1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

5.1. PRELIMINARY ANALYSIS OF THE L 2 (2n) CASE 633

Also ifs is the involution in (a'), then GA(U) = S(s)D_ and U = (zD+), so U = Vv.

In particular [D_, Z] = 1, so Z :::; Z(X-). If n = 2 mod 4, then T :=:; S(s),

so Z = U is D+-invariant; hence X = (H, DL) :::; Na(Z), contrary to 02 (X) = 1.

Thus n = 0 mod 4. Finally D+ is faithful on Vv, so applying 3.1.6 with TD+, Vv in

the roles of "Mo, V" as before, either 02(X+) of. 1 or q(D+T/0 2 (D+T), Vv) .:=:; 2. In

the latter case, as T/Gr(Vv) is cyclic and m(Vv/Gvn (t)) ~ n/4 fort E T-Gr(Vv),

n = 4 or 8. Thus (4) holds. D

LEMMA 5.1.7. (1) Na(Baum(T)) '.SM.

(2) Let HE H*(T, M) and set K := 02 (H). Assume [Z, H] of. 1. Then

(i) L = [L, J(T)].

(ii) K = [K, J(T)].

(iii) Either 02( (NL(T n L), H)) of. 1, or [V, L] is the S5-module for LT~

S5, and Z(H) = 1.

PROOF. We first prove (1). Let S := Baum(T). If J(T) :::; Gr(V), then

(1) follows from 5.1.2. Thus we may assume J(T) f:. Gr(V), so by 5.1.2, either

V/Gv(L) is the natural module for Lor [V,L] is the A 5 -module. In the former

case, Sn LE Sylz(L) by E.2.3.2, so (1) follows from 4.3.17.

Therefore we may assume that [V, L] is the A5-module. As [V, J(T)] of. 1,

we conclude from E.2.3 that LT~ S 5 , S = J(T) ~ E 4 is generated by the two

transvections in T, and (ZL) = [V,L] EB Gz(L). We may assume V = [V,L].

Assume that Na(S) f:. M; then no nontrivial characteristic subgroup of S is

normal in LT as M = !M(LT). Hence by E.2.3.3, L is an A 5 -block, so V =

02 (L) :::;I M. Let Q := 02 (LS). It follows using C.1.13.b that Q = V x Qc, where

Qc := Gs(L).

For any 1 of. S 0 :::; S normalized by LT, we have Na(So) :::; M = !M(LT), so

Na(S) f:. Na(So) by our assumption. Thus Hypothesis C.6.2 is satisfied with L, S,

T, Na(S) in the roles of "L, R, TH, A". Therefore by C.6.3.1 there is g E Na(S)

with Vg f:. Q. As V :::;I M, g f:. M.

Suppose that Qc f:. Qg. Since [Qc, Vg] n [V, Vg] :S Qc n V = 1, from the

action of S on V and hence on Vg, we conclude that Qc and V induce distinct

transvections on vg. Thus as IS : Qgl = 4, s = QcVQg. Let x E [Qc, Vg]#;

then x E Qc :::; Ga(L), so as M = !M(L) by Theorem 4.3.2, Ga(x) :::; M, so

V :::; 02 (Ca(x)). Since Qc induces a transvection on the A5-module Vg for Lg,

GL9s(x)QcQg/QcQg ~ S3, so V :S 02(GL9s(x)QcQg) = QcQg, contrary to V

and Qc inducing distinct transvections on vg.

Therefore Qc :::; Qg. Hence

(Qc) :S (Qg) = (QbVg) = (Qb),

so <I>(Qc) = <I>(Qb). Thus as <I>(Qc) ~ LT and g f:. M = !M(LT), <I>(Qc) = 1 =

<I>(Q).

Next we claim we can choose g so that S = QQg. If not then Q n Qg is a