1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

15.3. THE ELIMINATION OF Mr/CMf(V(Mf)) =Sa wr Z2 1123

LEMMA 15.3.9. V < 02(Y), so that Y ~ A4 x A4.

PROOF. Assume V = 02 (Y), or equivalently that Y ~ A 4 x A 4. By 15.3.7,

Y ~ M. As Aut(A4) ~ 84, Aut(Y) ~ 84 wr Z2 with CAut(Y)(V) = Autv(Y).

Thus YCM(V) = Y x CM(Y). Since Z has order 2 and CT(Y) :::l T, CM(Y) has

odd order; thus CM(Y) = 1 as F*(M) = 02 (M). Therefore M ::::; Aut(Y), so as

M ~ ot(2) by 15.3.2.1, we conclude M ~ 84 wr Z 2. But this is ruled out by

Theorem 13.9.1. D

15.3.2. Uniqueness theorems for Y and 02 (Cy(V;)). Our first main

goal, in Theorem 15.3.45 in this subsection, is to show that M = !M(Cy(Vi)8); to

do so, we first show that M = !M(Y8). We prove the two results simultaneously,

adopting a suitable hypothesis to cover both cases, and eventually establish the

common uniqueness result in 15.3.44.

Thus in this subsection, we assume:

HYPOTHESIS 15.3.10. Either

(1) Y+ := Y, or

(2) M = !M(Y8) and Y+ := 02 (Cy(V1)).

Let

H+ := H(Y+8, M) ={IE H(Y+8): Ii. M},

and write H+,* for the mimimal members of H+ under inclusion. As our goal

is to show that M = !M(Y+8), we will assume H+ is nonempty, and derive a

contradiction. Given IE H+, define Mr :=Mn I, Ur := (Zr), I* := I/Cr(Ur),

and R := 02(Y+8).

LEMMA 15.3.11. Assume IE H+. Then

(1) 8 E 8yh(I).

(2) Cr(Ur)::::; Mr.

(3) If case (2) of Hypothesis 15.3.10 holds, then Y+ = 02 (YnI) and Nc(Vi)::::;

M 2: Na(Y+) for i = 1, 2.

(4) Nr(Y+) =Mr.

(5) Either:

(i) Y+/0 2 (Y+) is Eg or 3i+^2 , Y+8/02,(Y+8) ~ 83 x 83, R = 02(Y8) :::1

31

and R = Cs(V2).

(9) Nr· (Y-t') = Mj <I*, and Y-t' =f. 1.

(10) Either

(ii) Y+ = [Y+, J(8)], so Y-t' ::::; J(I)*.

(11) J(I)* i Mj.

(12) If L::::; I with [V, 02 (Y n L)] =J 1 and Li M, then no nontrivial charac-