1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

1160 15. THE CASE .Cr(G, T) = 0

15.4.10, X::; CM;(V(Mi)) =:Hi, so as Hi::; Ca(Z) and X :::1 Ca(Z), X :::1 Hi.

Thus 02 (X) ::; 02 (Hi), so Xi is of order 3 for each i, and Yl ~ E3e for some

e ~ 1. Notice e ::; m 3 (M 1 ) ::; 2, so as X is not normal in M 1 , e = 2. Now

T ::; Ca(Z) ::; NM 1 (X), so as AutM 1 (Yi1) ::; GL 2 (3) and Y1 = (XM^1 ), it follows

that 02 (AutM 1 (Yl)) ~ Z3. Therefore Y1 = XX 1 , where X1 := 02 (Xo) and

X 0 is the preimage in Y 1 of Cy1(P 1 1 ) for P E Syl 3 (M1). Thus X1 E X, so by

15.4.10, X 1 ::; H 2. As H 2 ::; Ca(Z), X^2 and Xr are normal in Hi,. Therefore

02 (H'ff.)::; CH?(Y 12 ), so as m3(H2)::; 2, Y 12 = D1(03(H'i,)). Hence Yi is normal in

M 2 , and Y 1 is normal in M1 by definition, contrary to the simplicity of G. This

contradiction completes the proof of Theorem 15.4.8.

In the remainder of the subsection, we collect some useful consequences of

Theorem 15.4.8.

LEMMA 15.4.11. For each ME M(T):

(1) 02(M) = CM(V(M)).

(2) M is maximal with respect to :S. In particular, there is no unique maximal

member of Jvl(T) under :S.

(3) [V(M), J(T)] =/= 1.

PROOF. First CM(V(M)) ::; Ca(Z) = T by Theorem 15.4.8, so as V(M) is

2-reduced, (1) holds. Now if M :S M 1 E M(T), then

M = CM(V(M))(M n M1) = 02(M)(M n M1)::; M1

by (1), so M = M1. Thus Mis maximal in M(T) under :S, so since IM(T)I > 1

by Hypothesis 15.4.1.2, (2) holds. Then (3) follows from (2) and 15.4.6. D

Define Y to consist of those groups Yin B(G,T) U ~(G,T) such that Y =

[Y, J(T)]; we show Y is nonempty in the next lemma. Set S := Baum(T) and

E := D1(Z(J(T)).

LEMMA 15.4.12. Let M E M(T). Then Y n M =/= 0. Further for each Y E
YnM:

(1) y :::1 M.

(2) M = Na(Y) = !M(YT).

(3) For suitable s(Y) = 1 or 2, Y/0 2 (Y) ~ E 3 s(Y) and m([V(M), Y]) = 2s(Y).

(4) SE Syb(YS).

(5) R2(YT) = [V(M), Y] EB Ey, where Ey := Cn 1 (z(o 2 (YT)J)(Y) and E =

Ey EB C[v(M),YJ(S).

(6) Either

(i) s(Y) = 1, YT/02(YT) ~ 83, and IE: Eyl = 2, or

(ii) s(Y) = 2, YT/02(YT) ~ Ot(2), Y = Y1Y2 with Yi/02(Yi) ~ Z3,

[V(M), Y] = V1 EB Vi, where Vi := [V(M), Yi] ~ E4 for i = 1, 2, Yi = [Yi, J(T)] is

S-invariant, and IE: Eyl = 4.

PROOF. Set M := M/CM(V(M)). By 15.4.11.3, [V(M), J(T)] =/= 1, so as Mis

solvable, we conclude from Solvable Thompson Factorization B.2.16 and A.1.31.1

that X := [03(M), J(T)] and its action on V(M) are described in (3). Let X be

the preimage in M of X and Yo:= 02 (X).