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6 The Loewy length of projective modules
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,
idempotent (ii) I f (iii) I f
(iv)
of
e
I
f
:
-
S
I
For any i d e a l
I = Re
then
R,
?=
I f o r a l l n 2 1.
R/I
f o r any i d e a l
In particular,
R.
R-+
J(FG)% = 0 .
J(FG).
is a l e f t i d e a l of
i s semisimple, then SO i s
R
such t h a t
be an a r t i n i a n r i n g .
R
i s semisimple and
R
n
is
V,L(V)
which i s f i x e d i n t h e following, we s e t
p
t (G) = L(FG) , t h e nilpotency index of Let
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
of
f o r some
R.
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 (i) A s s u m e t h a t
proof.
R
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)
1.3. LEMMA. (Schanuel's Lemma).
R-modules and let P ,P 1
and hence
as the fletter module of
determines the isomorphism class of
2
V.
As
we shall see below, V
R(V1.
Let R
be an artinian ring, let V,V ,W 1
be projective R-modules such that the sequences
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
2
defined by
2
be
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
P ( V ) 8 P'
and Kerf
- -
and P
1
0
2
Ri(V1
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'
be two projective covers of
be the corresponding exact sequences.
R1 ( V )
%
Then
'.
ti) Let P
Proof.
1
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
and define the i-th socle
(I4 =
s
0,
V
=
R
for all n 2 1
1 Put
(V1 = SOC(V1
Si(Vl of V by
s7,.( v1 /si-l(V1 If
-
V an R-module.
Let R be an artinian ring and let
s
n
sp
=
R then we put S.(R) z
= SOC ( V/SiF1
S.(V1
=
(i2
(V)1
and refer to Si(R1
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.
The F-linear extension of the map g of the group algebra FG
.
g-l,g E G, is an anti-automorphism
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
(iii) If a* :
W*
--+
c W
2 -
1
V and V*
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,
W are FG-modules and
.
1
then
0. E
Hom(V,W),
F
defined by
then the map
is an FG-homomorphism if and only if so is a. (iv) If CY E Hom(V,W)
FG
and
E Hom(W,X),
then
v+a
x-
0-
In what
B
El+
is an exact sequence of FG-modules, then so is
(BCY)* 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
Ig(Trf)l ( W )
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 .
.
If€
w;If(W2)
W*
V*/W
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*
and only if so is
If
CY.
1.8. COROLLARY.
Let
is an FG-homomorphism if
then
FG
= (gfl (CY(U1) = f(g-l(a(u1)
(cc*f) ( g - h
=
V,f E W*
Hom(V,W),
CY E
la*(gfllCul
€
=
Since
(W1/W2)* is isomorphic to the FGsubmodule
(iii) and (iv).
for all v
into w*.
= (lrfrf, cg-lW,
= (gf, ( W ) = f(g%
and in the isomorphism
morphisms.
V*
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)
Hom(W*,V*). Since, by Lemma 1.7, F the assertion follows. 1.9. LEMMA.
CY
E Hom(V,w)
FG
F
i f and only if
F-spaces)
onto 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
2
then V* = V
is a direct decomposition of
v,'
V; 8
V
into FG-submodules
v;
is irreducible (indecomposable) if and only if
.
(indecomposablel
V*
v1
is irreducible
PRELIMINARY RESULTS
(I) It i s a standard fact that n
Proof.
cn
~g(n(t,))l for all
t,
E V,f
=
is an F-isomorphism.
n ( u ) ( g - l f ) = (g-lf)( u ) = f ( g v )
-
cf)
= IV(~V)I
is an FG-isomorphism.
TI
1
1
1
2
V* = V @ V
(ii) The direct decomposition
Since
I
and g E G,
V*
381
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)*)
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
(iii) Since W
:
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
x
h
E
P,U)
By Lemma 1.10, it suffices to verify that 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
[si+l(FGI/si(FG)1 * = J(FGIi
(iii
Proof. Thus,
Put J
=
383
J(FG).
as FG-modules
'+'
/ ~ ( ~ ~ )
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. FG
I (lHIG.
1.15. LEMMA.
F
FG
It is obvious that dimV = dim Hom(V,F).
F
F
F
Now apply Theorem 2.4.9 for the case Let
P be
a
X
.
If H = 1, then = 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.
384
CHAPTER 6
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)
.
= dim (P/S’*lP)
F
F
Then
.
-
dim (P/JiP)
F
= dim Hom(P/JZ+lP,FG)
F FG
=
- dim
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, a s FG-modu l e s
(V/W,
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 Tensoring with
w-+ v 4 v/w-
0
w e o b t a i n an e x a c t sequence
P
0 But
4
w+
4 PQ
F
v+
P 8 F
i s p r o j e c t i v e , hence so i s
P 8 (V/W,+ F
0
Cv/w1
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.
P
1.18. U E M M A .
H be a subgroup of
Let
be an FG-module.
P8
G,
gt
8 (ui0
rn
Q
m,
.1
,1 4
1g j
t
G k, 1 G
:
gt
prove t h a t , given
{uil and
5
=
2 =
{u.} J
F
g t @ (ui @ V,.) a r e bases of
g,h,h E H,
gz Then w e have
G
SO
=
be F-bases of
g ,g2,...,gk f o r H
in
Cg,
(gt 8 Ui) @ gtVjl
G.
V, Then
G
(U 8 VH)
F 1 G t
.
k,
z 8 gt v j
Q 24.1
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
onto 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 -
u E U,V E V
that
ggt
U and
j G n, is an F-basis of
Q (Ui Q v j l -
forward t o v e r i f y t h a t f o r all
ggt
m, 1
(U @ VHIG
d e f i n e s an F-isomorphism of
Write
21...rVn
UG ‘8 y, choose t h e elements P n. Then t h e mapping
f
that
V ,V
as FG-modules
F
Choose a l e f t t r a n s v e r s a l
3 As an F-basis of
1G f
and
u , u p , . . .,u
Let
IJ
V
U be an FH-module, and l e t
let
= r/G Q v
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,
and
PG(V)
As usual, we w r i t e
V.
R(V)
and
L(V)
p > 0,
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.
denote t h e Loewy length and t h e H e l l e r module of
V,
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
Thus
V.
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
388
CHAPTER 6
and t h e r e f o r e
NOW
,(iiil .
apply Lemma 2.1(il Let
.
V be an FG-module and l e t
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.
Let
2 E Z(FG)
,
then
ann(3:) V
i s a submodule of
V be an FG-module such t h a t
V.
VN is a p r o j e c t i v e FN-module.
Then
Proof.
W e may harmlessly assume t h a t
FN @ f o r some FN-module
Thus
vi
I(N)V
E FN.
Then
N +v
= 0
.E
I"N1
... @ I(N)).
V
if
a s required.
V
E
I(N)V ,
2.4. LEMMA.
+
+ N Vi
u
X
u
V and w r i t e
E
for a l l
= 0
by Lemma 3 . 3 . 1 2 ( i i ) .
V = P(1 1 G
Let
Fix
i f and only i f
i f and only i f
2
@ FN = V @
i n which c a s e
X,
V n (I(N)@
=
..,
Hence
1)
=
v1
But
i.
E ann(N+) V
f
... + vn,
+ N vi
= 0
i f and only
B
and l e t
be t h e Loewy length of
n
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
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
Proof. 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 )
a contradiction.
But I(G)e ? J ( F G ) e
Let V be an FG-module on which N
2.5. LEMMA.
FGe, so
of
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)
and
N
Since
H = G/N.
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
.?
is p r o j e c t i v e as an FG-module, so by Lemma 2.2
P/I(N)P
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 Thus
P i s p r o j e c t i v e , hence so is PI and s o , by Lemma 2 . 3 , N*J(FG)n-l*P # 0.
3.1.9
But, by Lemma 2 . 4 ,
5 S(FG)"-l.
+
ann(N 1.
P
E J(FWrn-l while by Proposition
=
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
=
Hence
J(FGIm'n-2P and t h e r e f o r e
N'
I(h')P
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)
=
F
But Q/JQ
dim(JiQ/J'+lQ)
(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
FN
(by Lemma 2.9)
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 ) tive over FH trivial.
k - 1, where
k = maxRi.
and conjugation action o f
N
i
Note that P ( V )
is projec-
H
on each factor 1(iV)’/I(NIi+’
is
Hence, by Lemma 1.17, @
i
where X
+
(P,(V, 8 I w i / I ( N l i + l )
runs over the FH
(and hence FGI-composition factors of FN.
R = m a d (P,(V)
X
Since PH(V 8 F
XI
@(PH(V,8 X) X
2
F
is a summand of P,(V)
8XI
Thus
8 XI 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
Vi denote the FG-module
( I ( N h ( i d + 3Q
F
where G
for some subgroup
v
.
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 Therefore, for a l l
FN.
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
g l ,...,gk
a left transversal for N
Then it is immediate that 8' i s
algebras.
V.
Note that
in G.
d
FH
(#)N
and the map
.
)
Then
unique element of End(P) FG is an injective homomorphism End(p)
8 I-+ 8'
VG
4
FG be a left transversal for N in H and
Let g1,g2, ...,gS
Hence the map
VG
:
G onto End(V
)
FH
Proof.
to
H
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
the inertia group of an irreducible FN-module = (il J ( F G ) V = FG-J(FHI~IJI
(ii) L ( f i ) Proof. which case
J(E)~P
Y and let E
=
for all n
1
G, let H be
End(fi).
FG
Then
is equal to the nilpotency index of J(End(fl) I . We may take
fi
=
FGe.
V
FH
=
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
are G-conjugate, say f , = g fig.
and f i
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 Multiplying ( 3 ) on t h e l e f t by
= ( F G e ) (eJ(FG)e)
J(FG)
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 (using ( 4 ) with
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
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(#).
FH
for a l l
n 2 1
Then
(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
is equal to the nilpotency index of F ( H / N ) .
By Theorem 3.4.21,
F"LY/N)
End(#)
FH
for some
Hence the first assertion follows from Theorem 3.3(ii).
c1 E
Z*(H/N,F*).
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.
are
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$, it follows t h a t
r
2,
- -
a
If G-submodule of
W"
=
Let
2,
W"
-$W) E
G
and l e t
CinfBl ( z , y l = infB(x,y) = 1 f o r a l l
KinfB
N
so t h a t
3 . 7 . LEMMA.
z,g E N
I n what follows we w r i t e
identifiable.
W = 0.
and so
B E Z2(G/N,K*), l e t infB be t h e element of
Note t h a t
e(z
F i n a l l y , suppose t h a t
V = $(U)
+
( u - $ ( u ) ) EW +PIrr.
and t h e r e s u l t follows.
be a normal subgroup of
N
W" n
such t h a t
V
Then, by t h e above,
v
Thus
a
i s a K G-homomorphism.
$
Assume t h a t Accordingly,
401
T(N)
Let
N
K
be any f i e l d .
For any
Z2(G,K*) defined by
B(zN,yNl and hence
I(N)
and
KN
{;1I #] M E N}. G,
let
B E Z2(G/N,K*)
a = infa.
(il
KaG-I(N)
(ii) I f
is an i d e a l of
KaG
such t h a t
i s a normal p-subgroup of
N
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 ,
Thus f
(g E G )
we have
is a surjective homomorphism of F-algebras.
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
f o r some
c o n s t i t u t e a K-basis f o r
1,
By (i), KO"G.I(N)
rn 2 1.
I(N),
i s an i d e a l of
and t h e r e f o r e
= I(N)KO"G
Furthermore, t h e above e q u a l i t y shows t h a t t h e n i l -
KaG.I(N) c - J(K'G).
Thus
7, n #
403
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
be a normal p-subgroup of
N
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
(ii) I f
N
and l e t
p
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
Hence
d hP
= 1,
and so by (71,
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 ,
=
a s F-spaces
S 63 P G * I ( N )
(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 Then
g E G,
t h e r e e x t s t unique
F* by
h : G{ i l g E G}
,
homomorphism.
+
Y(x,y)
(ii)By ( i ) ,r e p l a c i n g
a = inf8.
that 3.9.
let
N
Let
$I
defined by E
f(z) 2
=
t(g)n(gl.
= h(g)g,g€G.
determines an F-algebra
=
we have
G,
Hence
Y
=
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
Throughout t h i s s e c t i o n ,
p.
-I
be a normal p'-subgroup
Loewy length of
istic
and
g
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.
-
y = a(6h)
-g
Now apply Lemma 3.1.
COROLLARY.
p > 0,
x,y
with
n(g) E N
FYG with x u = Y(x,g)xg and
B(xN,yN).
=
and
a ( t ( g ), n ( g ) ) , s e t
FB(G/N)
Thus, given
proving t h a t
=
i s an F-basis of
f : F aG
Hence t h e map
h(g)
tCg1 E T
405
denotes a f i n i t e group and
G
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.
(0 G
H
LEMMA.
< G n-1,O
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
xij G j G
m-l),
W
be FG-modules.
=
(J'V/Ji+'V)
where
Set
n
@ (jW/>+lh')
F
J = J(FG).
Then
=
L ( V ) ,m = L ( W )
and
CHAPTER 6
406
Proof.
Put
Y.. $3
=
8 $W.
V'J
Then
F
3 Y
-
'ij
.+Yi
i+l,j
,j+l
and
+
Y . ./(Yi+l 23
NOW
let
Y
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
where
X
'
=
c
y xij - i +@j = tYzj/(Yi+l,j + Yi,j+ll- xR/xk+l
i+j=R
=%
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
be a normal subgroup of
N
W be an FG-module and s e t
L(V) Proof.
Put
I(M)' = Si*FM,
M
=
where
=
= J(FG1,
WN)
G
FM
F
ni.
-
G
.
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
such t h a t
W 8 (lNl
t(G/N1 + L(W)
and view
G/N J
V
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
GROUPS OF P-LENGTH
2
407
a s required. 4.3.
G
Let
LEMMA.
be a p-nilpotent
f i e l d of c h a r a c t e r i s t i c grc p
(i) I f
P
P(m
1
fl
and
L(P(V))
Proof. and
wG
P(V)
That
X
Y.
3
XN s YN
4.4.
=
t(H/N)
THEOREM.
p > 0 and l e t
has a normal p-subgroup
Q = 0 ,(H),
and l e t
P
(il
(iil
P,(VI
zz
(iiil
If
did,
then
S
then
Let
x
t(H/N)
X such t h a t XN
W.
Then
W.
WG
P(X) 2 P(Y)
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
=
N n H
and
NH
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
in
V Q , where
H.
WC
L(PG(V))2
complement i n
(iv)
G.
of
H
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
Y
hence
F
=
There e x i s t s a unique i r r e d u c i b l e FG-module
(ii
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
V
F
N = 0 ,(GI
where
H
W
and l e t
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
G L ( P G ( V )1 = (n-d) (p-1) + L ( ( V H ) 1 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,
F
wc
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
=
1
>
UH
i s i r r e d u c i b l e and
U
Assume t h a t
S
Then
X
C H.
-
VH,
U such t h a t
Y
X
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
in
S
and p u t
and
H
= ( uH ) " w 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.
F
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
by Corollary 3 . 2 . 5 ,
t(!P/QI where
pk = IT n S I .
dimU F Thus
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
=
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
4.5. LEMMA.
Let n
FG
n(y)
FH
-+
X- z E
=
NH and N n H
= 1
Then Kern =FG.I(N)
be the natural projection.
J(FH) = n(J(FG)I .
and, by Lemma l.l(iii), we have FG*I(N)
for some y € JtFH1.
r € FG.I(N) + J ( F H ) .
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 )
G with G
Then
more, by Proposition 3.l.l(il, =
(
F
This completes the proof of the theorem.
is the identity map on FH
then n ( x ] = y
1
6l3 [I(N)i/I(N)i+ll i20
@
of G.
:
-
L ( ( V H )G
be a normal p-subgroup of
for some subgroup H
proof.
+
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
=
t(S)
+ t(N1
-
1
(iil
6l3 (I(N)i/I(N)i+l) is a completely reducible FG-module, where G acts by
(iii)
C xqs +Q
$20 conjugation.
Proof.
- E xq CFQ
E I(N)'"
(i) * lii):
algebraically closed.
for all i 2 0, all x
I(N)'
and all s E S.
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 )
E
+ 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
q
= 1~1-l
e
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 ) lates J(N)i/l(N)i+l
for all i 2 0 (s-1)(
x
.
Therefore
and all s
c
q)z =
- c
zqs
*Q
+Q
also annihilates
(S-1) 1 4
SfQ
annihi-
Since for any z E I(N)
S.
E
x
But
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’
= kx(C1l
,k
= 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
GROUPS OF P-LENGTH 2
411
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
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,
idempotent (ii) I f (iii) I f
(iv)
of
e
I
f
:
-
S
I
For any i d e a l
I = Re
then
R,
?=
I f o r a l l n 2 1.
R/I
f o r any i d e a l
In particular,
R.
R-+
J(FG)% = 0 .
J(FG).
is a l e f t i d e a l of
i s semisimple, then SO i s
R
such t h a t
be an a r t i n i a n r i n g .
R
i s semisimple and
R
n
is
V,L(V)
which i s f i x e d i n t h e following, we s e t
p
t (G) = L(FG) , t h e nilpotency index of Let
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
of
f o r some
R.
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 (i) A s s u m e t h a t
proof.
R
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)
1.3. LEMMA. (Schanuel's Lemma).
R-modules and let P ,P 1
and hence
as the fletter module of
determines the isomorphism class of
2
V.
As
we shall see below, V
R(V1.
Let R
be an artinian ring, let V,V ,W 1
be projective R-modules such that the sequences
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
2
defined by
2
be
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
P ( V ) 8 P'
and Kerf
- -
and P
1
0
2
Ri(V1
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'
be two projective covers of
be the corresponding exact sequences.
R1 ( V )
%
Then
'.
ti) Let P
Proof.
1
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
and define the i-th socle
(I4 =
s
0,
V
=
R
for all n 2 1
1 Put
(V1 = SOC(V1
Si(Vl of V by
s7,.( v1 /si-l(V1 If
-
V an R-module.
Let R be an artinian ring and let
s
n
sp
=
R then we put S.(R) z
= SOC ( V/SiF1
S.(V1
=
(i2
(V)1
and refer to Si(R1
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.
The F-linear extension of the map g of the group algebra FG
.
g-l,g E G, is an anti-automorphism
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
(iii) If a* :
W*
--+
c W
2 -
1
V and V*
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,
W are FG-modules and
.
1
then
0. E
Hom(V,W),
F
defined by
then the map
is an FG-homomorphism if and only if so is a. (iv) If CY E Hom(V,W)
FG
and
E Hom(W,X),
then
v+a
x-
0-
In what
B
El+
is an exact sequence of FG-modules, then so is
(BCY)* 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
Ig(Trf)l ( W )
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 .
.
If€
w;If(W2)
W*
V*/W
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*
and only if so is
If
CY.
1.8. COROLLARY.
Let
is an FG-homomorphism if
then
FG
= (gfl (CY(U1) = f(g-l(a(u1)
(cc*f) ( g - h
=
V,f E W*
Hom(V,W),
CY E
la*(gfllCul
€
=
Since
(W1/W2)* is isomorphic to the FGsubmodule
(iii) and (iv).
for all v
into w*.
= (lrfrf, cg-lW,
= (gf, ( W ) = f(g%
and in the isomorphism
morphisms.
V*
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)
Hom(W*,V*). Since, by Lemma 1.7, F the assertion follows. 1.9. LEMMA.
CY
E Hom(V,w)
FG
F
i f and only if
F-spaces)
onto 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
2
then V* = V
is a direct decomposition of
v,'
V; 8
V
into FG-submodules
v;
is irreducible (indecomposable) if and only if
.
(indecomposablel
V*
v1
is irreducible
PRELIMINARY RESULTS
(I) It i s a standard fact that n
Proof.
cn
~g(n(t,))l for all
t,
E V,f
=
is an F-isomorphism.
n ( u ) ( g - l f ) = (g-lf)( u ) = f ( g v )
-
cf)
= IV(~V)I
is an FG-isomorphism.
TI
1
1
1
2
V* = V @ V
(ii) The direct decomposition
Since
I
and g E G,
V*
381
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)*)
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
(iii) Since W
:
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
x
h
E
P,U)
By Lemma 1.10, it suffices to verify that 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
[si+l(FGI/si(FG)1 * = J(FGIi
(iii
Proof. Thus,
Put J
=
383
J(FG).
as FG-modules
'+'
/ ~ ( ~ ~ )
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. FG
I (lHIG.
1.15. LEMMA.
F
FG
It is obvious that dimV = dim Hom(V,F).
F
F
F
Now apply Theorem 2.4.9 for the case Let
P be
a
X
.
If H = 1, then = 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.
384
CHAPTER 6
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)
.
= dim (P/S’*lP)
F
F
Then
.
-
dim (P/JiP)
F
= dim Hom(P/JZ+lP,FG)
F FG
=
- dim
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, a s FG-modu l e s
(V/W,
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 Tensoring with
w-+ v 4 v/w-
0
w e o b t a i n an e x a c t sequence
P
0 But
4
w+
4 PQ
F
v+
P 8 F
i s p r o j e c t i v e , hence so i s
P 8 (V/W,+ F
0
Cv/w1
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.
P
1.18. U E M M A .
H be a subgroup of
Let
be an FG-module.
P8
G,
gt
8 (ui0
rn
Q
m,
.1
,1 4
1g j
t
G k, 1 G
:
gt
prove t h a t , given
{uil and
5
=
2 =
{u.} J
F
g t @ (ui @ V,.) a r e bases of
g,h,h E H,
gz Then w e have
G
SO
=
be F-bases of
g ,g2,...,gk f o r H
in
Cg,
(gt 8 Ui) @ gtVjl
G.
V, Then
G
(U 8 VH)
F 1 G t
.
k,
z 8 gt v j
Q 24.1
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
onto 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 -
u E U,V E V
that
ggt
U and
j G n, is an F-basis of
Q (Ui Q v j l -
forward t o v e r i f y t h a t f o r all
ggt
m, 1
(U @ VHIG
d e f i n e s an F-isomorphism of
Write
21...rVn
UG ‘8 y, choose t h e elements P n. Then t h e mapping
f
that
V ,V
as FG-modules
F
Choose a l e f t t r a n s v e r s a l
3 As an F-basis of
1G f
and
u , u p , . . .,u
Let
IJ
V
U be an FH-module, and l e t
let
= r/G Q v
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,
and
PG(V)
As usual, we w r i t e
V.
R(V)
and
L(V)
p > 0,
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.
denote t h e Loewy length and t h e H e l l e r module of
V,
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
Thus
V.
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
388
CHAPTER 6
and t h e r e f o r e
NOW
,(iiil .
apply Lemma 2.1(il Let
.
V be an FG-module and l e t
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.
Let
2 E Z(FG)
,
then
ann(3:) V
i s a submodule of
V be an FG-module such t h a t
V.
VN is a p r o j e c t i v e FN-module.
Then
Proof.
W e may harmlessly assume t h a t
FN @ f o r some FN-module
Thus
vi
I(N)V
E FN.
Then
N +v
= 0
.E
I"N1
... @ I(N)).
V
if
a s required.
V
E
I(N)V ,
2.4. LEMMA.
+
+ N Vi
u
X
u
V and w r i t e
E
for a l l
= 0
by Lemma 3 . 3 . 1 2 ( i i ) .
V = P(1 1 G
Let
Fix
i f and only i f
i f and only i f
2
@ FN = V @
i n which c a s e
X,
V n (I(N)@
=
..,
Hence
1)
=
v1
But
i.
E ann(N+) V
f
... + vn,
+ N vi
= 0
i f and only
B
and l e t
be t h e Loewy length of
n
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
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
Proof. 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 )
a contradiction.
But I(G)e ? J ( F G ) e
Let V be an FG-module on which N
2.5. LEMMA.
FGe, so
of
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)
and
N
Since
H = G/N.
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
.?
is p r o j e c t i v e as an FG-module, so by Lemma 2.2
P/I(N)P
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 Thus
P i s p r o j e c t i v e , hence so is PI and s o , by Lemma 2 . 3 , N*J(FG)n-l*P # 0.
3.1.9
But, by Lemma 2 . 4 ,
5 S(FG)"-l.
+
ann(N 1.
P
E J(FWrn-l while by Proposition
=
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
=
Hence
J(FGIm'n-2P and t h e r e f o r e
N'
I(h')P
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)
=
F
But Q/JQ
dim(JiQ/J'+lQ)
(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
FN
(by Lemma 2.9)
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 ) tive over FH trivial.
k - 1, where
k = maxRi.
and conjugation action o f
N
i
Note that P ( V )
is projec-
H
on each factor 1(iV)’/I(NIi+’
is
Hence, by Lemma 1.17, @
i
where X
+
(P,(V, 8 I w i / I ( N l i + l )
runs over the FH
(and hence FGI-composition factors of FN.
R = m a d (P,(V)
X
Since PH(V 8 F
XI
@(PH(V,8 X) X
2
F
is a summand of P,(V)
8XI
Thus
8 XI 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
Vi denote the FG-module
( I ( N h ( i d + 3Q
F
where G
for some subgroup
v
.
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 Therefore, for a l l
FN.
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
g l ,...,gk
a left transversal for N
Then it is immediate that 8' i s
algebras.
V.
Note that
in G.
d
FH
(#)N
and the map
.
)
Then
unique element of End(P) FG is an injective homomorphism End(p)
8 I-+ 8'
VG
4
FG be a left transversal for N in H and
Let g1,g2, ...,gS
Hence the map
VG
:
G onto End(V
)
FH
Proof.
to
H
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
the inertia group of an irreducible FN-module = (il J ( F G ) V = FG-J(FHI~IJI
(ii) L ( f i ) Proof. which case
J(E)~P
Y and let E
=
for all n
1
G, let H be
End(fi).
FG
Then
is equal to the nilpotency index of J(End(fl) I . We may take
fi
=
FGe.
V
FH
=
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
are G-conjugate, say f , = g fig.
and f i
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 Multiplying ( 3 ) on t h e l e f t by
= ( F G e ) (eJ(FG)e)
J(FG)
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 (using ( 4 ) with
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
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(#).
FH
for a l l
n 2 1
Then
(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
is equal to the nilpotency index of F ( H / N ) .
By Theorem 3.4.21,
F"LY/N)
End(#)
FH
for some
Hence the first assertion follows from Theorem 3.3(ii).
c1 E
Z*(H/N,F*).
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.
are
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$, it follows t h a t
r
2,
- -
a
If G-submodule of
W"
=
Let
2,
W"
-$W) E
G
and l e t
CinfBl ( z , y l = infB(x,y) = 1 f o r a l l
KinfB
N
so t h a t
3 . 7 . LEMMA.
z,g E N
I n what follows we w r i t e
identifiable.
W = 0.
and so
B E Z2(G/N,K*), l e t infB be t h e element of
Note t h a t
e(z
F i n a l l y , suppose t h a t
V = $(U)
+
( u - $ ( u ) ) EW +PIrr.
and t h e r e s u l t follows.
be a normal subgroup of
N
W" n
such t h a t
V
Then, by t h e above,
v
Thus
a
i s a K G-homomorphism.
$
Assume t h a t Accordingly,
401
T(N)
Let
N
K
be any f i e l d .
For any
Z2(G,K*) defined by
B(zN,yNl and hence
I(N)
and
KN
{;1I #] M E N}. G,
let
B E Z2(G/N,K*)
a = infa.
(il
KaG-I(N)
(ii) I f
is an i d e a l of
KaG
such t h a t
i s a normal p-subgroup of
N
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 ,
Thus f
(g E G )
we have
is a surjective homomorphism of F-algebras.
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
f o r some
c o n s t i t u t e a K-basis f o r
1,
By (i), KO"G.I(N)
rn 2 1.
I(N),
i s an i d e a l of
and t h e r e f o r e
= I(N)KO"G
Furthermore, t h e above e q u a l i t y shows t h a t t h e n i l -
KaG.I(N) c - J(K'G).
Thus
7, n #
403
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
be a normal p-subgroup of
N
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
(ii) I f
N
and l e t
p
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
Hence
d hP
= 1,
and so by (71,
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 ,
=
a s F-spaces
S 63 P G * I ( N )
(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 Then
g E G,
t h e r e e x t s t unique
F* by
h : G{ i l g E G}
,
homomorphism.
+
Y(x,y)
(ii)By ( i ) ,r e p l a c i n g
a = inf8.
that 3.9.
let
N
Let
$I
defined by E
f(z) 2
=
t(g)n(gl.
= h(g)g,g€G.
determines an F-algebra
=
we have
G,
Hence
Y
=
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
Throughout t h i s s e c t i o n ,
p.
-I
be a normal p'-subgroup
Loewy length of
istic
and
g
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.
-
y = a(6h)
-g
Now apply Lemma 3.1.
COROLLARY.
p > 0,
x,y
with
n(g) E N
FYG with x u = Y(x,g)xg and
B(xN,yN).
=
and
a ( t ( g ), n ( g ) ) , s e t
FB(G/N)
Thus, given
proving t h a t
=
i s an F-basis of
f : F aG
Hence t h e map
h(g)
tCg1 E T
405
denotes a f i n i t e group and
G
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.
(0 G
H
LEMMA.
< G n-1,O
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
xij G j G
m-l),
W
be FG-modules.
=
(J'V/Ji+'V)
where
Set
n
@ (jW/>+lh')
F
J = J(FG).
Then
=
L ( V ) ,m = L ( W )
and
CHAPTER 6
406
Proof.
Put
Y.. $3
=
8 $W.
V'J
Then
F
3 Y
-
'ij
.+Yi
i+l,j
,j+l
and
+
Y . ./(Yi+l 23
NOW
let
Y
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
where
X
'
=
c
y xij - i +@j = tYzj/(Yi+l,j + Yi,j+ll- xR/xk+l
i+j=R
=%
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
be a normal subgroup of
N
W be an FG-module and s e t
L(V) Proof.
Put
I(M)' = Si*FM,
M
=
where
=
= J(FG1,
WN)
G
FM
F
ni.
-
G
.
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
such t h a t
W 8 (lNl
t(G/N1 + L(W)
and view
G/N J
V
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
GROUPS OF P-LENGTH
2
407
a s required. 4.3.
G
Let
LEMMA.
be a p-nilpotent
f i e l d of c h a r a c t e r i s t i c grc p
(i) I f
P
P(m
1
fl
and
L(P(V))
Proof. and
wG
P(V)
That
X
Y.
3
XN s YN
4.4.
=
t(H/N)
THEOREM.
p > 0 and l e t
has a normal p-subgroup
Q = 0 ,(H),
and l e t
P
(il
(iil
P,(VI
zz
(iiil
If
did,
then
S
then
Let
x
t(H/N)
X such t h a t XN
W.
Then
W.
WG
P(X) 2 P(Y)
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
=
N n H
and
NH
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
in
V Q , where
H.
WC
L(PG(V))2
complement i n
(iv)
G.
of
H
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
Y
hence
F
=
There e x i s t s a unique i r r e d u c i b l e FG-module
(ii
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
V
F
N = 0 ,(GI
where
H
W
and l e t
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
G L ( P G ( V )1 = (n-d) (p-1) + L ( ( V H ) 1 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,
F
wc
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
=
1
>
UH
i s i r r e d u c i b l e and
U
Assume t h a t
S
Then
X
C H.
-
VH,
U such t h a t
Y
X
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
in
S
and p u t
and
H
= ( uH ) " w 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.
F
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
by Corollary 3 . 2 . 5 ,
t(!P/QI where
pk = IT n S I .
dimU F Thus
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
=
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
4.5. LEMMA.
Let n
FG
n(y)
FH
-+
X- z E
=
NH and N n H
= 1
Then Kern =FG.I(N)
be the natural projection.
J(FH) = n(J(FG)I .
and, by Lemma l.l(iii), we have FG*I(N)
for some y € JtFH1.
r € FG.I(N) + J ( F H ) .
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 )
G with G
Then
more, by Proposition 3.l.l(il, =
(
F
This completes the proof of the theorem.
is the identity map on FH
then n ( x ] = y
1
6l3 [I(N)i/I(N)i+ll i20
@
of G.
:
-
L ( ( V H )G
be a normal p-subgroup of
for some subgroup H
proof.
+
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
=
t(S)
+ t(N1
-
1
(iil
6l3 (I(N)i/I(N)i+l) is a completely reducible FG-module, where G acts by
(iii)
C xqs +Q
$20 conjugation.
Proof.
- E xq CFQ
E I(N)'"
(i) * lii):
algebraically closed.
for all i 2 0, all x
I(N)'
and all s E S.
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 )
E
+ 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
q
= 1~1-l
e
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 ) lates J(N)i/l(N)i+l
for all i 2 0 (s-1)(
x
.
Therefore
and all s
c
q)z =
- c
zqs
*Q
+Q
also annihilates
(S-1) 1 4
SfQ
annihi-
Since for any z E I(N)
S.
E
x
But
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’
= kx(C1l
,k
= 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
GROUPS OF P-LENGTH 2
411
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