1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

1126 15. THE CASE .Cr(G, T) = 0

LEMMA 15.3.16. {1) WL E R 2 (LR) n R 2 (LRY+) and L+ is quasisimple.

{2) m 3 (L) ~ 1, YL i= l, R acts on L, L = [L, J(R)], and L is described in

C.2.1.3.

PROOF. By 15.3.15, m 3 (L) 2: 1 and we can appeal to the results in section

C.2. As m 3 (L) 2: 1, YL f= l, R acts on L, and 02(LR) :SR by 15.3.12.1.
As L/0 2 (L) is quasisimple and 02(LR) :S R = GT(V), WL E R2(LR) n
R 2 (LRY+)· As R acts on Land Li Mr, Lis described in C.2.7.3. By C.2.7.2,
L = [L, J(R)]. As Gc(V)::::; M but Li Mr, L+ i= l, so as L/02(L) is quasisimple,
soisL+. D

LEMMA 15.3.17. One of the following holds:

{1) m 3 (L) = 1 and L/0 2 (L) ~ L2(2n), n even, or L3(2).
(2) m3(L) = 2 and Y+ ::::; ()(J) = L.
{3) m3(L) = 2 and L* ~ SL3(2n) with n even.
PROOF. By 15.3.16, m 3 (L) 2: 1, YL f= l, and Lis described in C.2.7.3.
Suppose first that m 3 (L) = 1. Then from the list in C.2.7.3, L/0 2 (L) ~ L 2 (2n)
or L 3 (2m), with m odd. Now as SE Syh(I), SL= Sn LE Syl2(L) and SL acts
on YL since Y+ is S-invariant. It follows that n is even if L/0 2 (L) ~ L2(2n), and
that m = 1 if L/0 2 (£) ~ L 3 (2m). That is, (1) holds in this case.

So assume m 3 (L) = 2. Then (3) holds if L* ~ SL 3 (2n) with n even; otherwise

()(I) = L by A.3.18, so that (2) holds. D

In the remainder of the proof of Theorem 15.3.13, we successively eliminate the
various possibilities in C.2.7.3 given by 15.3.16.

LEMMA 15.3.18. L is not an L 2 (2n)-block.

PROOF. Assume otherwise. Then n is even by 15.3.17, while by 15.3.16, YL i= 1

and R normalizes L.

Let Lo := (L^8 ), so that So :=Sn Lo E Syh(Lo) and Mo :=Mn Lo ~ YL.

Then M 0 is an overgroup of S 0 in Lo, so Mo is contained in a unique Borel subgroup

Bo of Lo, and hence Bo is Mr-invariant. Therefore as Bo is solvable, Bo= Mo by

15.3.15.3. Then as Y+ :::J Mr by 15.3.11.4, we conclude that Y+ induces inner

automorphisms on Lo/02(Lo). By 15.3.12.2, Y+ = YLYc with IYLl3 = 3 = 1Ycl3,

and Y+/02(Y+) ~ Eg. As So is Mr-invariant, 80 ::::; 02(Y+S) = R; hence RL =

RnL E Syb(L), and RE Syl 2 (LR). Thus as V :S Z(R) and U(L) = [WL,L] since

Lis a block, WL = GwL(RL)U(L) by B.2.14, so that

V::::; Gu(L)(R)GwL(L). (*)

Now if [V, YL] = 1, then by (*), we have V :S Gu(L)CR(L)(YL) = GR(L),

since U(L)/Gu(L)(L) is the natural module for L 2 (2n). But then L ::::; Gc(V) ::::;

M, contrary to L i M. Therefore 1 i= [V, YL] is a B 0 -invariant subgroup of

[Gu(L)(RL), YL], so [V, YL] = Gu(L)(RL) by (*) and the structure of coverings

of the natural module. in I.2.3. But for b E Bo - R, Gu(L) (b) = Gu(L)(L), while

02 ,^3 (ML) ::::; Gc(V) ::::; Gc([V, YL]) by 15.3.2.1, so we conclude that (B 0 nL)/ RL is a

3-group, and hence n = 2. Thus as [V, YL] = Gu(L)(RL), [V, YLJ/G[v,YLJ(YL) ~ E4.

As V = V 1 E9 V2 where V 1 and V2 are the only Y+-invariant 4-subgroups of V,

Gu(L) (L) = 1 and we may take V2 = [V, YL], and hence YL ::::; GM(V 1 ) = M 2. Then

by(*), Vi::::; GwL(YL), so as Li M, also Nc(V1) i M. Hence by 15.3.11.3, we

are in case (1) of Hypothesis 15.3.10, where Y+ = Y. Then Y/0 2 (Y) ~ E 9 , so