THE MAIN LEl1t-'i.A

NAG(P)

VI. THE MAIN LEl1t-'i.A

In

this

section we prove

the

major res ult of this

thesis.

It

is the lemma which makes everything work. The familiar technique of reducing

a

minimal counter example is used. The pattern was set in E. Shult's work (8). The object is to reduce the minimal counterexample so that the character lemmas of sections Ill and IV may be applied.

The result is carried out for a general class two odd p group. Only section IV prevents us from getting the strong result.

In the beginning reduction steps are fairly complete.

As arguments are repeated often they become shorter with use.

Suppose that A is a group and CIA!, r)

= l

for

a

primer. Assume that pis a prime and (!Al, p)

=

l, p1r.

Suppose that AR is a group with extra special normal r subgroup R.

We

assume that A is irreducible on R/D(R) and trivial on D(R) with A

0

= ^CA(R). Suppose that

Q

is GF(p) or the rational field and Sis a primitive !AR.Ith root of unity and k

=

Q(S). Assume that V is an irreducible

k[A..~)

module which is nontrivial on D(R) and Cy(A

0 )

= v. ^W

^e

^also

suppose that for any n I exp A, n

~

re

+ 1

for any

c if r

2c+l I IRI.

(*) A

₁

is called a

(*)

group if VIA contai ns the

trivial A module for any R, V, and ^A~ A

1 satisfying the above conditions.

By (III. 11) abelian groups are

(*)

groups. By (IV. 11) an odd p group P is a

(*)

group if all of its irreducible representations are cyclic or the central

product of an extra special p group with a cycli c group of exponent p or p • 2 That is, all irreducible r epresentations

are subgroups of extra special groups. Note that subgroups and factor groups of

(*)

groups are

(*)

groups. In

particular, all class two groups of exponent p are

(*)

groups. We do not know if all those of exponent p 2 are

(*)

groups since there is a class two group P of order p6 of exponent p2

where P/Z(P) is of rank two and exponent p2 and P has a faithful irreducible character. This group is not covered in section IV.

The lemma is stated on the next page.

(VI. 1) Theorem: Suppose ~ A ~ ~ p group -2.f class

_s

2 for

£££

p. Suppose A l2, ~ (*) group. Assu."!le ~ AG

~!!solvable group with normal subgroup

G

^where

(!Al, !GI)

= 1. Suppose~

!GI =

^qmq₀

f2E.

~prime q

f

p (m ~ 0)

~ (q,q

0)

=

^{l. Assume k = Q(S) where Q =} GF(q) .2E the rational field~

S 1.2.

^!! primitive 1Alq₀ ~ of unitv. Sunoose V

l l

~ k[AG] module faithful 2£ G. Assu ... me ~

~

i) V ~ ~ ~ of eauivalent irreducible k[AG] rrodules ii)

! !

exp A

=

^pa ~ ^pb ~ re+ l

f£E.

^{l ,S b < a}

~any primer~~

r 2

c+ll IGI.

1) Cv(A)

1-

(0) 2E 2) Cv(A')

=

^(O) 2!:

3) Cv(A')

1-

(0) implies there

l l

cycl ic D

-

< A~

a) CV(A'D)

=

⁽⁰⁾ ~ b)

CG(A'D)

>

CG(A').

The proof begins on the next page.

We

assume that

(VI. 1)

is false

and

choose a

counterexample (A, G, V) minimizing IAI + !GI +dim V. So we have the following:

l ' ) Cv(A) = (0) and 2') Cv(A') :f: (0) and 3 I) for any cyclic D ,!S A

a•) Cv(A¹D) =f: (O) or

b') CG(A¹D)

'f:.

CG(A^{1) .}

(VI.

2)

V

!§.. ~ irreducible l<[AG] module.

Here V = v1

+ •..

^{i- Vt. is a sum} of equivalent irreducible k[AG] modules. Hence, (A,G,V

1) is a counter- example if and only if (A,G,V) is also. So t = l .

(VI.

3) VIG is a multiple of a single irreducible A G

i l \ . b - - - - - - 0

module

i..2.£

^{every A}₀ ^{l::i. A.}

1!!

particular, VIG

i&

homogenev~.

Suppose not. By (V. 6) there is A

0 _!S A

1 ^l::i. A of pri~

index p so that

•••

where the Ui are irreducible A

1G modules and V NAG

u

1 1AG.

Let Gi

=

^{ker[G - > Aut Ui), Qi= G/Gi.}

Clearly (A

1

,Q

₁

,u

₁^{) satisfies the} hypotheses of

(VI. l).

Hence,

(VI. 1)

holds in this

case

by induction.

Now

vlA ~ u

₁1AGIA

~A u

₁ 1A

IA.

l

l) So by (V. 2) Cu (Al) = (O) if and only if .Cy(A) = (O)~

l

2)

Also

by (V. 2)

we have,

< o) :f cu (

A

1 f\ A• ) =

c

(A• )

l u1

since A1

.2:

A'

;S

Cu (Al').

l Hence we find

3) there is D ;:: A

1

cyclic so that a") CU (A1¹D)

= ^(O)

^and

l

b")

CG (A1

¹

D)

^>

CG (Alt)•

-1 -1

Using the fact that A1 ^~A'

.2:

A1^{1, from} we get

al)

CU (A'D)

<

cu

^(A₁ ^{• D) = (}

o).

1 1

And CG - l

(D) .2: CG

- l

(A1

¹

D) -

^>

CG

-1 (Al') . ^->

C,..,

-1 ^1..:1 (A') bl)

CG (A'D)

> CG (A').

- l - l

au)

so

Choose l = -nl,

. ^.. . ^,

1\p as cos et representatives of Al ⁱⁿ

A.

We may arrange the ^j{'. so that u. = _it"i ul •

it7"

^l.

Further, x _{e: Al} acts upon ui as x ^l. = "Ti. -lxl\". acts upon

l. l.

and upon G. as x 1t· ¹ acts upon

Q

₁ with the isomorphism

-1 -~:l

1\'i l.

~> y G

1

^of

Q ₁

onto

Q _{1 ..}

Since

A' ;:: A

1

^and

A'

A

A, A'D

^A A we get

ai) CU. (A¹D)

l.

= (0) and

bi) CG.

- i

(A¹D) > CG. (A¹) . - i

So finally

a) CV(A¹D) =

(O)

and

b)

CG(A'D)

> CG(A') ^by (V.

7).

Therefore, VIA G is homogeneous.

0

(VI. 4) ~every A

0 <A~~ Cy(A₀)

:f

(O).

Suppose A

0

<A

and Cv(A

0)

= ^(O).

^{Hence we}

^may

^choose

A0 ~ A

1 A

A and A1 ^{< A since A is} nilpotent, and Cv(A1)

= ^(O).

Clearly A1' ~A'. So Cv(Ao') ~ Cv(A¹⁾ ~

(0).

So by (VI. 3), VIA G is homogeneous. Hence, using inGuction,

1

we may apply (VI. 1) to (A₁, G, V). From the foregoing, i t is clear that we have

3) a I) Cv(A1¹D) = (O) and b') CG(A1¹D) > _CG(A1') for a cyclic D ~ A1 • So

a) Cv(A'D) ~ cv<A

1•n)::: (O) CG(D) > _CG(A1¹D) ~ CG(A1^{1) ~} CG(A¹⁾ or

b) CG(A'D) ~ CG(A').

Hence the conclusion.

(VI. 5) A!§.

faithful .Q.!!

V.

B.l'ld

Suppose not. Let A

0

=

^{ker [A ~> Aut V] • Since G}

is faithful and V is an irreducible AG module we must have [A0 ,

G]

⁼ l. Hence (VI. 1) applies to (A/A

0 , G, V).

Let ~

=

^A/A₀ ^and

^t:.

₁

=

^A1A₀ ^/A₀ ^{• Then} _A1

=

^{A1 so clearly}

CV(~)

=

^{Cv(A) = (O) and CV(~•) = Cv(A') ~ (0). so there is}

Q ~ ~ cyclic with

a")

cv<t:.'

Q) =

(O)

^and

b") CG(t:,' Q) > CG(a•) • Let D

-

< A be cyclic such that DA

0/A

0 = Q. Then since

A'Ao/Ao

=

/!',

a) Cv(A

¹

D) = ^(O).

And CG(D)

~

CG(A'D)

~

CG(A') so b) CG(A

¹

D)

~

CG(A

¹) .

Therefore A is faithful on v.

Choose M

<

G as a maximal AG invariant subgroup of G.

The group G/M is an irreducible A module, where the action, for x

e

A and

~M e

G/M, is

-l (xlrx- 1 )M.

x(nM)

=

nx M =

From each A orbit on G/M choose a representative

^tt

1M. ^So that ll'lM,

•••

, 7tmM form a complete set of A orbit

representatives. By (V. 8) we may choose iri, i

=

l, . . . ^,

"Jr. -l = 'tt -l -

=

A()A

^l.

Af\(AM)

i -

Ai.

By

choosing A conjugates of

1\

1

=

l, ••• , -rrm we get a

complete set of coset representatives of M in G; rr ₁

=

l,

• • • ' ltm' • • • ,

l\"':l

AA

(AM)

J =

Aj'

j

= ^l,

^•••

, e. Further, A permutes the

l \ j

if we specify for x s A that,

x(~jM) = lfj(x)M.

Now v I G is homogeneous. Therefore, v IM= vi+ ••• + ^vf

with homogeneous components Vi. Further, G is transitive on the Vi's and M fixes each one. That is,

f

divides

IG/MI.

(VI. 6)

.ll

^f

1

¹

!llim

f = e = IG/MI ~~Vi may ~

numbered~ that A fixes V₁, itiVl = Vi, ~A permutes

!h2

Vi exactly

.!!. ll.

permutes ~ n₁•

Consider the permutation representation

f>

of AG on the Vi's • Now M is in the kernel of

f>.

Further G f'\ ker r/, is a proper AG invariant subgroup of G containing M, so i t must be M. Since G/M is abelian, Gf\ker

f>

is the

subgroup of G fixing every v

1• And now f = e = IG/MI.

But ~ is a transitive representation of A(G/M) given on the cosets of some subgroup B of order IA(G/M)j/e =

IA!. So Band A are Hall IAI subgroups of A(G/M). Hence they are conjugate in A(G/M). In other words, the

representation is given on the cosets of A. Therefore A fixes, say, v

1• Setting Vi= n₁v₁, for x e A we get -1

xvi=

x(~ivl)

= (xlri.x-l)vl

=~ix

^{vl =}

~i(x)vl

^{= vi(x)•}

So this step is complete.

(VI.

7) l !

f ~ l

!llim !2£.

^~ (A,AM) coset representa- tives

1'ri =

^l, ••• , 1( _{m - -}we have

vJA "VA

Li~lti>

^{vllA IA,}

i

~

^V

~AG

Vl(AM)jAG.

Since AM stabilizes vl and lstab(AG,Vl)I = IAGl/e

=

IAMI

we have AM= Stab(AG,V

1). Now M ~AG so V ~AG

Vl(AM)jAG.

By the Mackey Decomposition we get

vlA ::YA Vl(AM) IAGIA Cl.A ci:l$ 1\iVll 1t'-l IA

. (AM) 1f\ A

-1

since

(AM)...,.i

n A = ^{CA (1\'i)} = ^Ai.

Remark: If v

1

1A. contains the trivial Aj module then

J

contains the trivial A module by (V. 2). So

(0)

implies that Cv (A.)

= (0)

for each

j =

l,

l J

, m.

(Hence also for j

=

1, ••• , e.) Let

~ =

ker[A

--->

Aut G/M] •

(VI.

8)

l l v

₁

1Ar1

^{~ !l£.!}

^contain

^~

^trivial Ar1

submodule

~ ^f =

1. (i.e.

VIM~

homogeneous.)

• • •

Suppose

v

1

^1~

does not contain the

trivial~

submodule. Now ~ ~ AG since [~,G] ,::: ^{M and} ~ ~ A. So v 1 1Ax1'

1

is a homogeneous A.M1- module by (VI. 3) and (VI. l) applies to

(A.i~,M/M

1

^,v

1

⁾

^where

M1 = kar[M - >

Aut v

₁) . by

induct ion.

By assumption Cy

^{(~) =}

^(O).

l

Next

~

,::: Aj for every j. So

~·

,::: Aj ()A'

for

every

j.

If

Cv

1

^{(~') =}

^{(O) then Cy}

¹

^(Ajf\A

^{1) =}

(0) for every j.

Hence

by

(V. 2)

Cv I IA (A')

= (O) for every j. Hence l Aj

So we must have

Cv

^{(~1) ~}

(O).

l

This means that when we apply

(VI. 1)

^to

(~,M/M

1

^,V

1

⁾

we have

3) a")

Cv

(~¹0) =

(O)

l

and

b") CM/M

(~'O) ~

CM/M

(~').

1 1

Set Mi = ker[M - > Aut

Vi), !:li

= M/Mi . Now

~'D ~~so ~'D is centralized

by

every 1'1 • Hence conjugation of

~'D by ni

¹ fixes '\r'D elementwise.

Therefore,

So

by

(V. 8)

That is,

And

Again, since every 1ri. centralizes ~·

Cv(~'D) =

(O).

That is,

a)

Cv(A

¹

0)

~ CV(~¹0)

= ^(O).

Hence f = 1.

(VI. 9)

.Y. _A/P;v1

1&, abelian ~ f = l.

If f ~

l

then

A

is cyclic and irreducible on

G/M.

If

~i ~

n 1

⁼ l then the orbit [n~ I x e

A}

is faithful on

A/~. Hence i t is the regular representation of A/~.

That is,

Ai =

^~· By the remark preceding (VI.

8)

and (VI. 8) we are done.

(VI. 10)

1f

A/~ =

A

^~ nonabelian ~ f

=

^l.

Now

G/M

is a.n r group for a prime r. But ~ is a class two p group which is faithful and irreducible on the

GF(r)

module

G/M.

So we apply

(II.

7) to get a 7\~M which

.L

is fixed by no element of

A#·

In other words,

C~(n 1 ⁾

^{= 1}

or CA(TTi) =Ai= ~· So again by (VI. 8) and the remark preceding it we get f

=

^l.

Under the hypothesis of

(VI. l)

this means

VIM

is homogeneous and f = l.

Now

G/M

is an r section.

So

by

(V. 9)

we may choose an r Sylow subgroup R

0 of G fixed by A. Next let R be chosen in R

0 minimal such that i) R is A invariant, and ii) RM = G.

We eventually show that R is extra special.

Next consider VIAM =

v

₁

+ ...

+Vt where the Vi are homogeneous components. Since VIM is homogeneous, each

Vi is faithful and is a multiple of a single fixed

irreducible

M

module. Since Cv(A') ~

(O)

^we may choose v1 so that Cy (A¹) ~ (0). Clearly Cy(A) = (0) implies

l

Cy.(A) = (O), i = l, ••• , t. So we apply (Vl. l) to

l.

(A, M, v

1) and obtain D $ A cyclic so that a") Cy (A¹

D)

=

(O)

and

l

b") ~(A'D) ,2:: CM(A').

(VI •

11)

l l

A 12, abelian ~ CA (M) = A*

rf

l .

In this case, A¹ = l so ~(A¹D) = CM(D) ,2:: CM(A') = M.

Hence D $ CA(M). But Cv (A'D)

=

^{Cv (D) = (O) so}

l l

1

~

D

$A*.

(VI. 12)

CA(M) =A*~

1.

We may assume that A is nonabelian. Let U be a homogeneous component of v

11A'DM• Since VIM is

homogeneous,

U

is faithful on M. Now (A'D)'

= ¹

^{since A}

is class two, A' $ Z(A), and D is cyclic.

= (0), Cu(A'D) = (O). Further, cu[<A'D) •]

Since Cy (A¹D) l

=

u.

^{So in}

applying

(VI.

l) to (A'D, M,

U)

we get

3)

and D

1

^{$ A'D}

cyclic so that

Also since

a*)

c ₀

((A¹D)¹D

1 ]

= Cu(D

1 ) =

^(0),

we have D1 ^~ l. Hence

n

₁ ^{$ CA(M) = A*.}

(VI.

13) A*() _.°'M ^A

=

1 and CG(A·lc) = _- M.

Suppose A*

n '\I

^{= A}₀ ^{:f l . Now A}₀ ^{A A so we may take}

A1 = Z(A)nA

0 ~ 1 since A is nilpotent. We know that A*

centralizes Mand~ centralizes G/M. Hence by (V. 7) A 1 centralizes A and G. So A1

s

^{Z(AG). But} Vis irreduci al e so A1 is cyclic and acts as scalar multiplication on V by

(VI. 5).

Hence CV(A

1 )

⁼

(O).

By

(VI. 4)

Al= A. But then A is cyclic and

a) Cv(A¹A)

=

^{Cv(A) =}

(O)

^and

b)

CG(A)

= ^G

^~

^CG(A

¹⁾

= ^G.

Hence A*r\ ~ = 1. But then A*~/~ A A/~

1

^so

A*Ay/~n Z(A/A~) :/: 1 and CG/M(Aic) = l.

Hence CG(A*) = M.

(VI. 14)

~Em choose

R

~~RS CG(M),

R

is extra special, and RA AG. Further, D(R) SM, D(R)

S

CG(AG).

Now G = NG(M). But M = CG(A*) so by (V. 10) G =

CG(M)CG(A*) = CG(M)M. Now CG(M) is A invariant so R may be chosen so that RS CG(M).

Let R1

=

^{Z(R). Now} _R1

S

^{CG(M) so} _R1 ^{~ Z(G),} since Ri.~ = G. Further VIG is homogeneous and faithful so R1 is cyclic and acts as scalar multiplication on

v.

In

particular, because AG is faithful, R1

s

^{Z(AG). So R}₁ ^<

Mand R1

S

CG(AG).

By the minimal choice of R we must have M\\R = D(R)

as the unique maximal A invariant normal subgroup of R.

Let R

0 be any characteristic abelian subgroup of R. Now R/D(R) _~A G/M so if Ro <R then Ro ^~ D(R). But C (A*)

R ^~

D(R) and so Z(R) = Rl < R. Hence Ro ^~

D(R).

^{But then} Ro ^{~ M.} We already know that R <

0 - CG(M) ('\ M

=

^{Z(M) and VIM}

is homogeneous. So R

0 ~ Z(R) = R

1 and R

0 is cyclic. So R is the central product of a cyclic r group and an extra special r group. By the minimality of R, this means R is extra special.

Finally, R ~ CG(M) normalizes itself and is normalized by A. Hence R b AG.

(VI.

15) VIR ~homogeneous; CV(~) ~ (O).

Here VIG is homogeneous. So, since RA G, v!R

is completely reducible and the homogeneous components are permuted transitively by M since MR = G. But M centralizes R so vjR is homogeneous.

Suppose next that cv<Ar1) ~ (O). Now~ centralizes Further, ~ A A.

G/M "'A R/D(R), so i t centralizes

R.

iL

Hence _CV(~) is a k[AR] submodule of v. Let v < Cv(AM)

0 - l

be an irreducible

k[AR]

submodule. Since Z(R)

=

^D(R)

~ Z(AG) i t acts as scalar multiplication (nontrivially) on V hence also on V

0 • Further, on V

0 , A is represented as A/An· Now ~< A since Cv(A) = (O). Therefore V₀ is a k((A/~)PJ irreducible module. Also A/~ is faithful and

irreducible on

R/D(R).

Now

!RI

= r

2

c+l

I !GI.

Further, by hypothesis, pb

I

re + 1 for any e

~

c and any pb

~

exp A. Hence we may apply the fact that A is a (*) group to find that (O) ~ Cv (A)~ Cv(A). But Cy(A) = (O).

0

Hence Cv(i\i) = (O).

(VI. 16) (VI. 1) holds.

By (VI. 12) A*~ 1. And by (VI. 13) A*(\Ar~ = l.

Hence

Ai

_{1 <A.} So by (VI. 4) CV(~) ~ (O). This contradicts (VI. 15). Therefore (VI. 1) holds.

We now curtail the hypotheses on k.

WI. 17) Corollary:

1£

(VI. 1) ~ may assume that k

l l

any subfield

2£. Q(S). 1£l

particular, ~ may~

k = GF(q).

Suppose U is a homogeneous K[AQ)module satisfying all of (VI. 1) except that KS

Q(S)

is a subfield of

Q(S).

Let K(S) = k =

Q(S).

Then k is a finite extension of K.

Let

g

= ~(U. Let V be any irreducible l<[AG] submodule of

Q.

Then V is a K[AG] module isomorphic to m copies of U for some integer dividing the degree of the extension

[k:

r<].

We apply the theorem to (A,G,V). Suppose

v

~K[AG]

,u e ...,.---e--u

_{1 •}

m

It is clear that

Cv(L) !:!K(AG] lCU(L) 0 ••• e

Cu(L~

v

m

for any L ~ A. Also G is faithful on V since i t is on U.

The

two isomorphisms give (VI. 17).

(VI. 18) Corollary: Sunpose ~

!,n

(VI. 17) conclusion 2) arises. ~ i.§,

2)

CV(A') = (O).

~ there

i l

¹

t: ^D

^~A' ~ a)

CV(D) =

⁽⁰⁾

im£

b)

CG(D) =

^G.

Now VI

_A

1

G

= V

1 : r

^{••• +Vt where the Vi are (in the}

case of

(VI.

l))

homogeneous components. Let G. _i

=

^{ker[G - > Aut V..),} _{• - i} ^G.

=

^G/G _i ^{. •}

Then we apply (VI. l) to

(A•,g

₁

,v

₁^{). Since}

A

¹¹

=

^{1, and}

CV (A') = (O)

we get

by

(VI. 1) a cyclic D ~A' so that 1

a')

CV (D) = ^(O)

1

and b')

CG (D) = !h·

-1

Now D

_~A'

-

<

Z(A).

So

a)

CV(D) =

^{(0) and}

b)

CG(D) =

^G.

Remark: Again i t is no trouble to extend this by the argument of (VI. 17) to the field K < k •

.

-