1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

6.2. IDENTIFYING M22 VIA £ 2 (4) ON THE NATURAL MODULE 687

PROOF. Assume Zs = V n U. We begin by arguing much as at the start

of the proof of 6.2.11, except this time V* has order 2 by 6.2.11. By 6.2.10.5,

1 -/= [V*' U] ::::; vnu = Zs of order 2, so that V* is generated by a transvection

on U with center Zs. As U = (Zff) and Zs = [U, V*], U = [U, K*] by G.6.2. As

V ~ T, V ::::; Z(T), so G.6.4.4 shows that K* 9:'. Ln(2), 2 ::::; n::::; 5, 86 , or 87; and

by G.6.4.2, U is the natural module or the core of the permutation module for 86.

In each case K = NGL(u)(K), so H = K. Next by 6.2.9:

But if H is Ln(2) with 3 ::::; n ::::; 5, then V is not normal in CH·(Zs). Thus

H = K 9:'. L2(2), 85, or 87. In each case, V 1- 02 (H), so in particular

Vi 02 (X), where X := 02 (Mz), and hence V > V n X. By 6.2.12.1, L acts on

02 (0^2 (H n M)0 2 (L)) = X, so L acts on V n X. Therefore as Lis irreducible on

v, vnx = i.

Suppose first that H* is 85 or 87. Then there are x, y E H such that I :=

(Vx, VY) ::::; Mz and I* 9:'. 83. Then vx i Na(VY), but Cvx(U) ::::; Na(Z~) ::::;

Na(VY) by 6.2.l; so as V*x has order 2, Nvx(VY) = Cvx(U) is of index 2 in vx.

Similarly IVY: Nvy(Vx)I = 2, so as I is not a 2-group, 02 (1)::::; 02 (1)::::; X and

IZ(02(I))I = 2 by 6.2.6. But as x, y E Gz, z::::; vxnvY = Z(02(I)), so z::::; vnx,

contrary to the previous paragraph.

This contradiction shows that H* 9:'. 83 , so H* = (V*, V9*) for g EH - Mand

IVHI = IH: Mzl = 3. Thus vH ::::; (V, V^9 ), so that K = (V, V^9 ). Therefore case

(2) of 6.2.6 holds with K 9:'. 83 /Qg, and Z = V n V9 = Z(P), where P := 02 (K).

Notice as Zs ::::; P ~ H that U = (Zff) ::::; P. Then R := CH(F) ::::; CH(U) =

QH by 6.2.12.2. Also as case (2) of 6.2.6 holds, Nva(V) = P 9:'. E4, Cp(V) =

P n V, and P i L. Therefore T = P(f;, where t E T n L acts nontrivially

on P. Thus t is nontrivial on P/(P n V), so that t ¢:. V since [P, VJ ::::; P n

V. Therefore as Nout(P)(K*) 9:'. 83 x 83 and H = KT, we conclude H/R 9:'.

83 x Z2 and Cr(V) = VCR(V). Now R = PCR(P) as Inn(P) = CAut(P)(F) by

A.1.23. But CR(P) ::::; CR(V n P) = CR(V) by 6.1.10.2, so as CR(P) ~ H, CR(P)

centralizes (VH) = K. Therefore CR(P) = CR(K), so R = PCR(K) and CR(K)::::;

CR(V). Thus CR(V) = CR(K)Cp(V) = CR(K)(P n V), and hence Cr(V) =

VCR(V) = VCR(K) = VCR(P). Then [P, Cr(V)] = [P, V]::::; V, so as L = [L,P],

[L, 02 (LT)] ::::; V, and hence L is an L 2 (4)-block. Now (Cr(V)) ::::; Cr(L) = 1