1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

708 7. ELIMINATING CASES CORRESPONDING TO NO SHADOW

LEMMA 7.7.9. (1) If case (a) of G.2. 7.3 holds, then K is an A1-block.

(2) [L^0 ,TO' nK] i 02(L^0 *).

PROOF. Suppose case (a) of C.2.7.3 holds, where K is ax-block. Suppose first

that K is an L 2 (2n)-block, and either n > 2 or K < K 0 • Let Y be a Cartan

subgroup of Kon M. An argument in the proof of the previous lemma shows that

Ye =I= l, and supplies a contradiction. Thus K =Ko is a block of type L2(4), A5,

or A1.

Suppose next that K is a block of type A 5 or L 2 (4), and let YE Syh(MnKL^0 ),

Ye := Gy(V), YL := Y n L^0 , and YK := Y n K. By 7.7.4.1, IY : Yel ::::; 3. As

NK(YK) i M, 7.7.4.2 says that YK =I= Ye, and hence Y = YKYe = YLYe as

IY : Yel ::::; 3. Then

v_ = [V,L^0 ] = [V, YL] = [V, YK]::::: u n K::::: 02(K).

But as K is of type A 5 or L 2 (4), m(0 2 (K)/Z(K)) = 4 = m(V_), so V_Z(K) =

02(K). This is impossible, as Q n K E Syl 2 (K), and Q centralizes V but not

02 (K). This establishes (1); in particular K is not a xo-block.

Assume that [L^0 *, T 0 n K*]::::; 02(L^0 *). Then Ton K::::; Cr 0 (L^0 /02(L^0 )) = Q,

so Q E Syb(KoQ). Therefore K is a x 0 -block by C.2.5, contrary to the previous

paragraph. Thus (2) holds. D

LEMMA 7.7.10. L^0 ::::; K, and hence To::::; NH(K), so that K =Ko.

PROOF. We may assume L^0 i K, and it suffices to derive a contradiction.

Since 1 =/= [V, L^0 ] ::::; [U, L^0 ], we have L^0 * =/= 1. We will appeal frequently to the

fact that L^0 * is normal in M'Jr, and hence is normalized by MK :=Mn K, with

L^0 /02(L^0 ) of order 3.

Inspecting the groups listed in C.2.7.3 and appealing to 7.7.9.1, either m 3 (K) =

2 or K* ~ SL 3 (2n) with n odd. In the former case we apply A.3.18, and A.3.19

when K* ~ SL 3 (2n) with n even; we conclude that K is the subgroup of H gen-

erated by all elements of order 3 so that L^0 ::::; K, and the lemma holds in this

case.

Therefore we are reduced to the case where K* ~ SL 3 (2n) with n odd, and MI

is a maximal parabolic. Assume L^0 i K 0. Then [L^0 *, Mk]::::; L^0 *nMI::::; 02(Mk),

so as CAut(K•)(Mk/02(MJ<:)) is a 3'-group, [L^0 ,KJ = 1, contrary to 7.7.9.2.

This contradiction shows L^0 ::::; K 0. As we are assuming L^0 i K, we must have

K <Ko= KK^8 for s .E To - Nr 0 (K). Hence K* ~ L3(2) by A.3.8.3. As L^0 f:. K

and T 0 acts on L^0 , L^0 * is diagonally embedded in K 0. But the Sylow group T 0

acts on no such diagonally embedded subgroup with Sylow 3-subgroup of order 3,

completing the proof of the lemma. D

As L^0 ::::; K, L^0 ::::J MK-Hence as L^0 /0 2 (L^0 ) is of order 3, K is not L 3 (2n)
with n > 1 odd. Similarly if K ~ SL 3 (2n) with n even, then L^0 = Z(K*), so

that [L^0 ,K] = 1, contrary to 7.7.9.2. Thus n = 1 in case (g) of C.2.7.3.

Assume we are in the subcase of case (e) of C.2.7.3 where K* ~ Sp 4 (4) and

MI is a maximal parabolic. Then as L^0 * ::::J MI, L^0 * = 02 , 3 (Mk)). But then

[L^0 , TI] ::::; 02(L^0 ), contrary to 7.7.9.2.

Thus we are left with the subcase of case (a) of C.2.7.3 where K is an A 7 -block,

or one of cases (b)-(d), case (e) with K* ~ A 6 , case (f), case (g) with n = 1, or
case (h).