semisimple.
be a full set of nonisomorphic irreducible FG-modules.
k E {l, ...,r } ,
l p i s an irreducible constituent of
from Nakayama reciprocity (Theorem 2.4.101 that each constituent of
V / J ( F G )V .
Let
{Vl,.
. . ,Vr}
Since for all SOC((V~)~), it follows
Vk
is an irreducible
Hence
for some positive integers Assume
by way of contradiction that J(PG1V # 0.
ducible submodule W which is a copy of
Vj
by (81, there is a surjective FG-homomorphism
Then J(FG)V
for some j E (1
has an irre-
,...,r } .
Hence,
368
CHAPTER 5
is a nonzero element of
A
conclude t h a t
A
4CV)
such t h a t
C_J(FGIV.
Invoking Lemma 4 . 1 5 , we
is n o t semisimple, a c o n t r a d i c t i o n .
As an a p p l i c a t i o n of t h e r e s u l t above, we now prove t h e following group-
theoretkc f a c t . 14.17. COROLLARY. (Saksonov (1971)). of
G,
let
P E Syl (GI
P
d
and l e t
r be t h e number of p-regular c l a s s e s
Let
G.
be t h e number of double ( P , P ) - c o s e t s i n
P
Then
dp2 r Proof.
F
Let
by Theorem 2.3.2,
be an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c there e x i s t s exactly
i r r e d u c i b l e FG-modules.
V
say
P,
,..., V p ,
p.
Then,
of nonisomorphic
Keeping t h e n o t a t i o n of Theorem 14.16, w r i t e .r
f3bkVk
SOCV
(9)
k=1 Then we have
bk = i(Vk,S0CV) =
...,r} .
k E {I,
for a l l
irv,,v,
= i(Nk)p, lp)# 0
Since
P
i(V/S(FG)V, SOCV~ =
C akbk k= 1
(11)
we conclude t h a t P
d
= d i d 2
F
Z akbk 2 F , k=1
(12)
a s required. 14.18. of Let
COROLLARY.
G, E
let
P
(Saksonov C1971I1.
E Syl
P
(G) and l e t d,
be t h e a l g e b r a i c closure of
nonisomorphic i r r e d u c i b l e EG-modules. alent:
(iI
dp
= P
Let
r be t h e number of p-regular c l a s s e s
be t h e number of double ( P I P ) - c o s e t s i n F
and l e t
{V
,.. . , V r 1
be a f u l l s e t of
Then t h e following c o n d i t i o n s a r e equiv-
G.
369
P-RADICAL GROUPS
(iiil
G
...,
dinP(Vkl = ] P I for all k E {1,2, r}. E It follows from (lo), Put V = ( l p I G and A = End(V).
is p-radical and
Proof.
(i) * (ii):
FG
A
(111 and (12) that
=
Hom(V/S(FG)V, SocVl
and
FG a k = bk = 1
for all k
It follows from ( E l and (91 that
V/JiFGJ V
E
{1,2,
SocV and hence A
...,rj
End(SocV).
But
FG Socv is completely reducible, so A 14.16,
G is p-radical.
is semisimple and therefore, by Theorem
V
Hence, by Corollary 4.6,
=
SocV and the required
assertion follows by virtue of (91. (ii1 * (iiil:
By Corollary 4.6, G
is p-radical, while by Lemma 4.13,
,...,
r1. dimP(Vkk) = IPI for all k E {1,2 E (iii) =. ( i l : If uk = did'(.Vkl, then (uk/lPI1= 1 for all E Since G is p-radical,
k E 11,...,PI.
by virtue of Theorem 4.14.
Remark.
Owing to Saksonov (19711, all groups satisfying the equivalent con-
ditions of the above corollary are solvable. 14.19. LEMMA.
Let
A
=
Mn(Fl
and let $,$ E A* = Hom(A,F) be such that (A,$)
F and
(A,$]
are syrrrmetric algebras.
Then there exists a nonzero
in F
such
that for all a € 4
$(a1 = hJl(al
Proof.
f
Owing to Lemma 3.3.2(11,
f ( a ) (x) = $(xu)
and
are isomorphisms of (A,A)-bimodules. automorphism.
Let A o
:
A+
A*
g(al (x1
Hence g-lf
be the opposite ring to A
and
g
: A
=
$(xu)
:
A
-+
A
* A* where (x,a
( a @ a )a = a aa 1
2
1
2
A)
is an (A,A)-bimodule
so that A
is an
A @Ao F
module via
E
CHAPTER 5
370
A 8 A" i s a F A i s an i r r e d u c i b l e A 8 A"F such t h a t (g-lf)( a ) = ha f o r
A 8 A"-modules.
g-lf is i n f a c t an automorphism of
Then
F simple algebra with
F
a s a s p l i t t i n g f i e l d and
0#
mod'dle, we conclude t h a t t h e r e e x i s t s all
a
E
A.
Hence
E F
f(a1 = Xg(a) f o r a l l a
E
A.
Since
In p a r t i c u l a r , f o r a l l
a 6 A
we have
a s required. 1 4 . 2 0 . LEMMA.
Let
e s g be an idempotent of
e =
Then, f o r any p-
FG.
SEG singular
ec+ =
o
C e - = tr(.eCfl = 0. SEC f o r any p-singular c E CL(G).
C E CL(G1,
Proof.
n
Choose a p o s i t i v e i n t e g e r Owing t o Lemma 2 . 2 . 1 ,
element.
such t h a t f o r a l l
ePGe
i s simple, then
n g E G , gp
i s a p'-
we have
n n
n e
Furthermore, i f
C e i g p (modIFG,PGl1
= ep
SEG O ur choice of
n ensures t h a t f o r a l l g
C e SEC
Therefore, by Lemma 2.2.2,
Now assume t h a t t h e algebra division ring
D.
If
E
Hence w e may assume t h a t
=
eFGe
0.
C, e
Since
t s t h e c o e f f i c i e n t of
.4
C-l
C
l
Fe
'j
+
eFGe +. F
EZ
D,
i s a maximal s u b f i e l d of
eFGe =
in
i s p - s i n g u l a r , we a l s o have
eFGe
i s simple, say
g
M (D1
for a suitable
then
..
with matrix u n i t s
e... 23
The map
*
C X . . e . . + +CX..
i,j i s such t h a t
i s a symmetric algebra.
i
A
E F
such t h a t
c = Xe,
eC+ = eC+e E Z(eFGe) , w e have
tr(ac1
eC+ = pe
=
$(a)
$7,
On t h e o t h e r hand, by Lemma
(eFGe,tr) is a l s o a symmetric algebra.
3.3.11,
exists
(eFGe,$)
ZJ 23
Thus, by Lemma 14.19,
a E eFGe.
for a l l
f o r some
p
E
F.
there
Since
I t follows t h a t
371
P-RADICAL QROUPS
= +(,.ice =
But e
+e
= +(eC
11
t r ( i C + e11 1 =
.
is an idempotent of F G ,
11
required.
=
11
tr(C+e c ) 11
htr(ellc+)
tr(e
so
11
c+
and thus eC+
= 0
=
0, as
We are now ready to prove Let c
14.21. PROPOSITION. (Tsushima (1971b)).
G including 1.
be the sum of all p-elements of
Then
n
c ei
e2 =
i=1
where the
ei
are all block idempotents of FG
Proof. Let e
be a block idempotent of
c2e
by Theorem 3.6.41ii),
FG
of defect zero.
by Lemma 14.20, ce 14.22. LEMMA.
‘i i
1 2
...ik
e
FG
whose defect i s not zero.
Now assume that e
= 0.
FGe
By Theorem 3.6.4(i), =
of defect zero.
c2e = e ,
and so
is a block idempotent of
.
is a simple F-algebra.
proving the assertion.
Let G be a permutation group on a set
x
= {1,2
i ,i2, .. .,ik
be the stabilizer of the points
,...,n ) ,
in X.
Hence,
and let
Then the
following conditions are equivalent: (i) For every i , j E X, p
divides the order of
n
= o (ii) ( c PG-G;)’ k=1 Proof. (i) * (ii): Fix
i,j
E
{1,2
,...,n)
Gij
and let
S
be a coset decomposition of order of
By hypothesis, p
Gj over Gij.
G i j and
++
G.G. = 2.3 =
+ +
G.C. .(x 2.23
\G..IG:(z
=o
z3
1
+
... + xS) +
Then,
... + xS)
divides the
372
CHAPTER 5
g E G
for a l l
and a l l
,...,n].
f,d
(ii)* ( i ):
If
Therefore ( t i 1 holds.
E {1
n ( C FG*G;)* = 0 ,
i , j E {1,2,.
then given
.. ,n),
k=1
C.x z r n n Gixn
Since
0
=
rn # n , p
for
divides t h e o r d e r of
Gij,
a s required..
With t h i s information a t our d i s p o s a l , we now e s t a b l i s h t h e following r e s u l t . Assume t h a t a group
14.23. THEOREM. (Motose and Ninomiya ( 1 9 7 5 b ) ) . p-solvable and p - r a d i c a l , and l e t Then
F
be an a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
G has a n o n t r i v i a l i n t e r s e c t i o n .
Proof.
Let
d e f e c t zero.
G
Note a l s o t h a t , s i n c e
i s p-solvable,
SocCFGl =
FG
Thus
FG
Soc(FG1 = FGe
i s p-radical,
P E Syl (GI and l e t G
U Psi
=
we have
, 8,
I' =
(13)
I, be a c o s e t decomposition of
as a permutation group on t h e s e t
we see t h a t t h e s t a b i l i z e r of
psi
.
t h a t (13) holds i f and only i f So t h e theorem i s t r u e .
by v i r t u e
= 0
i=1 G
of
has no blocks of d e f e c t zero i f and only i f
n
Remark.
Then, by
1.
C FG* (5'') SESylp (GI
FG. (5'')
P.
G
On t h e o t h e r hand, because
P Regarding
including
i s equal t o t h e sum of a l l block idempotents of
e2
of Corollary 3.5.7.
by Theorem 4.4.
G
be t h e sum of a l l p-elements of
c
Proposition 14.21,
over
p.
FG has no blocks of d e f e c t zero i f and only i f each p a i r of Sylow pasub-
groups of
Fix
i s both
G
g
is
-1
gi Psi.
-1
i
Psi n gj-'p".
3
x
= @,Pq
, . . .,Pgn),
Invoking Lemma 1 4 . 2 2 ,
# 1
for a l l
G
we infer
i , j E {1,2
,...,n}.
According t o Tsushima (19861, Okuyama has proved t h a t any p - r a d i c a l
group i s p-solvable. As has been previously i l l u s t r a t e d , t h e r e
supersolvable and not p - r a d i c a l . supersolvable group p-radical?
i s a p-nilpotent group which i s not
I t i s t h e r e f o r e appropriate t o ask:
The negative answer t o t h e above question w a s
provided by Saksonov (1971) who e x h i b i t e d a supersolvable group of o r d e r which is n o t p-radical.
Is any
108
A simpler example, due t o Passman, i s found i n Bedi
373
p-RADICAL GROUPS
(1979) and will be considered below.
G be a s p l i t extension of
Let
M(31 by a group of automorphisms of o r d e r 2 :
I
G = < a ,b ,c,d a3=b3 = ~ 3 = d 2 = ,b-’ab=uc 1
Then
G is
of o r d e r
G’ =
G, we have
c
5
and
c
i s c e n t r a l , w e have
Frobenius group with
P
as a complement.
G-P
= 2.
and thus
G = G’P Since
p
F be an a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
Let
By t h e d e f i n i t i o n of
1
and i s supersolvable s i n c e it has an i n v a r i a n t s e r i e s
54
with c y c l i c f a c t o r s .
1
,a-’ca=c ,b- cb=c,d-’ad=Q2 ,d- bd=b2,d-’cd=c >
P
where
N G (PI # P
=
and thus
G i s not a
Invoking Theorem 3.13.4, we conclude
that
I t t h e r e f o r e s u f f i c e s t o show t h a t
nFG-I(P~)
is c e n t r a l
@G To t h i s end, p u t
of o r d e r 2 .
Therefore
I
Given
=
I
5 ~ ~ ( 2 - 1=) 0 ,
and s i n c e
g
there exists
~(x-ll
Id,ad,bd}5 X
X
nrlG*I(@I and l e t
SEG LEX
so
be t h e s e t of a l l elements of
E G
I(x-~) =
G = ,
o
fl =
such t h a t
for a l l
we s e e t h a t
G
x =
E
X.
<X>.
<X>
Hence, by
C (x-lIFG
&X Applying (141, we conclude t h a t
r a s required.
I.I(G1 = 0
and t h u s , by Lemma 3.11.1,
- FG* (G+I = FG+ c
5 z( F G I
and hence
Since
Proposition 2.1.10,
I(G) =
G
This Page Intentionally Left Blank
375
6 The Loewy length of projective modules The a i m of t h i s c h a p t e r i s t o provide some g e n e r a l information p e r t a i n i n g t o t h e Loewy l e n g t h of p r o j e c t i v e modules.
A s a p r e p a r a t i o n f o r t h e proof of t h e main
theorem (Theorem 4.6), we e s t a b l i s h a number of important r e s u l t s of independent interest. J(FG)
These r e s u l t s a r e d i r e c t l y connected with t h e nilpotency index of
t o be examined i n t h e next c h a p t e r .
1. PRELIMINARY RESULTS
G
Throughout t h i s s e c t i o n ,
p > 0.
istic
modules and
t h e Loewy length of
V,
p
R
e
(iii) I f
(iv)
R
of
I
:
R.
R-+
J(FG).
-
I
For any i d e a l
I = Re
then
?=
I f o r a l l n 2 1.
R/I
f o r any i d e a l
In particular,
S
R,
is a l e f t i d e a l of
i s semisimple, then SO i s
f
J(FG)% = 0 .
such t h a t
be an a r t i n i a n r i n g .
i s semisimple and
idempotent (ii) I f
R
n
is
which i s f i x e d i n t h e following, we s e t
t (G) = L(FG) , t h e nilpotency index of Let
V,L(V)
For any FG-module
i.e. t h e smallest integer
F i n a l l y omitting r e f e r e n c e t o
(i) I f
a f i e l d of c h a r a c t e r -
A s u s u a l , a l l modules a r e assumed t o be f i n i t e l y generated l e f t lG denotes t h e t r i v i a l FG-module.
1.1. LEMMA.
F
denotes a f i n i t e group and
I
f o r some
R.
of
i s a s u r j e c t i v e homomorphism of r i n g s , then
of
R,
J(R/I)= ( J ( R ) + I ) / I R
(i) A s s u m e t h a t
proof.
i s semisimple.
completely r e d u c i b l e , by P r o p o s i t i o n 1.6.2. ideal
J
of
idempotent of
R. R
If
1 = e
such t h a t
1
+ e I
2
= Re
with
.
e
Since
Hence
E I
R
is artinian,
.# = I @ J
and
e
E
J,
Ip
is
f o r some l e f t then
e
i s an
376
CHAPTER 6
(ii) Let J/I = J ( R / I l .
J" 5 I,
is nilpotent, say
J/I
Since R
for some n
> 1.
By ( i ) ,
J"
and therefore J / I
= J
and hence
0, as required.
=
(iii) Since f ( J ( R ) ) is a nilpotent ideal of S , f ( J ( R ) ) ZJtSl.
hence S/f(J(R)
But S/f(J(R)) is isomorphic to a factor ring of R / J ( R )
.
1.2. COROLLARY. Let N
=
Let f
:
be a normal subgroup of a group G.
Then
F(G/N) be the natural homomorphism.
FG'
Then
FG*I(N) and so the required assertion follows by virtue of Lemma l.l(iv)..
Let R
be an artinian ring, let V
jective cover of
where
.
is semisimple, by virtue of (ii)
(iv) Direct consequence of (iii).
proof.
we certainly have
Hence, by Corollary 1,6.17(ivI, it suffices to verify that
J(S/f(J(R)))= 0.
Kerf
R/I
t s artinian, so i s
Q(V!
is a superfluous submodule of PCV)
R-modules and let P ,P
V.
As
we shall see below, V
R(V1.
Let R
1.3. LEMMA. (Schanuel's Lemma). 1
and hence
as the fletter module of
determines the isomorphism class of
be an artinian ring, let V,V ,W 1
be projective R-modules such that the sequences
2
f
0
4
w-
P 1 . v-
0
4
v-
P 3
f
v
-
0 0
Then
v1 @ P 2 'L w2 @ P 1 Proof.
be a pro-
Then, by definition of P(Y), there is an exact sequence
V.
We shall refer to Q ( V )
are exact.
be an R-module and let P ( V )
Let X be the submodule of P @ P 1
defined by 2
be 2
377
PRELIMINARY RESULTS
If
P f3 P 2 -
71 :
P1 is the projection map, then IT(P
x n KerIT =
{(o,z2) Iz
I
E Kerf
) =
2
P
and
w2
Hence we have an exact sequence
o-w-x-+P-o Since P
X
z
is projective, we infer that
W @P 1
and the result follows. 2
Let R
1.4. LEMMA.
'
w2
8P 1
.
By the same argument,
be an R-module and let P be a
be an artinian ring, let V
projective R-module. (i) V
determines the isomorphism class of R(V)
v
(ii) Assume that f : P-
P for some R-module P
1
P ( V ) 8 P'
and Kerf
and P 1
be two projective covers of
- Ri(V1
be the corresponding exact sequences.
R ( V ) f3 P 2
P
while by Lemma 1.9.2, 2
1
P
2
.
P.+ z
V-
V
and let
ti
0
= 1,2)
By L e m a 1.3,
= R2(V)
8 P1
Hence, by the Krull-Schmidt theorem,
R 2 (8).
(ii) By Lemma 1.3,
Kerf f3 P ( V )
for some R-module P ' .
9
R(V) 8 P and, by Lemma 1.9.2(ii), P = P ( V ) @P'
Hence
and so, by the Krull-Schmidt theorem, Kerf
1.5. LEMMA. if
R(V) 8 P'
2
0
R1 ( V )
%
Then
'.
ti) Let P
Proof.
is a surjective homomorphism.
R(V)
a P'.
Let lG be the trivial FGmodule.
'
Then P ( 1G) = FG
if and only
G is a p-group. Proof.
FG/I(G)
Consider the exact sequence 0
4 I(G)
--+
lG, it follows from the definition of P ( lG )
only if I ( G )
5 J(FG).
If G
FG -+
FG/I(G)--+ 0.
that P(lG) = FG
is a p-group, then J(FG) =
Since if and
I(G) by Corollary
CHAPTER 6
378
Conversely, if I ( G ) C_J(FG1 then each g-1
3.1.2.
,gE
G, is nilpotent.
.
The desired conclusion now follows by virtue of
n
n
sp
=
and define the i-th socle
(I4 =
s
0,
V
=
R
Put
(V1 = SOC(V1
Si(Vl of V by
s7,.( v1 /si-l(V1 If
for all n 2 1
1
V an R-module.
Let R be an artinian ring and let
s
-
R then we put S.(R) z
= SOC ( V/SiF1
and refer to Si(R1
S.(V1
=
(i2
(V)1
2)
as the i - t h socZe of
For any ideal I of R, we put I" = R.
R.
Let R
1.6. LEMMA.
be an artlnian ring.
Then for all i 2 0
s ~ ( R= ~r ( J ( R I i ) By Proposition 1.6.29, Soc (R1
Proof.
=
r ( J (R)1
.
Assume, by induction,
that 'i-1
Since Si(R)/Si-l(Rl
(R1
= r ( J ( R 1i-ll
is a completely reducible R-module, we have
(R1 Si cR1
J
5 SiVl
(R)
and hence
J(R)~S~W5 s Thus Si(R1
5 r[J(RIi1 .
~ ~ ) ~ (R)-= ~ o s ~ - ~ i
Conversely, assume that J(R) z = 0 for some z E R .
Then J(R)x and therefore
(Rx
+
5 r(J(Rlic1) = Si-1 ( R )
Si-l(R1l/Si-l(R) is a completely reducible R-module. RX
proving that Let
F.
+ Si-l(R) ZSi(R)
x E Si(R1, as required.
V be an A-module, where A
Recall that the dual module
Thus
9
is a finite-dimensional algebra over a field
V* = Hom(V,F) is a right A-module with the
F
PRELIMINARY RESULTS
action of A
on
In the case A A-module
V
=
V*
FG
379
given by
there is a simple device which allows us to make any right
into a left A-module in the following natural way.
g-l,g E G, is an anti-automorphism
The F-linear extension of the map g of the group algebra FG
.
Therefore, if
W is a right FGmodule, then W
becomes a left FG-module under the action
V*
V,
In particular, for every left FG-module
is a left FG-module under the
action
We shall refer to the left FG-module follows, for any (left) FGmodule
V , V*
contragradient of
V.
1.7. LEMMA.
V be an FG-module
Let
(i) If W
is a submodule of
is a submodule of (ii) If W
c W
2 -
(iii) If
V*
V*
as the contragradient o f
V.
denotes the FG-module which is the
V, then
such that W*
are submodules of
V*/W V,
1
.
then
1
V and
W are FG-modules and
0. E
Hom(V,W),
then the map
F a* :
W*
--+
V*
defined by
is an FG-homomorphism if and only if so is a. (iv) If CY E Hom(V,W)
and
E Hom(W,X),
then
v+a
x-
(BCY)*
FG 0-
In what
B
El+
is an exact sequence of FG-modules, then so is
0
=
CY*B*.
If
CHAPTER 6
3 80
Proof.
(i) Let tn(gf)I
71
(W)
w E W,fE V*
for a l l
morphism with Kern (ii) Note that
of W;,
=
be the restriction mapping of
into w*.
=
Ig(Trf)l ( W )
= (lrfrf, cg-lW,
= (gf, ( W ) = f(g%
Since
and g E G, it follows that is a surjective FG-homo1 1 W Thus 'W is a submodule of V* and w* v*/W .
.
(W1/W2)* is isomorphic to the FGsubmodule
If€
w;If(W2)
W*
V*/W
and in the isomorphism
(iii) and (iv).
1'
-
01 l
i
V21/W11.
this corresponds to
The required properties are standard for F-spaces and F-homo-
It therefore suffices to verify that a*
morphisms.
V*
and only if so is
If
CY.
Hom(V,W),
CY E
is an FG-homomorphism if
then
FG = (gfl (CY(U1) = f(g-l(a(u1)
la*(gfllCul
(cc*f) ( g - h
=
for all v
€
V,f E W*
1.8. COROLLARY.
Let
and g E G.
V and b/
=
= f(a(g-lu))
g(a*f) ( V )
The converse is true by a similar argument.
be FG-modules.
Then (as
Proof. The map
c1 k-+a*
provides an isomorphism of Hom(V,W)
F-spaces)
onto
F Hom(W*,V*). Since, by Lemma 1.7, F the assertion follows. 1.9. LEMMA.
CY
E Hom(V,w)
i f and only if
FG
a* E Hom(W*,V*),
FG
V be an FG-module
Let
(i) The canonical mapping
71
:
V - + (V*)*
defined by
is an FG-isomorphism (ii) If V = V @ V 1
and
Vz,
(iii)
V
is a direct decomposition of 2
then V* = V
v,'
V; 8
V
into FG-submodules
is irreducible (indecomposable) if and only if
.
(indecomposablel
v1
v; V*
is irreducible
PRELIMINARY RESULTS
(I) It i s a standard fact that n
Proof.
cn
~g(n(t,))l
=
381
is an F-isomorphism.
n ( u ) ( g - l f ) = (g-lf)( u ) = f ( g v )
-
Since
cf)
= IV(~V)I
I
for all
t,
E V,f
and g E G,
V*
is an FG-isomorphism.
TI
1
1
1
2
V* = V @ V
(ii) The direct decomposition
of the F-space V*
is well known.
The desired conclusion now follows from Lemma 1.7(i). (iiil
V*
If
is irreducible, then (V*)*
is irreducible, then by Lemma 1-7(i).
is irreducible by Lemma 1.7(il.
is irreducible by (i).
Hence
V*
For any g E G,
If V
is irreducible
Properties (i) and (ii) immediately imply that
composable if and only if so is 1.10. LEMMA.
V
V*
is inde-
V.
let @gE WG)*
be defined by
Then the map
is an FG-isomorphism. Proof. Since
{@glgE GI
at least an F-isomorphism.
the result follows. 1.11. LEMMA.
(il
V
Let
is obviously an F-basis for V * ,
the given map is
Taking into account that for all g , h
.
=
E
G
@kg
V be an FG-module.
is projective if and only if so is V*
(ii) If
V
(iiil If W
is irreducible, then
V* zz Soc(P(V)*)
is a completely reducible FGmodule and
V
is irreducible, then
Hom(W,P(V)*) 2 Hom(W,l"*'*l
FG Proof.
(i) Assume that
v
FG is projective, say
FG@ for some FGmodule
W.
... @ FG
2
Vcr) F/
Then, applying Lemma 1.9(iil and Lemma 1.10, we have
382
CHAPTER 6
proving that V*
is projective.
there is a surjective homomorphism f
(ii) By definition of P(V1,
f*
Hence, by Lemma 1.7(iv), and Lemma 1.9(iiiI, Since Soc(P(V)*) (iii) Since W
P(VI*
The converse is a consequence of Lemma 1.9(i).
:
i s
Y*-+
P(V)*
:
P(V)-
is an injective homomorphism.
projective indecomposable and
V*
V.
By (i)
is irreducible.
is irreducible (Theorem 3.3.8), the assertion follows. is completely reducible, Hom(W,P(V) *I FG
5:
Hom(W, Soc(P(V)*) FG
.
Now apply (ii). Given x = CX$
E FG, let
z*
=
cx
g
g-1
Then, as has been previously observed, the map of FG.
x I-+-
x* is an anti-automorphism
Thus, if I i s a left ideal of FG, then
is a right ideal of FG. 1.12. LEMMA.
Let
I c I be left ideals of FG.
Then
1 -
r 1'
-
=RR(II
and
as FG-modules. Proof.
99 E FG.
Fix X = CX
By Lemma 1.10, it suffices to verify that
h
x
E
P,U)
if and only if Cx $I
9 9
E I
1
To this end, we first observe that given y = Cy g E FC, g
Hence Cxg@g E
I
1
if and only if
tr(xy*) = 0
.
for all y E I
The desired conclusion is therefore a consequence of Theorem 3.3.6(iiI.
PRELIMINARY RESULTS
i
1.13. LEMMA. For all (I) Si(FG)
*
0
FG/J(FGIZ
as FG-modules
[si+l(FGI/si(FG)1 * = J(FGIi
(iii
Proof. Thus,
383
Put J
=
J(FG).
'+'
/ ~ ( ~ ~ )
Then
and so
= J
as FGmodules
ji
for all i 2 1.
= J'
by Lemma 1.12 and Theorem 3.3.6(vII ( J ~ I R' ( ~J i 1 = r ( Si 1
(1)
(i) We have [by Lemma 1.7(ii)) [since
(FG)
= 01
(by (1)) =
Si(FGI
(by L e m a 1.6)
Hence the required assertion i s a consequence of Lemma 1.9(11
.
Now apply Lemma l.g(i1. 1.14. LEMMA.
Let
V be an FGmodule.
Then
dimV = dim Hom(V,FGl
F Proof.
F
It is obvious that dimV = dim Hom(V,F).
F
F FG
I (lHIG.
1.15. LEMMA.
FG
Now apply Theorem 2.4.9 for the case Let
P be
a
X
.
If H = 1, then
F = 1.
projective indecomposable FGmodule.
Then for all
i 2 0, Hom(P/J(FG)'P,FG)
FG Proof.
We may assume that P
Hom(P,Si (FG))
FG =
FGe for some primitive idempotent e of FG.
CHAPTER 6
384
Assume that f
:
P - +FG
t s an FGhomomorphism.
if and only if f(P)2 Si(FG)
- S(FG)’P Kerf 3 Assume that Kerf
>_ J (FG)”P.
Then
f(J(FG)ie) so f(e) E Si(FGI
by Lemma 1.6.
so
= 0,
f(J(FG)’P)
= J(FG)if(e)
=
0
Hence
f(P1 Conversely, assume that f ( P )
Tt suffices to show that
=
FGf(e1
5 Si(E%).
5 Si(FG)
Then, by Lemma 1.6, J(FG)’f(P)
=
0
and
as required.
We have now accumulated all the information necessary to prove the following useful result. Let P = P(Y1
1.16. PROPOSITION.
module
V.
Then for all i
be a projective cover of an irreducible FG-
a0
dim (S(FG1iP/JIFGIC*lPl = dim HomU(FG)i/J(FGl i+l,V*)
F Proof.
F FG
Put Si
=
S2.(FG1 and J
dim(JiP/Ji*’P1
= J(FG)
.
Then
-
= dim (P/S’*lP)
F
F
.
dim (P/JiP)
F
= dim Hom(P/JZ+lP,FG)
- dim
F FG =
F
Hom(P/JiP,FG) FG (by Lemma 1.14)
dim Hom(P,Si+ll -dim Hom(P,Si) F FG F FG (by Lemma 1.151
Hom(f’,Si+l/Si) F FG
= dim
= dim Hom((Si+l/Si)*,P*) = =
(by Corollary 1.8)
F FG dim Hom((Si+l/Si)*,v*) (by Lemma l.ll(iii)) FG i i+l (by Lemma 1.13 (ii)) dim H o m W /J ,V*) F FG
as asserted. m For later use, we next record,
1.17. LEMMA.
Let P be a projective FG-module and let
position factors of an FG-module
V.
Then
V 1 , V 2,...,Vn
be com-
PRELIMINARY RESULTS
385
n P Q V Z @PQVi Proof.
a s FG-modules
i=1
F
I t s u f f i c e s t o show t h a t f o r any submodule
PQVZPQW@PQ F
F
W
of
V,
(V/W,
a s FG-modu l e s
F
To t h i s end, consider t h e n a t u r a l e x a c t sequence
0
w-+ v 4 v/w-
0
w+
4 PQ
v+
P 8
F
P
0
w e o b t a i n an e x a c t sequence
P
Tensoring with
4
P 8 (V/W,+
F
0
F
Cv/w1
P8
Csee P r o p o s i t i o n 2.5.7 (ii)) F t h e above sequence s p l i t s and t h e r e s u l t follows.
But
i s p r o j e c t i v e , hence so i s
H be a subgroup of
Let
E M M A . 1.18. U
be an FG-module.
G,
gt
8 (ui0
u , u p , . . .,u
Let
IJ
= r/G Q v
,1 4
3 As an F-basis of
1G f
Q
m,
and
rn
1g j
t
G k, 1 G
5
V ,V
21...rVn
m, 1
G
:
gt
Cg,
that
{uil and
2 =
{u.} J
g t @ (ui @ V,.) a r e bases of
ggt
=
g,h,h E H,
gz Then w e have
SO
=
G
(U 8 VH) F 1 G t
.
k,
z 8 gt v j
We a r e t h e r e f o r e l e f t t o UG @ V . F g E G, f(gz) = gf(r). Using t h e f a c t
and
U and
V,
r e s p e c t i v e l y , it is s t r a i g h t -
that
ggt
Then
Q 24.1
u E U,V E V
forward t o v e r i f y t h a t f o r all
Write
(gt 8 Ui) @ gtVjl
G.
V,
onto
F prove t h a t , given
in
j G n, is an F-basis of
Q (Ui Q v j l -
(U @ VHIG
d e f i n e s an F-isomorphism of
U and
be F-bases of
g ,g2,...,gk f o r H
UG ‘8 y, choose t h e elements P n. Then t h e mapping
f
as FG-modules
F
Choose a l e f t t r a n s v e r s a l
.1
V
U be an FH-module, and l e t
let
F
respectively.
Thus
Then
(U 8 VHf Proof.
.
Q Cui Q v j ’
=
g , Q Chu. 0 hv z
.)
3
386
CHAPTER 6
f(gzI = (gz
Q
hu;I
Q (g
6
hv .I 3
=
(q,h Q u .I 8 (gshv .I 3
3
and, on t h e o t h e r hand,
gf(z) SO
=
(ggt Q U i ) Q (99v
t i
(gsh 8 u .) Q ( g hv
) =
z
5
.)
3
t h e lemma i s t r u e .
2. THE LOEWY LENGTH OF PROJECTIVE COVERS
G
Throughout t h i s s e c t i o n , F
denotes a f i n i t e group,
a f i e l d of c h a r a c t e r i s t i c
cover of t h e FG-module Finally,
L(V)
V.
R(V)
and
p > 0,
and
PG(V)
As usual, we w r i t e
N
a normal subgroup of
( o r simply
P(V))
G,
a projective
lG f o r t h e t r i v i a l FG-module.
V,
denote t h e Loewy length and t h e H e l l e r module of
respectively. Our a i m is t o examine t h e s i t u a t i o n where t r i v i a l l y on
V.
Thus
V
V
is an FG-module and
N
acts
can be a l s o viewed a s an F(G/N)-module v i a
and w e compute t h e Loewy lengths o f t h e corresponding p r o j e c t i v e covers and
PG/N(Vl
PG(v)
Let IT :
be t h e n a t u r a l homomorphism.
V,
we have
Let
V be an FGmodule.
FG
+
F(G/NI
G/N
Then, by t h e d e f i n i t i o n of t h e a c t i o n of
on
Since FG*I(NI = I(N)*FG
w e have
FG*I(NIV = I(N)V
Thus I ( N ) V Therefore
i s a submodule of
V/I(NIV
t r i v i a l l y on
V,
V
such t h a t
can be viewed a s an
then
N
a c t s t r i v i a l l y on
F(G/Nl-module.
I(N)V = 0 and so
V/I(N)V.
Moreover, i f
N
acts
THE LOEWY LENGTH OF PROJECTIVE COVERS
387
(2)
V
(i) If
FG/FG'I(N),
=
(ii) A map f : homomorphism. (iii) If f morphism
act trivially on FG-modules V
Let N
2.1. LEMMA.
V-
then
W
V
'1 F(G/N)
and
kr.
as F(G/N )-modules
is an FG-homomorphism if and only if
f is an F(G/N)-
Furthermore, the kernels of both homomorphisms are the same. :
M
4
V8W
is an FGisomorphism, then f is also an F(G/N)-iso-
.
(iv) J(FG) V = J(F(G/N)) V
(i) Assume that V = FG/FG*I(N).
Proof.
algebras, we certainly have
dimV F
is at least an F-isomorphism. G/N,
=
Since F(G/N)
dimF(G/Nl.
FG/FG*I(N)
as F-
Hence the map
F
Since this map obviously preserves the action of
it is in fact an F(G/N)-isomorphism.
(ii) Direct consequence of the definition of the action of
G/N
on
v.
(iii) Direct consequence of (ii) (iv) Let
TI :
FG -+ F(G/N)
be the natural homomorphism.
By Lemma l.l(iii)
,
and so by (1) we have
as required.
2.2. LEMMA.
Let
V be a projective FG-module.
Then
V/I(N)V
F (G/N)-module. Proof.
By hypothesis, FGe... 8 F G s V 8 X
for some FGmodule
X.
Hence I(N1FG 8
... 8 I(N)FG
I ( N ) V @ I(N)X
is a projective
CHAPTER 6
388
and t h e r e f o r e
NOW
,(iiil .
apply Lemma 2.1(il
.
V be an FG-module and l e t
Let
3:
E FG.
Then we p u t
01
ann(3:) = { V E V ~ X U=
V I t is c l e a r t h a t i f
2.3. LEMMA.
2 E Z(FG)
,
then
ann(3:) V
VN is a p r o j e c t i v e FN-module.
V be an FG-module such t h a t
Let
V.
i s a submodule of
Then
Proof.
W e may harmlessly assume t h a t
FN @
X,
f o r some FN-module
I(N)V
Thus
vi
E FN.
Then
i f and only i f if
V
E
V
I(N)V ,
2.4. LEMMA.
N +v
= 0
.E
I"N1
2
... @ I(N)).
+ N Vi
X
u
V and w r i t e
E
= 0
by Lemma 3 . 3 . 1 2 ( i i ) .
V = P(1 1 G
+
u
Fix
i f and only i f
a s required. Let
@ FN = V @
i n which c a s e
V n (I(N)@
=
..,
for a l l
Hence
1)
i.
=
v1
But
E ann(N+) V
f
... + vn,
+ N vi
= 0
i f and only
B
and l e t
n
be t h e Loewy length of
V.
Then
( i ) SocV = G V (ii)
G+ E J ( F G ) ~ - ~
(iii) J(FG) V =
Proof.
(i) L e t
FGe/J(FG)e
2
Soc(FGe1
lG.
f o r some
I(G)V
lG.
e
Then If
0 # 1 E F.
be a p r i m i t i v e idempotent of
V i s i d e n t i f i a b l e with
0 # x E Soc(FGe), then
gx =
2
Thus soc(FGe) = FG
a s asserted.
FGe
+
+ (FGe)
= G
FG
such t h a t
and, by Theorem 3 . 3 . 8 ( i i i ) , for a l l
t
g E G, so x = XG
389
THE LOEWY LENGTH OF PROJECTIVE COVERS
is a
(ii) By hypothesis, J(FG)n-le
nonzero submodule
FGe, so
of
G+ E J(FGln-’e C J(FG)n-l.
+
(iii) If I(G)e = FGe, then soc(FGe) = G (I(G)e) = 0 , I(G)e
is a proper submodule of FGe.
maximal submodule of FGe.
(i)
P(VIN
modules
V;
Hence I(G)e
= J(FG)e
=
t2
a
is projective.
for all
<,
Then
for some irreducible FGP(V)
Since
lN for some i E {1,2 Hence Soc(X.)
z
1
N
=
P(If
)
9.. . @ P ( V n )
Write
Since P ( V I N
By Theorem 3.3.8(iii),
N acts trivially on Soc(P(V)).
are G-conjugate.
... @ V
is irreducible.
where the Xi are indecomposable FN-modules.
2
II
acts trivially.
Then, by Theorem 1.9.3,
1.
and thus we may assume that V
that Soc(x.)
is a
J(FN1 * P ( V )
and some
module, each Xi
V
is a projective FNSocP(V),
and hence
5
(SOC(P(V)))~ SOC(P(V)~I, it follows
,...,
7711.
But, by Lemma 5.3.9, all the
Soc(PN(lN))
as required.
(ii) Direct consequence of (il and Lemma 2.4(iii).
for all
<,
so X. 2
.
We are now ready to prove the following result. 2.6. THEOREM. (Lorenz (1985)l.
Let
V be an FG-module on which N
trivially.
L(PG(V)) = L(PG,N(V) 1
(vl If G/N
Thus
for some e 2 1
Write P(VI/J(FG)P(V) = Ifl @
(i)
and J(FG)e
as required.
ePA,(lN)
(ii) I ( N )* P ( V ) Proof.
But I(G)e ? J ( F G ) e
Let V be an FG-module on which N
2.5. LEMMA.
a contradiction.
i s a p’-group, then
acts
xi
PN (1N )
CHAPTER 6
390
L ( P G ( r n I = L(PN(lN)I Proof.
(i) S e t
P
=
PG(V)
H = G/N.
and
N
Since
a c t s t r i v i a l l y on
V,
it follows from ( 2 ) t h a t
Thus, by Lemma 2 . 1 ( i i ) , we o b t a i n an exact sequence of FH-modules
Moreover, by ( 3 ) and Lemma 2 . l ( i v ) , we have
But
.?
P/I(N)P
is p r o j e c t i v e as an FG-module, so by Lemma 2.2
i s p r o j e c t i v e as
Thus, by (41 and ( 5 ) ,
an FH-module.
PH(V,
= P/I(N)P
a s required. (ii) P u t
n
=
L(PH(V1) and
rn = L(PN(lN1I.
Since
P,(V)
1
P/I(N)P,
i t follows
from (5) t h a t
J(FGI~-'.P ~frw)-P Now
P i s p r o j e c t i v e , hence so is PI and s o , by Lemma 2 . 3 ,
I(h')P
=
+
ann(N 1.
P Thus
N*J(FG)n-l*P # 0.
3.1.9
But, by Lemma 2 . 4 ,
5 S(FG)"-l.
E J(FWrn-l while by Proposition
Hence
J(FGIm'n-2P and t h e r e f o r e
N'
=
J(FGlrn-lJ(FG)nclP# 0
L ( P ) 2 mtn-1.
(iii) A s s u m e t h a t
J(FG1 .J(FN)
= J ( F N )*J(FG).
Keeping t h e n o t a t i o n of (ii), we
have
by v i r t u e of (51.
Applying Lemma 2.5, we deduce t h a t
J(FGln*P 5 J ( F N ) * P and t h a t
391
THE LOEWY LENGTH OF PROJECTIVE COVERS
We now claim that for any
J(FN)% = 0
(7)
J(FGlnkP C - J(FNlkP
(8)
k 2 1,
If sustained, the required assertion will follow from ( 7 ) by taking k
=
m.
In
(by induction hypothesis) (since J (FG)' J ( F N ) = J (FN)' J (FG)) (by (61)
as required.
(iv) Assume that N
Then L ( P (1 ) I
is a p'-group.
N
N
=
1
and hence by (ii)
L(PG(V1) s L(PG/N(V)1 Since J(FG)'J(FN) = J(FN)'J(FG)
=
0,
it follows from (iii) that
L(PG/N(v))
L(PGCVI) as required. (v) Assume that G / N
Then L(PG,N(V'l)
is a p'-group.
L(PG(V)1 By Proposition 3.1.8(ii),
=
1 and hence by (ii)
p LCPN(lN1)
J(FG) = FG*J(FM
J(FG1 *J(FM
=
=
J(FM FG and hence
J(FIv1 * J ( F G ) .
Applying (iii1, we conclude that
as required.
.
L(PG(V) Q L(PN(lN))
As an application of Theorem 2.6, we now prove
2.7. THEOREM. (Willems (1980)).
Let V
a KerV is p-solvable of order p m,a
be an irreducible FG-module such that
a O,(p,m)
L(P(V1)
= 1.
aCp-11 + 1.
Then
CHAPTER 6
392
Set K = Kerv.
We argue by induction on the order of G.
Proof.
( K ) = 1 then a = 0 and there is nothing to prove.
0
P'IP
N = 0 , ( K ) # 1.
P
Assume that
Then, by Theorem 2.6(iv) , L ( P ( V ) ) = L(PG/N(V)).
assertion follows by the induction hypothesis. Then, by Theorem 2.6(ii),
If
Thus the
Finally, assume that N = 0 ( K ) # L
P
we have
Using the
where
IN
.
This completes the proof of the theorem. 2.8. COROLI~\RY. (Wallace (1968)). U
p m , ( p , m ) = 1.
Then
Let
G be a Psolvable group of order
a(p-1) + 1.
v=
lC, we get
t(G) 2 L(P(lG))
> a(p-1) +
Proof. Applying Theorem 2.7 for
1,
as required. 2.9. LEMMA. Let
V be an FC-module. Hom(V,lG) FG
Proof.
An F-linear map f : V-+
Then HO~(V/I(G)V,lG)
F lC is an FG-homomorphism if and only if
I(G)V5 Kerf.
Thus Hom(V,lG) is a subspace of Hom(V,l ) consisting of those G FG F f E Hom(V,LG) for which Kerf 2 I ( G ) V . Since the latter subspace is isomorphic
F
to Hom(V/I(G)V,lG) , the result follows. F Next we establish the following important result. 2.10. THEOREM. (Alperin,Collins and Sibley (1984)). which N
acts trivially.
Let V
be an FG-module on
View FN as an FG-module via conjugation of
G
on €%
39 3
THE LOEWY LENGTH OF PROJECTIVE COVERS
Then for all
i ? 0 we have FG-isomorphisms P ~ / ~ ( 8 Y )J(FN) '/ICN) J(FN)
F where PG,,fl (V)
P (v)/J(FN) ~ '+'pG
i s viewed as an FGmodule by letting N
convention, J(FN1 Proof.
= J (FN)'
=
FN)
sct trivially.
.
By Theorem 2.6(i)
and Lemma 2.5(ii) and 2.1(ii)
PG/N ( V ) Put P = P,(V) ,J = J(FN)
5
PG( V )/ J (FN)PG( V )
-
as FG-modules
(9)
and consider the map
I
P/JP 8 J ~ /(IN I ji
J~P/J~+$
(10)
H ys + Ji+'P
(z+JP) 8 (y+I(N)J')
This i s obviously well defined, F-linear and surjective. B
(v)
Moreover, for a i l
'G, g((r+JP) 63 (y+I(N)Jil
63 @yg-l+I'(N)Ji)
= (gM1
i s mapped to
Thus the map (101 is a surjective FG-homomorphism,
In view of ( 9 ) , we are left
to verify that the modules in (10) are of the same F-dimension.
Q
Put
= P (1 )
N
N
and observe that, by Lemma 2.5(11, we need only show that dim(Q/JQ 8 J i / I W Ji)
=
dim(JiQ/J'+lQ)
F But Q/JQ
(11)
F
lN, so dimQ/JQ = 1 and therefore
F dim (Q/JQ 8 Ji/I ( N )Ji ) = dim (Ji/I! N ) J i
)
F
F
= dim HomWZ/I(IVIS
*
i ,lN)
i/Ji+l ,1,)
= dim Hom (J
F
(by Lemma 2.9)
FN
which equals the right-hand side of (111, by virtue of Proposition 1.16.
9
Returning to the discussion of Loewy lengths, we now prove
2.11. THEOREM. (Loren2 (1985)).
Let N
be a normal p-subgroup of G
be an FGmodule on which N acts trivially. conjugation of G
on N.
Then
and let
View FN as an FGmodule via
CHAPTER 6
394
where X
runs over the FG-composition factors of
Proof.
Set P = P ( V ) , H
G
=
G/N
FN.
ki = L(I(N)iP/I(N1iflI , i > 0.
and
By
Theorem 2.10,
and hence
If rn
C
ki, then
T(N)i+lP
JIFG)mItN)iP
= annpT(N1
t( N ) 4 - 1 ,
where the latter equality follows by virtue of Corollary 4.1.7, since P over FN.
is free
It therefore follows that
+
I ( N I ~ ( ~ ) - ~ - ~ . S ( F G ) ~ . ~ o( N ) ~ P and thus
Therefore, L ( P ) 2 t ( N )
+
k - 1, where
k = maxRi.
Note that P ( V )
i tive over FH trivial.
and conjugation action o f
on each factor 1(iV)’/I(NIi+’
is
Hence, by Lemma 1.17, @
(P,(V, 8 I w i / I ( N l i + l )
i where X
N
is projec-
H
runs over the FH
@(PH(V,8 X) X
2
F
(and hence FGI-composition factors of FN.
R = m a d (P,(V)
Thus
8 XI
X Since PH(V 8 F
XI
is a summand of P,(V)
8XI
we also have
F L 2 L(PH(V8 X) F
as required.
2.12. THEOREM. (Lorenz (1985)).
Let N
be a normal p-subgroup o f
G, let V
THE LOEWY LENGTH OF PROJECTIVE COVERS
be an irreducible FG-module and assume that G = N H with N
fl H =
1.
For each i
> 0,
let
for some subgroup
Vi denote the FG-module
( I ( N h ( i d + 3Q
v
F where G
.
i+l acts by conjugation on I ( N ) ’ / I ( N )
G (I) Pc(V) ( l H )8 PH(V), where N acts trivially on PH(V). F G (ii) L ( P G ( v ) ) > L(PH(v)) + L ( ( V H ) 1 - 1 G t ( N ) - 1 + maxL(Vi) (iiil L ( ( V H ) i In particular,
G
L((VH)
if and only if all
V.
) =
t(N)
are completely reducible. n
Because X
is projective, it follows that
(ii) If n = L ( p H ( V ))
,
then by (i),
.
proving (ii)
(iii) We first observe that
Thus if m < ni = L ( V i ) ,
then
x 1 PG ( V )
as
required.
395
H of G
CHAPTER 6
396
Here the latter equality follows from Corollary 4.1.7, since
FN.
Therefore, for a l l
i2
(VHIG is free over
0
Since the last assertion is obvious, the result is established.
.
3. THE LOEWY LENGTH OF INDUCED MODULES
Throughout this section, F finite group.
As
denotes a field of characteristic p > 0 and
usual, all modules over a ring R
G a
are assumed to be left and
is an R-module and E = End(W) , then W will also R be regarded as an E-module via If W
finitely generated.
@w 3.1. LEMMA.
Let W
=
Re
=
for all 4 E E , w E W
4tW)
for some idempotent e
of R
and let E
=
End(W). R
Then
W as an E-module is equal to the nilpotency
In particular, the Loewy length of index of J ( E ) . Proof. W
E W.
For each x E @Re, let f
By Proposition 1.5.6,
X
E
E be defined by f x ( w ) = Wx for all
the map
1 , is an anti-isomorphism of rings. J(eRe) = eJ(R)e and so
eRe-
E
5-
j-
Furthermore, by Proposition 1.6.35,
fx E J ( E I i
iv
J (E)
Finally, J(E)'W = 0 i€ and only if the E-module W eRe
i
if and only if x E ( e J ( R ) e )
.
Thus
= Re ( e(8) ~e )
(eJ(R1e)'
=
0.
i s equal to the nilpotency index of
Hence the Loewy length of J(eRe)
=
eJ(R)e.
Since
E", the result follows.
3 . 2 . LEMMA.
Let N
module and let H
be a normal subgroup of
be the inertia group of
V.
G,
let V be an irreducible FNThen each FH-homomorphism
397
THE LOEWY LENGTH OF INDUCED MODULES
€I:
8 I-+
VH
+
8'
VH
extends to a unisue FG-homomorphism 8'
is an F-algebra isomorphism of End(V
H
VG
:
G onto End(V
)
VG
4
.
)
FH
Proof. g l ,...,gk
FG be a left transversal for N in H and
Let g1,g2, ...,gS
a left transversal for N
Then it is immediate that 8' i s Hence the map
in G.
Then
unique element of End(P) FG is an injective homomorphism End(p)
8 I-+ 8'
d
FH algebras. to
V.
Note that
(#)N
and the map
FG
5 $.
is surjective and the result follows.
8.
G
End(V of FG FG I N isomorphic
+
(v
is the sum of all submodules of
Hence for any $ E End(pI, $(#I
extending
This proves that the given map
=
We have at our disposal all the information necessary to prove the following result in which 3.3.
L(p) denotes the Loewy length of the FG-module F.
THEOREM. (Clarke ( 1 9 7 2 ) ) .
Let N be a normal p'-subgroup of
Y and let E
the inertia group of an irreducible FN-module (il J ( F G ) V = FG-J(FHI~IJI = (ii) L ( f i )
=
G, let H be
End(fi).
Then
FG
J(E)~P
for all n
1
is equal to the nilpotency index of J(End(fl) I .
FH Proof. which case
We may take
fi
=
FGe.
V
=
FNe for some primitive idempotent e
Hence, by Lemma 3.1, the Loewy length of
module is equal to the nilpotency index of J ( E 1 .
of as
FN in an E-
Thus (iiI is a consequence of
(i) and Lemma 3.2. Write l = e + e 1
2
+...+
as a sum of primitive idempotents of FN with
e e
rn =
e
.
Then we have
J ( F G ) F = J(FG)e = FG(J(FG)el = FGe(J(FG)e)
+FGe2(J(FG)e)+...+FGem(J(FG)e)
as left FG-modules, where the sum is not necessarily direct.
(1)
For each aEeiFGe,
CHAPTER 6
398
let f a E Hom(FGei,FGe) be defined by f
FG
U
(XI
=
for all x E FGei.
za
Then, as
can be seen from Lemma 1.5.5, the map eiFGe + Horn (FGei ,FGe) FG
is an F-isomorphism. such that f
:
FGe, then there is an a E eiFGe
In particular, if FGei
FGei--+ PGe
is an FG-isomorphism.
For the sake of clarity, we divide the rest of the proof into three steps. Here we prove t h a t
S t e p 1.
J(FG)e = FGeJ(FG)e. it follows from (1) that we need only verify that
Since FGeJ(FG)e cS(FG)e,
eiJ ( F G ) e Let f i
5 eiFGeJ (FG)e
be the block idempotent of FN
Then f;
be the sum of G-conjugates of f i . Now if f
and fi
with eifi i s
(1 G =
e i’ 1 G i G rn,
We may therefore assume that f
1
1
1
f?
and hence
2 1
2 2 2
and let
(2)
a central idempotent of FG.
are not G-conjugate, then f ? f * = 0 eiFGe = e .f .f? F G P f e
i G rn)
= 0
-1
and f i
are G-conjugate, say f , = g fig.
Then -1 (g e i g ) f ,
=
g
-1
(eifilg = 9-le.g z and FNe are in the same block
and so the irreducible FN-modules FN(g-leig)
FNf,.
But N
FN(g-’eig)
is a p‘-group, so FNe FGe
2
FG(g-’eig)
By the foregoing, there is an a E eiFGe isomorphism. y E FGe.
Hence there is a b E eFGei
Therefore sub
=
z
for all
IC
2
and hence
FGei
such that f
:
FGei-
such that fi’(y) in FGei.
Thus
ei = eiab = ( e .a)b = ab and so for any c E eiJ(FG)e ,
c
= e . c = (able = a ( b c ) E eiFGeJ(FG)e 2
This proves (2) and hence the required assertion.
=
FGe
yb
is an FG-
for all
TBE LOEWY LENGTH OF INDUCED MODULES
S t e p 2.
Here we prove that
for a l l
J ( F G l n F = J(E)"#
399
n
>
1.
By Lemma 3.1, it s u f f i c e s t o v e r i f y t h a t
J(FG)ne The case
=
>
for a l l n
( F G e ) (eJ(FG)e)n
n = 1 being proved i n S t e p 1, w e argue by induction on n.
1
So assume
that
k J(FG) e
= ( F G e ) (eJ(FG)e)
J(FG)
Multiplying ( 3 ) on t h e l e f t by
J(FGIk+le
for a l l
k
for a l l
k
n
(3)
n
(4)
gives =
whereas multiplying (3) on t h e r i g h t by
k+l
(J(FG)e) J(FG1e
G
gives
k ( J ( F G ) e ) ( J ( F G ) e ) = ( F G e ) (eJ(FG)eIk+l f o r a l l k 4 n
(5)
Thus we have
J(FG)~+'~= (J(FG)~)~+' = ( J (FG)e )
( J (FG)e )
= (J(FG)"e)
(J(FGle1
( W e ) (eJ(FG)e)n+l
=
proving t h a t (31 holds f o r
S t e p 3.
k
=
(using ( 4 ) with
(using ( 4 ) with
k
=
n)
k - n - 1)
(using (51 with k = n )
n + 1
we nou complete t h e proof by showing that
"fl = FG-J ( F H )"#
J(FG)
We keep t h e n o t a t i o n of Lema 3.2 and p u t
E
= End(#).
for a l l
n 2 1
Then
FH
(by Lemma 3.21
as required.
.
CHAPTER 6
400
Let N
3.4. COROLLARY.
algebraically closed field of characteristic p
let F be an
G,
be a p-nilpotent normal subgroup of
and let V be a principal inde-
n
composable FN-module.
Then
LCFsJ) is equal to
the nilpotency index of
(8) I.
J (End
FG
Our proof of the equality J ( F G ) n F
Proof.
only on the fact that if N
=
J ( E ) n V G in Theorem 3.3 relied
is a pr-group then any two principal indecomposable
FN-modules in the same block are isomorphic.
Since the latter property also
holds under present hypothesis (see Corollary 3.10.101
.
for all n ? 1.
S(FGInf = J ( E I n f l of Lemma 3.1.
3.5. COROLLARY.
deduce that
The desired conclusion now follows by virtue
Further to the assumptions and notation of Theorem 3.3, assume
is an algebraically closed field of characteristic p .
that F exists
, we
Z2(H/N,F*) such that L ( f i 1
c1 E
is equal to the nilpotency index of
Furthermore, if for all q # p
F'(H/NI.
cyclic, then Proof.
L(PI
Then there
the Sylow q-subgroups of H/N
are
is equal to the nilpotency index of F ( H / N ) .
By Theorem 3.4.21,
F"LY/N)
End(#)
for some
c1 E
Z*(H/N,F*).
FH Hence the first assertion follows from Theorem 3.3(ii).
The second assertion is
a
a consequence of the first and Lemma 3.4.9.
G
We close by providing a sufficient condition under which L ( V I the nilpotency index of FP, where P Theorem 3.3.
i s
is equal to
a Sylow p-subgroup of the group H
in
To achieve this, we need to establish some preliminary results
concerning twisted group algebras. 3.6. LEMMa.
Let K
the order of G. Proof. reducible.
be an arbitrary field whose characteristic does not divide
Then, for any c1 E Z 2 ( G , K * ] ,x"lG
is a semisimple K-algebra. V
It suffices to verify that every K'G-module Assume that F/
is a submodule of
V.
over K, its subspace W has a complement in V ,
Since
is completely
v
is a vector space
say
V=W@W'
Let 0
:
V+
W
be the projection map, and let $ : V
-+
V be defined by
THE LOEWY LENGTH OF INDUCED MODULES
u E V and y E G,
Bearing i n mind t h a t f o r a l l
w e deduce t h a t
h'"
i s an
V E
v.
-1 -1 --1 E kr. Then, f o r any z E G , v E W and so v) = z v. -1 z e z v = U and $ ( V l = 0 . S e t t i n g W" = Ker$, it follows t h a t
r
2,
- -
a
If G-submodule of
V
Then, by t h e above,
v
Thus
a
i s a K G-homomorphism.
$
Assume t h a t Accordingly,
W"
=
N
Let
W" n
such t h a t 2,
W"
-$W) E
G
infB(x,y) = 1 f o r a l l
N
3 . 7 . LEMMA.
z,g E N
F i n a l l y , suppose t h a t
V = $(U)
+
( u - $ ( u ) ) EW +PIrr.
T(N)
Let
N
be any f i e l d .
For any
Z2(G,K*) defined by
B(zN,yNl and hence
I(N)
I n what follows we w r i t e
so t h a t
K
and l e t
CinfBl ( z , y l =
KinfB
W = 0.
and so
B E Z2(G/N,K*), l e t infB be t h e element of
identifiable.
e(z
and t h e r e s u l t follows.
be a normal subgroup of
Note t h a t
401
and
KN
{;1I #] M E N}. G,
let
B E Z2(G/N,K*)
a = infa.
(il
KaG-I(N)
(ii) I f
N
is an i d e a l of
KaG
such t h a t
i s a normal p-subgroup of
KO"G.I(N) and t h e nilpotency index of (iii) I f
N
KaG*I(N)
G
KaG/KaG*I(N)
and charK = p,
2
K B (G/N)
then
5 J(KaG) i s equal t o t h a t of
i s a normal Sylow p-subgroup of
are
f o r t h e augmentation i d e a l of
i s t h e K-linear span of be a normal subgroup of
KinfBN
G
J ( f G ) = KaG*I(N)
J(U)
and charK = p ,
then
and l e t
CHAPTER 6
402
In partfcular, by (if), the nilpotency indices of SCpGl
and J(U1 coincide.
(i) Define a surjective K-homomorphism f : KaG
Proof.
B
-+
K (G/N)
by
f(5, = p Then, for all z , y E G ,
(g E G )
we have
is a surjective homomorphism of F-algebras.
Thus f
We are therefore left to
verify that Kerf = f G * I ( N ) . If n
E
N,
n-
then
iE
so I ( N )
Kerf,
be a transversal for N
elements
-
t,t E T.
and let S be the F-linear span of the
G
in
It will next be shown that
K'G
= S
+ KaG*I(Nl
and for this it suffices to verify that each
g
and n E N .
for some t E T
= tn
g=G =
proving that
9E
Fix
Kerf.
5 E
S
+
pG0r[N)
xi
E F,ti E
(6)
gE
S
+ KUG.T(Nl ,g
E G.
Write
Then
a-l(t,nltn
=
a-'(t,n)'i + u-l(t,n~t(n-i),
and hence ( 6 ) is established.
Then, by (61, z 3:
where
and thus'
5 Kerf
A?G*T(NI Let T
5 Kerf
= A
can be written in the form
... + XnTn+y
t1 +
1
T, 1 G i G n , and y E K ' G * I ( N ) .
Because y
E
Kerf,
we
have
f(z) = A T N + 1
which implies that 1 Kerf
K%*I(N)
(ii) Fix n
E
=
... =
A
n
1
... +
= 0 and so
t N
n n
x
E
=
0,
KaG - l ( N ) .
This shows that
and hence that Kerf = K D " G * I ( N ) , as required. N
and write
(N(
=
pd
for some d 2 1.
Since a(z,y) = 1 for
THE LOEWY LENGTH OF INDUCED MODULES
n-
Bearing i n mind t h a t t h e elements we conclude t h a t
I(N)" = 0
KaG , so KaG*I(N)
potency index of
c o n s t i t u t e a K-basis f o r
1,
rn 2 1.
f o r some
= I(N)KO"G
KaG.I(N) c - J(K'G).
Thus
7, n #
403
By (i), KO"G.I(N)
I(N),
i s an i d e a l of
and t h e r e f o r e
Furthermore, t h e above e q u a l i t y shows t h a t t h e n i l -
J ( W ) = I(N) coincides with t h a t of
KaG.I(N).
(iii) D i r e c t consequence of ( i ) ,(ii) and Lemma 3.6.
3.8. PROPOSITION.
Let
N
be a normal p-subgroup of
p
i c a l l y closed f i e l d of c h a r a c t e r i s t i c
6 E Z2(G/N,F*)
(i) There e x i s t s
N
(ii) I f
and l e t
such t h a t
i s a normal Sylow p-subgroup of
c1
G,
c1 €
F
G, l e t
be an algebra-
z2( G , F * ) .
i s cohomologous t o
ir.fB.
then
= F'(G/N)
(a)
F%/J(F"G)
(b)
The nilpotency index of
Proof.
(i) By Lemma 3.4.9,
FaG
i s equal t o t h a t of
J(FN1.
we may harmlessly assume t h a t f o r a l l x,y E N
Fix
g E G,n E N
and w r i t e
pa
=
(NI
.
(7)
Then
----1 = A 9ng-l
gng k E F*
f o r some
and so by (71,
d Hence
hP
= 1,
so
h = 1 and t h u s c--
gng Clearly
F'G.I(NI
of t h e set
1;
i s a l e f t i d e a l of
- iln E N } ,
-
-1 =
gng-l
a
F G.
FaG*I(N1
I(N)
i s t h e F-linear
(8)
span
p G * I ( N ) i s t h e F - l i n e a r span of t h e s e t
(;(;lg I E) G n W e claim t h a t
Because
f o r a l l g E G,n E N
i s an i d e a l .
E NI
To s u b s t a n t i a t e our claim, we need only
404
CHAPTER 6
as claimed.
T be a t r a n s v e r s a l f o r N i n G containing 1, and l e t S be t h e F-
Let
{ilt E TI.
l i n e a r span of
We now prove t h a t
FaG Given
and so
t
E T
and
F'G*II(NI
x,y E N ,
=
S 63 P G * I ( N )
a s F-spaces
(9)
we have
is i n f a c t t h e F-linear span of
Observe t h a t t h e l a t t e r s e t combined with
T c o n s i s t s of
IGI
elements.
The
equality
now proves ( 9 ) , Setting
t N = Z + FaG.I(N),
algebra with t h e elements e x i s t unique
i t follows from (91 t h a t
{ G I t E 7')
t E T , n E N with
a s a basis.
t t = tn. 1
2
Given
Setting
6 ( t l I , t 2 N ) = a ( t l, t 2 ) a - l ( t , n ) we have
and so
FO"G/FO"G.l(N)
i s an
t ,t2E T , t h e r e 1
F-
GROUPS OF P-LENGTH 2
Given Define
g E G,
t h e r e e x t s t unique
h(g)
F* by
h : G-
=
tCg1 E T
405
and
a ( t ( g ), n ( g ) ) , s e t
n(g) E N
y = a(6h)
{ i l g E G}
,
i s an F-basis of
f : F aG
Hence t h e map homomorphism.
+
Y(x,y)
proving t h a t
a = inf8.
3.9.
let
N
Let
$I
G,
f(z) 2
t(g)n(gl.
= h(g)g,g€G.
determines an F-algebra
=
we have
Y
Hence
=
i n f @ a s required.
F be an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c
v.
If
P
G
of
E Syl
P
H
and l e t
be t h e i n e r t i a group of
(HI assume t h a t PN
i s equal t o t h e nilpotency index of
a H.
Then t h e
.
J(FP).
Apply Corollary 3.5 and P r o p o s i t i o n 3.8 ( i i l (bl
4. GROUPS OF p-LENGTH
.
2
G
Throughout t h i s s e c t i o n ,
p.
E
be a normal p'-subgroup
Loewy length of
istic
=
by a cohomologous cocycle, i f necessary, we may assume
c1
an i r r e d u c i b l e FN-module
Proof.
g
Now apply Lemma 3.1.
COROLLARY.
p > 0,
x,y
defined by
B(xN,yN).
=
(ii)By ( i ) ,r e p l a c i n g
that
FB(G/N)
Thus, given
and
-
-g
FYG with x u = Y(x,g)xg and -I
Then
with
denotes a f i n i t e group and
F a f i e l d of character-
All conventions and n o t a t i o n s adopted i n t h e previous s e c t i o n remain
i n force. Our aim i s t o provide circumstances under which t h e i n e q u a l i t y of Theorem One of our r e s u l t s w i l l prove t h a t t h i s i s always
2 . 1 2 ( i i ) becomes an e q u a l i t y . t h e case i f 4.1.
H
LEMMA.
is p - n i l p o t e n t with elementary a b e l i a n Sylow p-subgroups. Let
V and
W
be FG-modules.
=
(J'V/Ji+'V)
xij (0 G
< G n-1,O
G j G
m-l),
Set
n
@ (jW/>+lh')
F where
J = J(FG).
Then
=
L ( V ) ,m = L ( W )
and
CHAPTER 6
406
Proof.
Y.. $3
Put
=
8 $W.
V'J
Then
F
3 Y
-
'ij
.+Yi
i+l,j
,j+l
and
+
Y . ./(Yi+l 23
NOW
Y
let
=
1
C Y.. i + j = i $3
for
R G n+m-l.
0
O = Y
c
C Y
= x.. w
Yi,j+l)
n+m-l - n+w2 -
Then
''*
CY0=V8W F
and t h e canonical map
?
YR
yiji+j=R y i e l d s a s u r j e c t i v e homomorphism
y xij - i +@j = tYzj/(Yi+l,j + Yi,j+ll- xR/xk+l =%
i+j=R where
X
'
=
c
X...
It follows t h a t
i+j=R ' 3
thus proving t h e a s s e r t i o n . 4.2. Let
LEMMA.
Let
N
be a normal subgroup of
W be an FG-module and s e t
V
=
G
WN)
G
such t h a t
W 8 (lNl F
L(V) Proof.
Put
I(M)' = Si*FM,
M
=
where
G/N J
t(G/N1 + L(W)
and view = J(FG1,
FM
ni.
.
G/N
i s a p-group.
Then
1
a s an FG-module v i a
F M S (1 ) N
G
.
Then
and
r(M1i / r ( ~i+l) = ni for suitable integers
-
G
iG
In t h e n o t a t i o n of Lemma 4.1, we t h e r e f o r e have
2
GROUPS OF P-LENGTH
407
a s required. 4.3.
LEMMA.
G
Let
be a p-nilpotent
f i e l d of c h a r a c t e r i s t i c
H
grc p
V
W
and l e t
P
1
fl
L(P(V))
and
=
There e x i s t s a unique i r r e d u c i b l e FG-module Proof.
and
Y
wG
P(V)
That
Y.
3
XN s YN
4.4.
=
t(H/N)
THEOREM.
p > 0 and l e t
has a normal p-subgroup
H
and l e t
P
(il
(iil
P,(VI
G.
of
Q = 0 ,(H), zz
(iiil
If
did,
then
S
then
Let
x
t(H/N)
X such t h a t XN
W.
W.
WG
P(X) 2 P(Y)
Then
X with XN
and
W
Finally, the equality
N
Let
Let
be an a l g e b r a i c a l l y closed f i e l d of
V be an i r r e d u c i b l e FG-module. G
with
W
F
=
NH
N n H
and
W
G
1 f o r some p - n i l p o t e n t
=
be an i r r e d u c i b l e submodule of
T be t h e i n e r t i a group of
Assume t h a t
V Q , where
H.
in
WC
L(PG(V))2
complement i n
vN,
i s a p a r t i c u l a r case of Corollary 3.5.
(Lorenz (1985)).
characteristic
subgroup
i s a submodule of
The e x i s t e n c e of an i r r e d u c i b l e FG-module
follows from Theorem 3.4.5 and Lemma 3.4.9.
L(P(V))
W
i s a p a r t i c u l a r case of Lemma 3 . 4 . 2 6 ( i ) .
be i r r e d u c i b l e FG-modules with
X
hence
be an a l g e b r a i c a l l y closed
be an i r r e d u c i b l e FN-module with i n e r t i a
is an i r r e d u c i b l e FG-module such t h a t
P(m (ii
F
N = 0 ,(GI
where
(i) I f
p
group, l e t
G
t(T/Ql + L ( I V H )
)
-
1 with e q u a l i t y i f
T n S has a normal
S E syl (H)
P
i s elementary a b e l i a n o f o r d e r
n
p
and
pd
is t h e p-part of
F G L ( P G ( V )1 = (n-d) (p-1) + L ( ( V H ) 1
(iv)
If
VQ i s i r r e d u c i b l e , then L(PG(V))2 t ( S )
+ t(N) - 1
and e q u a l i t y holds i f and only if t h e FG-module
is completely reducible.
Here w e view
I(N)i/IUV)i+l
as an FCmodule by l e t t i n g
408
G
CHAPTER 6
a c t by conjugation.
(il
Proof.
and ( i i ) . By Lemma 4 . 3 ( i )
L ( 8 l = t(T/Ql.
Invoking Theorem 2.12,
, we
PH(Y) 2
have
#
and
we i n f e r t h a t
L ( P G ( V l ) 3 t(T/Q) + L ( ( V H ) G )
-
1
and
= (lHfQ 8 =
PG(V,
wc
F
U such t h a t
By Lemma 4 . 3 1 i i l , t h e r e e x i s t s a unique FT-module induced module component.
X
=
>
U
UH
i s i r r e d u c i b l e and
Assume t h a t
S
C H.
X
Then
-
VH,
H
Q
W.
2
The
s i n c e both have a common FQ'l' nS
i s a normal complement f o r i s normal i n
U
S
in
and p u t
and
= ( uH ) " w x
Y X
X
Thus w e have
I t follows from Lemma 4 . 2 t h a t
LIPG(V1l Since
G/
(iiil
Assume t h a t
p a r t of
dimV.
2
T/Q,
5'
t(G/)
+ LC(VH)G l
-
1
t h e required a s s e r t i o n follows.
is elementary a b e l i a n of o r d e r
By assumption on
S, T
pn
and l e t
pd
be t h e p-
n S has a normal complement i n S,
and
F by Corollary 3 . 2 . 5 ,
t(!P/QI where
pk = IT n S I .
k =n -d
(ivl
By ( i i ) ,w e have
k(p-1)
+
1
U be a s i n t h e proof of (ii)so t h a t
Let
i s n o t d i v i s i b l e by
dimU F Thus
=
p
and hence t h e p-part of
dimV
F
.
which proves t h e a s s e r t i o n by applying (ii)
equals
U
H
VH.
Then
p d = IH/Tl.
409
GROUPS OF P-LENGTH 2
L(PG(V)l
V
Let N
Let n
proof.
This completes the proof of the theorem.
of G.
FG
:
n(y)
NH and N n H
= 1
Then Kern =FG.I(N)
J(FH) = n(J(FG)I .
5 J(FG).
Thus
2
-
for some
FG-I(N) 5 J(FG1 and therefore x
E
Z
E
,TI
Further-
Now, if x € J(FG),
y E FG*I(N) and so
Conversely, let x E FG*I(NI + J(FH1.
and hence r(z] = n ( Z l
some y E J ( F H ) X- z E
we have FG*I(N)
for some y € JtFH1.
r € FG.I(N) + J ( F H ) .
=
be the natural projection.
and, by Lemma l.l(iii),
more, by Proposition 3.l.l(il, =
G with G
Then
FH
-+
is the identity map on FH
then n ( x ] = y
1
(
be a normal p-subgroup of
for some subgroup H
-
L ( ( V H )G
6l3 [I(N)i/I(N)i+ll i20
@
F
4.5. LEMMA.
+
t(S)
G L((VH) 1 2 t (N) with equality if and only if
and, by Theorem 2.12(iii),
is completely reducible.
=
Then
J(FG1.
T(X) =
y
for
Hence
J(FG).
We are now ready to establish the final result of this section. 4.6. THEOREM. CLorenz (19851).
G
=
NH and N n H
complement S ,
=
G with
be a normal p-subgroup of
for some Frobenius group H
1
where
Let N
Q and S are p '
with kernel Q
and
and p-groups, respectively.
Then the
following conditions are equivalent: (i)
t(G1
(iil
=
t(S)
+ t(N1
-
1
6l3 (I(N)i/I(N)i+l) is a completely reducible FG-module, where G acts by
$20 conjugation.
(iii)
C xqs +Q
Proof.
- E xq
E I(N)'"
for all i 2 0, all x
E
I(N)'
and all s E S.
CFQ
(i) * lii):
algebraically closed.
By Corollary 3.1.18, we may harmlessly assume that F By taking
v=
lG
t(G) 2 L(PG(V1l and so L(PG(V)) = t ( S )
+ t(N) -
1.
(ii) * (iii): owing to Theorem 3.7.7,
in Theoreq 4.4, we have t(S)
+ t(N) -
1
Now apply Theorem 4.4(iv).
our assumption on H
implies that
is
410
CHAPTER 6
S(FH) = I ( S ) e where
c
e
q
= 1~1-l
SfQ Set X
FG*I(N) and Y
=
= I(S)e.
Then, by C1) and Lemma 4.5.
S(FG)
=
X+Y
(2)
i+l By hypothesis, J(FG) annihilates each I(N)i/l(N) .
i
I(N)i/I(N)i+l,so Y annihilates each I ( N )
.
x
But
Therefore
also annihilates
(S-1) 1 4
annihi-
SfQ lates J(N)i/l(N)i+l
for all i 2 0 (s-1)(
x
and all s
c
q)z =
- c
zqs
*Q
+Q
Since for any z E I(N)
S.
E
zq
SfQ
the required assertion follows. (iii) * (i): An easy calculation shows that for all i 2 0 e * I ( S ). 1 ( 8 ) i e Set 8
=
t(N)
+
t(S)
-
1.
be written as a product of then a
=
+
eI(N)’+le
=
e * l ( N ) i l ( S ) e+ eI(NIi*le
In view of ( 2 1 , we have to show that if c1 E FG
G
factors each of which belongs to either
involved in a.
t(S). of c1 = 0 .
kz
X
can or
Y
0.
We argue by descending induction on the number
that
(3)
C
t(N1.
If
&,
2 t(N1,
then
c1 E
Xt(N’
,k
= kx(C1l
= 0.
Then the number of factors from Y
of factors from
x
We may therefore assume involved in
c1
is at least
Let n
= n (a) denote the length of the longest consecutive subproduct Y Y Clearly, if n 2 t ( S ) then consisting entirely from factors in Y .
Y
So
belongs to
assume that n < t ( S ) . Then a contains a subproduct which either an Y n YX’Y or to Y ’XiY(i > 0 ) . We examine the first case, the second
being entirely analogous.
Now
n
Thus we have c1 =
c1 1
+
c1 2
with c1 ,a2 E J(FG)
9“
,
but
kz(a2) > kx(u)
and
411
GROUPS OF P-LENGTH 2
Q (a) = Q (a) ,n (a1 > n (a). By induction, we deduce that a = a
x
x 1
hence
a
= 0.
Y 1
Y
This proves the theorem.
1
2
=
0
and
This Page Intentionally Left Blank
413
7 The nilpotency index Let G be a finite group, let F be a field of characteristic p be the nilpotency index of J ( F G ) .
Our aim in this chapter is to provide some
important information on the number t ( G ) . bounds for t ( G ) ,
and let t ( G )
After examining an upper and lower
we investigate circumstances under which the equalities hold.
Almost all of the results presented pertain to p-solvable groups since virtually A number of interesting examples is also
nothing is known in the general case. provided. 1. SOME RESULTS ON p-SOLVABLE GROUPS
Throughout this section, G denotes a finite group and p write
GL(n,p)
and S L ( n , p )
group of degree n
a fixed prime.
We
for the general linear group and the special linear
over the field P
P
of p-elements, respectively.
Our aim
is to record some group-theoretic results required for subsequent investigations. First we record the following conventions. of
If M
5N
are normal subgroups
G, we set
so that
‘G/M (N/W Expressed otherwise, CG(N/MI where the image of g E G
Let @
and consider
V
p
=
CG(N/M)/ M
is the kernel of the homomorphism
is the automorphism of N/M
nM 1.1. LEMMA.
=
G
-+
Aut(N/M),
given by
W g-lngM
be the Frattini subgroup of a p-group
G,
let
(G:@) = p
G / @ to be a d-dimensional vector space over the field
elements (see Lemma 4.3.1).
Then
L?
P
d
of
414
(il
CHAPTER 7
C
=
1
,x2 ,...,xn > i f and o n l y i f G = < z ,x
particular, i f
1
,...,x > ,
@zl,@x2,...,0xn. In
i s spanned by
n 2 d.
then
G can be generated by e x a c t l y d elements.
(ii) The group
{x1 ' z 2 ,...,xd proof.
2
V
G
generates
(i) If
,...,Oxd 1
{ax
i f and only i f
G = <x ,x2,...,zn>, then
V
The subset
i s generated by
Ox l , . . . , @ x n .
d
elements, w e
Since a d-dimensional v e c t o r space cannot be spanned by less than
X
=
> d.
n
have
.
{xl,. . ,x 1 ,
G = <X,O>,
we have
{Oxl
,...,Ozd 1
Conversely, i f
G = <x
@z,...,Oxd
t h i s case As
1.2.
Let
and
,...,xd > ,
@
t i ) we get
then by
G/O
can be excluded
G=
,...,Xd>.
(Z
,...,O X d > .
V =
In
v. '
AuttG)
and t h e f a c t o r group
d
I
divides
pm
V = C/O. of
U
G
Since
@
Aut(G)/I
Thus
.
n (~'-1) i=l
i s a c h a r a c t e r i s t i c subgroup of
induces an automorphism
i s obviously a homomorphism f o r
d e f i n i t i o n , we have (ii) Fix a b a s i s
( y I , y ,,..., y d ) - -
Clearly we g e t
Oxi
G
- d(d+ll/2.
every automorphism
=
t h e s e t of a l l automorphisms of
I i s a p-group of order l e s s than p 'n-d'd.
( i ) Put
f(U1
and l e t
GL(d,p).
IAut(G1
Proof.
G,
invariant.
i s a normal subgroup of
The subgroup
rn = nd
I
Denote by
i s isomorphic t o a subgroup of
$(@xi)
then by ( i ) ,w e have
be t h e F r a t t i n i subgroup of a p-group
IGI = p".
I
(i) The set
all
V,
i s a b a s i s of
which leave every element of
U-
O
an a p p l i c a t i o n , we prove t h e following r e s u l t of P. H a l l (1933).
(G:O) = pd
where
since
a r e l i n e a r l y independent and so form a b a s i s of
1
THEOREM.
(iil
G = <X>
so
Setting
G.
from any generating s e t of (ii) I f
.. .,OXn > .
=
V
Conversely, assume t h a t
V.
i s a b a s i s of
1x1
I
= Ker
=
for a l l
p (n-d)d.
<,
into
of Aut(V)
V.
The map
2
GL(d,p).
By
and ( i )follows.
{Ox l , . . . ,Ox 1 of d such t h a t
Aut(G)
f(U)
G,
y
V
iE G
over and
F
Let
P'
Qi
=
Oxi
Because any element
t h e group
$
X be t h e t o t a l i t y of for a l l of
i
E {1,2
1 satisfies
I a c t s on X by t h e rule
,...,d ) .
SOME RESULTS ON P-SOLVABLE GROUPS
(Y,#Y,
, . . . I
Ydl
-
Let S be an orbit under this action. the stabilizer I(Y1 of an element Y
$(yil
=
y
i
=
=
and
trary orbit, each orbit is of length
,...,y d ) .
(y,,y2
is the index of
If $ E I(Y1, then
Because G = < y ,y,
Hence I(Y) =
1.
JIQdl)
I...,
Then the length of S
for all i E {1,2 ,...,d}.
1.1, we get $
(W,1
415
IS1 = 1 1 1 .
,...,yd>
Since S
1 1 1 divides
(11; hence
by Lemma is an arbi-
1x1.
This shows
that I is a p-group whose order is a divisor of p (n-d)d d(d-1)/2 is a consequence of the fact that IGL(d,p1 I = p
The last statement n lz (pi-l).
i=1
For future use, we need an important property of the upper p-series of a psolvable group, known as the Hall-Higman's lemma (Hall and Higman (1956)).
This
is recorded in 1.3. LEMMA.
Let G be a p-solvable group and let
.
,
subgroup of Op ,p(GI/Opt (G)
@/o ,(GI P
be the Frattini
Then
In particular,
(0 (G11 C_ 0 (GI if 0 ,(GI = 1
(bl
C
(c)
The group G/O
P
G P
P'IP
P (G) is isomorphic to a subgroup of Aut(0
P'rP
(G)/@)
and 4ut(0
(G)/@)
P'SP
0
P'rP
Put K = C (0
(GI C_ K.
G P'tP
(GI/@) and observe that, by Lemma 4.3.1,
By the definition of
p-group, so that, if K
3
0
P'rP
0
(G) of order prime to p .
0
LG)/Op,(G1
P'PP
P'IP
GL(d,p)
(GI/@I
where pd = Proof.
2
(GI, K
0
P'rP
P'rP
(GI cannot be a nontrivial
must contain an element g
Then g
of order prime to p.
(G), K / O
not in
induces an automorphism of But an automorphism of a p-group which is
the identity modulo the Frattini subgroup has order a power of p (Theorem l.Z(ii)). Hence g
induces in 0
P'rP
(G)/@ an automorphism which is not the identity,
contrary to the definition o f
K.
Thus K = 0
P'rP
(GI, as required.
416
CHAPTER 7
W e c l o s e t h i s s e c t i o n by recording some standard group-theoretic f a c t s required f o r subsequent i n v e s t i g a t i o n s .
The p r o o f s of t h e s e f a c t s a r e r a t h e r
involved and t h e r e f o r e w i l l be omitted. I t w i l l be convenient t o introduce t h e following terminology.
i s invoZved i n
K
a group
H
of a subgroup 1.4. THEOREM.
p
= 3
provided
P
K
i s isomorphic t o a homomorphic image
G.
of
G
Let
p
be a p-solvable group and assume t h a t e i t h e r
G,
i s a Sylow p-subgroup of
5
or
G.
SL(2,3) is not involved i n
and
(i) I f
G
W e say t h a t
P
then every normal a b e l i a n subgroup of
Op, ,p(G).
is contained i n
G
(ii) I f a Sylow p-subgroup of
i s of c l a s s a t most
2,
G
then
has p-length
1.
Proof.
( i i l i s an immediate consequence of ( i l .
p,
For a given prime
z
P
x z!
and
P
(il
abelian of o r d e r
we d e f i n e
p
If
p 3 and exponent G
Property
t o be t h e semidirect product of SL(2,p) on z!
P
i s odd, then t h e Sylow p-subgroups of
G = Q d ( 3 ) , then
Proof.
Qd(p)
w i t h r e s p e c t t o t h e n a t u r a l a c t i o n of
SL(2,pl
1.5. THEOREM.
(ii) I f
.
For t h e proof of (i) r e f e r t o Gorenstein (1968,p.2341.
&d(pl
Let
1.6. THEOREM.
i s of 3-length 2.
CG(P n 0 (GI P'rP
II (GI P
Proof.
5 Opr,p (GI
,
Z (P) 5 Op ,p (GI
and, i n p a r t i c u l a r ,
See Gor n s t e i n (1968,p.228) Let
Sylow p-subgroup
G.
P If
p
.
A
and Huppert (1967,p.691)
be an odd prime, and l e t
such t h a t
P of c l a s s
G.
C(P)
1.7. THEOREM.
involved i n
be t h e p-length of
P
and
(see a l s o Koshitani (1982a,p.33)).
G be a p-solvable group with Sylow p-subgroup
II (GI
and l e t
(ii)
a r e non-
For t h e proof of ( i l , r e f e r t o Glauberman (1971, Example 1 1 . 4 ) .
Gorenstein (1968,p.203)
(il
zp.
p.
Property (ii) is an easy e x e r c i s e based on Glauberman (1971,pp.32-331
c(P)
x
0 (GI # 1, G P
G
be a f i n i t e group with a
i s p-solvable and
i s an a b e l i a n normal subgroup
P,
then
Qd(p) is not A
20
P ' IP
(G).
SOME RESULTS ON p-SOLVABLE GROUPS
417
See Glauberman (1986, Lemma 6.31 and Gorenstein (1986, Theorem 8.1.3)
Proof.
( f o r a s h o r t proof r e f e r a l s o t o Koshitani (1982a, Lemma 2 ) .
1.8. THEOREM.
(Hall-Higman).
p.
(il
I? = p
n
P
g be an element of
Let
g
nomial of
in
,
be a p-solvable group of l i n e a r transform-
0 (G) = 1 a c t i n g on a v e c t o r space over a f i e l d
a t i o n s i n which
teristic
G
Let
V
(X- U p ,
is
G
of o r d e r
F
of charac-
Then t h e minimal poly-
where e i t h e r
or
n
n
(ii) There e x i s t s an i n t e g e r
q
pn.
G
and t h e Sylow q-subgroups of
p
such t h a t
a r e nonabeltan.
-
1 i s a power of a prime
I n t h i s case, i f
n
is t h e
least such i n t e g e r , then
n
n-n
O(p O - 1 ) G P c p n
p Proof.
See Gorenstein (1968, p. 359).
For f u t u r e use, w e a l s o record t h e following g e n e r a l group-theoretic 1.9. THEOREM.
pi
where t h e Then
A
Let
A
a r e d i s t i n c t primes, and l e t
has a complement i n
Proof.
p # 2
Let
Here
Dn
n
G
>2
Mn(pl
Q for
1.10. THEOREM.
D
and
n
i s defined f o r
n2 3
P
Let
i=1 of G/A.
P..
h a s a complement i n a l l
n
b 4
p = 2
if
p
i s odd,
p = 2
p
= 2
P is
and and
n
for
2n
.,Q
.
and
n
>
3
n > 3,
let
2n.
S ( n ) be
defined by
p
n
which c o n t a i n s a
Then
isomorphic t o
= 3,
n > 3,
Finally, for
be a nonabelian p-group of o r d e r
n- 1 c y c l i c subgroup o f o r d e r p
(iii) I f
A
s i x npi ’,
be t h e d i h e d r a l and g e n e r a l i z e d quaternion groups of o r d e r
t h e semidihedral group of o r d e r
(ii) I f
2
=
as follows:
and
(i) I f
be a Sylow p.-subgroup
Pi/A
i f and only i f
IG/Al
with
See Huppert (1967, p. 124).
Recall t h a t t h e group
if
G
be an a b e l i a n normal subgroup of
facts.
then then
P
Mn(P)
i s isomorphic t o
D3
or Q3
P i s isomorphic t o M n ( 2 ) , Dn , Qn or S(n)
CHAPTER 7
418
Proof.
i s s a i d t o be regular, i f f o r any
P
A p-group
x1 ,x2,... ,xn i n
1.11. THEOREM.
<x,y
>'
P
Proof.
i s of exponent
and
P
p,
Throughout t h i s s e c t i o n ,
B
P
then
p > 0.
i s a block of
FG,
G
i s regular
P
then
i s regular
.
i s r e g u l a r , then
P
is abelian.
.
t
2. UPPER AND LOWER BOUNDS FOR
p,
i s l e s s than
P
See Huppert (1967, p.3221.
characteristic
there e x i s t
P be a p-group.
Let
p = 2
(iii) I f
x,y E P
such t h a t
( i ) I f t h e nilpotency c l a s s of
(ii) I f
m
See Gorenstein (1968, g.1931.
F
denotes a f i n i t e group and
As u s u a l ,
t(G1
then we w r i t e
an a r b i t r a r y f i e l d of
i s t h e nilpotency index of
t(B)
J(FG).
f o r t h e nilpotency index of
If
J(B).
We s t a r t by recording some p r o p e r t i e s which a r e immediate consequences of previously e s t a b l i s h e d r e s u l t s .
2.1.
PROPOSITION. (Passman (1970a), Tshushima (1967), Wallace (1968) 1 .
be a p-solvable group of o r d e r
Proof.
2.2.
p rn,
Apply Proposition 3.1.9 Let
LEMMA.
group.
a
B
where
p 1, m and a >
and Corollary 6.2.8.
be t h e p r i n c i p a l block of
FG,
1.
Let
G
Then
.
where
G
i s a p-solvable
Then
t ( B 1 = tCG;/O , ( G I ) P Proof.
D i r e c t consequence of P r o p o s i t i o n 3.1.20.
.
The f i r s t e q u a l i t y i n t h e r e s u l t below i s contained i n a work of Clarke (1972). 2.3.
PROPOSITION.
and l e t
B
Let
G be a p-solvable group of p-length 1, l e t P E Syl ( G )
P
be t h e p r i n c i p a l block of
FG.
Then
t(G1 = t ( P ) = t ( B )
UPPER AND LOWER BOUNDS FOR $(GI
Proof. p-length
419
0 , ( G ) P / O ,(GI E Syl ( C / O , ( G I I
Note that P
P
P
P
and G/O , ( G I
P
is of
P
Hence, by Lemma 2.2, it suffices to prove that * ( G I = t ( P ) .
1.
By
Corollary 3.1.18, we may harmlessly assume that F is algebraically closed. Furthermore, by Proposition 3.1.8(ii), we may also assume that G
is p-nilpotent.
The desired assertion is now a consequence of Morita's result (Corollary 3.10.10). An alternative proof, which avoids the use of Proposition 3.1.8(ii) and Corollary
m
3.10.10, can be obtained by applying Corollary 6.3.9.
2.4. PROPOSITION. (Koshitani (1977b), Tsushima (1978bII.
a
group of order p m , and only if P Proof.
and let P E Syl (GI.
i s cyclic.
t(G)
Then, by Proposition 3.16.20,
Hence, by Proposition 2.3,
t ( C 1 = pa.
Conversely, suppose that t ( G I and @/M
Then
P
=
pa if
is cyclic.
Assume that P
p-length 1. 4.3.2,
where p ) m ,
Let G be a p-solvable
=
pU
.
the Frattini subgroup of H/M.
a-r J (FGI
t ( G ) = t(P1
Put
M
and therefore, by Theorem
o , ( G I ,H P
=
is of
G
=
o ~ , , ~ ( G,pr ) =
IH/MI
Then, by Proposition 3.1.9,
5 FG* J (FH)
and so
Hence, by Propositions 2.1 and 2.3,
t(H)
=
pr , it follows from Theorem 4.3.2 that H/M H/O
is cyclic of order p.
is a subgroup of Aut(H/@I. length 1.
Hence P
H/M
Since H/M
1 Aut t H / G ) I
Thus G/H
is a pr-group,proving that G
=
is of order
Hence, by Lemma 4.3.1,
Therefore
p - 1 and, by Lemma 1.3, G/H is of p -
is cyclic as asserted. a
am, u
p-leng.th 1 and o f order p
if the Sylow p-subgroups of If
= p ,
is cyclic.
2.5. PROPOSITION. (Motose and Ninomiya (1975aI).
Proof.
r
t(H/M1
P E Syl ( G I , P
G
>
1, p ) m.
Let G
be a p-solvable group of
Then t(G) = a(p-1) + 1 if and only
are elementary abelian. then t ( G 1 = t ( P )
by Proposition 2.3.
Now apply
Theorem 4.3.2. 2.6. PROPOSITION. (Motose and Ninomiya (1975a1, Wallace (1968)).
Let G
be a
420
CHAPTER 7
U
p-solvable group of order p m , a 2 1, p
1 m.
Then the following conditions are
equivalent: t(G1 = 3
(il
(ii) pa = 3
Proof.
or pa = 4 and the Sylow 2-subgroups of (i)
(iil
:
Let P
C
are elementary abelian.
be a Sylow p-subgroup of G.
By Proposition
1.1,
and p = 2.
This implies that a = 1 and p = 3 or that a = 2 G
since P
is of p-length 1
t(G) = tlP)
In both cases
is abelian (see Proposition 3.16.20). Thus, in case p
by Proposition 2.3.
U
= 4
,
13
Hence
is elementary
abelian by Proposition 2.4. (iii) =. (il : By the foregoing, G
fs of p-length
Now apply Propositions
1.
2.5 and 2.4,
Let G be a finite group of order Pam, a ? 1, p solvable.
k
m, not necessarily p -
It will next be shown that the inequalities
+
t ( G 1 P a(p-11
1 and
given in Proposition 2.1, no longer hold.
t IG)
pa
In order to accomplish this, we first
prove tHo auxiliary assertions. 2.7. LEMMA.
idempotents
Let 1 of
FG.
=
e +
... + e n ,
t ( G 1 = max{R
Since FG = F G e l @
respectively.
, . . . , & , I < maxb
... @ FGe, 1
Since kZ
< Ti
be the Loewy
Then
... T,I
... @ J ( F G )k en
k
,.. .,&,I.
and ri
and J ( F G e i l = J ( F G ) e i
J ( F G ~= ~J V G ) e t~ Hence t [ G ) = max{k
let Ri
For each i E { 1 , 2 , . . . , ? 2 1 ,
and composition lengths of F G e Z ,
Proof.
where the eZ' are primitive orthogonal
for all
,
we have
for all k 2 1
i, the assertion
follows. 2.8. LEMMA.
Let C = (c..) , 1 6 i , j 6 k ,
be the Cartan matrix of FG.
ZJ
k t(G)
max
I
C c..}
lea 3'11 zJ
Then
UPPER AND LOWER BOUNDS FOR
...,FGek
proof.
Let FGel,FGe2,
FG-modules.
t(G)
421
be all nenisomorphic principal indecomposable
Then, by the definition of
C, cij
is the multiplicity of the irre-
FGei,
as a composition factor of ducible FG-module FGe./J(FG)e 3 j Hence, the composition length of FGei is equal to
Now apply Lemma 2.7.
il + c i 2 +
.
c
2.9. EXAMPLE. (Wallace (1968)).
p
1 m,
i
G k.
... + c ik
There exists a group
of order pam, a
G
1,
such that t ( G 1 < a(p-ll
Let G = A
Proof. Since
1 4
,
5
+ 1
the alternating group of degree 5 , and let p
= 5.
I G ( = 60, we have
a(p-ll
+
It will be shown that t ( G ) 4 4. ally closed.
1 = l(5-11
+
1 = 5
We may harmlessly assume that
F is algebraic-
Then it is well known (see Brauer and Nesbitt (1941)) that the
Cartan matrix of FG is
Invoking Lemma 2.8, we conclude that t(G) G 4 as required. 2.10. EXAMPLE. and let F
Let q
be a power of a prime with q
be a field of characteristic.2.
(i) The Sylow 2-subgroups of G (ii)
G.
l(mod 41, let G-PSL(2,q)
Then for some a
are dihedral of order '2
2
t ( G ) > 2a
In particular (by Proposition 2.11, of
.
t(G)
3
t(P)
where
P
is a Sylow 2-subgroup
Indeed, property (i) is a consequence of a general group-theoretic fact
(see Gorenstein (1968),p.418). ically closed.
To prove (ii), we may assume that F
Owing to Erdmann (1977, Theorem 2(a),p.667),
projective indecomposable module Loewy length L ( V )
of
V
V
there is a
in the principal block of FG
is given by
is algebra-
such that the
422
Hence, by Lemma 2.7
as required. Our next aim is to prove that t ( G 1 G dimJCFGl
1
f.
F and to provide necessary and sufficient conditions under which the equality holds. In what follows, we choose a complete set FGe
1
...,FGek
,FGe 2 ,
of nonisomorphic principal indecomposable FG-modules, where trivial FG-module.
C = Cc..), 1 G i , j G k , 23
We denote by
FGel/JCFG)el is a the Cartan matrix of
FG . 2.11.
With the notation above,
LEMMA.
k t ( G ) G max { Z c . . } G max i < j~ = 1 'J i
+
{dimJ(FG)ei
~
F~
1) G dimJ(FG)
k
+
1
F
The inequality on the left was established in Lemma 2.8.
Proof.
k C c..
j=1 '3
is equal to the composition length of N e i ,
we have
k
proving that
k max { C c , . ) 9max{dW(FG)ei 1 ~ j=1 q '3 F
+
11
The inequality on the right being obvious, the result follows. 2.12.
G.
THEOREM. (Motose and Ninomiya (1975a)).
.
Let P be a Sylow p-subgroup of
Then the following conditions are equivalent:
(i) t(G1
=
dimJ(FG)
+ 1
F (ii) G = P is cyclic or Proof.
(i) * (ii):
1
+
G is a Frobenius group with cyclic complement P.
Assume that dimJ(FG1 F dimJ(FG) F
=
+
1 = t(G).
max {dimJ(FG)ei 1
+
1)
Then, by Lemma 2.11,
t(G1
UPPER AND LOWER BOUNDS FOR
Since dimJ(FG)el 2 IPI
-
423
(by equality (11 in the proof of Theorem 3.8.41, it
1
F follows that for all i # 1
J(FG) = J(FGlel and J(FG)e. = 0
z
B
This implies that the principal block module.
Hence, by Proposition 3.8.10, B
3.1.20,
3
FP.
that B = FGe
1
Since
1
and hence that JCB)
F G
and so, by Theorem 3.8.11, t(G)
=
G
FG
contains only one irreducible Thus, by Proposition
is p-nilpotent.
is the only nonzero idempotent of
dimJ(FG1
But
of
=
=
J(FG)e
dimJ(FP1
Thus
IP/ - 1
=
F =
J(FG).
=
FP, it follows
P or G is a Frobenius group with complement p .
t(P) by Proposition 2.3, so t(P1
=
IPI
and hence, by Theorem 4.3.2,
P is cyclic. (ii) * (il: By Theorem 3.8.11, t(P)
=
IPI
.
as asserted.
dimJ(FG1 = ]PI
-
1, while by Theorem 4.3.2,
F Hence, by Proposition 2.3,
t(G]
.
=
t(P)
=
IPI
=
dimJ(FG1 + 1
F
We next quote the following result, whose proof is almost identical to that of Proposition 3.1.8.
For details refer to Passman (1970a) or to Karpilovsky (1985).
2.13. PROPOSITION. (Passman (1970a)l. index n
and let
F be an arbitrary field.
J ( I P G )5~ F'G*J(F%I
(i)
Let N
be a normal subgroup of Then, for all
J(F%)
of
Z2(G,F*),
C_ J(FO"G)
(ii) If n # 0 in F, then J(FaGI = flG*J(F'Nl J(FOLG) and
c1 E
G
=
J(F'N).F'G.
In particular
have the same nilpotency index.
Repeating the argument of Proposition 3.1.9, we immediately derive 2.14. COROLYLRY. (Passman 11970al).
Let Iv
that G/Iv
Z2(G,F*).
is p-solvable and let
c1 E
be a normal subgroup of If
I G/NI
=
G
such
npa where
then
In particular, if G
is p-solvable of order npa
where
(n,p) = 1, then
( n , p ) = 1,
CHAPTER 7
424
J@'QG)pa 2.15. PROPOSITION.
p > 0, let N
Let F
P
0
be an algebraically closed field of characteristic
and let { e l ,
= 0 ,(GI
=
...,e
S
}
be a full set of representatives
of Gconjugacy classes of block idempotents of FN. let Gi
be the inertia group of
a E Z2(Hi,F*1
e
i and let H i
=
For each
Ei/N.
iE
{1,2,.
.. , s } ,
Then there exists
such that
i
c1
t ( G ) = max { t ( F i H i ) } 1~2% cl.
c1
i
2
Proof.
.
is the nilpotency index of J ( F 'Hi).
where t ( F H . )
Direct consequence of Morita's Theorem (Theorem 3.10.9).
2.16. COROLLARY.
Further to the assumptions and notation of Proposition 2.15,
n is p-solvable and let p
assume that G
Hi, 1
i
4
s.
be the order of Sylow p-subgroups of
Then
ai. t ( P Hi)
pni
and
1%-
Proof.
Apply Corollary 2.14 and Proposition 2.15.
Throughout this section, F finite group and t ( G ) Assume
that G
.
is an arbitrary field of characteristic p ,
G
a
is the nilpotency index of J ( F G 1 .
is a p-solvable group of order Pam, p
1, m,
U
> 1.
Then, by
Proposition 2.1, we have
Furthermore, by Proposition 2.5, if
G
if and only if the Sylow p-subgroups of
is of p-length G
1 then t ( G ) = u(p-1) + 1
are elementary abelian.
this section is to provide further information on the structure of t ( G 1 = u(p-1)
+
1.
Our aim in C
with
We first illustrate that Proposition 2.5 is no longer true
without the assumption that G
is of p-length 1.
Let
G
G = S
be t h e symmetric group of degree
4
p = 2.
and l e t
4
i s a s o l v a b l e group whose 2-length i s g r e a t e r than
ObvLously
1 and whose Sylow 2-sub-
I t w i l l be shown t h a t
groups a r e n o t elementary a b e l i a n .
t ( G ) = 4 ( = a(p-1)
+
1)
Actually t h i s f a c t can be immediately deduced from L o r e n t z ' s theorem (Theorem However, r a t h e r than appeal t o t h e g e n e r a l r e s u l t , we g i v e a simple ad
6.4.6).
hoc argument which i s a p p l i c a b l e t o t h e p r e s e n t very s p e c i a l s i t u a t i o n .
X of
f o r any subset
G,
X+
we w r i t e
f o r t h e sum of elements i n
3.1. PROPOSITION. (Motose and Ninomiya ( 1 9 7 5 a ) ) .
i
s t a b i l i z e r of
2.
E {1,2,3,4)
and l e t
F
G =
Let
s4 '
A s usual,
X.
let
Gi be t h e
be an a r b i t r a r y f i e l d of c h a r a c t e r i s t i c
Then
+
(i) J ( F G ) = FG+
N
where
FG*I(N)
i s t h e Klein's f o u r group contained i n
G.
(ii) t ( G ) = 4
Proof. G
nN
(i)
Since
N
i s a normal 2-subgroup of
G
G = NG
with
and
it follows from Lemma 6.4.5 t h a t
= 1,
J(FG) But
G
S3
=
J(FG1)
J ( F G 1 = FG:
and so
+
FG.I(N)
by v i r t u e of Example 3.7.9.
This proves
(i).
(ii) We f i r s t show t h a t
+
+
+ + GigGj
- 1
GigGj Since
i # j,(C;)'
=
for a l l
= 0
+ + Gi Gg c j l
g
=
6G:
= 0.
i , j E {1,2,3,4)
and a l l
G? G+
it s u f f i c e s t o v e r i f y t h a t
Assume t h a t
i # j
and w r i t e
.ti
g E G = 0.
+ G+i ( k , k ) ,
G:(k,i)
=
Gl(k,!L,i)
and
G+i ( R , i )
=
G:(k,i,k)
and so
+ +
+
G . G . = Gi(l+(k,L) 2 3 = 2Gi
+ ( k , i ) + ( k , k , i ) + (R,i) + ( k , i , k ) )
+ 2G:(k,sil
+
2Gi(k,i)
= o proving (1). Since
I(fl)'=
FN + ,
it follows from ( i l t h a t
I n case
{1,2,3,4) = { i , j ) U { k , k l .
Then
Gi+
(1)
CHAPTER 7
426
I
+ I(B)G:
= FG*N+
(G:)'
But, by (l),
=
G+*ICN)C:
0 and
Since
= 0.
we t h e r e f o r e conclude t h a t
t(G)
Hence
+
4 = 3(2-1)
'
and so, by Proposition 1.1, t ( G ) = 4 .
1
t(G) = a(p-1)
We next provide f u r t h e r circumstances under which
G
t h a t t h e Sylow p-subgroups of
a r e elementary abelian.
+
1 implies
The following
preliminary observation w i l l c l e a r our path. 3.2.
Let
LEMMA.
P E Syl ( G I .
p P
If
P
G
be an odd prime, l e t
be a p-solvable group and l e t
p3 and exponent
i s of order
p2,
then
G
i s of p-length
1.
Proof.
P
G
i s of p-length
P
(GI 1 , w e may assume t h a t
E Sylp(G/Op
since
C / o ,(G1
If
i s p-solvable.
X/Y
Assume t h a t
2
1,
X. QY/Y
p.
P
We f i r s t show t h a t
Q
Then
Q/(Q
Qd(p1 i s
p
2
Q
(G:XI
and
not involved i n
[PI
G
A
= p3, P/Z(P)
of
P
5 0P'nP ( G I ,
0 (G) # 1 P
i s not involved i n
G
Y 4 X.
and some
G
IYI .
Let
and so
P.
Q
X/Y,
C. By
p 3 and
Q be a Sylow p-subgroup of On t h e o t h e r hand,
a contradiction.
A
and so, by Theorem 1 . 7 ,
5 0P'rP ( G )
Thus
f o r any
P.
is a b e l i a n .
which i s a normal a b e l i a n subgroup of and t h e r e f o r e
k
p
Since
&d(p) a r e nonabelian of order
i s a Sylow p-subgroup of
abelian normal subgroup Since
1
is a Sylow p-subgroup of
n Y)
Qd(pI
Qd(p) f o r some subgroup X of
Hence
G.
(GI = 1 i n which c a s e
0
Theorem 1 . 5 ( i ) , t h e Sylow p-subgroups of exponent
then so i s
P.
proving t h a t
Thus, i f
3:
E P,
then
z E < Z ( p ) ,x>
Hence, by t h e foregoing,
G
has p-length
1.
'
z E 0
P'IP
(G)
GROUPS G
t(G)
WITH
=
U(p-1) + 1
427
We are now ready to prove the following result established by Motose (1980) and by Koshitani (1982a) for p # 2.
for p = 2
Let G be a p-solvable group with a metacyclic Sylow p-subgroup
3.3. THEOREM.
P of order pa, a > 1. t(G)
+
a(p-1)
=
1 if and only if P
If P
Proof.
If p = 2, assume that
is elementary abelian, then G
assume that t ( G 1 = a(p-11 is of p-length
+
Then
+
is of p-length
1 by Proposition 2.5.
1
Conversely
By Proposition 2.5, it suffices to verify that
1.
Let G
1.
$ Sq.
is elementary abelian.
(Proposition 3.16.20), so t ( G ) = a(p-1)
G
G/Op,(G)
be a counter-example of minimal order.
If
0 , ( G I # 1, then by Proposition 2.1,
P
a(p-1) and so t[G/O
P
,(GI
) =
hypotheses as G, case
1
=
t (GI 2 t (G/O
+
1.
a(p-1)
P
C/o ,(GI
b
10 (GI1 = p ,b 2 1.
P
If t ( O ( G I )
P
1
satisfies the same
P
Thus we must have
Thus t ( O ( G I )
P
=
0 , ( G I = 1, in which
P
3
b(p-1) + 1, then by
b(p-1) + 1, in which case 0 (GI
elementary abelian by Proposition 2.5.
P
> a [p-1) +
(GI )
and Proposition 2.1,
a contradiction.
0 (G)
P
But the group
a contradiction.
0 (GI # 1, say
Theorem 4.1.9
+
But P
is metacyclic, hence so is
p2.
and therefore 0 (G) is elementary abelian of order
by Lemma 1.3,
C
G
is
P
P (0 (GI) = 0 ( G ) .
Furthermore,
Thus G/O (GI is isomorphic to a subgroup P P P Because G is not of p-length 1, the group P is nonabelian.
of Aut(0 ( G 1 ) .
P
We now distinguish two cases. Case 1.
Assume that p = 2.
[C/O ( G ) l G 6.
P
Since G
By the foregoing, IPI 2 8, 10 ( G ) I
P
is not of p-length 1, we have
0 (GI = C (0 ( G ) ) is elementary abelian of order 4 P G P fully on 0 (GI by conjugation.
\ G I = 24,
and C/O (GI
P
4
and
in which case acts faith-
P
Let S = < a >
be a Sylow 3-subgroup of
G and 0 (G)
P
=
{l,z,y,Z!j}.
Since
CHAPTER 7
428
-1 (0 ( G I 1 = Op(G) I we may assme that a-'za = y and a ya = q. Consequently, G P N ( S ) n 0 ( G ) = 1 and (N(S1 1 = 6 by Sylow's theorem. We conclude therefore G P G that G is a semidirect product of 0 ( G ) and N(S) and so N(S) is the autoC
P G G Since the latter obviously implies that G z
0 (GI.
morphism group of
s , we
P derive a desired contradiction.
Case 2.
We now assume that p
If 0 CG1
is odd.
P
is cyclic of order p ,
G/O (GI is a p'-group, contrary to the assumption that G
is not of p-length
P
is elementary abelian of order p2.
Thus 0 (GI
P
IAut(0 (GI J
P
( P / o (G)
we have
P
I
=
1 or
I
= IGL(2,pJ
p.
.
length 1, a contradiction.
=
1.
Since
p(p-11 '(P+l),
is nonabelian and metacyclic, so P
But
and exponent p2.
of order p 3
1
then
It follows from Lemma 3.2 that G
and G / O p , ( G )
In view of Theorem 3 . 3 , the case where p = 2 separate treatment. For t h e r e s t of t h i s section
F
is
is of p -
S4
deserves a
is asswned t o be algebm-
i c a l l y closed. 3 . 4 . THEOREM. (Motose (19801).
e ,e 1
2
,. . .l e t
let Gi
Let p = 2, let G/O , ( G )
be all block idempotents of
be the inertia group of e
t(GJ =
I
FO
P
,( G ) .
P
IH$[
= 8
5
if
4
otherwise
Proposition 2 . 1 , that t ( G 1 Sylow 2-subgroups of Hil
>
4.
,(G1 .
and let
. ., t ) ,
Then
? - P
...,tI
for some i E t1,2,
Since the order of Sylow 2-subgroups of G
Proof.
S
For each i E {I,.
and put Hi = G . / O
i
"=
is 8, it follows from
zni
is the order of
...,t } .
Then each n
On the other hand, if
then by Corollary 2.16
n. Assume that and hence
kI
(H
= 24
lHil #
8
t ( G ) = 4.
and
Suppose that
for some k .
for all ak E Z2(Hk,F*). t(G) = t(S4).
t ( G ) G max I 2 KiGt lHil # 24 for all
Then Hk
2
'1
i
E {l,
[ H i [ # 8 for all i E {l,
S
and so, by Lemma 3.4.9,
...,t},
but
FakHk
Hence, by Proposition 2.15 and Corollary 2.16,
Thus t ( G ) = 4
by Proposition 3.1.
i
FS4
2
G WITH
GROUPS
1H.I = 8
F i n a l l y , assume t h a t
a. F 'H.
8
ai
t ( H . 1 = 5 by Theorem 4.3.4.
implies t h a t
Hence
t(G1 = 5
and by Lemma 3.4.9,
.
a s required.
429
i E {1,2 ,...,t 1 .
f o r some
d i h e d r a l group of o r d e r
E Z2(Hi,F*1.
+ 1
= U(p-1)
FHi
Hi
Then
is a
for a l l
By t h e foregoing, t h i s
W e now p r e s e n t two examples which i l l u s t r a t e t h e following f a c t s :
G
There e x i s t s a group
ti)
and a c e n t r a l p'-subgroup
fl such t h a t
t ( G ) # t(G/fl1 G
(ii) If
i s a s i n ( i 1 , then
t ( B ) # t (G)
where
B
Assume t h a t
p = 2
and l e t
is t h e p r i n c i p a l block of
FG . 3.5.
EXAMPLE. (Motose (19801).
group of o r d e r 3.
(121-'d(12)
that
G
Denote by =
d-l
and
t h e s e m i d i r e c t product of
A
by
.
Then
and l e t
a c t s t r i v i a l l y on
be a c y c l i c S4
such
t(G) = 4
by
Theorem 3.4. 3.6. EXAMPLE. (Motose ( 1 9 8 0 ) ) . c y c l i c groups of o r d e r
f
Then
3.
=
2
=
L e t a homomorphism
f
z2 x z2
x
and
S4
and
into
s3.
with r e s p e c t t o
S3+
:
can be regarded as a homomorphism of
a s e m i d i r e c t product of
M
p
Assume t h a t
GL(2,31
GL(2,31,
,
be
be defined by
since
s4
Let
G
f.
Then t h e following r e l a t i o n s
be t h e s e m i d i r e c t product of
hold:
-1 a ca = c Let
, a-lda
be a p r i m i t i v e 3rd-root of
E
e Then
=
e
cd
1 in
=
c
and
b-ldb =
6'
and put
= (1+Ec+E2c2)(l+d+d')
i s a block idempotent of
FM
a-lea # a Hence, i f
, b -1cb
such t h a t and
G ( e ) i s t h e i n e r t i a group of
Theorem 3.4 and Example 3.5, we d e r i v e
e,
b-leb = e then
is
I G ( e ) / M ( = 8.
Thus, by
CHAPTER 7
430
Furthermore, i f
t(BI
= t(Sbl.
3.7.
Remark.
B
FG,
is t h e p r i n c i p a l block of
then by Lemma 2.2,
t ( B 1 = 4.
Hence, by Proposition 3 . 1 ,
Similar examples can be constructed f o r an a r b i t r a r y prime
p
( s e e Motose ( 1 9 8 2 ) ) .
t(G1
4. COMPUTATION OF
Throughout t h i s s e c t i o n ,
WITH
M[p)
M(p)
denotes an e x t r a - s p e c i a l p-group of order
E Syl ( G )
P
odd, defined i n Sec. 2 of Chapter 4.
Note t h a t , by Corollary 4 . 2 . 4 , t ( M ( p ) ) =4p-3.
The information obtained w i l l be important f o r subsequent i n v e s t i -
subgroup.
Note t h a t i n case
i s of p-length
p = 3,
case
1
(Theorem 1 . 4 ) .
1.
with
M ( 3 ) E Sy13(Gl
(G)
E Syl
P
t ( G ) = 4p-3.
Hence, by P r o p o s i t i o n 2.3,
G
Mlp)
with
As w e s h a l l see below, t h e case where
c r u c i a l f o r t h e computation of from now on we assume t h a t
In
which a r e
F
t(GI
with
M(3) E Sy13 ( G I .
G
=
&d(3) is
For t h i s reason,
i s a f i e l d of c h a r a c t e r i s t i c
3
and begin by
G = Qd(3).
a t t a c k i n g t h e case where Recall t h a t t h e group
M(31
~ ( 3 = )
i s defined by
z3 = y 3
=
1,
P
= 3c,
ye
=
x y , 2 = x>
M = SL(2,31 a c t s n a t u r a l l y on t h e elementary a b e l i a n group
where
U is regarded as a vector space over t h e f i e l d F
group
Q d ( 3 ) i s defined t o be a semidirect product of
t h i s action.
U
3
of and
3 elements.
The
M with r e s p e c t t o
I t w i l l be convenient t o f i x t h e following n o t a t i o n :
H
= Qd(3)
a = Note t h a t
G
any p-solvable group
The most important example of such a group i s t h e group
(see Theorem 1.51.
The group
5
p
t h e r e e x i s t 3-solvable groups
not of 3-length Qd(3)
a s a Sylow p -
M(p)
Our aim i s t o extend t h i s r e s u l t t o p-solvable groups having
gations.
p3, p
[: i],
b =[!j
:]
and
c =
y]
in
M
=
SL(2,3)
COMPUTATION OF
t(G)
WITH
M(p) E Syl (G) P
431
Moreover, we put
where FT
T over F
is the group algebra of the group
(see Lemma 4.l(iv)
below]. 4.1. LEMMA.
With the notation above, the following properties hold:
f is a central idempotent of FM
(i)
(iil a 2 f = -f and fuf (iii)
'I
=
-f(u+u2) = fu2f
is central in FM.
(iv) ( ( a 0 b ) f ) '
=
0.
In particular, C
Hence
is a cyclic group of order
for all u E b'
3
T 2 = C2(l-Uob)f
consisting
of
and
commutes with T3 =
f
(a0b)f.
which implies that T
P-linearly independent elements.
-
(f-~)u(f-~) = ( ~ - T ) G ( ~ - T+) T+
(v)
(vi) Q+~Q+=
(u+-~)o+
for all u E U - 1
(vii) f(aobl3cQ' = G(y2-y)Q' (viii) (f-T)uM+ E C3M+ (ix) T+u&+(l-c) E C'M'
+ + TuM
(x] (xi)
+ + =MuT
calculation.
and
Q+x ( a 0 b ) f =
Q+(y-y2)z
and M+u(f-T) E M'C3 and
(l-c)Q'uT+ E M+C3
for all u E U for all u E U for all u E U
= O
+vQ T + ~ ( ~ - T ~ (i-c)
Proof.
for all u E U
= ( i - c ) ~ + v ( f - ~ ) u=~ o +
for all u , ~ E ) U
The validity of (il , (iil , (iv) and (v) follows by straightforward Note also that ( x ) is an easy consequence of (ix).
(iii) Invoking (i) and (ii), we derive
CHAPTER I
432
A s i m i l a r argument shows t h a t
(vi)
Because
Tb = T
and
're =
T.
Q i s both t r a n s i t i v e and r e g u l a r on U - 1 , Q+ u Q+
=
C
u4Q+ = ( U + - l ) Q +
&Q Applying ( i i ) ,w e have
(vii)
A s i m i l a r argument proves t h a t
(viii)
Invoking ( i l
, (iiil,
Q + ~ ( a o fb )= Q+ (y
- y2),
and ( v i i ) , w e derive
and a s i m i l a r argument shows t h a t
Because (ix)
Q is t r a n s i t i v e on
U-1,
t h e a s s e r t i o n follows.
Owing t o ( i ) ,( i i i ) ,and ( v i i l it follows t h a t
S i m i l a r l y w e have
we have
433
Since
Q is
U-1,
t r a n s i t i v e on
t h e a s s e r t i o n follows.
(xi) The proof of (ix) t o g e t h e r with (vii) and (x) ensures t h a t
A s i m i l a r argument shows t h a t
Because
Q
i s t r a n s i t i v e on
.
p r o j e c t i v e space follows.
U-1
and
M/ = PSL(2,3)
{ h , x 2 } , {y,y2}, { q , z 2 y 2 } , { s 2 y , q 2 2 }
,
i s t r a n s i t i v e on t h e the desired assertion
With the n o t a t i o n above,
4.2. LEMMA.
and
J(FH) = A + B + C Proof.
Note t h a t
4.l(i), ( i i i ) , FT
J(FS)Q+. t a t i o n s of F-dimension
Since
M, 10.
J(FS)Qi
is c e n t r a l i n {1,2,3}
where
is a n i l p o t e n t i d e a l of FM.
Thus
J(FM)
FM
and t h a t , by Lemma
c o n t a i n s both
J(FT1FM and
is t h e s e t of degrees of a l l i r r e d u c i b l e F-represen-
is a s p l i t t i n g f i e l d f o r
M,
the radical
J(FMI
On t h e o t h e r hand, one e a s i l y v e r i f i e s t h a t t h e s e t
is of
434
CHAPTER 7
consists of some F-linearly independent elements in J ( F T ) F M establishes the first assertion. natrual map FH--+ F(H/U) = FM
+
J(FS)Q+.
This
Taking into account that the kernel of the
is contained in J(FH) , we obtain J ( F H ) = A+B+C.
The next lemma is crucial.
4.3. LEMMA. (i) C'
=
The following properties hold: and C5 = 0
U'FM
and A 3 = 0
(ii) A 2
=
(iii) B'
5 T+FH + (J@-T) C' ( ~ - T I F H 5 T+C2(f-r)FH + (f-t)C2T+FH + T'C4
(iv) B 3 (v) B5
=
M'FV
0
(vi) BA'B
=
o
(vii) B 3 A
=
AB3 = 0
(viii) B*AB' Proof.
=
o
(i) Straightforward
/ii) Applying Lemma 4.l(vi), we derive
and
as asserted.
(iii) By Lemma 4.1(i) ,(iii), we derive
J@-T
is central in FM.
Applying Lemma 4.l(v),
COMPUTATION OF
tCGZ
WITH
M(pl
Syl
P
(GI
as required. (ivl
Invoking Lemma 4 . l ( v l t o g e t h e r with ( i )and (iii), we o b t a i n
a s desired. (vl
Applying ( i ), ( i i i l , ( i v )
and Lemma 4 . 1 1 ~ 1 , i t follows t h a t
a s required. (vil
By Lemma 4 . l ( v i i i l and ( i i ) ,we have
as a s s e r t e d .
435
CHAPTER 7
436
c - c3c2
+
0
+
c5
=o S i m i l a r l y w e have (viiil
Applying ( i i i l and Lema 4 . l ( i x l , (XI,we i n f e r t h a t
a s required. 4.4.
B 3 A = 0.
.
Proof.
H
If
LEMMA.
=
Qd(3I
and
p
then
= 3,
t ( H l = 9.
Applying Lemma 4.3, we s e e t h a t
(A+B15 c - FHmA3
+
FH*AB3
+
FH*B3A
+ B2AB2 + B 5 + FH*BA2B
=o Since
C5 = 0 ,
proving t h a t Put
x=
i t follows from Lemma 4 . 2 t h a t
t(Hl 1
9.
+ x + x2
. and
V =
together with Lemma 4 . 2 , we o b t a i n
1
+y +
y2.
Applying Lemma 4.1(1)
-
(iv)
437
and
whence
By Lemma 4.2, J ( F H ) *
contains XT'VT'.
Applying (11, we have
thus completing the proof. We have now come to the demonstration for which this section has been developed.
The following result is essentially due to Motose (1984a) who
established the crucial case where p = 3. 4.5. THEOREM.
Let G be a p-solvable group, p
P isomorphic to M(p1.
Then t ( G 1 = 4p- 3
odd, with a Sylow p-subgroup
438
CHAPTER 7
Proof.
G
Assume t h a t
t(G)
Proposition 2.3,
4p-3.
=
1.
is of p-length
1 . 4 ) , so we may assume t h a t
p 2 5
If
p = 3
“hen, by Corollary 4.2.4
G
then
G
and ‘chat
i s not of p-length
F
we may a l s o assume t h a t t h e f i e l d
Corollary 3.1.18,
t(G) # 9
A s s u m e by way of c o n t r a d i c t i o n t h a t
i s of p-length
and
1 (Theorem
By
1.
i s a l g e b r a i c a l l y closed. G
and l e t
be a counter-example of
minimal o r d e r . We set
W
0 ,(G), G
=
U
= G/W,
=
O (G )
1
3
3
1
and
P g
Let
g
and
= PWW / 1 PI G
G /U
=
2
1
1
g E G
be t h e images of an element
G
in
G2,
and
respec-
tively.
G
W e f i r s t observe t h a t
I u11 IP I
= 3,
33.
=
IG,I
then
IU
If
G2
IAut(U )
I
1 .
=
33,
By Theorem 1 . 6 ( i ) ,
xl , y l X
morphism
and
G
G+
e
.
PI
V
Q
5 Q.
. =
CG:X)
Then
IVI
If
Sylow 2-subgroup we must have
Q
1
1
G
Q
of
=
P1 , G
and so
If
contrary t o the f a c t t h a t
i s of 3-length 1, again a
3’
i s elementary a b e l i a n of o r d e r
U1
U
and so
<
G 2,
G
U
SL(2,3) n G and so
k R > , k 2 2,R 2 0.
,c y 1
1
1
=
1
t ( G ) = t(X1
SL(2,3).
.
<x , y > 1
under t h e n a t u r a l homo-
X and hence G 2
i s normal i n then
1
U 1 preserving t h e same r e l a t i o n s a s those
2
G =
<x
=
Hence we may assume t h a t
by Proposition 3.1.8
5 SL(2,3). V E Sy12(G2),
Hence, i f
has a normal Sylow 3-subgroup
lU1 I = 3‘.
If
SL(2,31 s i n c e
c
contrary t o t h e f a c t t h a t
Let
x
and
The l a t t e r implies t h a t
The group then
P
=
denote t h e inverse image of 2
(ill,
1
U 1 contains
We may r e p l a c e g e n e r a t o r s of
Let
U
then
by Lemma 1.3.
1
U1
and so
2
=
Aut(U )
Aut(u 1 = GL(2,3).
i s a subgroup of
between
I
Thus w e may assume t h a t
contradiction. and
i s contained i n
IVI > =
2
4,
[:
then and
V
5Q
=
P
,
contains the
.
Thus
= SL(2,3).
be a Sylow 2-subgroup of
GI such t h a t
U Q = 0 1 I
3 r 3
,(GI).
Because
CHARACTERIZATIONS OF GROUPS G
nU
WITH
1
it follows from Theorem 1.9 that G
439
9 U =1,
and Q
= 1
a- 1 t(G) = p +p- 1
1
U l and SL(2,3).
is a semidirect product of 1
Taking into account that SL(2,3) cAut(U 1 that G
I
IG
and any
c(
=
E
GI
is a quaternion group or a cyclic group and
23*33, it follows from Lemma 3.4.9 that for any subgroup H
FH
(2)
Let { e s ] be the set of all block idempotents of FW group of
e
Assume that W # 1.
in G. G
subgroups of =
and G/W
t(G/n = t(Is/W) < t ( G )
are isomorphic to M(3). then I s
of Gl
Z2(H,P*), EP"H
9
U 1 , we deduce
= Qd(3).
Since every 2-subgroup of since
acts naturally on
are isomorphic. for a l l
B
Then
and let I,
IG/Wl
q
IGI
be'the inertia
and the Sylow 3-
Hence, by assumption
Is/W
such that the Sylow 3-subgroups of
If the Sylow 3-subgroups of rs/V
are of order
Q
3*,
is of 3-length 1 [Proposition 3.16.20), so by Proposition 2.3 and t(Is/W) < 9.
Corollary 3.2.5,
Invoking (2) and Proposition 2.15, we therefore
conclude that
Thus W = 1 and
a contradiction.
G
= Qd(31,
contrary to Lemma 4.4.
This
completes the proof of the theorem.
5. CHARACTERIZATIONS OF GROUPS
G
WITH
t (GI
=
pa-l+p
-1
Throughout this section, G denotes a finite group and F an arbitrary field of characteristic p . order p a , (il
Recall that, by Theorem 4.3.4, if P
is a noncyclic group of
then the following conditions are equivalent:
t (PI = pa-'
+p - 1
(ii) t ( P I > pa-' (iiil P
contains an element of order p
a- 1
Our aim is to generalize the above fact to p-solvable groups. restrict ourselves to the case where p 2 3 .
In future we shall
The reason for this is as follows.
CHAPTER 7
440
p = 2
Let
.
= S
Then t h e Sylow 2-subgroups of
and hence s a t i s f y condition ( i i i l .
8
order
G
and l e t
G
a r e d i h e d r a l of
On t h e o t h e r hand, by Proposition
3.1.
t ( G 1 = 4 # 2'
p
Thus f o r
2,
=
+
2-1
Theorem 4.3.4 cannot be extended t o a r b i t r a r y p-solvable groups.
The following r e s u l t i s of independent i n t e r e s t and is t h e key t o t h e s o l u t i o n of our problem. 5.1.
THEOREM.
(Koshitani (1983),(1985a1, Motose (19851).
group with a Sylow p-subgroup
p = 2 Set
M
=
,
S
G/@
and l e t
t(G)
2 pa-1
0 , ( G I ,N = 0
P'PP
P
@/O ,CG)
and
@/oP , ( G )
be t h e F r a t t i n i
and t h a t
G
i s n o t of
is cyclic
P
4
(G), and
pr =
(N/M(.
Owing t o Corollary
F is a l g e b r a i c a l l y closed.
w e may harmlessly assume t h a t
3.1.18,
pa
be a p-solvable
Then
1.
Proof.
of o r d e r
Assume t h a t
0 (GI/Op,(G). P'nP
subgroup of p-length
P
G
Let
For t h e sake
of c l a r i t y , we d i v i d e t h e proof i n t o f o u r s t e p s . Here we prove that
Step 1.
that
is cyclic.
O/M
imply t h a t
G
Since
i s of p-length
G/N
is
isomorphic t o a subgroup of
(GL(2,pl 1 2.
=
p(p-11 (p2-11,
JlFGlP
and
the f i r s t assertion w i l l
B y Proposition 3.1.9,
a-r
GLf2,p)
we have
5 FG*J(FN)
and hence
oz Thus
If
t (N) > $-l
N/M
since
which c a s e
iV/O
is a subgroup of
t(N/M)
r-1
5F G ~ J C F N ) ~
and hence, by Proposition 2.3,
i s c y c l i c , then
G/N
a-1
JLWP
< pr
i s of order Aut(N/Ol
p
and hence
Thus
(Lemma 3 . 1 ) .
by Proposition 2.4.
< t(N/M) < pr
is of p-length N/M
1
i s noncyclic, i n
I t follows from (1) t h a t e x a c t l y
one of t h e following holds:
pP1
G
and
r
3
G
CHARACTERIZATIONS OF GROUPS
t ( G ) = pa-' + p
WITH
-1
441
or
N/M
N/@
I f ( 2 ) h o l d s , then by Theorems 4.3.4 and 1.10,
p2
order
i s cyclic.
@/M
and
GL(2,p)
Aut(N/@)
@/M
and
We now prove t h a t
Step 2.
p = 2
or
i s of o r d e r
p
x
~
o
f o r some
P
G/O
follows from Theorem 1.9 t h a t
p
is a subgroup of
T
=
e ,e ,...,e 1
2
of
e
in
i
T
where
be a l l block idempotents of
G,
1G
i
G
s.
FM
U. E
2
2.
GL(2,3)
and l e t
p
3.
It
and
G/N.
=
N/@
k
Gi
on
k
3
x
k
3
.
3 x k 3
Let
be t h e i n e r t i a group
Then, by P r o p o s i t i o n 2 . 1 5 , w e have
"i
H . = G./M, z z
Invoking
i s t h e s e m i d i r e c t product of
1
".
U.
G/N.
is a s e m i d i r e c t product of
t(G) = max { t ( F Hi) l G S where
~
so t h a t , by S t e p 2 ,
G L ( 2 , 3 ) with r e s p e c t t o t h e n a t u r a l a c t i o n of
and
1
p E {2,3).
p # 2
Assume by way of c o n t r a d i c t i o n t h a t
G/@
€
and i s conjugate t o
i s t h e minimal polynomial of every p-element o f
The aim of t h i s s t e p is t o prove that
Hence
i s a subgroup of
3.
=
G/N
Theorem 1.8, we t h e r e f o r e d e r i v e t h a t S t e p 3.
G/N
is cyclic.
[: y] (X-1)'
(3)
i s elementary a b e l i a n of
Thus i n any case
I n view of S t e p I, every p-element of
Thus
p2
is elementary a h e l i a n of o r d e r
t ( F 'Hi)
and
Z2(Hi,F*I
i s t h e nilpotency index of
J ( F 'Hi). Let
S be t h e s e m i d i r e c t product of
t h e n a t u r a l a c t i o n of homomorphism and l e t
[Hi:(Li
n Hi)] G 2 ,
SL(2,3)
L
i
on
k
3
x
k x k 3
a3 .
and
3
Let
be t h e i n v e r s e image of
SL(2,3) with r e s p e c t t o
f : G/M--t G/@
f(H.)
n S.
be t h e n a t u r a l
Since
it follows from Lemma 3.4.9 and Proposition 2 . 1 3 ( i i ) t h a t U.
= max { t ( L i n Hi)] lG<s 1GG (Li n H i ) / ( ( @ / M ) n Hi) i s isomorphic t o a subgroup of
t ( G ) = max
NOW
Ui
=
{ t ( F 'H.11
Theorem 4.5 and P r o p o s i t i o n 3.1.9, and Proposition 4.1.8, we d e r i v e
t(Ui)
< 9.
S and, by
Hence, applying Theorem 4.1.9
CHAPTER 7
442
= 3r
Completion of the proof.
Step 4.
@/M
By Step 1, t h e group group of G/O
< t(G)
3r = 3‘-’
contrary t o t h e f a c t t h a t
G/N
we have
GL(2,2l,
GL(2,2)
on
G/N
Since
i s a sub-
GL(2,Z). Applying Theorem 1 . 9 , w e deduce t h a t
=
Z2 x Z2 and
i s a semidirect product of
n a t u r a l a c t i o n of
s of 2-length 2 .
i s c y c l i c and G
Z
x
2
Z .
CL(2,2) with r e s p e c t t o t h e
Thus
2
G/Q
and t h e r e s u l t
S
follows.
p be an odd prime and l e t
Let
n 2 3.
Mn(p)
Recall t h a t t h e group
is
defined by
n-1
n-2
5.2. THEOREM. (Koshitani (1985al).
Let
p
P be a Sylow p-subgroup of
solvable group and l e t
G
be a p -
G
be an odd prime, l e t
p
of o r d e r
U
with
U
2 2.
or
Mu@)
Then t h e following c o n d i t i o n s a r e equivalent:
t (G)
(i)
=
pa-’
+
pa-’ G
(ivl
where
B
i s t h e p r i n c i p a l block of
FG
< t t G ) < pa
has p-length
1 and
P
i s isomorphic to e i t h e r
>3
for
a
(v)
t ( P 1 = pa-1
(vi)
-1
+ p - 1,
(ii) t ( B ) = pa-’
(iiil
p
4.
Z a-l P
x Zp
p- 1
p Q - l < t ( P 1 < pa
(vii) (viii)
P
i s noncyclic but has a c y c l i c subgroup of index
p
P i s isomorphic t o e i t h e r Z a-l x Zp o r Ma@) f o r u >
Proof. and 1.10.
Conditions (vl Hence, i f
G
-
( v i i i ) a r e equivalent by v i r t u e of Theorems 4.3.4
i s of p-length
equivalent by Proposition 2 . 3 .
3
P
1,
then conditions (i)
-
(viii) are
We a r e t h e r e f o r e l e f t t o v e r i f y t h a t each of t h e
c o n d i t i o n s ( i l , ( i i l , ( i i i l and ( v i i i ) imply t h a t
G
i s of p-length
Now C i ) obviously implies ( i i i l and C i i i l implies t h a t
G is
1.
o f p-length
1,
G
CHARACTERIZATIONS OF GROUPS
by v i r t u e of Theorem 5.1.
t(B)
Lemma 2 . 2 ,
i s of p-length
i s of p-length
G
which case
0 (G) # 1 since G
Suppose t h a t
Q
Let
G
Theorem 1 . 5 ( i I .
Q
z
z
$-2
P c 0 (GI. - P’tP
X
Y
and
X/Y.
x
a
i s of o r d e r
P,
Hence
G
2
Ma@)
i s of p-
G
then
a 2 3.
f o r some
To
0 ,(G) = 1 i n P
Y 4 X
Q
X/Y
and
i s metacyclic s i n c e
p3
and exponent
p,
p
P
5 0P ’ Q
CGl.
4 P.
Since
Clearly
$
Qd(p).
i s meta-
by v i r t u e of
Therefore, by Theorem 1 . 7 , i f
A
then
and hence
P
P
such t h a t
Then
This i s a c o n t r a d i c t i o n .
a n a b e l i a n normal subgroup of
zp’
x
we may harmlessly assume t h a t
has subgroups
On t h e o t h e r hand,
is of p-length
P
i s p-solvable.
be a Sylow p-subgroup of
cyclic.
z
2
Assume t h a t
1,
P
P
If
1 by Proposition 3.16.20.
prove t h a t
C/O , ( G I
and by
1.
F i n a l l y assume t h a t ( v i i i ) . holds. length
443
t ( B 1 > pa-’
Then
Hence, by Theorem 5.1,
P
G
1 and t h e r e f o r e
Assume t h a t ( i i l holds.
t(G/O r(G)l.
=
t ( G ) = pa-l + p - 1
WITH
A
is
E Z(P),
a P
and thus
1 as d e s i r e d .
i s of p-length
W e c l o s e t h i s s e c t i o n by i l l u s t r a t i n g t h e f a c t t h a t t h e converse of Theorem 5 . 1 does n o t hold.
Cil
G/Q
S
4
,
Namely, we g i v e examples of groups
@/02,(G1
(ii) G/Q 1 S4, @ / / 0 2 r ( G l
z2 2 z
t(G1 = 9 > 2
and
G
such t h a t
4-1
t ( G 1 = 7 4 24-1
and
The above examples w i i i i n d i c a t e t h e d i f f i c u l t y i n c h a r a c t e r i z i n g a l l 2-solvable
G
groups
with
t ( G ) > 2a-1
and t h e underlying f i e l d Q =
Let 0
and
T
F
where
2a
is t h e o r d e r of Sylow 2-subgroups of
i s of c h a r a c t e r i s t i c
< a , b l a ” = 1, b 2 = a 2 , bab-’
W
&.
2.
be t h e quaternion group and l e t
=
be automorphisms defined by
o(a) = b ,a@) = ab, ~ ( 0=)a-’, Then
G
U 3 = 1, -r2 =
1, TUT-’
= U2
be t h e s e m i d i r e c t product of
R =
and so
Q
R
and
~ t b =) a-’b
i s isomorphic t o
with r e s p e c t t o t h i s a c t i o n of
Then we have
02,(W)= 1 ,
o p
=
Q
i s t h e F r a t t i n i subgroup of
v/
2
s
s3
oz(w)
. R
Let
on
CHARACTER 7
444
For the r e s t of t h i s sectson, we assume that p
closed.
=
2
und F
is algebrae'cally
The latter condition does not impose any restriction, by virtue of
Corollary 3.1.18.
We need the following preliminary result.
5.3. LEMMA. CMotose (19851). Proof.
Put X
=
+
R 9FV
t(W) = 7.
With the notation above, and
Y
=
J(FQ)FW
.
Since J ( F Q ) F W C_ J ( F W ) ,
it
follows from Lemma 4.2 that J(FW1 = J(FR'R)FV + J ( F Q ) F W
= X + Y We shall show that X 3 = 0.
Since Y 5 J(FW)'
=
in C.
0 this will imply that
(X+Y17 = 0
+ + + -- 0 for + + + R xR yR = 0 for all
It suffices to verify that R CCR y R follow if we show that
=
all x,y E &, z,y E C = Q
Since {ll, { a 2 } and C are all orbits of &
and the latter will
- {l,a21.
Fix x , y
under the action of R,
we have
R + ~ R + ~ R=+ c+R+~R+= ( C + ~ ~ R + =
(a2+11 (a+b+ab)'R+
= o proving that t ( W I Since R+xR+
=
7.
C+R+
for x E C and a 2 + 1 and
Q
+
are central in F W ,
also have
This implies that t ( W ) > 6
5.4. EXAMPLE. (Motose (1985)1.
and hence that t ( W ) = 7 . Keeping the notation of Theorem 5.1, we shall
we
CHARACTERIZATIONS OF GROUPS G
+p - 1
t (G) = pa-'
WITH
2
z2
and
In what follows, we shall regard this as a homomorphsim from
w
to
present an example of a group G Consider the homomorphism R-
Let
M
=
G
G/Q
with GL(2,3)
,
IS
@/02,(GI
445
t ( G ) = 9 > 8.
defined by
GL(2,3).
denote the semidirect product of the elementary abelian group
x
Then clearly root of
M
of order
G/@
= 0 f(G),
and let e
1
3'
W with respect to the above homomorphism.
and
z
S b and @ / M S
be a block idempotent of
.
Let
FM
be a primitive 3rd
E
defined by
e = (1+Ee+~~e')(~+d+d') Then
# e
U-leu
and
-1 T eT = e
and so the inertia group
G(e)
of
e
in G
is
given by G(e) = < T , Q > * M
{el
Let
=, 'M
e
=
,...,e
e, e
let Hi = G(ei)/M, 1 =Si G 9. c1
i
E
Z2(Hi,F*1
1
be the set of all block idempotents of
Owing to Proposition 2.15,
a
t(F
and
there exists
such that
a. { t ( F 'H.)) z lGX9 a. is the nilpotency index of J ( F "Hi). t(G)
where
FM
iH.)
=
Since
a. t ( F 'Hi)
=
t(Hi)
by Lemma 3.4.9, we have
t(G1 If Hi 3.1.9.
max It(Hi)l lG59
contains no Sylow 2-subgroups of On the other hand, t ( H
contains an element T b
5.5.
=
EXAMPLE.
) =
of order
(Motose (1985)).
t(W1
G, = 7
= 7
< 8.
t(H.1
G 8
by Lemma 5.3.
by Proposition Since
8, we also have
Again, we preserve the previous notation.
aim is to present an example of a group G t(G1
then
with
G/@
S
,
Q/02 (GI
2
z2
Our and
CHAPTER 7
446
Consider the homomorphism
+ GL(2,31
and regard it as a homomorphism from W
defined by
Let G be the semidirect
to G L ( 2 , 3 ) .
product of the elementary abelian group M = < c > x < d > Then clearly M = 0 ,(GI, G / @
to this homomorphism.
,...,e
{e
= M+,e
or
*M
S
and W and @/M
be the set of all block idempotents of FM.
}
with respect 2
z .
Let
Then G(ei) = G
and so
by virtue of Proposition 2.15 and Lemma 3.4.9.
by Proposition 3.1.8(ii)
Now
and Corollary 4.2.4(ii).
t(W1
Thus
= 7
by Lemma 5.3 and
t(G)
=
t ( W 1 7 7, as
required.
“hroughout this section, G characteristic p .
is a finite group and
F is an arbitrary field of
We know, from Theorem 4.3.6, that if
P is a group of order
pa which is not elementary abelian, then the following conditions are equivalent: (il
t ( P ) = ( a + l )(p-1) + 1
(iil t ; P ) < (a+2)(p-1) + 1 (iii) There exists a central element z
of order p
in P
such that P / < z >
is
elementary abelian. Our aim is to extend this result to p-solvable groups under certain conditions. Recall that the prime p
is said to be a Fermat prime if p = 2n + 1 for some
n a 1 (the latter, of course, implies that n is a power of 2). 6.1. LEMMA. (Motose (1984d)). group P
of order p
regular.
,a
2 1.
If t ( G ) < (a+2) (p-1)
2-subgroups of Proof.
a
C/O , ( G I
P
Let G be a p-solvable group with a Sylow p-subAssume that G is of p-length
+
1, then P
a 2 and P is
is a Fermat prime and the Sylow
are nonabelian.
Owing to Proposition 3.16.20,
.?
is nonabelian.
Hence, by Theorem
p
l.ll(iii), assume t h a t
i s odd.
,
0 ( G ) = 1. P
U
0 (GI, @ P
=
w e may harmlessly
(Lemma 2.21,
P
Put
IUI = PSI S 2 1 .
write
t ( G ) 3 t(G/O ,(GI)
Since
U and
t h e F r a t t i n i subgroup of
By Theorem 4.1.9, w e have
t(G)
z t ( U ) + t(G/U) -
1
while by Proposition 2.1,
Hence w e deduce t h a t
I t follows, from Theorem 4.3.6,
<x,U>'
r e g u l a r and
5U
for a l l
(XU)P =
u
that
X E
I s o f exponent
P,
p.
Because
P
is
it follows t h a t
2
for a l l
x
E
P,u E U
Hence we have
..+x+l = u2-I ... uxu
;Cp-l+. Iorall
X E P
and
...,x,l
sum of endomorphisms
GL(U/@).
Sylow p-subgroup
P
of order
Sylow 2-subgroups of ( i ) t ( G ) = a(p-1)
P
+ 1 or
+
i s of p-length
P/
Let
G
>
1.
pa,a
G/O ,(GI
(ii) t ( G ) < (a+2) (p-1)
that
U.
of
x-p(,lp and
= 1
(1)
$-l+...+x+l
By Lemma 1.3,
G/U
isthe
.
i s a subgroup of
Invoking (1) t o g e t h e r with Theorem 1.8, t h e r e s u l t follows.
6.2. THEOREM. (Motose (1984811.
(iii) G
X -s s u s = x ux
where
U E U ,
=
be a p-solvable group with a r e y l a r If
p
i s n o t a Fermat prime o r t h e
a r e a b e l i a n , then t h e following are e q u i v a l e n t :
t(G) =
(a+l)(p-1) +
1
1 1 and
P
has a c e n t r a l element
z
of o r d e r
p
such
i s elementary a b e l i a n
In p a r t i c u l a r , by Proposition 2.5,
t ( G ) = a(p-1)
+
1 i f and only i f
P
is
elementary a b e l i a n . Proof.
By Lemma 6.1,
our assumption on
G
forces
G
t o be of p-length
1.
CHAPTER 7
448
Hence, by Proposition 2.3,
t ( G ) # a(p-1)
+
t(G)
If P
t(P1.
=
1 by Proposition 2.5.
Hence in this case (i), (ii) and (iii)
In case P
are equivalent by Theorem 4.3.6.
not elementary abelian, then
is
is elementary abelian, then (i),
(ii) and (iii) are always true, again by Proposition 2.5.
This completes the
proof of the theorem.
Let G
be a finite group and F
ODD.
an arbitrary field of characteristic p . U
, a?
is a group of order p
to Theorem 4.3.5, if P
p
t ( G ) =pa-’,
7. CHARACl‘ERIZATIONS OF GROUPS G WITH
Owing
1, then the following
conditions are equivalent:
t ( P ) = pa-’
(i)
(ii) P = M ( 3 )
or
P = Z xjz X Z 2
2
2
Our aim in this section is to extend this result to p-solvable groups where p The case p
odd.
= 2
seems formidable for the following reason.
is
Assume that
p = 2 and consider the following three groups: G G
1
2
G3
= Z x z 2
= S
4
x Z
x Z
2
2
x Z
3
x Z x Z 3
3
xz!
3
3
is the group constructed in Example 3.6.
These three groups have the same order
2333.
Furthermore, by Corollary 3.2.5,
Proposition 3.1 and Example 3.6, we have
t(Gl) = 4
t(G3) = 5
t(G ) = 4
The above shows that there exist two groupsI namely
GI
and
GpI
with the same
nilpotency index whose Sylow 2-subgroups have different structures. hand, there are two groupsI namely
G2
and
GB,
On the other
with different nilpotency indices
whose Sylow 2-subgroups have identical structure.
This explains why we shall
restrict ourselves to the case p 2 3.
7.1. LEMMA. Then
t(G) =
Proof.
Let G
pa-1
be a p-solvable group of p-length
if and only if P
M(3)
or
Direct consequence of Theorem 4.3.5
P
z
U
1 and of order p ,a x Z2 x Z2
.
and Proposition 2.3.
1.
CHARACTERIZATIONS OF GROUPS G
Let G
7 . 2 . THEOREM. (Motose (1986all.
of order pala > 1,p odd.
subgroup P and P
2
If p
=
and P
3
IGI.
argue by induction on
2
has p-length at least 2 .
Then t ( G ) = pa-'
if and only if p = 3
=
Furthermore, by Corollary 3 . 1 . 1 8 ,
N/@
group of G L ( N / @ ) , where
N/@
2
M ( 3 ) , we
Then G/N
i s
(GI
, pr
is not cyclic. G/N
and hence
is of p-length
I N/MI
and
is isomorphic to a sub-
is regarded as a vector space over
cyclic of order p
=
P (Lemma 1 . 3 ) . P Indeed, if
is a p'-group,
2.
G L ( 2 , 3 ) acts naturally on the elementary abelian group E
of
Let T
order 9.
and P
P'tP
P
contrary to the assumption that G Every subgroup
by Theorem 4.5.
we may also assume
Put M = 0 ,(GI I N = 0
has p-length at least two, the group N/M
i s cyclic, then
9 = 33-1
To prove that p = 3
by the Frattini subgroup of N / M .
Since G
t(G)
Owing to Lemma 7.1, we may harmlessly assume that G
i s algebraically closed.
let @/M
GL(2.3)
be a p-solvable group with a Sylow p-
M ( 3 ) , then
t ( G ) = pa-'.
Conversely, assume that
N/M
449
M(3).
Proof.
that F
t ( G ) = pa-1, p ODD
WITH
(respectively, Sl
(respectively, S L ( 2 , 3 1 )
be the semidirect product of E
with respect to this action.
of
and
We divide the
rest of the proof into a number of steps.
P I
Step I .
of
Here we prove that if
GL(2,31
(hence a
=
J ~ F N P#
r + l l and
r-1 Assume that J ( F N l p # 0.
]@/MI
=
0, 3r-2
then p
=
3,
C/N
is a subgroup
.
Then, by Proposition 2.3,
It follows, from Theorem 5.2, that
r- 1 + P - l
t(N1 = p and that N / M
I
z
r-l x
z
P
P
or N / M
minimal number of generators of N / M
2:
Mr(p)
for 1 2 3 .
is two, we conclude, from Lemma 4 . 3 . 1 ,
N/@
is an elementary abelian group of order p'-
Now
G/N
GL(2,p)
is a subgroup of
Since in both cases the
In particular,
I @/MI
=
that
pr-2
GL(N/@) = G L ( 2 , p ) , and every element of order p
in
is conjugate to
[A' ?]
for some 0 # A e P p
CHAPTER 7
450
(X-1
Thus
is the minimal polynomial of every element of order p
)*
Since p
is odd, Theorem 1.8 implies that p = 3 .
Step 2.
Here we prove that
lQ/MI
8-2 and
=
p = 3,
+
t ( N I G 3p1
r- 1
i s a subgroup o f
GL(2,3) (hence a = r + l ) ,
2.
Then t ( N ) G pr-'
= 0.
Assume that J ( F N I p
G/N
in G / N .
and, by Proposition 3 . 1 . 9 ,
a-r S(FGIp
But, by hypothesis, t ( G )
-
=
pa-'
FG*J(FN)
and so, raising both sides to the power of
1 yields
Thus we must have
Since p
I Q/Ml=
implies that N / M
is odd, Theorem 4.3.5 N/Q
3r-2,
is elementary abelian of order
M(3). 3'
Hence P G/N
and
=
p
=
3,
is a subgroup of
GL(2,3).
Invoking Step 1, we deduce that p
IQ/MI
= 3r-2
Step 3.
.
G/N
G/@
is a subgroup of
3'-l
+
is a semidirect product of N / @
and
G/N
T.
Let f
:
G/M+
G/Q
be the natural homomorphism, let L
4.1.9
Since U
Then
2
>
(H:V).
together with Proposition 3 . 1 . 8 ( i i ) , we derive
Set U = V/((F/M) n H )
< 9.
and
,
3r = 3a-1 = t ( G ) = t ( f H I
t(U)
2.
By Proposition 2.15, there exists a block idem-
be the inverse image of f ( H ) n S and let V = L n H . Invoking Lemma 3.4.9
and
G/@ = S.
of FM such that for H = G(e)/M
for some ci E Z2(H,F*).
GL(2,3)
is a subgroup of t(N)
Reduction t o the case where
hence G / @
3,
Furthermore, by (1) and ( 2 1 , we also have
By Step 2 and Theorem 1 . 9 ,
potent e
=
=
t(V)
and observe that, by Theorem 4 . 5 and Proposition 2 . 1 ,
is isomorphic to a subgroup of S ,
and Proposition 4.1.8
that
it follows from Theorem
CHARACPERIZATIONS OF GROUPS G
WITH
where the latter equality follows from Step 2. Since
t ( U ) = 9 and M(3J
2.1 and 2.4 that U
=
group of
S.
of order 4.
Step 4 . Let f
, p ODD
>_ @/M
Thus H
t ( U ) = 9.
v#
Hence a 3-Sylow subgroup
Then, by induction, t ( V )
G/M.
Thus we may assume that N
has 3-length at least 2, so G/N
=
T
3
is a 3-
Thus G / N = SL(2,3J since G/N
must contain an element
contains an element of order
3.
as required.
Completion of th e proof. :
FG+
F(G/@)
=
FS
be the natural homomorphism.
Then, by Lemma 4.2 and
the proof of Lemma 4.4,
where
and
451
U = G I @ is a subgroup of S which contains a Sylow 3-sub-
But G
Hence G/@ = S ,
a-1
and so D is isomorphic to P.
3ai1 = '3
is isomorphic to M(3).
and
p
is of exponent 3, it follows from Propositions
Assume that either M # 1 or
group, V = G
=
contains a Sylow 3-subgroup of S.
D of V is of order 271@/MI
implies that D
t(G)
Z
such that Z 5 = 0.
is a right ideal of FS
Since
it follows that
JWG) = f-l(Z1 + J ( F N ) F G On the other hand, it is easy to see that
(f-'(Z)
1
5 Kerf
= J(F0)FG
and so
Taking into account that
t(N)
+ 2
(Step 2 1 , we derive
CHAPTER 7
452
Thus F = 2
Theorem l . S ( i ) ,
.
@ = 1, which by S t e p 3 implies t h a t
and
P
M ( 3 ) and t h e r e s u l t follows.
G = S.
Hence, by
453
8 Radicals of blocks Let
B
be a block of a group a l g e b r a
FG
d.
with d e f e c t
d
motivates t h i s chapter i s t o discover r e l a t i o n s between
t ( B ) of
index
J(B).
on lower bound f o r
The problem t h a t and t h e nilpotency
The f i r s t two s e c t i o n s provide some g e n e r a l information
t(B)
and upper bound f o r
t(Z(8)). The r e s t i s based on t h e
Fong correspondence and t h e Khshammer' s s t r u c t u r e theorem f o r blocks of p s o l v a b l e groups.
The l a s t s e c t i o n provides a number of important a p p l i c a t i o n s .
G
Among o t h e r r e s u l t s , w e prove t h a t i f and
t ( B ) = pd
i f and only i f
1. A LOWER BOUND FOR
t( B )
istic
p > 0.
denotes a f i n i t e group and
Given a block
index of t h e Jacobson r a d i c a l As u s u a l , w e w r i t e
t h e exponent of integers.
A,
G,
CR(G1
B
J(B)
B
of
and exp(G)
Given a s u b s e t
F a f i e l d of character-
6(B)
and
G, n E
x+=
=
Cx fix
alaPn E (if
n
xp
1.1. LEMMA.
Let
=
N
and a finite-dimensional F-algebra
n E No and l e t A
f o r some
m2 I
tA,AI}
X = n
rP
a,
IXE
then by d e f i n i t i o n 'X
be a finite-dimensional F-algebra. and
T ( A ) = u Ti(A) i=0
(ii) There e x i s t s
rn E No such t h a t
= 0)
XI
a0
( i ) [ A , A ] 5 T n ( A ) , !?',(A) 5 T n + l ( A )
and
denote t h e s e t of a l l nonnegative
~ 1 2E I A , A I I
{a E
G
No
m
TJA)
B.
f o r t h e d e f e c t group of
we w r i t e
T ( A ) = {a E
d
6( B ) .
f o r t h e s e t of conjugacy c l a s s e s of Let
X of
d(p-1) + 1 Q t (B)< p
FG, we w r i t e t ( B ) f o r t h e nilpotency
of
respectively.
then
is cyclic.
6(Bl
I N TERMS OF THE EXPONENT O F
G
Throughout t h i s s e c t i o n ,
i s p-solvable,
Trn(A) = T ( A ) .
454
CHAPTER 8
Proof. (i) Direct consequence of the property
x
[A,AI
E
P
implies
E [A,AI
established in the proof of Lemma 2.2.1.
...,aS
(ii) Choose a basis u l , u 2 ,
iE
for the F-space T ( A ) .
there exists m : 2 1 such that u?
{1,2,. ..,s}, S
2 =
Z Aiui with hi
F, we have for rn
E
i=1
rn
s C
2
5 ",(A)
For any p-subgrour, D C E
Hence, given
0 (mod I A , A l )
and hence that T,(A)
of
5 D.
6(C)
Ck(G) with
<
[A,A].
E
,...,rn 1 ,
maxim
=
i
mrn
:A
i=1
This proves that T ( A )
For each
m
=
A.
G, let ID(G) be the F-linear span of all c+,
Observe that, by Lemma 2.2.5,
ID (GI is an ideal of
n
Z (FG). 1.2. L E M .
(i) I,(G)l
Let D be a p-subgroup of Is
Ex g 9
E
1 [FG,FGI 51D(G)
FG with
a Z(FG)-submodule of
(ii) Given x =
G.
T(FG) and n E No with q
=
p
n
2 exp(D),
we have
1 xq E ID(G) Proof. (i) Assume that y E ID(GI1 ideal of
and
Z
E
Z(FG).
Since ID(C) is an
Z(FG), we have for all t E ID(G)
z t E I,(O
It follows that
tP((yz)t)= t r ( y ( z t ) ) = 0 1 proving that yz E ID(G)
to Lemma 3.3.9(iil,
.
Hence ID(G)
1
is a
for all t E I,(G)
Z(FG)-submodule of FG.
Z(FG) = [FG,FGll and so, by Theorem 3.3.6(1), 1 [FG,FGl = Z(FG)
Bearing in mind that ID(G)5 Z(FG),
we derive
I D ( G ) ' >- ZIFG)'
= [FG,FGI
as required. (ii) Assume by way of contradiction that t P ( X q C + ) # 0
for some
cE
Ck(G)
Owing
t(B)
A LOWER BOUND FOR
with and
S(C) C - D.
+G
tr(xS
)
q 2 exp(8(C)l,
Then,
# 0
455
C
so by Lemma 3 . 9 . 2 ( f f ) ,
f o r some p - r e g u l a r s e c t i o n
tr(xs+)=
S
G.
of
i s p-regular
Because
x
@-lxg
# 0
w e have 3.3.13, Let
x 9 T(FG), A
rn
1.3. THEOREM.
(ii) I f
T(B).
D = d(B)
Put
prn < exp(D1,
E
xE
T(B)
rn
T(A)
T(A).
B = B ( e ) be a block of
E
as the l e a s t
No.
FG
and l e t
n
6(B)
I t s u f f i c e s t o show t h a t
T ( B ) = Trn(B)
then
ID(G) and so Z ( B 1
Hence, by Lemma
Owing to Lemma 1.1, t h e r e e x i s t s
i s t h e exponent of
and l e t
prn 2 exp(D), then
=
Let
n Then p
To prove ( i ), assume t h a t
e
T (A)
such t h a t
(Kiilshammer (19821).
be t h e index of Proof.
n
G.
We now d e f i n e t h e index of
T (A) = ?'(A).
nonnegative i n t e g e r
of
So t h e lemma i s t r u e .
a contradiction.
be a finite-dimensional F-algebra.
rn E No such t h a t
(i) I f
S-'
f o r t h e p-regular s e c t i o n
T ( B ) # T,(B)
p" 2 exp(D)
rn
q = p
and l e t
= Z(B)e LID(G).
.
By P r o p o s i t i o n 2.2.6,
Hence, by Lemma 1 . 2 ( i i ) ,
for a l l
w e have
tr(xqZ( B 1 ) proving t h a t
xq
E Z(B1
L
=
[B,B],
tr(xqID(G1) Consequently
=
3:
0,
E Trn(B1
and t h e r e f o r e
T(B1 = Trn(B). To prove ( i i l
, assume
that
4
m
= p
< exp(D1
and w r i t e
e = C e a
SEG g. z(FG1Z(FC(D)) together with Theorem 2.2.4, G c o n t a i n s a p-regular element g E CG (D). Choose an element
Applying t h e Brauer homomorphism
we infer that
d
E
d-1
Suppe
D of maximal o r d e r and a p-regular element k i s n i l p o t e n t and
dh = h d .
Thus
W e are t h e r e f o r e l e f t t o v e r i f y t h a t To t h i s end, we f i r s t note t h a t
(d-l)he
E
CG(D) with g-l
= k.
Then
i s n i l p o t e n t and so ( d - l ) k e E T ( B ) .
(d-llhe 9 Trn(B).
CHAPTER 8
456
and t h a t , by our choice of
8g-l
must be p-singular.
r e g u l a r and thus
q
d,
and
y
If
8#
E Suppe
tr(d4g-ly) = 0.
Since
1.
,
dg
=
gd,
t h e element
y
then by Theorem 2 . 2 . 4
is
p-
The conclusion i s t h a t
But then
and hence
a s required.
8
As an a p p l i c a t i o n o f Theorem 1.3, we now prove t h e following r e s u l t .
1.4.
THEOREM.
D
group (i) I f
(Kiilshammer (1982)).
q = pn
and l e t
f o r some
B
Let
n
E
4 < e x p ( D ) , then t h e r e e x i s t s
=
B(e1
be a block o f
FG
with d e f e c t
No. 2
E J(B)
with
xq 9 I B , B ] .
In
particular,
(ii] I f
Proof. with
yq
Therefore
q = exp(D),
then
z4 = 0
( i ) By Theorem 1.3,
[B,B].
y = k
for a l l
T(B) #
z E J(Z(B)).
T (B)
and so t h e r e e x i s t s
Applying Theorem 3 . 3 . 1 0 ( i ) , w e also have
+
D
f o r some
k4 E [ B , B ] , we deduce t h a t
a s we wished t o show
k E [B,B]
and some
D
E JCB).
y E T(B)
T ( B ) = [B,BI + J ( B ) . Because
457
Owing t o Theorem 1 . 3 , we have
(iii)
for all
g E G
fact that
z E J(Z(B)).
and a l l
Applying Lemma 3 . ? . 2 ( i I
i s a Z(FG)-module, w e d e r i v e
[B,B]
0 = tr((gz)%+) for a l l
together with the
C E CR(G).
Thus
= [tr(gz(C1/%+)14
tr(FGz(C1")')
0 and so
=
= 0
Z ( c q +
C E CR(G)
for a l l
By Theorem 3 . 9 . 3 , w e i n f e r t h a t
o
z 4 ~ +=
cE
f o r a l l p-regular
CL(G)
with
6 (c) 5 D G
Since each
CE
CR(G)
with
C
5 Suppe
i s p-regular and
6(C)
5D
(Theorem 2 . 2 . 4
G and P r o p o s i t i o n 2.2.6)
it follows t h a t
zq = 24, = 0 T h i s completes the proof of t h e theorem. 1.5. COROLLARY.
B.
of d e f e c t groups of Proof.
be a block of Then
Suppose t h a t
G.
sylow p-subgroup of
t ( B ) 2 pn-'
Then
G
P
pn,n > 0 ,
and l e t
+
be t h e exponent
.
1.
pn be t h e exponent of t h e
i s a b e l l a n and l e t
n =
o
for a l l
z E JCFGI.
Apply Theorem 1 . 4 t i l ) .
1.7. COROLLARY.
If
t
is t h e exponent of Sylow p-subgroups of
zt = Proof.
FG
D i r e c t consequence of Theorem 1 . 4 ( f ) .
1.6. COROLLARY.
Proof.
B
Let
o
G,
then
z E J ( Z (FG)
for a l l
Apply Theorem 1 . 4 ( i i ) .
We next i l l u s t r a t e t h a t Theorem 1.4U.i) need not be t r u e f o r
z E J(B).
The
following two observations w i l l c l e a r our path.
1.8. LEMMA.
(i)
Let
(R1 x R 2 x
modules.
R,RI,R2,...,R n be r i n g s R G x R G x x Rn ) G
...
and l e t
... x RnG
G,GI,G2
be groups.
a s r i n g s and
R
x
... x Rn-
CHAPTER 8
458
(ii) R(G x G 2 ) 1 (RGIlG2 as rings and R-modules (iii) For any
k
>
1, M (RIG 4 M (RG) as rings and R-modules k k
(i) Put S
Proof.
projection map.
=
R
1
Denote by
(7
Kerfi = 0
:
i
... x Rn
x
2
71
n Then
x R
SG-
and let fi
RiG
Ri be the
: S+
fi'
the ring homomorphism induced by
i
and KerTIi = (Kerfi)G, 1
n.-
Consider the map
i=1
1
... x RnG (z) ,... (33 1 -.
R G x R2G x
SG-
z C-L
(z) ,TI
(71
l i
,TIn
2
is a homomorphism of rings and 5'-modules.
It is clear that IT
n n n KerTI = n Kerni = n (Kerfi)G = t n Kerfi)G
i= 1
i=l TI
is injective.
To prove that
,...,ynl
and write
yi = Cr .g gs
with r
gi
be such that f . ( r i = r 1c 2 g gi'
E
... x
E R G x R G x
Ri, 1 c i
$G n .
TGr
gl
9
0,
i=1
is surjective, fix
TI
(yl,y2
=
Since
For any g E G,
n.
Then
RnG
Ti(Crggl =
let P
9
E S
yi, 1 G i G n , and thus
,... , ~ , l ,
(Y,,Y,
=
as required. (ii) By looking at the chain of rings R
-
is a free R-module with
(g1,g21
{ g l g 2 1 g lE G 1 ,g 2
5 RGl 5 ( R G l I G 2 , we E
G21
as a basis.
infer that
(RG1)G2
Thus the map
g l g 2 induces an isomorphism R ( G l x G2)-
of R-modules.
(RGl)G2
Since this induced map obviously preserves multiplication, (ii) is
established. (iii) Let e . . be the n x n 23
elsewhere, and let ei zero entry elsewhere.
J,g
matrix with (i,jl-th entry
be the n x n
M k I R ) G - + Mk(RG) of R-modules. preserves multiplication.
A
and
induces an isomorphism
routine calculation shows that this map also
So the lemma is true.
1.9. LEMMA. Let G = H x P , where k FH (Oil for suitable k,ni i=1 "i
nM
matrix with (i,j)-th entry g E G
e..gW e 23 i , j ,g
Then the map
1 and zero entry
P
is a p-group and H is a p'-group.
and division rings Di.
Then
Write
459
M,
(ii) Each
(Dip) is indecomposable
i (iii)
P i s t h e d e f e c t group of any block of
(i) D i r e c t consequence of Lemma 1.8.
Proof.
D.P
(ii) I t s u f f i c e s t o show t h a t each
D.P.
zero c e n t r a l idempotent of
p.
characteristic
e
(ii) L e t C
C_
Suppe.
i s indecomposable.
e E Z(Di)P
Then
Hence, by Theorem 2.2.4,
C
By Theorem 2.2.4, of
5H
e,
e
FG
be any block idernpotent of
P i s t h e d e f e c t group
Thus
FG.
=
as required.
be a non-
a s required.
C E Ck(G1
and l e t
and hence
e
Z(Di) i s a f i e l d of
and
1
Let
.
P is
be such t h a t C.
t h e d e f e c t group of
I t i s now an easy m a t t e r t o show t h a t Theorem 1.4(111, need not be t r u e f o r
z E J(B). (Kilshammer (19821 I .
1.10. EXAMPLE.
P
=
X
verify t h a t
F ( H x PI
FH
21
be an elementary a b e l i a n group of o r d e r
4.
let
Bo
has a block
has a block B
J(FP) =
Then
be an a l g e b r a i c a l l y c l o s e d f i e l d of
H be a nonabelian group of o r d e r
2,
characteristic
F
Let
I(P)
1
M (Fp). 3
and so
J(B1
M3(F1. Let
and l e t I t i s easy t o
Hence, by Lemma 1.9, t h e group algebra
FP.
T ( P ) be t h e augmentation i d e a l of
M3(I(P)).
Identifying
J(B)
and
M3(I(P)),
we see t h a t
g-1
0
h-1
0
0
0
0
0
0 z =
but
' 2
# 0.
.
2 . AN UPPER BOUND FOR
t(Z(B1)
Throughout t h i s s e c t i o n , istic of
2.1.
p.
denotes a f i n i t e group and
F
a f i e l d of character-
Our aim i s t o provide an upper bound f o r t h e nilpotency index
J ( Z ( B ) ) , where
LEMMA.
G
Let
B
P
i s a block of
t(Z(B))
FG.
be a p-subgroup of
G
and l e t
H be
a subgroup of
C
such
CHAPTER 8
460
that
D,
C(P) G let p
5 H C_ NIP). :
e
where each
Z(FH)
PG wfth d e f e c t group
be t h e Brauer homomorphism, and l e t p ( e ) = e
FH.
i i s a block idempotent of
Z G Proof.
C ,C
Let
I (G) D
let
e be a block idempotent of
Let
Di
If
+ . .. + e n' ei f
is a d e f e c t group of
5 D.
D.
then
G Z(FG1-
1
,...,Ct
be t h e F-linear
Proposition 2 . 2 . 6 ,
L E Ck(H)
L
and
x E G.,
f o r some
2
that
5 Ci
+ + C1 , C p , .
span of
IDCG)
i
6 ( C i ) C D and G We know, from Lemma 2.2.5 and
,Ci.
Z(FG1
e E ID(G1 I
e.
containing
If
w e conolude t h a t
C I XEG
I (HI DxnH
with
then
E {1,21.,.,t},
P(e) E
Takinq i n t o account t h a t
..
i s an i d e a l of
f o r some
Because
CR(G)
be a l l elements i n
(HI D"~H Z(FH1
is an i d e a l of
ei = e i p ( e ) ,
and t h a t
we see t h a t
e E Hence, by Lemma 3.16.5,
D . C Dx
%a
2.2.
C1
5D,
nH
as required.
G,
of
and t h e r e f o r e
H
let
El.
be a l l d i s t i n c t p-regular c l a s s e s of
Zi
(i)
c1,c2,. ..,zr
be t h e image of
iE
=
PC(P1,
and l e t
For a l l
Proof. As
i
For each
Ci under t h e n a t u r a l homomorphism G -
a r e a l l d i s t i n c t p-regular c l a s s e s of
,...,rl. c E {c ,...,cr}, C 5 C(P1 G C E { C ,-..,C 1, 6 ( C l / P r
H/P
.. , r } ,
E {1,2,.
G/P.
and
I cil
Then =
lcil
{1,2
(ii) For a l l
E C.
G n)
9
P be a normal p-subgroup
Let
let
x
x E G
f o r some
G
,...,C
(iiil
i E I DxnH (HI
i
(1 Q
G
LEMMA.
for a l l
e
C I CHI x€GD%H
Let
x
IGI = pnq, i s a p'-element
where
and
P
6(C)
i s a d e f e c t group of
( p , q ) = 1.
and as
5
Fix
C E {C
-
C.
,...,C
}
and l e t
461
g E G,
For any
let
cii(a1
Let
By (ii), we have
H/P.
=
where
h-lxh
Then
E Cz(Z1.
k,
element, say of o r d e r
=
and
d = 1.
P
We next show t h a t
CHCS)/P
=
5 CH(xl
d3c
(1)
d
is a p-element, it follows t h a t
d E P.
f o r some
(h-lxh)k = xkdk
x is
Since
a p'-
dk
=
and (11 i s e s t a b l i s h e d .
H (2)
5 C*(Z).
C,(rl/P
and t h e r e f o r e
=
h E C
Thus
G/P.
in
xd
1=
and hence t h a t
g
be t h e image of
Since (1)
implies ( i i i ) ,we a r e l e f t t o v e r i f y ( i ) .
ICI
To t h i s end, we f i r s t observe t h a t
(E:Cz(G) 1 X ,X
Assume t h a t
Then
-1
yx,y
x1
=
3:
1
d
and
.
yx2y
E
H
{c ,...,cr}
f o r some
commute.
X
Thus
Y
Finally, l e t
y
E
yx y-ld
-1
=
2
y,
o r d e r of
E Y.
y m E P.
then
+
ipn
such t h a t
X2,
and
z
d E P.
jm = 1.
Then we
ma71
C E {Cl
Let
y
xi
E
56(x2)
X i =1,2. i'
and so
we have
are distinct. of
H/p
C(P1.
and l e t
If G there are integers
Because (p,m) = 1,
E
m
i
is the
j
and
Hence
i s t h e h i g h e s t power of
element.
,...,Cr-
assume t h a t
. n .
pn
P
a r e p'-elements,
proving t h a t t h e
y As
and l e t
By ( i t ) ,
yx y - l
and
z2,
=
denote a conjugacy c l a s s of p'-elements
be such t h a t
-
=
(H:CH(~))
a r e such t h a t
y E H
Since
=
s i n c e by (1)
=
,...,Cr 1
=
p
.n
p yJm E p p
(2)
G , yipn
d i v i d i n g t h e o r d e r of
is a
p'-
.n be such t h a t
yzp
c.
E
Then, by ( 2 1 ,
Y
=
?
and t h e r e s u l t follows.
2 . 3 . LEMMA.
Let
P be a normal p-subgroup of
G,
H
let
=
PC(P1
and l e t
G
n
:
of
FG+ FH
F(G/P1 be t h e n a t u r a l homomorphism. Di be a d e f e c t group of
and l e t
For each proof.
bi, 1 G i
,...,~
( 1b a r e a l l d i s t i n c t blocks of
iE
,...,n},D./P
(i) n ( b l ) , n ( b 2 )
(iil
Let
(i) L e t
{1,2
b ,b 1
ei
,...,bz
be a l l blocks
n.
F(H/P)
i s a d e f e c t group of
4 = n(Z(FHFH)) and l e t
2
n(bi).
be t h e block idempotent of
FH
462
CHAPTER 8
bi, 1 4 i
contained i n
4
n.
By Proposition 3.1.1,
and t h e r e f o r e , by Lemma 3.4.16, idempotents of
T.
idempotents of
F (H/P).
of
ci
C',Cr, 1
2
1
...,T ( en
2
be a l l p-regular c l a s s e s of
a r e a l l d i s t i n c t p-regular c l a s s e s of
(iil
e
Let
FH
be a block idempotent of
r e g u l a r c l a s s e s of
H
such t h a t
Suppe =
and l e t
c
suppn(e) = C r U where
Cl denotes t h e image
H/P.
Hence, by Theorem
1
,.. U Ct.
U
be a l l p -
C ,C ,,..,Ct 2
Then
... u c; 4 G/P.
Ci under t h e n a t u r a l homomorphism G
i s t h e image of
CI
If
F(H/P).
T contains a l l c e n t r a l kdempotents of
2.2.4,
H.
T contains a l l c e n t r a l
G/P, then by Lemma 2 . 2 ( i ) ,
under t h e n a t u r a l homomorphism G-+
...,C f
are a l l d i s t i n c t primitive
I t t h e r e f o r e s u f f i c e s t o prove t h a t
C ,C f...,Cp
Let
lT(e 1 ,
i s a nilpotent ideal
KerT
The
d e s i r e d conclusion now follows from Lemma 2 . 2 . ( i i i l . 2.4. THEOREM. (Okuyama (1981)).
Let
B
=
B(e)
be a block of
FG
with d e f e c t
d
Then
proof.
=
Z(FG)e,
k k 2 1, J ( Z ( B ) ) ' = J ( Z ( F G ) ) e .
all
If
Z(B)
We have
d = 0
is true for write
D
J(B)
then
d
=
0
=
so
J ( Z ( B 1 ) = J ( Z ( F G ) ) e and t h e r e f o r e f o r
Thus we need only v e r i f y t h a t
0 (Theorem 3.6.4)
and t h e r e f o r e
and w e argue by induction on
f o r a d e f e c t group of
B.
d > 0.
S ( Z ( F G ) ) e = 0.
Thus ( 3 )
I n what follows, w e
For t h e sake of c l a r i t y , t h e rest of t h e
proof i s divided i n t o t h r e e s t e p s .
S t e p 1. H = C(P). G
Let Let
g E D be an element of o r d e r p n , n > 0,
let
P
p : Z(FG) -+ Z(FH) be t h e Brauer homomorphism.
=
and l e t
I t w i l l be
shown t h a t
By hypothesis, t h e r e e x i s t s
C E C&(G)
with
C 5 Suppe and 6 (C) = D.
Hence
463
H
0
C # @ and so, by Theorem 2.2.3, p(e1
=
sum
natural homomorphism. F(H/P)
Ker'll = FH(g-11,
is at most d .
Let
n(f)
Then, by Lemma 2.3,
with defect at most
d-n,
+ em
Fix f E { e
of block idempotents of FH.
Lemma 2.1, the defect of f
Therefore we may write
.-.
+ e2 +
e 1
as a
# 0.
p(e1
,...,e 1
TI
and observe that, by
m
: FH+
F(H/P)
be the
is a block idempotent of
since the defect of f is at most d.
Since
it follows from (3) by induction hypothesis that
d-n
f 5 FH(g-1)
J ( Z ( F H 1 IP
Since pd - 1 2 pd-n (pn
- 11,
(5)
raising both sides of (5) to the power of pn
-1
yields
i=o
where the last equality follows by virtue of Lemma 3.11.2.
proving (4) by virtue of Step 2. Zp,
=
,... ,S r
Sl,SZ
Let 21
I: FS: i=1
p(J(Z(FG1)
.
5J ( Z ( F H 1 ) .
be all p-regular sections of
d
be a p-regular element of G
such that 3cg = gx. the coefficients of ficients of
z
G and let
We next prove that J(z(FG)~P -le
Let x
Thus we have
and
g
c
z
(6)
- Pr
It suffices to show that, for any z 3c
and leg
in
Z
are equal.
are constant on conjugacy classes of
G
This is so since the coef-
G.
We may harmlessly assume that either x E Suppz or xg E Suppz. a
defect group of e .
Then, by Lemma 2.2.5
n
of order p d in J(Z(FG))' -'e,
a p-element of
and Proposition 2.2.6,
Let D be
ID(G) is an
CHAPTER 8
464
ideal of
Z(FG) containing e .
x E Suppz or xg
group of
D.
Since
Thus we may assume that g belongs to a defect
,H
Let P = < g >
e.
z E T,(G).
it follows from the definition of ID(G) that g
E Suppz,
belongs to a conjugate of
In particular, we must have
C(P) and
=
P :
Z(FG)
Z(FH)
f
the Brauer
G
homomorphism.
Since x,xg E H,
same as the coefficients of x
the coefficients of x
and zg P(Z1
in P(z).
and xg
in z
are the
But, by (4),
pn-li E FEI( c g 1 i=o
and so the coefficients of x
Step 3.
and xg
Completion of the proof.
in
are equal, proving ( 6 )
p(Z)
Owing to Theorem 3.3.14, we have
J(Z(FG))Zp, = 0 Hence, multiplying both sides of (61 by
J(Z(FG)) we obtain d
cPCZ(FG)lP e
= 0
proving ( 3 ) and hence the result. 2.5. COROLLARY.
G
Let
a
be a group of order p m, ( p , m ) = 1.
Then
a J(Z(FG1Ip Proof.
Let e ,e r . . . , e 1
2
= 0
n be all block idempotents of FG.
Then, by
Theorem 2.4, a
S(Z(FG))P ei
a
a
Since J(Z(FG)lp = J(Z(FG)IP e l A
+
...
=
...,nl
for all i E {1,2,
0
a + J(Z(FG)IP e n ,
the result follows.
.
weaker version of Corollary 2.5, namely the equality
a+lJ(Z(FG))( p "'(p-') was established by Passman (19801. 3.DEFECT GROUPS OF COVERING BLOCKS.
Throughout this section, G and B
denotes a finite group, N
F an arbitrary field of characteristic p > 0.
i s a block of
group of b .
FG
covering a block
a normal subgroup of
G
Our aim is to prove that if
b of FN, then 6 ( B )
f?
N is a defect
465
DEFECT GROUPS OF COVERING BLOCKS
Let B = BCel
3.1. LEMMA.
b = b(f1
covers b .
tively, such that B (i) V
and
be blocks of FG and FN, respec-
Then the following conditions are equivalent:
is an indecomposable FNrmodule such that V
UN for
B.
U lying in
some projective indecomposable FG-module (ii) V
is a summand of
is a projective indecomposable FN-module such that V
lies in bg = g - l b g
for some g E G. Let X
Proof.
Let f * be the sum of all distinct G-conjugates of f.
(ii), respectively. Then
Y denote the classes of FN-modules satisfying (i) and
and
is a central idempotent of and f*V = V
f*U = U for U E B there is a g
so V E Y
E
G such that
and thus X
To prove that Y
YV
L
5 X,
U E B,
WN.
is a summand of
Then
so L
71
for
V
V,
so
=
E
X. V
is a summand of
(see Lemma 3.10.1).
Because
E
bg.
first assume that
V
E
Hence
V is indecomposahle,
Clearly, V
is projective,
V f E X.
V'
If
is such that L = V/J(FN)V
Y
is a summand of
Let
71
:
uN
for the
No and
then for a suitable r E
r Q - + J(FNl U be a projective cover of J(FN)pU.
But Q E B ,
(&/J(FN1QIN.
Vi/J(FN)Vi for some i and V
Now assume that
V E Y and V'
E
as well.
V
z
E
=@
Vi, all Vie X,
X. Then
.
Then
if
V and V '
V/J(FN)V is a composition
Applying the above argument twice, we derive Q
Since obviously 'V E X
least one
1
QN
so we may assume they have a composition
factor Q/J(FN)Q in common (for some & E Y ) factor of Q
hence
X are in the same block.
are linked (see Proposition 1.10.14),
X2 Y.
e
induces a surjective homomorphism
and therefore L
V E X.
=
5 Y.
is a composition factor of some projective module
FG and ef*
V
E
x
and X
.
is nonempty, X
E
X and
contains at
V in each block b g , so X contains all V in all b g , that is
So the lemma is proved.
We are now ready to prove the following result.
466
CHAPTER 8
3 . 2 . THEOREM. (Knb;rr (19761I.
Let F
be an arbitrary field of characteristic
p > 0, let N be a normal subgroup of
covers b .
and FN, respectively, such that B then D
nN
G and let B and b be blocks of FG If D i s a defect group of B ,
is a defect group of b .
Proof.
6CB)
We denote by
6 ( b ) defect groups of blocks B
and
Let X and Y be the classes of FN-modules satisfying (i1 and
respectively.
We know, from Lemma 3.1, that
(ii), respectively, in Lemma 3.1.
Theorem 3.16.14, there exists an irreducible module M E b where Vz(M) denotes a vertex of M . Therefore M
i s a summand of
UE B
FG-module
and b ,
. Owing
Now M
(U/J(FNIU)N
2
V/J(FN)V
x
=
Y.
By
such that ' U s ( M ) = 6 ( b I , G for some V E X = Y.
for some projective indecomposable
to Lemma 3.16.13, the indecomposable FG-module
U/J(FN)U
is FN-projective, so by Lemma 3.16.1,
On the other hand, by Proposition 3.10.6, there i s a block
6(B]
and
C
1:
SO
6(Bl)
and
6(b) = N n 6(Bll.
B
of
FG such that
Hence
G
6(b) = 6 ( B ) n N , G
as asserted.
4. REGULAR BLOCKS.
Throughout, G
denotes a finite group, H
field of characteristic p > 0, classes of
G and, for each C
is a block of FH, we write potent of FH
lying in b
associated with b .
usual, CRCG)
As E
a subgroup of
CR(G) ,6(C)
G and F an arbitrary
denotes the set of conjugacy
is a p-defect group of
b = b [ e , h ) to indicate that e
and
h
C.
If b
is a block idem-
the irreducible representation of
Z(FH)
We denote by
the natural projection so that T(c+)
Note that
71
=
(c n H I +
for all C E CR(GI
i s a surjective F-homomorphism, but need not preserve multiplication
467
REGULAR BLOCKS
Let
b=b(e,AI
be a block o f
bG i s defined.
LEMMA.
bs
5s 5 G
being a r i n g
corresponding t o
bG o r
i s defined and e i t h e r one of
AG
proving t h a t
b(e,A)
(bSIG i s defined, then so is
C E CR(G1,
2
1
' 1Z ( F S )"1,Z (FHI
we have
.
Hence = (lor
bc
= AoTI =
)on
2
AG
1
( b S ) G i s defined.
i s defined i f and only i f
hG a r e t h e i r r e d u c i b l e r e p r e s e n t a t i o n s of
Z(FG)
Since
(ASIC
a s s o c i a t e d with blocks
b G , r e s p e c t i v e l y , t h e a s s e r t i o n follows.
(ii) By hypothesis,
AG
is an i r r e d u c i b l e r e p r e s e n t a t i o n of
sition 2.2.6, there exists
x(L+) # 0
Proposition 2.2.6,
a s asserted.
=
6 ( b J C_ b ( b G ) . G
(ASf
(bSIG and
b
Consider t h e sequence of n a t u r a l p r o j e c t i o n s
TI = II 71
and t h e r e f o r e
and l e t
.
(bSIG= bG
(i)
Then, f o r a l l
G
be a chain of subgroups of
Z( F G )
Hence
i s an
6 ( b ) always denotes a d e f e c t
I n what follows
bG i s d e f i n e d , then
Proof.
and
FG
1'
r f t h e map
FH.
t h e o t h e r and (ii) I f
H
Let
be a block of If
hen.
b.
group of
(il
=
bG t h e unique block of
homomorphism), we denote by
4.1.
hG
and l e t
Z(FG) (which is e q u i v a l e n t t o AG
i r r e d u c i b l e r e p r e s e n t a t i o n of
and say t h a t
FH
.
f o r some
C E CE(GI
with
6(Cl = 6(b
L E CR(H)
with
L
6 (b) c 6 (L)
i7
G
.
5 c.
)
Then
Z(FG).
such t h a t
6(L)
+ A (C ) # 0 .
56(C) G
Thus
By Propo-
G
and, by
468
CHAPTER 8
For f u t u r e use, we next record 4.2. LEMMA. of
(il
FG
and
B
N
Let
FN,
respectively.
V E B,WE b
W
and
B
Furthermore, i f
be blocks
Then t h e following conditions a r e equivalent:
FG
b,
( i )* (ii) :
W,
and
r e s p e c t i v e l y , such
VN.
i s a submodule of
covers
V
and FN-modules
V
then f o r any i r r e d u c i b l e FG-module
V N
t h e r e e x i s t s an i r r e d u c i b l e submodule of Proof.
b
and
b
covers
(ii) There e x i s t i r r e d u c i b l e
that
G, and l e t B
be a normal subgroup of
B
Write
Ee # 0.
B(E)
=
B,
in
b.
belonging t o
b = b ( e ) so t h a t , by Lemma 3.10.2,
and
If
e
a n n i h i l a t e s a l l i r r e d u c i b l e submodules of
f o r any i r r e d u c i b l e FG-module
V
in
( i )i s equivalent t o
B,
then
e
annihilates a l l irreducible
Ee
B
and so
VNI
B.
FGmodules i n
Hence t h e same i s t r u e f o r
S(B).
potent i n
Since J ( B 1
E
Ee Ee
i s n i l p o t e n t , we conclude t h a t
i s an idem=
0.
Thus ( i )
implies ( i i ) , ( i ):
(ii)
Assume t h a t ( i i l holds.
Then, by C l i f f o r d ' s theorem,
VN = g W C B 1=
g ,g2,...,g,
EeV
=
f o r some
eV = W
Assume t h a t
containing
-1 e = g elg
.. . @ egn W
b
and l e t
g E G.
g-lW i s an i r r e d u c i b l e submodule of
4.3. LEMMA.
H.
If
Let
C ( P ) C_ H
G
then
:iomomorphism with r e s p e c t t o Proof.
eW
and
=
W
and so
( d i r e c t sum of FN-modules)
VN
B
and by
b
covers
w=9-b V N
b = b ( e , X ) be a block of
2 N(P), c
V
b
= 1
b (e 1
)
B.
t h e block of
1
and hence by Lemma 3 . 1 0 . 2 ,
I t follows t h a t
e(g-lW) = g-le so
=
V be any i r r e d u c i b l e FG-module i n
Then, by t h e above
f o r some
EV
and hence t h a t ( i i ) implies ( i ) .
covers
W.
r4
By hypothesis,
W any i r r e d u c i b l e submodule of
Denote by FN
B
kr @
eg
Ee # 0
This proves t h a t
G.
in
... CB gn
P.
XG
= X.p,
#
o
belonging t o
FH where
In particular,
I t s u f f i c e s t o show t h a t f o r each
and l e t
b. P be a p-subgroup of
p : Z(FG)-+
b
G
,
Z(FH) i s t h e Brauer
i s defined.
C E CR(G),
REGULAR BLOCKS
469
XG(C+I = ( l o p ) CC*)
n
C
To t h i s end, w r i t e
fl H =
Ci with each
U
i=1
n(C
+)
=
Ci i n
Ck(H1.
Then
n +
c ci
i=1 and
and so we need only v e r i f y t h a t
xcc3 z Because e i t h e r
c2. -c
L E Ck(H) n suppe
x(cc n c G ( p ) ) +
for a l l
i E {l,.
Ci n CG(P) = 8, we may assume t h e l a t t e r .
C (P) o r G
6(L) = 6(b).
be such t h a t
6(Ci) c 6(L1
Hence
=
Then
P
5 6(L)
h(C)
and s o , by P r o p o s i t i o n 2.2.6,
=
since
0,
Let
P a H.
as r e q u i r e d .
H N
Let
be r e g u l a r w i t h r e s p e c t t o
4.4.
of
Let
LEMMA.
FG.
(i)
G.
be a normal subgroup of
N
N
be a normal subgroup of
is said t o
G
and l e t
B
=
B(E,A)
be a block
B i s r e g u l a r with r e s p e c t t o N
b
of
FN
(iii) B = b'
f o r some block
b
of
FN
Proof.
( i )* ( i i ):
Let
Owing t o Lemma 3.10.4,
such t h a t
G
b = b(e,pl
:
for a l l
FN
which i s covered by
y E FN,
C
h (Z*( F N ) )
z E Z* ( F N )
.
From t h e d e f i n i t i o n of
p
G , we
hence
G
C E CL(G) i s such t h a t
be a block of
p (Z*(FA4 1-
Jt(!.~(r) 1 = h(x1 f o r a l l
p (y) = p ( y )
B
which i s covered by
t h e r e e x i s t s an F-algebra isomorphism
Jt
If
FG
of
Then t h e following c o n d i t i o n s a r e e q u i v a l e n t :
f o r every block
have
B = B(E,h)
.
if
( i i ) B = bG
B.
A block
.. , n l
$ ( p (r))=
A (x)
N,
p (C
then
G +
G J t ( P ( z l ) = A(zl
for a l l =
+
0 = h(C ).
x E Z*(FN)
Hence for a l l
z E Z(FG)
410
CHAPTER 8
and so pG
E i s a block idempotent of bG and thus B
=
b
G
Because p ( E l = h ( E ) = L
Z(FG),
is an irreducible representation of
G
.
(ii) =* (iii): Obvious
G
(iii) * (i): Let b = b ( e , p ) be a block of FN and
such that B = b ,
are equivalent and so it suffices to show that p
C E CR(G1
.
c
with
N.
the result follows.
G +
(c
)
= 0
Then l.I
G
for all
The latter being a consequence of the definition of p
G
,
We have now come to the demonstration for which this section has been developed. Let N
4.5. THEOREM. of FG
and F N ,
(i) If B
be a normal subgroup of
G and let B
and b
be blocks
respectively
is regular with respect to
N, then B covers b if and only if
G
B = b . (iil If D
is a defect group of B
and
CG(D)
5N ,
then B
is regular with
n
respect to N ,
and B
(iii) The block bG
=
B’
is defined if and only if there exists a block of FG
is regular with respect to N
G
E N
for some normal p-subgroup P
Furthermore, by Lemma 3.10.2, if
then bG = bG 1
2
(il
Proof.
if and only if bl If B
assume that B = b potent
E
G
of FG.
covers b ,
and b
bl and b
.
are blocks of F N ,
then B = bG by Lemma 4.4(ii).
G
Conversely,
for some block idem-
for all r E Z*(FN) , we infer, from
Because p (r) = p ( x )
(ii) Write B = B ( E , X )
covers b . and assume that C E C!L(G)
then by Proposition 2.2.6,
D
5 6(C)
is such that C
But then g E N
and so C
ZN,
N.
If
which means that g E C G ( D )
G
some g E C.
is regular
are G-conjugate.
G If b = b ( e , u ) , then B = B ( E , p )
L e ma 3.10. 4, that B
x(C+) # 0,
of G, then B
In particular, bG is defined and is the only block of FG
with respect to N. covering b .
is the unique block of FG which is regular
and covers b .
with respect to N C (P)
which
b.
and covers
(iv) If bG is defined, then bG
(vl If
B of FN which is covered by B .
for any block
a contradiction.
for
471
TAE FONG CORRESPONDENCE
bG is defined, then bG is r e g u l a r (Lemma 4.41 and covers b , by
(iii) I f
B
v i r t u e of ( i ) . Conversely, assume t h a t and covers (iv)
b.
Then, by ( i l ,
(v)
B = b
and
Write
G
B = B(E,X)
56(B)
bG and so bG is defined.
C E CR(G)
and l e t
6(B)
.
5 6(C).
P
and hence
a contradiction.
which i s r e g u l a r
B
FG
be any block of
which i s
bG
.
then by Proposition 2.2.6,
G, P
FG
Then, a s w e have seen i n (iii), bG is r e g u l a r ,
b.
r e g u l a r and covers
b
=
bG i s defined and l e t
Assume t h a t
covers
B
i s a block of
56(C). G
P
Because
N.
C
be such t h a t
If
A(C+) # 0 ,
i s a normal p-subgroup of
By t h e argument of (ii), we deduce t h a t C
5N ,
G
5. THE FONG COFfRESPONDENCX. G
Throughout t h i s s e c t i o n ,
F
and
If
H
i s a block of
G,
i s a subgroup of
FN,
then
G
a normal subgroup of
an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c
6 = B (E )
If
N
denotes a f i n i t e group,
p
3
0.
denotes t h e i n e r t i a group of
G(E)
E.
then Z
T G , H : CFG ( H I
(FG)
i s t h e r e l a t i v e t r a c e map. 5.1.
LEMMA.
Let
module such t h a t of
FH,
then
Proof.
H
FH
1T : FG'
8
{1,g2, . . . , g n}
V@
lies i n
V
FH
FG
and
Owing t o Proposition 1.10.15,
put
X =C-
(C n H I .
C
nH = C Then
V
b
l i e s i n t h e block
.
H
afforded by
in
V
.
(1)
J,
G.
Let
I$
and
F,
respectively.
and
be
it s u f f i c e s t o v e r i f y t h a t
$.rr(C+) = W C + ) To t h i s end, w r i t e
If
( d i r e c t sum of F-spaces)
is a l e f t transversal for
t h e c e n t r a l c h a r a c t e r s of
b
G
be an i r r e d u c i b l e F H -
be t h e n a t u r a l p r o j e c t i o n and w r i t e
... @ gn
n
V" = V @ g where
V
and l e t
VG is an i r r e d u c i b l e FG-module.
bG i s defined and Let
G
be a subgroup of
U C 1
U 2
... U C,
for a l l for suitable
Ci
C E CR(G) E Ck(H),
(2) and
472
CARPTER 8
+2,
and so X
= 0,
Let H be a subgroup of
LEMMA.
5.2.
This proves ( 2 ) and hence the result.
by virtue of (1).
H contains the inertia group of W.
irreducible FN-module such that
W(H) and W ( G )
with N C_HC_G, and let W be an
G
Denote by
the sets of all irreducible FH and FG-modules, respectively,
whose restriction to N
W as a summand.
have n
(i) If
V
E W(H),
fl
V-
(ii) The map
?E
then
W(G1
induces a bijective correspondence between the isomorphism
classes of FH and FG-modules in W(H) Proof.
Let S denote the inertia group of
respectively.
W
and assume that the result
Then, for X E W(S1,
is true for S instead of H.
XI-+f
and W ( G ) ,
the maps
satisfy ti) and (ii), with respect to W ( S ) ,W(H) and
,pectively.
($IG
Because
Hence we may assume that S
WW). Because
V E
v
there exists a composition factor X Since XN
in particular, W of XN
XN.
Then V
W.
of
$ such that
is completely reducible, VN
i s a summand of
isomorphic to
and
W(S1 , W ( G ) ,
res-
Let U
2U
=
H. (#IH,
is an irreducible submodule of "
factor of XH.
9
XG, it follows that ti) and (ii) are satisfied
2
with respect to W(H) , W ( G ) . (i) Suppose that
X*
v
is a composition i s a submodule of
XN
i
denote the sum of all submodules
and, by Clifford's theorem, U
i s an
n
irreducible FH-module such that X that X
2
v".
V is irreducible, we conclude
Since
is also irreducible.
(ii) Assume that V ,V E W(H1 1
(
=
have
W as a summand.
V1 8 V:
are such that
2
(
and
=
V
2 1
fl
V
Let
be a block of FG
B
=
B(E)
Then
for some
VE
W(H).
(flH
(c),,
.
The latter being a conse-
be a block of FN, let H = G ( E )
covering B.
of FH with EeE # 0.
fi. 2
and we ace left to verify that each 2
quence of Clifford's theorem, the result follows. 5 . 3 . LEMMA.
2
V 2 8 V: , where the FH-modules V ; and Vi do not
Thus
U E W ( G ) i s isomorphic to
1
and let B
Then there exists exactly one block b
Moreover,
=
=
B(E)
b(e)
THE FCNG CORRESPONDENCE
(il
The r e s t r i c t i o n of
T G,H
algebras; i n p a r t i c u l a r ,
(iil
Z(b1
is an isomorphism
Z(B)
of F-
E = TG,H(eI
By r i g h t m u l t i p l i c a t i o n ,
(iiil
Z(bl
to
473
FGe
n
i s a f r e e b-module of rank
=
(G:H)
L e f t m u l t i p l i c a t i o n g i v e s an isomorphism
Mn(b)
B z End(FGel
of F-algebras
1
b Proof.
By d e f i n i t i o n ,
i s a c e n t r a l idempotent of
E
FH,
so we have a
decomposition E
where each
e
= e
1
2
i is a block idempotent of 0 # EE = EE'= Ee
b = b ( e ) of
t h e r e i s a block
i
E {1,2
,...,k).
transversal for
FH
H
in
G.
+
FH.
of
Ee = eE =
FG.
and
... + EekE,
put
'e
=
geg
Then, f o r a l l d i s t i n c t
=
'EVE
0.
=
e = ei
-1
,
T,
T
be a l e f t
we have
o
Er = T (el GrH
Hence
and l e t in
u,V
f o r some
i s a c e n t r a l idempotent
Observe a l s o t h a t
proving ( i i l Let
e
+
EeE # 0 and
with
'eve = 'e( uEv €1 v e since
ek
Since
E iEe E
g E G,
For a given
...
+ e +
x
. be an element of
FGE'
xFGe
with
=
0.
Then, f o r a l l
t E T,xte=O
and
o
=
c rtet-l = E T
X E ~ =
x
Thus l e f t m u l t i p l i c a t i o n g i v e s an i n j e c t i v e homomorphism FGE'-End(FGe)
b algebras.
Since FGE' = @ FGte ET
=
@ t(FGe),
ET
w e have dimFGE' = n'dimb F F and t h u s
= didn@)
F
of F-
CHAPTER 8
474
FGE'
2
End(PGe1
3
b In particular, E'
Now E E ' e = Ee # 0,
is a block idempotent of FG.
EE' # 0 and therefore E = E ' ,
SO
proving (iii).
Given z,y E Z ( b ) , we have
-
uzueveuy u,vET
tt
c
=
z y
ET =
since ' e v e = 0 for u #
x
u u z
and
V
e
'G,H'W'
u
z for
=
all
u E T.
Since for all
E Z(b),
TG,H(z)e
=
C
t
m
the map
TG,H: Z(b)
3
Z(B1
tt L z ee = z
=
ET
E T
is an injective homomorphism of F-algebras.
More-
over, this map is bijective since dimZ(b1 = dimZ(B1, by virtue of (iii). F F proves C i 1 . Finally, suppose that b'
=
b'(e')
is another block of
This
FH with Ee'E # 0.
Then
0 # Ee'€
=
c
tetE m ' = ee'
ET and e = e ' ,
thus completing the proof.
We are at last in a position to attain our main objective, which is to prove the following important result known as the Fong correspondence. 5.4. THEOREM. (Fong (19611, Reynolds (19631, Kiilshammer (1981(b)). algebraically closed field of characteristic p > 0, let N of G; let ti1 (iil
If b If
of FH, that b
B
= B(E)
be a block of FN
is a block of
B is a block of FG covering B, called the Fong correspondent of B covers B
and
B =
be an
be a normal subgroup
and let H = G(E).
FH covering 6, then bG
G b .
Let F
is defined and covers
then there exists a unique block in FH
(with respect to
In particular, if b l ,
...,b
8)
B b
such
are all distinct
475
THE FONG CORRESPONDENCE
blocks of FH covering B,
then 6',
... ,bf
are all distinct blocks of FG
8.
covering
(iiiI If b
b(eI
=
is the Fong correspondent of
B
=
B(E1
in FH, then
(el and e is the unique block idempotent of FH with EeE # 0 GtH (b) The restriction of T to 2 ( h ) is an isomorphism Z ( b ) Z ( B ) of FG,H algebras. (a) E
=
T
M,(b),
(c) B
where n
V h VG
(dl The map
=
(G
:
HI
induces a bijective correspondence between the isomorphism
classes of irreducible modules in b (eI The blocks b (f) B
and B
and B ,
respectively.
have a defect group in common
is the unique block of FN covered by
b.
is p-solvable and N = 0 ,(GI, then e
(g) If G
P
H is a defect group of b .
and a Sylow p-subgroup of
= E
In particular, by Step 1 in the proof of Theorem
3.10.9,
b
P
for some k 2 1 not divisible by
Mk(P(H/N))
(i) Write b
Proof.
covers 8 ,
Because b
submodule W
of
=
fl
summand.
(ii) Let B
VN belonging to 6,
W
is contained in H .
bG is defined and
fl
E
9-l) and
Invoking Lemma 5.2,
G
lies in b
has
.
we
W
as a
Therefore,
bG covers 8 . be a block of
such that W
FG covering 6 and let V be an irreducible F G there exists an irreducible FN-module W
Owing to Lemma 4.2, i s a summand of
contained in H , V
L
VN.
V
is defined and
Moreover, by Lemma 4.2, If EeE UG = 0 ,
Since the inertia group of
for some irreducible FH-module U
VG
N has W as a summand (Lemma 5 . 2 ) . then bG
Now, for any g E G, 9W E 6(g
Is an irreducible FG-module, whose restriction to N
module in B .
B
in b .
V
it follows from Lemma 4.2 that there exists an irreducible
Hence, by Lemma 5 . 1 ,
by Lemma 4.2,
and some aEZ*(ff/N,F*).
and fix an irreducible FH-module
b(e1
therefore the inertia group of infer that
p
b
then e E
lies in b
If b
G
covers 8 .
8= 0
and so
,
W is
whose restriction to
is the block of
FH containing U,
by Lemma 5.1, and hence B = b
Let
in
G
.
E be the block idempotent of B .
eEU = E e U
= EU =
0.
But then ~ k = r 0,
CHAPTER 8
476
a contradiction, and hence EeE # 0.
This proves (ti), by appealing to Lemma 5 . 3 .
(iii) Properties (a), (b) and (c) are direct consequence of Lemma 5.3. and b
(a), note that by (c) B ducible modules. where
have the same number of nonisomorphic irre-
f1 P fl2
Thus, it suffices to show that
Vi is an irreducible FH-module in b , i
implies
(VilN, i = 1,2.
since
w = gw
5.2 we have
V
%
1
the same defect.
E =
B1
56(B)
Hence
8@
H.
E
g-ll =
(f)N
and,
B(E)
Consequently,
(V1)N and (V2)N have
contains the inertia group of FJ2,
and, by (d) and Theorem 2.2.3,
6 ( b ) = b ( B ) , proving G
is another block of FN
= B1(E1)
h - l ~ h for some h E H.
conclude that
P
so by Lemma
2
G If
(f)N
such that
V , proving (a).
6(b)
By Lemma 4.1,
But H
then
2
f3
in
we infer that
E
for some g E G, and hence for g E
W 2 as a common summand.
1
($IN,
( V $ l N is a submodule of
8z f,
If
v2,
v1
= 1,2.
Owing to Lemma 4.2, there exists an irreducible FN-module Wi
Wi is a submodule of
To prove
E
1
=
E,
Because
proving (f)
6 ( b ) and
6(B)
have
(el.
covered by b ,
then by Lemma 3.10.2,
is a central idempotent of FH, we
E
.
Property (g) is a consequence of a general result to be proved in the next
.
section (see Proposition 6.4(iii) , (ivl) 5.5. COROLLARY.
p > 0,
Let
F be an algebraically closed field of characteristic
exists a subgroup H of defect group D
G containing 0 ,(GI P
and a block b
Then there
of FH
with
such that where n = ( G : H)
(i) J ( B ) P M n ( J ( b )1 ,
(ii) D
be a block of FG.
let G be a p-solvable group and let B
is a defect group of
(iii) t ( B ) = t ( b ) , where
B
t(B1
and D E Syl ( H I
P
and t ( b ) are the nilpotency indices of J ( B )
and J ( b ), respectively. (ivl b
I-
Mk(f(H/O
P
,(G))
for some
k
1 not divisible by p
and some
a E Z2(X/O ,(GI ,F*). P Proof.
Property (iiil
is a consequence of (i), while properties (i), (ii),
477
TEE FONG CORRESPONDENCE
(c), ( e l , (gl
and ( i v ) follow from Theorem 5 . 4 ( i i i l ,
.
Another u s e f u l observation is provided by 5.6. PROPOSITION. p > 0,
G
let
that
Z(FG)
E E
a block
B
Assume t h a t
1.
.
B
B
covers a block
5
with d e f e c t group
t(B)
t ( & ,ID1
=
B(E)
=
5
with
PO ,(GI P
of
such
0 (5)# 1 and P
with
such t h a t
101
and
5 E Syl
P
and S t e p 1 i n i t s p r o o f , we have
By Theorem 3 . 1 0 . 9 ( i i )
B
=
FG
be a block of
Then t h e r e e x i s t s a p-solvable group
FZ
of
Proof.
be an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c
be a f i n i t e p-solvable group and l e t
D #
d e f e c t group
F
Let
g
Mk
C f l (G/O P ,(GI 1
and
D E Syl (GI P k
f o r some
p
1 not d i v i s i b l e by
(2)
a E Z2(G/0
and some
P
(GI , F * ) .
It is a
standard f a c t ( s e e Karpilovsky (1985, pp.96,97)) t h a t t h e r e i s an e x a c t sequence
where
5-
Z
1-
FE.
f a c t o r of
Hence, by (11, B
Mk(FE) where
B1,
...,Bs
1k ,
p
= M k (E)
5,
-
B.
i s p-solvable, and of
o
P
(G/O
-
such t h a t
1
(3)
f(G/O
Mk(B1l x
... x M k ( B s )
FE,
we i n f e r t h a t
f o r some block
B
and
P
,( G ) )
Mk(F&
is a d i r e c t Since
(4)
FZ
of
%modules coincide.
then by ( 2 ) and (31, ID1 =
d e f e c t group of
x
(GI
i t follows from ( 4 ) and Proposition 1.5.3 t h a t t h e p - p a r t s of t h e
dimensions of i r r e d u c i b l e group of
P'
i s a d i r e c t f a c t o r of
a r e a l l blocks of
B Since
2
i s a c e n t r a l p '-subgroup of
Z
G/O
P
,(G))
101.
in
X 4
1 we have
5
If
X/Z
i s a Sylow p-sub-
0 (G/O , ( G ) ) # 1. P P
t(B) = t(E).
5
is a
Since
G
Hence, t h e i n v e r s e image
satisfies:
and
5
Hence, by Theorem 2.2.7,
Furthermore, by ( 4 ) , w e a l s o have
D #
'
i s a n o n t r i v i a l p-group
CHAPTER 8
470
Z
Since
-
G,
is a c e n t r a l p'-subgroup of
.
follows.
2.
P of
n o n t r i v i a l normal p-subgroup
X
we i n f e r t h a t
Thus 0
P
(2) #
= Z x
P f o r some
1 and t h e r e s u l t
6. THE KULSHAMMER'S STRUCPURE THEOREM. Throughout t h i s s e c t i o n , and
F
G
denotes a f i n i t e group,
N
an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c
FN,
a block of
G,
group of
G(E)
then
a normal subgroup of
p > 0.
denotes t h e i n e r t i a group of
= B(E1
is
H i s a sub-
If
E.
6
If
G
then
T G f H : CFG(HI i s t h e r e l a t i v e t r a c e map.
Given s u b s e t s
NG(X,Yl
ZVG)
---*
X,Y
of
FG,
w e put
n NG(Y)
= NG(X1
All t h e r e s u l t s of t h i s s e c t i o n a r e e x t r a c t e d from an important work of Kiilshammer (1981b).
6.1. LEMMA.
b
and l e t
B
Let =
=
B(E)
be a block of
FG
b ( e ) be t h e Fong correspondent of
( i ) For each d e f e c t group
6
of
6,
B
covering a block
B
= B(E)
d
FN
8.
with r e s p e c t t o
t h e r e i s a d e f e c t group
of
of
b with
6 = d n N (ii) I f
6 =
6
B and d i s a d e f e c t group of
i s a d e f e c t group of
d n N, then f o r each normal subgroup P of
y = ~ ( € 1 of Proof.
F(PN)
(PN:N) = (P:(Pfl N))
lie in
FN
block
y = Y(E)
where
H = G ( d ,
C5N
(bl
C
5P
t h e r e i s a block
.
5 SuppE
i s a power of
by Theorem 2.2.4.
t o o , and w e a r e l e f t t o v e r i f y t h a t
(a)
6
( i ) This i s a p a r t i c u l a r case of Theorem 3.2.
(ii) Since
F(PN)
with d e f e c t group P
d with
b with
of
F(PN).
such t h a t
Hence
E
p,
a l l block idempotents of
is a block idempotent i n
F(PN),
P i s a d e f e c t group of t h e corresponding
Owing t o Proposition 2.2.6,
there exists
C E CR(H),
479
THE K ~ ~ L S H A M M E R STRUCTURE ~S THEOREM
d
(c)
i s contained i n a d e f e c t group of
P
By ( b ) and ( c ) ,
PN = AN.
C.
a
i s contained i n a d e f e c t group
y
of
6
On t h e o t h e r hand, Theorem 3 . 2 i m p l i e s t h a t
=
and so
= Y(E),
A n N.
This g i v e s
the equality
P is
and t h u s 6.2. of
PROPOSITION.
FN
6
b with
6
TI :
B = B(E1
Let
FG
be a block of
8
i s a d e f e c t group of =
= YCE).
D n N.
P
Let
8
covering a block
b = b ( e ) be t h e Fong correspondent of
and l e t
Suppose
y
a d e f e c t group of
B
D
be a normal subgroup of
8.
with r e s p e c t t o
B
D i s a common d e f e c t group of
and
= 8(E)
with
and
6 C P , let
be t h e n a t u r a l p r o j e c t i o n and s e t H = G ( E ) . Then G There i s a block 6 ' = 6 ' ( e ' l of FN(P1 with d e f e c t group D and
FG-+
(i)
FC(P)
H n(e)e'
=
el.
(ii) For each block
D
d e f e c t group
and
6'
i n Ci), t h e r e are blocks
8'
= 8'(€'1,
E ' = TI(€)
,
B' = B ' ( G ' ) FN(PI
of
FN(P) G
of
with
6
with d e f e c t group
N such t h a t (a1
b'
(bl
TI(E)E' = E ' , Proof.
(6')'
6,
=
P.
(Ell
P.
Y'
F(NN(P)). If
g E N
If
Y
= Y(E)
F(PN)
of
= IT(€)
of
with
F ( f l p f l ( P ) ) = F(PNN(P))
i s a c e n t r a l idempotent of
E'
F(NN(P)) with
F(NpN(P)).
E'
Hence
€'El E
'
# 0.
= E
F(CN(P)).
By Lemma 6.1,
TI('€)
= 'TI(€)
On t h e o t h e r hand, f f
g
and, i n p a r t i c u l a r ,
is an element i n
N(P) with G
6'= 6' ( E ' )
Pn
'TI(€)
= TI(€) 'T(E)
Let
is a
E
and we have a block
Again, by Lemma 6.1, t h i s block has d e f e c t group
g E N ( P ) , then G
(PI.
with
= Y'(E')
Clearly
be a block of
c e n t r a l idempotent of of
(B'IG = B
and
It follows, from t h e Brauer correspondence (Theorem 3.9.21),
t h a t t h e r e e x i s t s a block with d e f e c t group
8'
with r e s p e c t t o
( i ) Owing t o Lemma 6.1, t h e r e i s a block
d e f e c t group
81 = 6
B'
fs t h e Fong correspondent of
=
6.
for a l l =
T(E),
CHAPTER 8
480
and so
E
=
Let
:
FG
'II
proving t h a t
'E,
-+
FCG(D) be t h e n a t u r a l p r o j e c t i o n .
Then
n r ( I r ( e ) )= ~ ' ( ef ) 0 a(e)
and
a(e)
e
=
~ ' ( e =] T f l e 1 where each
+
e
2
+
write
... + e
+
Then
n'(e 1
+
... + n ' ( e n )
~ ' ( e i~s ]a c e n t r a l idempotent i n F(CH(D)). This implies t h e b'
existence of a block I t follows t h a t
b'
e'.
=
b'(e')
b'(e')
=
b'
(ii) A s s u m e t h a t =
1
FYYHCPIl.
as a sum of block idempotents i n
T(e)e'
F(NH(P)).
i s a nonzero c e n t r a l idempotent of
=
F(NH(P)) w i t h n ( e ) e '
of
has d e f e c t group
b ' ( e ' ) i s a block of
D, proving
=
and T ' ( e ' ) g o .
e'
(i).
F ( N H ( P ) ) with d e f e c t group
D
and
Then we have
e r E f = e ' n ( e ] a ( a ) = e'.rr(ee) = e ' n ( e ) = e' B'
and we g e t a block
b'
=
and
=
B'(E')
of
F ( N G ( P ) ) with
b'(e'l
i s t h e Pong correspondent of
B' has
d e f e c t group
0.
= B'(E')
C of
By Theorem 5 . 4 ,
with r e s p e c t t o
By Lemma 4 . 3 , t h e blocks
defined and f o r each conjugacy c l a s s
(c n
B'
E r e ' € ' # 0.
(B'IG
and
G o r H , we have
CN,(P))+E~ + J ( Z ( B ~=I T ( C + ) E ' + J ( z ( B ~ ) )
or
(C n y H ( P ) ) + e '+ s ( z ( ~ ' =) )n ( C + ) e ' + J ( z ( ~ ~ I ) , respectively.
Now
.rr(e)e' = e r implies t h a t
( b r ) H = b.
( B r ) G = ( b ' ) G = bG = Bso
n(E)E' # 0
and t h u s
n(E)Ef = E',
proving ( i i ) .
Hence
R'=B'(E')
(b')H are
THE IdkSHAMMER'S STRUCTURE TEEOREM
The next r e s u l t provides an i m p o r t a n t reduction t o 6.3. PROPOSITION.
FN, and l e t b
of
6
Assume t h a t
b
b(e1
=
FG
be a block of
C,(PI.
8
P
Let
D
and
8
covering a block
B
be t h e Fong correspondent of
i s a d e f e c t group of
6 = D n N.
with
B = B(E)
Let
481
=
8.
with r e s p e c t t o
B
a common d e f e c t group of
CP
6
D with
be a normal subgroup of
B(E)
and
and p u t
H = G(E).
b
(i) There e x i s t blocks of
FNH ( P , b l 1
F ( C G ( P ) ) and
A
(a)
=
B
(bl
NG(P,BIl
=
b
bl
b
Fong correspondent of
=
1
b
1
1
P is a normal eel # 0.
1
1
=
b,
a
=
y = Y(E)
y
t h e r e i s a block
= y (E 1
a l(ell
a
E
= 1
A l (El)
B (E 1
of
1
1
= e
1
Since
, E I E e l # 0.
Because
# 0
= E Ee eE = E Ee e = E Ee I 1 1 1 1 1
of
F ( C G ( P I ) with
E 1
Ee E # 0 and a corresponding 1
1
FNG(P,B1).
S = C ( E ) n CG(P).
G i
ee
By Lemma 6 . 1 ,
1
E Ee 1 1
B
=
1
has d e f e c t
a r e a r b i t r a r y blocks of
F(CpN(P)) with
of
1
Eee
=
the
proving ( i l .
FCPN) with d e f e c t group P .
of
Since
i s a block
FNH(P,bl), r e s p e c t i v e l y , a s s t a t e d i n t h e theorem.
1
Set
H
and
1
Ee
=
Then
S
-H C
s i n c e f o r elements g E S
,
we have
gy = g ( ( y l ) P N )= ( S Y l ) P N= ypw = y 1
Let
c
1
)
such t h a t
By Theorem 5.4, w e may assume t h a t
b = b (e
t h e r e e x i s t s a block
A
1
of
1
( s e e Lemma 3.1.11)
6 ( e ) with r e s p e c t t o b = b (e 1
FflH(P,bl).
Assume t h a t
block
D
F ( C H ( P ) ) with
FflH(P,bl) l i e s i n F C H ( P 1
while by Theorem 4.5,
we g e t a block
1
w e may assume t h a t
bl(ell be a block of
=
every block idempotent of
F ( C H ( P ) ) and
1
NH(P,bl)CG(Pl
Let
a = a l ( e l ) of
1
B 1 = B (E )
with d e f e c t group
1
and
G.
subgroup of
(ti1
bl ( e l ) i n ( i l , t h e r e a r e blocks
of FNG(P,B CG (PI (bl) = B1
1
a = a (e
and
D.
(E 1
= A
1
=
H bl = b
F ( C H P l ) with
of
(1) Owing t o PropositTon 6.2,
Proof.
D,
bl(el)
with d e f e c t group
(ii) For each block
group
=
= c l ( f l ) be t h e Fong correspondent of
B
1
=
B ( F - 1 with r e s p e c t t o 1
1
482
Y1
CHAPTER 8
=
Y1(Ell and c
respect to Y
=
c,lf,)
= y1 (El).
the Fong correspondent o f
1
and for elements g E C,(P)
- S,
with
f, = f,.
We wish to show that
Assume by way of contradiction that f
b 1 = b1 (e 1 )
+ f2.
For elements g E S ,
we have
we have
Thus
contrary to
# 0.
Ee E
E
1
1
Hence f
=
f, and
1
CGP)
C,(P)
=
bl
c1
B1
=
It follows that
and A
1
= A (E 1
1
has defect group D.
NOW (a) follows from
bG = bG = B 1
B1
It follows from (a1 that N H ( P , b )CG(P1 5 NG(P,B1). tainment, assume that g
€
NG(P,B1l.
To prove the reverse con-
Then
O # g ( ~ E )= g ~ E 1 1
By Lemma 3.10.2,
1 1
there exists an element c E C G ( P ) with
= E
"E
1
particular, gc E H . of
Hence "5' = S and " c
= gccl (gcfl)
FS with 0 # gc(ElflEl)= E gcf 1
E
1 1
The uniqueness of the Fong correspondent implies that
gCf which in turn implies that
1
=
f
1
and "c
1
=
c
1
1
,
in
is a block
48 3
Let G
6.4. PROPOSITION.
a block of FG
FN with
be p-solvable, let hr = 0 , ( G I
with defect group D
natural projection.
such that B
Put P = D n O
EEZ(FG).
(GI
P'rP
6
covers a block
be
= 6(E)
andlet n : F G + F ( C G ( P ) l
of
bethe
Then
P is a Sylow p-subgroup of 0
(il
and let B = B ( E )
P
(G)
P'rP
such that
0 ,(NG(P1) = "(PI
P
and 0 ,(NG(D1l = NN(D1
P
(ii)
(iii) E
F(CN(P)l ,F(CG(Pll and F(NG(P)l.
is a block idempotent of
T(El
= E
D is a Sylow p-subgroup of G
(iv)
Proof.
(il
Theorem 2.2.4,
Choose a Sylow p-subgroup E
y
=
g-lDg.
P'IP
(G)
with P
of F(QN) has defect group Q.
= Y(E)
there is a defect group d
of 0
of
B
with Q
5 Q.
By
F(&N), and by Theorem 3.10.9,
is a block idempotent o f
corresponding block
that d
Q
=
d
0
P'rP
(GI.
the
By Lemma 6.1,
Let g E G
be such
Then
Pc Q
=
g-log n
o
(G) =
PrrP
g-lpg
and
Hence P
is a Sylow p-subgroup o f
0 ,(N P)l
C
P
6
ICG(Pl 5 PN.
0
P'rP
(GI.
It follows from Lemma 7.1.3
Taking into account that N G ( D )
5 NG(P),
that
(i) is
established. (iil As we have seen in (i), the block By Proposition 6.2, we get a block
y'
of F(PN) has defect group P.
y = y(e)
= y'(E'),
E'
= T(E),
Of
I" (P)) = F(PNN(P))= F(PCN(Pl), a block 6 ' = B ' ( E ' ) of F ( C N ( P ) ) and a block B' = B ' ( E ' ) of F(N(P)) with E ' E ' # 0. By Lemmas 5.3 and 3.1.11,
F(IV
G
E f E F(CG(p)) 5 F(PN) and even E' E F(PCN(P)). So we get E'
=
E'E' =
E'
and
484
CHAPTER 8
( i i ) follows
B1
(iii) L e t
=
B1
d e f e c t group of of
0
P'rP
B 1 (El) be a block =
FG
of
with
As i n ( i ),
B1 ( E l ) .
(GI and so w e may assume t h a t P 0 # T(E
so (ii)implies t h a t
= 1
n ( E 1 = IT(€).
n(E
El
# 0, and l e t
E
D1 n 0
P'tP
(G)
D n 0
=
P'PP
1
= IT@
E)
D
be a
1
i s a Sylow p-subgroup
(G).
I t is clear that
)IT(€),
Thus
1
0 # IT(E )IT(E) = n(EIE) 1
E
and (iv)
E,
= 1
.
proving ( i i i ) .
D i r e c t consequence of (iii)and Theorem 3.10.9(11).
The next observation is t h e so c a l l e d F r a t t i n i argument.
6.5.
N.
group of
Then
Proof.
g-lPg
N
Let
LEMMA.
G = I\IG(P1*N
g be an a r b i t r a r y element of
Let
i s a Sylow p-subgroup of
.
n E N.
some
required. 1
Let
+
a
4 X+
%
Y
-t
Then
and hence
5N
g-'Ng
and so
g-lPg = n - h
Hence, by Sylow theorem, G
for
g E NG(P)*N, a s
Z
be an extension of f i n i t e groups, where
1
8 i n t h e following way:
:
For any
Y-
Out
is not
z = AutZ/InnZ
y E Y,8(y)
X
ranges over those elements i n
3:
G.
This extension d e f i n e s a homomorphism
n e c e s s a r i l y abelian.
where
N.
gn-l E N (PI
I t follows t h a t
2
G and l e t P be a Sylow p-sub-
be a normal subgroup of
c o n s i s t s of a l l maps f o r which
=
+ Z , Z H , ' ; Z
y.
W e come now t o t h e main r e s u l t of t h i s s e c t i o n . 6.6.
(Kclshammer (1981b).
THEOREM.
G be a f i n i t e p-solvable group, l e t F
Let
be an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c
FG
block of blocks
B
= B (E 1
group (i)
G B
D =
with d e f e c t group 1
)
of
F ( C G ( P ) ) and
= A
A 1
H of
N (P,B )
G
and l e t
B = B(E)
Then t h e r e a r e a normal subgroup
1
and a subgroup
B
D.
p,
I
1
(E 1
of
FNG(P,B )
be a
P of
D,
with d e f e c t
1
such t h a t t h e following p r o p e r t i e s hold:
THE K~~LSHAMMER'S SPRUCTURE
485
THEOREM
is a Sylowp-subgroup of H
(ii) D
(iii) Po , ( H I
P
(iv) B
1
(v) H/O
= 0
P'rP
Mn(FCY(H/6
P
(H)
,( H I )
1 and some a E Z2(H/0 , ( H ) ,F*)
for some n
P
is isomorphic to an extension of P with the group NG(P,B1)/PC(P),
,(H)
P
G
where the corresponding homomorphism
is induced by the natural homomorphism NG(P,BI)-+ Au-. Put N = 0 ,(GI
Proof.
P
and let 8
We argue by induction on the order of
Case 1. Let b
S =
=
FN
be a block of
= B(E)
covered by
B.
G and distinguish two cases.
G(E) # G.
b ( e ) be the Fong correspondent of B
with respect to
5.4, we may assume that D is a common defect group of B there are a normal subgroup P o f
D, blocks b
=
6.
By Theorem
and b .
By induction,
b l ( e l ) of F ( C ( P ) ) and S
a
= a l ( e l ) of
FN (P,b 1 with defect group D, a subgroup H of N S ( P , b l ) ,
s
a natural number rn
1
and
CY E
Z2(E,F*I where
=
H/o P , ( H I ,
such that the
following conditions hold:
b
S
= b
D is a Sylow p-subgroup of H
is isomorphic to an extension of P where the automorphisms are induced by N s ( P , Put n = rn(G of FN(P,B
G
and choose blocks B
: S)
as in Proposition 6.3.
1
)
/C,(P)
= B (E ) 1
1
of
. F(C(P)) and A G
=
A1(EI)
Then (i) is a consequence of (i') and
1
Proposition 6.3, (iv) follows from Theorem 5.4, and by Proposition 6.3,
NG(P,B1)/C(P) N(P,bll/C(P) G S S which implies (v).
Case 2.
S = G(E) =
In this case, E
G
is a central idempotent of FG, so we may apply ~roposition6.4.
486
CHAPTER 8
P
Put
op ,, p (GI , H
D n
=
= N~ ( P I
,
F(CG(P)
FG
:
'TI
be t h e n a t u r a l p r o j e c t i o n .
and l e t
Denote by
B1
=
B (E 1
1
)
A
and
F(CG(P)) and F(NG(P)) with t h e block idempotent E
of
N (P1 = H = NG(P,B ) .
By Proposition 6 . 4 ( i i ) , w e have block
A
=
Al (El)
(iii) a r e now d i r e c t consequence of Proposition 6.4.
= A (E 1
t h e blocks
1
respectively.
= 7T(E),
By Proposition 6 . 2 , t h e
G
F(NG(P)) h a s d e f e c t group D.
of
1
P a r t s ( i ) ,( i i ) ,and By Lemma 6.5,
and
G/N
2
NG(P1/NN(P1
= N(P)/CN(P)
G
we have
From Lemma 7.1.3,
CG(P)
5 PN.
Thus (iv) and (v) a r e immediate conse-
quences of t h e isomorphism e s t a b l i s h e d i n S t e p 1 of Theorem 3.10.9.
This
completes t h e proof of t h e theorem.
G
Let
and
G
w e say t h a t
G
H.
on
I f a homomorphism ($ : G - + A u t ( H ) i s given,
be two groups.
acts on H
The a c t i o n
0
Let
H
($
@;
via of
is called f a i t k f u 2
G
be an a c t i o n of a group
product H H G
( o r simply
t h e homomorphism ($
H gG
G
if
on another group
i f the action
($
i s c a l l e d an action of Ker@ = 1.
H.
The semidirect
i s understood) c o n s i s t s of
($
( h , g ), h
a l l pairs
E
H, g E G ,
Usually, t h e subgroups
(H,lI
with t h e product defined by t h e formula
and
(1,G)
a r e i d e n t i f i e d with
H
and
G,
r e s p e c t i v e l y , so t h a t
H X G = HG, H 4 H X G for a l l 6.7. that
ghg-'
=
@ ( g )( h )
g E G, h E H .
COROLLARY.
P
and
=
D.
Further t o t h e assumptions and n o t a t i o n of Theorem 6.6, assume Then
THE K~SHAMMBR'S
n a 1
f o r some
H
Let
H/o
verify that
c1
P
,( H )
D xs D
i s a p'-group
H/Op
s.
and hence
,( H I .
P
S
D
Since
H
=
i s normal.
0 ,(H/o , ( H I ) = 1,
Because
S
f o r a s u i t a b l e p'-group
H/op, (HI
t h e Sylow p-subgroup of
a c t i n g f a i t h f u l l y on
and hence
P,
D
1
acts f a i t h f u l l y on
THEOREM. (Kkshammer (1981bl 1.
Let
D 4 S f o r some
H / o ,(HI P
Thus
D,
a s required.
FG
D.
with d e f e c t group
that
9
P
=
D.
G be a f i n i t e p-solvable group, l e t
be an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c block of
DO , ( H ) / O , ( H ) P P
i t follows from Lemma 7 . 1 . 3
P
D.
Theorem 6.6(v) i m p l i e s t h a t
W e c l o s e t h i s s e c t i o n by e x h i b i t i n g circumstances under which
6.8.
acting
By Theorem 6 . 6 ( i v ) , it s u f f i c e s t o
i s a Sylow p-subgroup of
is a Sylow p-subgroup of
cDs(D)5 D
S
where
be as i n Theorem 6.6.
By Theorem 6 . 6 ( i i ) ,
p'-group
E Z 2 ( D X S,F*),
D.
f a i t h f u l l y on Proof.
and some
487
STRUCTURE THEOREM
p
and l e t
B = B(E)
F
be a
A s s u m e t h a t a t l e a s t one of t h e following
c o n d i t i o n s hold: (a)
G has p-length
(bl
D
(cl
D is a T . I .
1
i s abelian
G
subgroup of
(i.e.
f o r each
g
E G,
either
g
-1
Dg
=
D or
S - l D g n D = 11 For a l l normal subgroups
(d)
BG
Then
P = D,
=
D
of
and a l l blocks
where
P i s a s i n Theorem 6.6.
f o r some
n 2 1 and some a E Z2(D N S , F * ) ,
f u l l y on
D.
Proof. (a)
If
G
BQ
= BQ(EQl
of
F(CG(&)
B , fl (&,B 1 = DCG(&) G Q
with
Q
Q
Hence, by Corollary 6.7,
where
S
is a p'-group
acting faith-
W e keep t h e n o t a t i o n of Theorem 6.6.
has p-length
1,
H
then
i s a l s o of p-length
1.
Hence, by Theorem
6.6 ( i i ) ,( i i i ) ,
DO , ( H ) P and t h e r e f o r e (b)
If
D
=
0
P'IP
( H ) = PO , ( H )
P
P = D.
i s a b e l i a n , then
D
5 CG(P).
I t follows from Lemma 7.1.3 and Theorem
488
CHAPTER 8
6.6. (iii) t h a t
D
CH(P)
(c)
Assume that
D is
and
D
If
= 1 = P.
5 POP ,(HI T.I.
a
P
= D.
subgroup of
G.
P # 1, then
DOp,( H I and so (d)
D
=
H
c_
and
5NG(P1
P
If
~NG(D1.
(H1 0 P'rP
=
Po
P
= 1,
then
H
= 0
Pf
(H)
Hence
, (HI
P.
In t h i s case we have
H = H n D C G ( P ) = DCH(P) C_ DO
(H)
P
by Lemma 7.1.3.
DO , ( H I = PO ,(HI
Thus
and
P
P
D
P.
=
Note t h a t i n t h i s case
S = 1.
even
?. APPLICATIONS.
G
Throughout t h i s s e c t i o n , characteristic
J(B) of
p > 0 and t ( B ) B
of a block
J(FG).
of
FG.
Given s u b s e t s
F
denotes a f i n i t e group,
t h e nilpotency index of t h e Jacobson r a d i c a l
As u s u a l , we w r i t e
X,Y
an a r b i t r a r y f i e l d of
of
FG,
t(G1
f o r t h e nilpotency index
we put
N ~ ( X , Y I= N (XI n N (YI G G 7.1.
Let
LEMMA.
B
a l g e b r a i c closure of group
D
be a block of
F.
FG
with d e f e c t group
Then t h e r e e x i s t blocks
B1,
D
...,Bk
and l e t of
EG
E be t h e with d e f e c t
such t h a t for a l l
m 2 1
In p a r t i c u l a r ,
Proof.
By Lemma 3.16.10,
we can w r i t e
E 8 B = B
F
f o r an i n t e g e r
k 2 1, where each B
Invoking Corollary 3.1.18,
i
@...
@Bk
i s a block of
we t h e r e f o r e derive
FG
with d e f e c t group
D.
APPLICATIONS
as required.
8
Let G
7 . 2 . THEOREM.
be a finite p-solvable group and let B
with defect group D. (a) G
has p-length
(b) D
is abelian
(c) D
is a T . I .
BG = B , IVG(Q,B 1
1
subgroup of G
=
Q
be a block of FG
Assume that at least one of the following conditions hold:
(d) For all normal subgroups Q
Q
489
of
D and all blocks B
Q
of F ( C G ( Q ) ) with
DC,(Q)
Then
Owing to Lemma 7.1, we may harmlessly assume that F
Proof.
ically closed. having D
Hence, by Corollary 6.7,
t ( B ) = t(Pff1 where
as a normal Sylow p-subgroup and CY E 2’ ( H , F * ] .
is algebraH
is a group
Thus t ( B l
=
t (D) by
virtue of Proposition 6.3.8. Let G be a finite p-solvable group and let B
7.3. COROLLARY. FG
with abelian defect group D
... x Z!
Z! x ? lx pnl pn2
, ni
be a block of
> 0, 1 < i
k.
Pnk
Then
t(B1 = 1 - k
k
+
ni
C p i=1
Direct consequence o f Theorem 7.2 and Corollary 3.2.5.
Proof.
8
To prove our next result, we require the following preliminary observation. 7.4. LEMMA.
Let B = B ( e ] be a block of F G ,
G
:
and let f
(i) f ( B )
=
FG-+
@
B 1
(ii) t ( B )
>
t(P)
Proof.
5
E { B 1 ,... ,B
F(G/P)
... d Bn +
t(Bi)
let P be a normal p-subgroup of
be the natural homomorphism.
for some n 2 1 and some blocks B 1
-
1 for all
iE
11,
...,n}.
Part (i) is a consequence of Lemma 2.3. n
)
and put
Then
t
=
t(5).
Let
7
=
of
F(G/P)
To prove (ii), fix
be the restriction of f
Then Ker? = Kerf n B
,...,Bn
(Kerf)@ = I ( P ) B
to B .
CHAPTER 8
490
F(J(B))= J(T(B)).
and, by Lemma 6 . 1 . 1 ( i i i ) ,
J(!ht-l
Since
# 0, w e d e r i v e
Then Ker? = I ( P ) B
J(B)t-l and so t h e r e e x i s t s
G.
s i t i o n of t)
9 I(P)FG
J(Blt-l
t) E
We can w r i t e
= FG*I(P).
-
tJ =
I(P)B.
Let
C 9.s. j=1 3 3
with
4
Thus we may assume t h a t
4
G = U g . P be a c o s e t decompoj=1 3 s FP. Clearly, j s1
9 I(P),
i n which case w e may
write s
'
=
Ccxx,cxEF SLEP
+
W p # 0.
I t w i l l next be shown t h a t
,
Cc2#O
and
6 P
Assume t h e contrary.
Since
+
WP = (Cg .s .)P+ = cg'. ( s .P+, 3 3 3 3 and s i n c e
S
.€"E
Fp
3
for a l l
j,
+
0 = SIP
c c = 0, 31EP
and so
w e have
s .P + 3
=
0 for a l l j .
Hence
c CXX)P+ = ( c c ) P + 6 P XEP
= (
a contradiction.
Thus
I@' # 0.
Since
by Lemma 3.11.2,
and s i n c e
e I ( P ) ~= I ( P ) he h a 0,
f o r any i n t e g e r
5 S ( B )h
we d e r i v e
wp+ E J ( B ) t + t ( P ) - 2 Thus
t(B)
+
t
t(P1
-
1,
as a s s e r t e d .
We a r e now ready t o prove 7.5.
THEOREM.
and l e t (i)
B
(Koshitani (19831, (1984)).
be a block of
d(p-1) + 1 < t ( B )
(ii) t ( B ) = pd
FG
Let
with d e f e c t group
pd
i f and only i f
D is c y c l i c .
G
be a f i n i t e p-solvable group
D of o r d e r p d .
APPLICATIONS
By Lemma 7.1, we may assume that F
Proof.
d = 0
then J ( B ) = 0
is algebraically closed.
If
by Theorem 3.6.4(1), in which case both assertions are
We may therefore assume that d > 0.
trivial.
pa
We first show that t ( B )
t(B1
with
By Corollary 5.5, we may assume that D
t(B)
Proposition 7.2.1,
<
p
t(G)
d
and hence, by Proposition 7.2.4, D then t ( B )
pa
=
and
.
=
pd
if and only if
is a Sylow p-subgroup of Furthermore, if t ( B )
=
pd
D is cyclic. G.
Hence, by
then t ( G ) = p
d
Conversely, if D is cyclic
is cyclic.
by Corollary 7 . 3 .
t ( B ) ? d(p-11
We are left to verify that on d
491
Since d > l ,
[GI.
+
We argue by double induction
1.
If G = 6 ( B ) ,
wehave p I I G I .
then B = F G
and the required assertion is true by Proposition 7.2.1. Let N = 0
P'
(G)
and let 6
=
B(E)
be a black of FN
Theorem 5.4,
FH, H = G(E),
has a block
t(B) = t(b).
If G # G ( E ) ,
then since
induction.
Hence we may assume that G
=
b with the same defect d IG(E)
I<
\GI
P
= 0
B*
has a block
H in which case
(X), [PI = p
P with defect d-r
Since d-r < d ,
r
and
and
with the same defect d
x
=
X/P.
x
t(5) -
1.
E
By
and
E Z(FG).
Then, by
such that 0 ( X ) # 1
P
$(P) = t(B).
and
By Lemmas 2.3 and 7.4,
t ( B * ) 2 t(P1 +
B.
we get the result by
Proposition 5.6, there exists a finite p-solvable group and FX
covered by
Let
FX has a block
By Proposition 4.1.8,
we get by induction that
Thus
as asserted.
8
7.6. THEOREM. (Koshitani (1985a). B
be a block of FG
with defect group D of order p
then the following are equivalent. (i) t ( B ) = pa-'
Let G be a finite p-solvable group and let
+p- 1
d
.
If p 2 3
and d
2,
CHAPTER 8
492
< t ( B ) < pd
(ii) pd-’
(iii) t (D) = pd-’ (ivl
t
p
-1
p d - 1 < t ( D l .: p d
(v) D (vi) D
is noncyclic but has a cyclic subgroup of index p is isomorphic to either Zpd-l
Proof.
for d
x Zp, or M d ( p )
3
By Lemma 7.1, we may assume that F is algebraically closed.
equivalence of (iii)- ( v i )
follows from Theorem 7.5.2.
The
The implication
(i) * (ii) being trivial, we are left to verify that (ii) implies (iv) and (iii) implies (i).
By Corollary 5.5, we assume that D
Assume that (ii) holds. by Theorem 7.5.1,
t(B) = t(D)
G
Since
is of p-length
and so pd-’
1.
t(G),
we have
t(G)
> p
d- 1
.
G.
Hence,
Invoking Theorem 7.2, we deduce that
d < t ( D ) < p , proving (Lvl
.
Assume that (iii) holds. Hence, by Theorem 7.2,
t(B1
i s a Sylow p-subgroup of
Then, by Theorem 7.5.2,
G
.
has p-length
t ( B ) = t ( D ) = pd-1 t p - 1, proving (i)
.
1.
We close this section by quoting the following interesting results, the proof of which is beyond the scope of this book. 7.7, THEOREM. (Koshitani (1977b1).
Let F
be an algebraically closed field of
FG with cyclic defect group D. if P E syl (GI is cyclic
P
- Ctl c
-
C
c c+l
...
...
c c
. . . , .c ct;
A
493
SURVEY OF SOME FURTHER RESULTS
8 . A SURVEY OF SOME FURTHER RESULTS.
Throughout this section, G denotes a finite group, F
p > 0, B a block of FG and tCB1
B.
=
for which
t(B)
3
.
Let D
We know, from Theorem 3.6.4, that t ( B )
J ( B ) = 0) if and only if D = 1.
J(B)
J(B).
the nilpotency index o f
We first examine those blocks B defect group of
a field of characteristic
=
1
denote a
(equivalently
Okuyama (1986) asserts that "it is true that
0 if and only if p = 2 and
ID1
= 2", which should read "it is true
that J ( B ) ' = 0 and J ( B ) # 0 if and only if p = 2 and
ID1
= 2".
However,
we were unable to find an explicit reference for this result. All information pertaining to the Brauer tree of a block from Feit (19821.
The following result describes all blocks
8.1. THEOREM. COkuyama (19861).
Let
B may be extracted B with e ( B )
F be an algebraically closed field of
characteristic
p > 0 and let D be a defect group of the block B of FG.
Then t ( B ) = 3
if and only if one of the following conditions hold:
(a) p = 2, D z
z2
x
z2
and B
(b1 p
is odd,
ID1
=
is isomorphic to the matrix ring over FD
where A4
Morita-equivalent to FA;,
p,
is the alternating group of degree
the number of irreducible FG-modules in B
(p-1)/2 and the Brauer tree of
= 3.
B
or is
4.
is p-1 or
is a straight line segment such that the
exceptional vertex is at an end point (if it exists). The principal block of the following groups satisfies conditions of Theorem 8.1 (see Okuyama (1986)1 : (i) G z
z2 x z2
(ii) G z S
P
or
GP A
or
CzA
P'
and p
where p
=
2
is odd.
The next result is of independent interest 8.2. THEOREM. (Okuyama (198611. characteristic 2, let projective cover of 2-subgroup of
Let F be an algebraically closed field of
lG be the trivial FG-module and let P ( 1G )
lc.
If the Loewy length of P(lG) is
then a Sylow
G is dihedral.
The next result provides a family of groups G of P U G )
3,
be the
is 3.
for which the Loewy length
CWTER 8
494
8.3.
(Erdmann (1977, Theorem 4 ) ) .
THEOREM.
q
power of a prime with
teristic Let
3(mod4],
and l e t
B
be a p-solvable group and l e t
We know, from Theorem 7.5,
F
G =
PsL(Z,q), where
q
is a
be a s p l i t t i n g f i e l d of charac-
P ( l G ) is
Then t h e Loewy length of
2.
G
C
Let
3.
be a block of
FG
of d e f e c t
d.
that
Owing t o Theorem 3.16.14,
B
t h e r e e x i s t s an i r r e d u c i b l e FG-module i n
which i s a d e f e c t group of
B.
Hence t h e i n e q u a l i t y
t(B)
d(p-1)
a v e r t e x of
+
1 can a l s o
be proved as a consequence of t h e following g e n e r a l r e s u l t . 8.4. THEOREM.
LNinomiya (19841333.
be a f i e l d of c h a r a c t e r i s t i c a vertex of
V
V
has o r d e r
p
n
p
3
,
i s g r e a t e r than o r equal t o
Let
0
G be a f i n i t e p-solvable group, l e t F V be an i r r e d u c i b l e FG-module.
and l e t
If
then t h e Loewy l e n g t h of t h e p r o j e c t i v e cover of nlp-1) t 1.
Consider t h e following chain
and assume t h a t
i f and only i f
F is
G
G
J ( Z ( F G ) ) i s an i d e a l of
FG.
FG
?
The e q u a l i t y
I t i s t h e r e f o r e appropriate
What a r e necessary and s u f f i c i e n t c o n d i t i o n s f o r
i d e a l of that
Then, by Theorem 3.11.5,
i s p - n i l p o t e n t with a b e l i a n Sylow p-subgroups.
holds i f and only i f t o ask:
a l g e b r a i c a l l y closed.
J(Z(FG))
t o be an
This problem was solved by Clarke (1969) under t h e assumption
i s p-solvable.
The following r e s u l t shows t h a t t h i s assumption i s
redundant.
8.5. THEOREM. (Koshitani ( 1 9 7 8 ) ) . characteristic vable.
p > 0.
If
Let
J(Z(FG))
F
be an a l g e b r a i c a l l y closed f i e l d of
i s an i d e a l of
FG,
then
G
i s p-sol-
A SURVEY OF SOME FURTHER RESULTS
495
Combining Theorem 8.5 with the results of Clarke (1969, Lemma 8 and Theorem), we obtain the following solution of the mentioned problem. Let G
8.6. THEOREM. (Koshitani (1978), Clarke (1969)).
let F
be a finite group and
be an algebraically closed field of characteristic p > 0.
is an ideal of FG
Then J(Z(FG))
if and only if one of the following conditions hold:
j IGI
(i) P (ii) G
is abelian is a p-nilpotent group with an abelian Sylow p-subgroup P, and G ' P
(iii) G
is a Frobenius group with complement P (iv) p
=
2, G
is 2-nilpotent, P'
Frobenius group with complement P
=
and kernel G'
Z(P)
has order 2
and
G'P
and kernel G' n Ozr(G), where
is a P
is a Sylow
2-subgroup of G. (v) G
has an elementary abelian Sylow p-subgroup P
L such that G
3 L 1 H, L/H
and kernel H, G/H
p - 111
acts on
+y
z-lya:
(vi) n L/H z P
and L
=
G'
is a Frobenius group with complement P
is a semidirect product of
=
2
transitively, and every x E
for all y E
=
P, L
and
2
such that
satisfies x E C,(P) "
z P,A
or
F-~II.
2 , G -has normal subgroups
where P
and normal subgroups H ,
H and
is a Sylow 2-subgroup of
L
such that
G
2L 3H
and
G, G ' = H*Z(P), Z(P) has order 2,
is a Frobenius group with complement P
and kernel H .
Although Theorem 8.5 is quite adequate for the purpose of characterizing all groups G
for which
J(Z(FG)) is an ideal of FG, it can nevertheless be
significantly improved.
Namely, the following result holds:
8.7. THEOREM. (Motose (1979)). characteristic p > 0.
Let F be an algebraically closed field of
If J(Z(FG)) is an ideal of FG, then
G'
is either a
p-nilpotent group or is a p'-group. The proof of the above theorem does not use Clark's characterization.
It is
based on the following two results which are of independent interest. 8.8.
THEOREM. (Kilshammer (19791, Motose (1979)).
sional indecomposable algebra over a field
F.
Let A
be a finite-dimen-
If J ( A ) = A * J ( Z ( A ) ) , then A
CHAPTER 8
496
is primary.
.
Moreover, if F is a splitting field for A ,
matrix ring over Z ( A ) .
8.9. THEOREM. (Asano (1961), Motose (19791).
indecomposable quasi-Frobenius algebra.
Let A
then A
is a full
be a finite-dimensional
Then the following conditions are
equivalent:
# 0
(i) J ( A )
.
(ii) e J ( A ) * # 0 for every primitive idempotent e (iii) R ( J ( A ) ) 5 J ( A ) * -
of A
The following lemma due to Motose 11983) is often useful in investigating the nilpotency index of J ( F G ) . Let B , I
8-10. LEMMA.
and J
be subsets of a ring A
which satisfy the
following conditions: (i) I A I = IBT (ii) I J = J I (iiil BJ Then
5J B
(JIAIn c -YIA.
Proof.
f
Moreover, if
The case n
=
= 0
then J I A
c J(A) -
1 being trivial, we use induction on n. !JIAIn
Assume that
5f I A
Then we have (JIAln+l C -YIAJrA
=
JnIArJA
=
YIBIJA
(since I A I = I B I )
=
~IBJIA
(since
IJ = JI)
(since
BJ
C
(since
IJ
=
c FIJBIA =
Y+’IBIA
c S+’IA, proving the first assertion. JIA
IJ = JI)
(since
A.
JI)
(since I A I = I B I )
Now assume that J
is a nilpotent right ideal of
- JB)
n
= 0.
Thus J I A c - J(A)
Then
= 0
and so
as asserted.
To illustrate an application of Lemma 8.10, we establish the following results. 8.11. THEOREM. (Motose (1983)1 .
Suppose that a group G
has subgroups H
and U
A
such t h a t
G
=
497
SURVEY OF SOME FURTHER RESULTS
UNG(H)U and H
5 NG(U).
Then
[ J ( F H ) u + F G ) ~ (=~ )o and
5J U G 1
J(FH) 'U proof.
i
A = FG, B = FN (H), I = {U 1 ,
Put
and
G
J = J(FH) and apply Lemma
8.10. 8.12.
THEOREM.
(Motose (1983)).
FN and l e t H
a block idempotent of
Proof.
A = FG, B
Put
N
Let
be a r i g h t t r a n s v e r s a l f o r
FH,
r=
H
in
G.
{el
and
3
S
A s i m i l a r argument shows t h a t
derive
J(FH1e LS(FG1
FHe
=
TBI
=
J(FH).
e be
let Then
{ u I , u p,..-,a 1
Let
Then
TAT = eFGe =
e.
be t h e i n e r t i a group of
=
G,
be a normal subgroup of
-1
FHeaieai u i i=l eFHe
=
=
IJ = J I
and
BJ
5 JB.
Invoking Lemma 8.10, we
a s required.
The next two r e s u l t s a r e easy a p p l i c a t i o n s of Theorem 8.11.
For t h e termin-
ology and p r o p e r t i e s of f i n i t e groups of Lie type we r e f e r t o C u r t i s (1970). 8.13. THEOREM. characteristic
U
(Motose ( 1 9 8 3 ) ) .
r
such t h a t
and an a b e l i a n r'-subgroup
B
Suppose t h a t
is a s e m i d i r e c t product of a normal r-subgroup
H
=
B n N.
8.14. THEOREM. (Motose (1983)).
p
be an odd prime d i v i s o r of
Let
Then
5 J (FG)
J (EH1 'U
let
G has a s p l i t (B,N)-pair of
q
be a prime power, l e t
q-1.
Then
G
=
SL(2,q) and
t ( G ) i s t h e p - p a r t of
q-1.
The n e x t r e s u l t i s a u s e f u l companion of Theorem 8.14.
8.15. THEOREM. characteristic
(Alperin (197911.
2.
Then
Let
G
=
SLC2,2nl
and l e t
F
be a f i e l d of
CHAPTER 8
498
To s t a t e our next r e s u l t , w e r e c a l l t h e following p i e c e of information ( s e e Broue and Puig (1980)).
Br
P '
. FG+
For any p-subgroup
FCG(P)
defined by
Br (g)
P
P of =
g
FG,
b of
C,(&)
g
C G ( P ) and
f ?
C F G ( P ) --+
BY
P
FCG(P).
(g)
0
=
B
For a block
( & , b ) c o n s i s t i n g of some p-subgroup
a B-subpair i s a p a i r
and some block
the natural projection
for
otherwise induces a homomorphism of F-algebras of
G
Q of
G
such t h a t
BPQ(lB)lb # 0 Let
F be an a l g e b r a i c a l l y closed f i e l d of c h a r a c t e r i s t i c p FG.
a block of
Then
f o r a l l B-subpairs G.
B
0 and l e t
NG(&,b)/CG(&)
is s a i d t o be n i l p o t e n t i f
( Q , b ) , where
3
B
be
i s a p-group
NG(&,bl denotes t h e s t a b i l i z e r of
(Q,b) in
The s i g n i f i c a n c e of a n i l p o t e n t block stems from t h e f a c t t h a t it i s iso-
morphic t o a f u l l matrix algebra over t h e group algebra of i t s d e f e c t group ( s e e Pliig ( 1 9 8 6 ) ) .
Whether t h e converse of t h i s i s t r u e i s not however known.
We
a r e now ready t o quote t h e following r e s u l t .
8.16. THEOREM. characteristic
(Okuyama (19811).
p
and l e t
Let
F
be an a l g e b r a i c a l l y closed f i e l d of
B = B ( e ) be a block of
FG
of d e f e c t
d.
Then
d
J(Z(FG)Ip - l e # 0 i f and only i f Let
t (G)
= 2
and
G
t(G)
B
i s n i l p o t e n t with a c y c l i c d e f e c t group.
be t h e nilpotency index of
J(FG).
have been c h a r a c t e r i z e d by Theorem 3.12.4.
t(G) = 4
(1982131 determined t h e s t r u c t u r e of
G
following hypotheses: (i) 0 ,(GI
O2 ,(G)
i s abelian i s metacyclic
with
and
G
f o r which
The case where
i s p-solvable i s s e t t l e d by Proposition 7.2.6.
r i a t e t o i n v e s t i g a t e t h e case where
(ii)
The groups
G
t(G) = 4
t (G)
=
3
It is t h e r e f o r e approp-
i s p-solvable.
Koshitani
under e i t h e r of t h e
A SURVEY OF SOME FURTHER RESULTS
0 ,(GI
(iii) The o r d e r of
i s n o t d i v i s i b l e by
499
3.
He a l s o obtained t h e following r e s u l t .
8.17. THEOREM.
(Koshitani (198233)).
P E Syl (G). P
If
t(G) = 4,
p
then
Let
G be a p-solvable group and l e t
= 2
and one of t h e following p r o p e r t i e s
hold:
P i s c y c l i c of o r d e r
(i)
(ii)
P
.
8
i s elementary a b e l i a n of order
S
G/02,(G)
(iii)
4
.
We c l o s e by quoting t h e following r e s u l t s .
8.18. THEOREM. (Motose ( 1 9 8 3 ) ) . be of order
pp
P i s r e g u l a r and t ( G )
is +elian,
i s elementary a b e l i a n .
(Ninomiya ( 1 9 8 1 ) ) .
characteristic
p
P
'b
Let
= ~ ( p - 1 )+ 1,
P
then
be an a l g e b r a i c a l l y c l o s e d f i e l d of
G be a p - n i l p o t e n t group.
and l e t
Fo ,(GI, l e t Pe
of
P
If
8
8.19. THEOREM.
e
P E Syl (G)
be a p-solvable group, l e t
F be a f i e l d of c h a r a c t e r i s t i c p > 0 .
and l e t
0 (G)/Opf,p(GI P'rPrP'
G
Let
For each block idempotent
denote a Sylow p-subgroup of t h e i n e r t i a group of
e.
Then t h e following conditions a r e equivalent:
G is
(il
(ii) I f 5
p-radical
e
then
ex
=
e
for all
[O ,(G1 ,Pel
E
P
e
(iii) I f
i s a block idempotent o f
8.20. COROLLARY.
(Ninomiya (19811).
Theorem 8.19, assume t h a t
G
Then
P
FO ,(G , P
then
P,
5 CFG (FOP, (GIe ) .
Further t o t h e assumptions and n o t a t i o n of
P (G) and t h a t p n g P i l = 1 f o r a l l
E Syl
g E G-N
G
(PI.
i s p - r a d i c a l i f and only i f one of t h e following c o n d i t i o n s hold:
Pd G
(il
(ii)
P.
FO ,(C) P
i s a block idempotent of
G has a subnormal subgroup ff which i s a Frobenius group with complement
8
8.21.
THEOREM.
element
g
(Tshushima ( 1 9 8 6 ) ) .
of o r d e r
p
in
G,
Let
I0 ,IG) ,gl P
G
be a p - r a d i c a l group.
is nilpotent.
'
Then, f o r any
CIIRPTER 8
500
Let F be an algebraically closed field of
8.22. THEOREM. CMotose (197711. characteristic p > 0, let U
be the F-dimension of the projective cover of the be the principal block of FG.
trivial FG-module and let B
If P E Syl CG)
P
is
cyclic, then
.
dimJGB1 = dimB(1- l / u l
F if and only if G
F
is p-solvable of p-length
8.23. THEOREM. miller (1971)).
Let
sn
1.
be the symmetric group of degree n.
Then lim(dimJ(FSnl/dimFSn) = 1
n
F
+ m F
8.24. THEOREM. (Koshitani (1982b)).
Assume that p
=
2 and
G is a 2-solvable
group with a Sylow 2-subgroup P which satisfies one of the following conditions: (l.1
Ciil
P is cyclic of order 4 P
is elementary abelian of order
(iii) G = S Then t(G)
=
4.
.
8
501
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On t h e indecomposable r e p r e s e n t a t i o n s of a c e r t a i n c l a s s of groups, Proc. London Math. SOC. ( 3 ) 10, 497-513.
STONEHEWER, S.E. [19691
Group a l g e b r a s of some t o r s i o n - f r e e groups, J. Algebra 13, 143-147.
SWAN, R.G.
119631
The Grothendieck group o f a f i n i t e group, Topology 2 , 85-110.
TSUSHIMA, Y. [1967]
Radicals of group a l g e b r a s , Osaka Math. J. 4 , 179-182
119681
A group a l g e b r a of a p-solvable group, Osaka Math. J. 5 , 89-98.
[1971a] On t h e a n n i h i l a t o r i d e a l s of t h e r a d i c a l of a group a l g e b r a , Osaka Math. J. 8, 91-97. [1971b] On t h e block of d e f e c t zero, Nagoya Math. J. 44, 57-59. I19741
On t h e e x i s t e n c e of c h a r a c t e r s of d e f e c t z e r o , Osaka Math. J.11, 417-423.
[1977)
On t h e weakly r e g u l a r p-blocks with r e s p e c t t o 14, 465-470.
I1978al On t h e p'-section 83-86.
0 '(G),
P
Osaka Math. J.
sum i n a f i n i t e group r i n g , Math. J. Okayama Univ. 2 0 ,
[1978b] Some n o t e s on t h e r a d i c a l of a f i n i t e group r i n g , Osaka Math. J . 15, 647-653. [1978c] On t h e second reduction theorem of P. Fong, Kumamoto J. S c i . 6-14.
(Math) 13,
[1978d] Problems on t h e r a d i c a l of a f i n i t e group r i n g s , Proc. 10th. Sympos. on Ring theory (Shinhu Univ. Matsumoto, 1 9 7 7 ) , Dept. of Math. Okayama Univ, Okayama, 116-120.
BIBLIOOWHY
518
TSUSHIMA, Y. I19791
Some notes on t h e r a d i c a l of a f i n i t e group r i n g 11, Osaka J. Math. 16, 35-38.
[1983]
A note on Cartan i n t e g e r s f o r p-solvable groups, Osaka J. Math. 2 0 ,
675-679. 119861
On p-radical groups, p r e p r i n t .
VILLAMAYOR, O.E.
119581
On t h e semisimplicity of group a l g e b r a s , Proc. Amer. Math. SOC. 9 , 62 1-627.
119591
On t h e semisimplicity of group a l g e b r a s , 11, Proc. Amer. Math. SOC. 9 , 621-627.
WADA, T.
I19771
On t h e e x i s t e n c e of p-blocks with given d e f e c t groups, Hokkaido Math. J . 6 , 243-248.
119811
Blocks with a normal d e f e c t group, Hokkaido Math. J. 10, 319-322.
[19831
On t h e number of i r r e d u c i b l e c h a r a c t e r s i n a f i n i t e group, Hokkaido Math. J. 1 2 , 74-82.
119851
On t h e number of i r r e d u c i b l e c h a r a c t e r s i n a f i n i t e group, 11, Hokkaido Math. J. 14, N02, 149-154.
WALLACE, D.A.R. P h i l . SOC. 54, 128-130.
119581
N o t e on t h e r a d i c a l of a group algebra, Proc. Camb.
[19611
On t h e r a d i c a l of a group a l g e b r a , Proc. Amer. Math. SOC. 1 2 , 133-137.
I1962al Group a l g e b r a s with r a d i c a l s of square zero, Proc. Glasgow Math. Assoc. 5 , 158-159. 11962b1 Group a l g e b r a s w i t h c e n t r a l r a d i c a l s , Proc. Glasgow Math. Assoc. 5 , 103-108. I19651
On t h e commutativity of t h e r a d i c a l of a group algebra, Proc. Glasgow Math ASSOC. 7 , 1-8.
[19671
The Jacobson r a d i c a l s of t h e group a l g e b r a s of a group and of c e r t a i n normal subgroups, Math. Z. 100, 282-294.
[19681
Lower bounds f o r t h e r a d i c a l of t h e group algebra of a f i n i t e p-solvable group, Proc. Edinb Math. Soc(2), 16, 127-134.
[1969]
On commutative and c e n t r a l conditions on t h e Jacobson r a d i c a l of t h e group algebra of a group, Proc. London Math. SOC. 19, 385-402.
119701
The r a d i c a l of t h e group algebra of a subgroup, of a polycyclic group and of a r e s t r i c t e d SN-group, Proc. Edinb. Math. SOC. V o l , 17 ( S e r i e s I1 P a r t 2 , (165-171).
119711
On commutative and c e n t r a l c o n d i t i o n s on t h e Jacobson r a d i c a l of t h e group algebra of a group 11, J. London Math. SOC. 4, 91-99.
519
BIBLIOGRAPHY
WALLIS, W.D. I19681
A reduction of t h e problem of semisimplicity, J . Algebra 10, 501-502.
WARD, H.N.
119611
Some r e s u l t s on t h e group a l g e b r a of a group over a prime f i e l d , Seminar on F i n i t e Groups an'd Related Topics, Harvard Univ., 13-19.
119681
The a n a l y s i s of r e p r e s e n t a t i o n s induced from a normal subgroup, Michig. Math. J. 15, 417-420.
WATANABE , A.
I19791
On Fong's r e d u c t i o n s , Kumamoto J. S c i .
m a t h ) , 13, 40-54.
I1982al R e l a t i o n s between blocks of a f i n i t e group and i t s subgroup, J . Algebra 78, 282-291. 11982bI p-blocks and p-regular c l a s s e s i n a f i n i t e group, Kumamoto J . S c i . , (Math), 1 5 , 33-38. WILLEMS, W. 119761
Bemerkungen z u r modularen D a r s t e l l u n g s t h e o r i e 111. I n d u z i e r t e und eingeschrgnkte Moduln, Arch. Math. (Basel) , 26, 497-503.
119781
%er d i e Existenz von Blscken, J. Algebra 53, 402-409.
119801
On t h e p r o j e c t i v e s of a group a l g e b r a , Math. Z.
1 7 1 , 163-174.
ZALESSKII, A.E. I19651
On t h e semisimplicity of a crossed product, S i b i r s k . Math. Zh. 6 , 1009- 1013.
[19701
On group r i n g s of s o l v a b l e groups, Izv. Akad. Nauk BSSR, s e r , Fiz-Mat., 13-21.
I19731
On t h e semisimplicity of a modular group algebra of a s o l v a b l e group, S o v i e t Math. 14, 101-105.
119741
The Jacobson r a d i c a l of t h e group algebra of a s o l v a b l e group i s l o c a l l y n i l p o t e n t , Izv. Akad. Nauk. SSSR, Ser. Mat. 38, 983-994.
ZASSENHAUS, H. I19491
The theory of groups, 2nd ed. Chelsea, New York.
52 0
Notation
Number Systems
the n a t u r a l numbers the rational integers t h e r a t i o n a l numbers t h e r e a l numbers t h e complex numbers t h e i n t e g e r s mod
ni
S e t Theory proper i n c l u s i o n L -
inclusion
I XI
t h e c a r d i n a l i t y of t h e s e t
x-Y
t h e complement of
Y
in
x x
Number Theory
a divides b
a does not d i v i d e b g r e a t e s t connnon d i v i s o r of t h e p - p a r t of
n
and
b
NOTATION
n
t h e p'-part
P'
521
n
of
Group Theory
F*
t h e m u l t i p l i c a t i v e group of a f i e l d
<X> 'n G x G 1
2
t h e subgroup generated by
X
t h e c y c l i c group of o r d e r
n
F
GI and G2
d i r e c t product of
H dG
H
D NS
t h e semidirect product of a normal subgroup subgroup S
X
t h e normalizer of
=
[G,Gl
D and a
X in G
t h e c e n t r a l i z e r of
G'
G
is a normal subgroup of
G
in
t h e commutator subgroup of
G
n Gi
d i r e c t product of
? Gi
r e s t r i c t e d d i r e c t product (or d i r e c t sum) of
Gi, i
E
I
iEI
ZEI
G
expG
t h e exponent of
Z (GI
t h e c e n t r e of
G
t h e p-part of
g
the p'-part
g
of
t h e set of a l l conjugacy classes of
G
t h e s e t of a l l Sylow p-subgroups of
G
t h e s t a b i l i z e r of t h e o r b i t of
x
t h e index of
H
z
in
G
t h e maximal normal p'-subgroup of t h e maximal normal p-subgroup of
G
G
Gi, i E I
NOTATION
522
Op(G/Opr(Gl) i n G
t h e image of
t h e general l i n e a r group of degree of p elements
n over t h e f i e l d
t h e s p e c i a l l i n e a r group of degree
n over F
P
P
=
{ g E GlgM E CGlM(N/M1
P
1
t h e n-th term of t h e lower c e n t r a l s e r i e s of
G
= < g n 1 g E G>
G
the number of double (P,Pl-cosets i n
t h e n-th term of t h e Brauer-Jennings-Zassenhaus series for G t h e n-th term of Lazard s e r i e s f o r =
G
n
(1
G
+rmn)
t h e weight of t h e F r a t t i n i subgroup of
G
n
synnnetric group of degree a l t e r n a t i n g group of degree
n
t h e group of a l l nonsingular transformations of t h e v e c t o r space TI t h e group of a l l
n x n nonsingular m a t r i c e s over R G
t h e automorphism group of
V
t h e automorphism group of an R-module =
ig l g E P
the
cE
CIICG)
p-length of
1
C
z
t h e semidirect product of t h e nilpotency c l a s s of
P
x
z
P
and
SL(2,p)
P
t h e d i h e d r a l group of o r d e r
2n, n > 2 2 n , n> 3
t h e generalized quaternion group of order t h e semidihedral group of o r d e r =
LEp
n- 1 = 1,
2n, n > 3 n-2
bP = 1, b-lab = al+P
t h e e x t r a - s p e c i a l p-group of o r d e r odd.
H
C H 2
HI
i s G c o n j u g a t e t o a subgroup of
>
p3 and exponent
HZ
p,p
NOTATION
H
= H
'G
HI
and H
523
are &conjugate
Rings, Modules, and Characters Z (Rl
the centre of R
A @ B
tensor product
charR
the characteristic of R
? Ri
direct sum
+I
direct product of rings FG
over F
the group algebra of G
the augmentation ideal of FG aug (x)
G'F
R0
the augmentation of
3:
the twisted group algebra of
G
over
F
the opposite ring of R the ring of all n x n-matrices over R the Jacobson radical of a ring R
J(V)
the radical of an R-module
SocR
the socle of a ring R
socv
the socle o f a module
V
V
the additive group of all R-homomorphisms V -
V
the n-th direct power of the composition length of the annihilator of
V
the projective cover of
I,
the unit group of a ring R the Loewy length of the Heller module of = Hom(V,F)
F
V
V
V
W
NOTATION
524
Si ( R )
t h e i - t h s o c l e of
R
si(v)
t h e i - t h s o c l e of
V
r
support of t r a c e of
z
V
t h e o u t e r t e n s o r product of
and
W
t h e inner t e n s o r product of FC-modules
U and
V
the induced module
t h e r e s t r i c t i o n of
vH lG o r 1
FG
vL AL
I d (XI
V
to
FH
t h e t r i v i a l FG-module =LQV
F = L 8 A F t h e sum of a l l two-sided i d e a l s of in X
FG
contained
V
G-conjugate of
c o n t r a g r a d i e n t module t h e intertwinning number
V
F
did F
dimension of
over
i n f (V1
i n f l a t e d module
ext(V)
extension of
XV
t h e c h a r a c t e r afforded by
V
gX
t h e c h a r a c t e r afforded by
'V
XG
t h e induced c h a r a c t e r
X+
=
V
c x fix A
t h e commutator subspace of d e f e c t group of
C E CL(G)
d e f e c t group of a block
B
t h e F-linear span of a l l 'C C E CL(C) and 6 ( C ) C_ D
with
G t h e F-linear span of a l l such t h a t D 3 6 (C)
C+ with
C E CR(C)
525
NOTATION
the F-linear span of a l l C" where C p-regular class of G with 6 ( C ) = D
t(GI
the nilpotency index of
A*
=
is a
J(FG)
Hom(A,F)
F FX
the F-linear span of X
1
=
{a E Al$(aX)
=
0
1
=
{a E AI$(Xu)
=
0
1
X
X1
01
= { a E A ~ U X= =
{a E AlXu
0 1
=
n =
{a E AlaP
for some n 2 11
IA,Al
E
the trace map
V
obstruction to the extension o f
to FG
the tensor product of representations the sum of all p-elements of KerX
=
{3:
KerT
=
Ix E FGIT(gccl
V
E FGIXv(g;c) = =
0 for all g E G I
o
the multiplicity of of w
n
R(FG1
=
FGI trtB
{a E
1
G including 1
=
for all g E
V
o
GI
as a composition factor
for all sufficiently
large n E N}
z
P'
P =
1 FS:
where S1,.
i=1
.. ,S r
are all p-regular
sections of
G.
trA
the trace
the matrix A
Kerb'
the kernel of the representation afforded by
KerB
the kernel of the block
of
B
n =
{;c
E FGlP
E IFG,FGl}
n = {z?
13:
E
XI n
= {g E
GI$
E
XI
a complement of =
J(A) in A
Z(FG) n Z(FN)
the inertia group of the block
b
V
NOTATION
526
e*
the sum of all &conjugates of
e
the relative trace map the Cartan matrix of
FG
the nilpotency index of J ( B ) the nilpotency index of J ( Z ( B ) ) the unique block corresponding to
ux ( V )
the vertex of 'I
6 (BI
the defect group of
B
the defect group of
e
xG
Cohomology Theory the group of all A-valued 2-cocycles of a coboundary the subgroup of a l l coboundaries =
Z 2 (G,A) / B 2 (G,A1
the cohomology class of obstruction cocycle = (infa)(x,yl =
a(xN,yN)
f
E . Z 2 (G,A)
G
527
Index
Action, 4, 486 faithful, 486 transitive, 5 Additive structure, 1 Algebra, 2, 52 definable over a subfield, 114 direct product of, 53 Frobenius, 123 homomorphism o f , 2 index of, 455 matrix representation of, 57 of finite representation type, 289 regular representation o f , 57 representation of, 56 separable, 114 symmetric, 123 uniserial, 285 Alperin, 392, 497 Annihilator, 27 of induced modules, 335 Anti-automorphism, 379 Asano, 496 Augmentation, 6 9 Augmentation ideal, 68 Azumaya, 36, 292 Bedi, 372 Bijective correspondence, 19 between the isomorphism classes of R and M (R)-modules, 19 betwee; the sets of ideals of R and M ( R ) , 20 bztween the a-representations and PG-modules, 145 Bilinear form, 123 associative, 123 nonsingular, 123 Bimodule, 44 Blackburn, 112 Block, 59 irreducible representation associated with, 59
nilpotent, 498 Block idempotent, 59 Brauer, 79, 81, 139, 192, 195, 200, 226, 421
Brauer correspondence, 223 Brauer homomorphism, 76 Brauer-Jennings-Zassenhaus series, 302 Brockhaus, 175, 193, 195 Broue, 498 B-subpair, 498 Burnside, 190, 225 Canonical injection, 88 Cartan invariants, 65 Cartan matrix, 65 Central character, 59,64 Centralizer, 4 Centre, 1 of a group, 4 of a ring, 1 of symmetric algebras, 133 Characteristic subgroup, 302 Characterization of defect group of a block, 2 7 1 defect of a block, 7 9 exponents of defect groups of blocks, 455
dimension subgroups, 308, 309 dimFJ(Z'(FG)) 217 elements of J(FG), 140, 177 H-injective modules, 95 H-projective modules, 9 4 J(R), 3 0 J ( V ) , 26 kernel of a block, 195 local rings, 36 p-groups with cyclic subgroups of index p , 417 p-radical groups, 357, 358, 365, 367 projective modules, 41,42 SOC V , 26 vertices, 103
.
528
INDEX
Clarke, 397, 418, 494, 495 C l i f f o r d , 141, 146, 149, 240 C l i f f o r d ’ s theorem, 141 Coboundary , 142 Cocycle, 142 Cohomology c l a s s , 143 Cohomologous cocycles, 143 C o l l i n s , 392 Commutative diagram, 40 Commutator, 301 Commutator subspace, 73 Component, 100 Composition s e r i e s , 8 equivalent, 9 factors o f , 8 length o f , 8 Conjugate of a module, 1 4 0 Conlon, 163, 167, 168 Conlon’s theorem, 163 Contragradient module, 379 Covering of a block, 229 Criteria for absence of blocks of d e f e c t zero, 372 algebra t o be Frobenius, 125 algebra t o be of f i n i t e r e p r e s e n t a t i o n type, 289 algebra t o be s e p a r a b l e , 114 algebra t o be symmetric, 126 algebra t o be u n i s e r i a l , 290 block t o be defined, 467, 471 block t o be r e g u l a r , 469 block t o be of zero d e f e c t , 180 block t o cover another block, 232 blocks t o be quasi-primary, 255 c e n t r e of a r i n g t o be indecomposable, 16 commutativity of J(FG), 260, 263, 265 commutativity of t h e p r i n c i p a l block, 257 complete r e d u c i b i l i t y of a module, 363 complete r e d u c i b i l i t y o f induced modules, 87 dim#(B) = dimFB(l-l/u), 500 dimpJ(FG) = dimJ(FG1 =
, 193 IGl (PI - 1. 198
each block t o be quasi-primary, 255 element t o be i n r a d i c a l , 2 7 , 30 equivalence of composition s e r i a s , 9 e x i s t e n c e of normal p-complement, 225 e x i s t e n c e of p r o j e c t i v e cover, 48 e x t e n d i b i l i t y of modules, 146 ( e , f ) t o be simple induction p a i r , 336 ( e ,f)t o be simple r e s t r i c t i o n p a i r ,
32# ,“wc
0, 338 t o be completely r e d u c i b l e , 341
( e p ) H t o be completely r e d u c i b l e , 344
eJ(FH)FG C FG*J(FH), 344 eJ(FH)FG FG*J(FH)e, 344 f a c t o r algebra t o be Frobenius, 130 f a c t o r algebra t o be symmetric, 130 f i e l d t o be p e r f e c t , 114 f i e l d t o be a s p l i t t i n g f i e l d , 55,56 Frobenius algebra t o be u n i s e r i a l , 291 FG t o be of f i n i t e r e p r e s e n t a t i o n type, 298 FG t o be l o c a l , 106 PG t o be l o c a l , 152 fW t o be completely reducible, 341
(g)G
t o be completely r e d u c i b l e ,
344H
fJ(FG) = fJ(FH)FG, 345 FG J(FH)M 2 J(FG)M, 346 FG t o be indecomposable, 1 1 2 group a l g e b r a t o be semisimple, 7 1 group t o be d e f e c t group, 278 group t o be Frobenius, 184 group t o be of p-length 1, 416, 426 group to be p - n i l p o t e n t , 244, 245 group t o be 2 - n i l p o t e n t , 248 group t o be p - r a d i c a l , 361 group t o be r e g u l a r , 418 group t o s p l i t over a normal subgroup, 417 homomorphism t o be e s s e n t i a l , 46 H t o be of p’-index, 352 H2(G,F*) t o be i n f i n i t e , 150 i d e a l t o be i n J ( R 1 , 3 1 i d e a l t o be Jacobson r a d i c a l , 29 i d e a l t o be n i l p o t e n t , 56 induced module t o be indecomposable, 168 induced module t o be p r o j e c t i v e , 99 i r r e d u c i b l e module t o be H-projective 352 isomorphism of p r o j e c t i v e modules, 49 isomorphism of t w i s t e d and ordinary group a l g e b r a s , 1 4 4 , 151
J(B)
J(B1
J ( F N ) B , 281 = B J ( Z ( B ) ) , 246 =
J ( B ) = I ( D ) * B , 282 J ( R ) t o be n i l p o t e n t , 31, 34 J ( A ) t o be p r i n c i p a l i d e a l , 292 J(FG)V = J ( F N ) V , 281 J(FG) t o be p r i n c i p a l i d e a l , 299 J(FG) = R(c), 175 J(FG) C FG*J(FH), 338 J(FG)’-= 0 , 250 J ( F G ) C Z(FG), 251 J(FG) FG*J(Z(FG),246 J(FG) C J(FH)FG, 338 J(FH) J ( F G ) , 338 J(FG)f-= FG*J(FH)f, 345 J(FG)e = FG.J(FH)e, 347 J ( F G ) e = J ( F H ) F G e , 347
INDEX
529
J ( F G 1 = FG.J(FH), 352 t ( G ) < (a+ ) (p 1) + 1, 447 J ( F G ) = JLFHIFG, 352 449 t(G) = pa-', J ( B 1 2 = 0 , 249 t ( G ) = U(p-1) + 1, 419, 427, 447 J ( F G 1 3 = 0 , 420 t(G) = pa, 419 J ( V ) = 0 , 25 t isted group algebra to be local, 152 J ( Z ( F G ) ) dto be an ideal, 495 to be completely reducible, 338 # 0 , 498 Z, E J ( V ) , 27 J(Z(FGI)P L(P(lG)I = 3, 494 Curtis, 79, 188, 209 lifting idempotents, 156 module to be absolutely irreducible,55 Dade, 154 module to be artinian, 5, 7 Defect, 77 module to be completely reducible, 12 of a block, 77 module to be finite direct sum of of a class, 77 irreducible modules, 23 Defect group, 77 module to be finitely cogenerated, 5 of a block, 77 module to be finitely generated, 6 of a block idempotent, 77 module to be flat, 46 of a class, 77 of covering blocks, 466 module to be free, 161 module to be injective, 43, 285 Dimension subgroups, 302 module to be injective hull, 288 Direct decomposition of rings, 15 module to be irreducible, 28 Direct power of a module, 19 module to be H-injective, 96 Direct product of p-radical groups, 360 module to be H-projective, 96 Division ring, 21 module to be noetherian, 6,7 module to be projective, 41, 42 Eilenberg , 126 module to be projective cover, 51 Equivalent representations, 57 modules to belong to the same block, 64Erdmann, 421, 494 modules to be linked, 63 Essential homomorphism, 46 module to be separable, 114 Essential submodule, 287 module to have composition series, 8 Exact sequence, 40 module to have finite decomposition, 7 Extension of ground field, 52 module to have maximal submodule, 6 External direct sum, 4 module to have zero radical, 25 to be completely reducible, 338 Feit, 493 Mu nilpotency of J ( R ) , 31 Fermat prime, 446 P ( l 1 = FG, 377 Filtration of 1 ( G ) , 304 G p-regularity, 196 Fixed-point-space, 186 ring to be full matrix ring, 18 Fong, 171, 474 ring to be indecomposable, 16 Fong correspondence, 474 ring to be local, 36 Fong correspondent, 474 ring to be simple artinian, 32 Fong's dimension formula, 170 ring to have finite decomposition, 17 Formanek , 177 Soc FG = FGc, 176 Frattini subgroup, 320 Frobenius, 183 submodule to be fully invariant, 14 Frobenius complement, 183 submodule to be radical, 25 submodule to be socle, 25 Frobenius kernel, 183 t ( B ) = pa-l+p-1, 442 Fully invariant submodule, 14 t C B ) > pn-' 4 - 1 , 457 Glauberman, 416, 417 t ( B ) = pd 490 Gorenstein, 183, 416, 417, 418, 421 t B pdL1+p-l, 491 Gow, 226 t C B ) < pd, 491 Green, 168, 218, 277 t ( B ) = /DI,492 t ( B ) = 3 , 493 Group, 4 t ( G ) = 4, 499 alternating, 5 composition series of 303 t ( G ) = dim#(FG) +1, 421 cyclic, 5 t ( G ) = t ( P 1 , 418 dihedral, 320 t ( G 1 = pa- +p-1, 442 elementary abelian, 303 pa- < t ( G ) < p a , 442 extra-special, 320 t ( G ) = (ail)Cp-1) + 1, 447 Frobenius, 183
?
&'
530
g e n e r a l l i n e a r , 413 generalized quaternion, 320 involved, 416 metacyclic, 318 multiplicative, 4 m u l t i p l i c a t i v e of a f i e l d , 4
M(p), 320
M ( p ) , 320 nylpotent, 301 nilpotency c l a s s o f , 301
PSL(2,q), 421 p-constrained, 112 p-element o f , 4 p '-element o f , 4 p-length o f , 283 p - n i l p o t e n t , 197 p-radical, 357 p-regular c l a s s o f , 4 p-regular element o f , 4 p-regular s e c t i o n o f , 138 p-singular c l a s s of, 4 p-solvable, 110 Q d ( p ) , 416 r e g u l a r , 418 semidihedral, 320 s o l v a b l e , 110 s p e c i a l l i n e a r , 413 symmetric, 5
S(n), 417 Group a l g e b r a , 67 augmentation i d e a l o f , 68 c e n t r e o f , 71 H a l l , 151 P. H a l l , 414, 415, 417 Hall-Higman's lemma, 415 Hall-Higman's theorem, 417 Hamernik, 255, 257 Height, 304 Heller module, 376 G. Higman, 415, 417 D. Higman, 94, 96, 225, 298 H i l l , 312 Hochschild, 225 Homogeneous component, 14 Homomorphism of a l g e b r a s , 2 Huppert, 112, 321, 416, 417, 418 M.
Ideal, 1 nil, 1 nilpotent, 1 Idempotent, 1 primitive, 1 c e n t r a l l y primitive, 1 orthogonal, 1 I i z u k a , 178 Inductive set, 3 I n e r t i a group, 141, 229 I n f l a t i o n map, 146
INDEX
I n j e c t i v e h u l l , 287 I n t e r n a l d i r e c t sum, 3 Intertwining number, 90, 363 I t e r a t e d s o c l e , 378 Jacobson, 114 Jacobson r a d i c a l , 22 dimension o f , 66 of d i r e c t sums, 25 of d i r e c t products, 31, 119 of f u l l matrix r i n g s , 31 of module, 22 of r i n g , 22 of submodule, 24 Jennings, 307, 308, 309, 311 Jennings' formula, 311, 321, 322, 3 3 0 Jennings' theorem, 307 Jordan-Hb;lder theorem, 9 Karpilovsky, 151, 423 Kasch, 298 Kawada, 226 Kernel of a block, 195 Kernel of a module, 195 K h a t r i , 343, 359, 360 Kneser, 298 Knorr, 277, 281, 335, 336, 343, 466 Koshitani, 240, 262, 318, 321, 325, 329,
416, 417, 419, 427, 440, 442, 490, 491, 492, 494, 495, 498, 499, 500 Krull-Schmidt theorem, 39 Kiilshammer, 134, 178, 207, 214, 246, 453, 455, 456, 459, 474, 484, 487, 495 Kiilshammer's s t r u c t u r e theorem, 484 Kupisch, 298 Lattice, 3 isomorphism, 3 homomorphism, 3 of submodules, 3 Lazard, 310 Lazard s e r i e s , 302 L e f t a n n i h i l a t o r , 127 Left transversal, 4 Lie product, 74 Linear equivalence, 143 Loewy l e n g t h , 284, 375 of induced modules, 400 of p r o j e c t i v e covers, 391, 407 Loewy s e r i e s , 284 Loncour, 119 Lorenz, 389, 393, 394, 407, 409, 425 Lower c e n t r a l s e r i e s , 301 Mackey decomposition, Malcev, 209 Maschke's theorem, 7 1 Matrix u n i t s , 17 Maximal element, 3 Maximal submodule, 3
lo1
INDEX
531
Michler, 195, 226, 235 Nilpotent element, 1 Ninomiya, 265, 266, 323, 329, 343, 357, Modular law, 3 Module, 2 360, 362, 372, 419, 422, 425, 494, absolutely irreducible, 54 499 artinian, 5 Noether-Skolem Theorem, 241 Normal closure, 348 basis of, 4 Normalizer, 4 completely reducible, 12 Normal p-complement, 197 composition length of, 10 direct product of, 3 N -sequence, 302 dual, 284 N!?mber of irreducible modules, 81 external direct sum of, 3 faithful, 28 Okuyama, 175, 372, 462, 493, 498 Orbit, 4 finitely cogenerated, 5 Outer tensor product, 116 finitely generated, 5 flat, 45 Partially ordered set, 3 free, 4 Passman, 108, 110, 310, 418, 423, 464 G-invariant, 141 p-complement, 107 head of, 90 Perfect field, 113 homogeneous, 14 Primitive ideal, 28 H-injective, 91 Principal block, 195 H-projective, 91 Projective cover, 46 imprimitive, 82 Projective matrix representation, 143 indecomposable, 3 Projective representation, 143 induced, 82 inflated, 149 completely reducible, 144 injective, 42 irreducible, 143 in a block, 61 a-representation, 143 Proper submodule, 3 irreducible, 2 Puig, 498 internal direct sum of, 3 linked, 63 Quasi-primary block, 253 left, 2 noetherian, 5 Regular block, 469 of finite length, 10 Reiner, 79, 188, 209, 241 principal indecomposable, 12 Relations between projective, 41 radical of, 22 Jacobson radical of a module and a reducible, 2 submodule, 24 Socle of a module and a submodule, 24 regular, 123 Relative trace map, 268 separable, 113 socle of, 22 Representation, 56 S-projective, 274 completely reducible, 57 uniserial, 285 faithful, 72 indecomposable, 57 unital, 2 Modular law, 3 irreducible, 57 Miller, 500 of algebras, 56 Morita, 105, 228, 238, 240, 255, 284, of groups, 72 292, 299, 419, 424, regular, 57 Morita's theorem, 240 underlying module of, 56 Motose, 119, 189, 265, 266, 315, 318,323,Restrictionof a module, 82 325, 326, 328, 329, 343, 351, 360, 362,Hetraction, 275 372, 419, 422, 425, 427, 428, 429, 43O,Reynolds, 179, 474 437, 440, 444, 445, 446, 447, 449, 495,Right annihilator, 127 496, 497, 499, 5 0 0 Right transversal, 4 Ring, 1 Nakayama, 126, 130, 240, 290, 292 associative, 1 Nakayama's lemma, 29 commutative, 1 Nakayama reciprocity, 90, 367 direct product of, 2 Nesbitt, 192, 200, 226, 421 homomorphism, 1 Nilpotency index, 118 identity element of, 1 Nilpotent conjugacy class, 218 indecomposable, 16
532
INDEX
Upper p-series, 282 local, 36 Upper bound for t(Z( B ) ) , 462 of endomorphisms, 2 Unit (left, right), 3 0 of endomorphisms of a completely Universal characterization of induced reducible module, 15 modules, 88 of endomorphisms of direct powers, 19 of matrices, 17 Vertex, 100 opposite, 1 Villamayor, 108 semisimple, 22 socle of, 22 Robinson, 214 Wallace, 189, 192, 193, 198, 250, 251, 260, 313, 392, 418, 419, 421 Ward, 154 Saksonov, 362, 365, 368, 369, 372 Watanabe, 178 Scalar matrix, 17 Wedderburn, 209 Schanuel's lemma, 376 Wedderburn-Malcev theorem, 209 Schur's lemma, 21 Second cohomology group, 142 Weight, 304 semidirect product, 486 Willems, 108, 154, 168, 391 Separable element, 113 Zassenhaus, 190 Separable extension, 113 Zorn's lemma, 3 Separable polynomial, 113 Short exact sequence, 40 split, 40 Sibley, 392 Simple induction pair, 336 Simple restriction pair, 336 Shoda, 240 Snider, 177 Socle, 22 of Frobenius algebras, 131, 134 of module 22, Source, 101 SDieael. 250. 251 Splitting field, 55 Splitting homomorphism, 40 Stabilizer, 4 Subgroup, 4 generated by a set, 4 Submodule, 14 essential, 26 fully invariant, 14 superfluous, 26 Subring, 1 Support, 68 of central idempotents, 76 Swan, 154 System of imprimitivity, 82 Tensor product, 44 inner, 97 of algebras, 117 of Frobenius algebras, 126 of modules, 44 of projective representations, 149 Thompson, 183 Trace map, 137 Transitivity of induction, 87 Tsushima, 139, 177, 246, 361, 363, 366, 367, 371, 372, 418, 419, 499 Twisted group algebra, 144
