S46 i2. LARGER GROUPS OVER F2 IN Cj(G, T)
so (P n UB)+ = Gp+(x).
Recall X = 02 (0 2 ,z(L)), so X::;; L::;; I::;; Na(P).
LEMMA 12.7.19. [P+,X] = 1.
PROOF. Assume otherwise. Take x E LnUB, and take y E U^9 to be the product
of 3 transpositions. Then [G[P+,xj(x),y] -=f.1 as y inverts X, so Gp+(x) -=f. Gp+(y),
whereas Gp+(x) = (U9 n P)+ = Gp+(y) since neither x nor y is a transposition.
This contradiction establishes the lemma. D
LEMMA 12.7.20. Lis a A 6 block, V = 02(M), and M =LT.
PROOF. As [P,X]::;; (Un UB)V by 12.7.19 and I centralizes (Un U^9 )V/V, it
follows by Coprime Action that [P, X] = V. Thus X = VY where Y has order 3,
so LT= VNLr(Y) by a Frattini Argument.
If Lis a A 6 -block then as Gr(L) ·= 1 by 12.7.12, 02(M) = V and M =LT
by 12.7.6.2. Thus we may assume that L is not a A5-block. Then 1 -=f. Zy :=
Di(Z(Nr(Y)) n 02 (LT), and Zy is in the center of VNr(Y) = T, so that Zy ::;; Z.
Let Vy := \Z~); then Vy E R'2(LT) by B.2.14 and Vy ::;; Ga(Y). As ·cr(L) = 1,
Gvy(L) = 1. Let Vo := VyV, so that also Vo E R'2(LT) by B.2.12. ·By B.4.2.8,
the unique FF*-offender in LT on Vis R 2 and m(V/Gv(R2)) = m(R2). Then it
follows from B.4.2 and B.3.4 that q := q(AutLr(V 0 ), V 0 ) 2:: 2, with equality only if
Vy/Gvy(L) is the A 6 -module (so that m(Vy) = 4 since we saw Gvy(L) = 1) and
either
(i) R 2 is an FF* -offender on Vy and hence Li centralizes Zy, or
(ii) There is a strong FF*-offender A in Ton V 0 with m(V/Gv(A)) == m(A) +1,
so that A= 02 (L 2 f') and again Li centralizes Zy.
However by 3.1.8.1, q ::;; 2, so indeed q = 2. Therefore Vy is the 4-dimensional
A5-module in which LiT centralizes a point, so as U = 02(Lif') by 12.7.18, U is
not quadratic on Vy, impossible as [Vy, U, U] ::;; [Vy n U, U] ::;; Vy n \z) = 1 using
- 7.15.1. D
We are now ready to complete the proof of Theorem 12.7.14.
By 12.7.20, Lis a A 6 -block, V = 02 (M), and M =LT. By 12.7.18, U ~Es, so
M /V ~ S 6. In particular M, and hence also T, are determined up to isomorphism,
so T is isomorphic to a Sylow 2-group of He. Thus J(T) ~ E 64. But by our remark
before 12.7.17, V n U = Vi[V, Ui] is of rank 4, so as U ~Es,
1u1=1u11un VI= 8·16 = 21 ,
so U ~ E54 and hence U is the preimage D~ of J(T) in T and LiT /U ~ 84.
As Tis Sylow in He, Gr(U) ::;; U, so as U induces GAut(U))(U) by A.1.23,
U = GH(U). Thus H ::;; Out(U) ~ ot(2). "Recall 02(H) = 1 by 12.7.15.1. As
V = [V,Li], [O(H), V] = 1 by A.1.26. Then since K = (V^0 z), K* centralizes
O(H). Further H = KLiT and LJ.T ~ 84 with V = 02(LJ.T). Now examining
Ot(2) for subgroups satisfying these conditions, we conclude H* is L 3 (2), A 6 , A1,
85, or I'L2(4). Next Ur:= Go(T) = Gt.Jnv(T) ~ E4 and Go(LiT) = f'i. Therefore
H is not A5 or 85, since those groups fix a point of of U, but K moves t. If H
is I'L 2 (4), then [U,K] is the A 5 -module for K, impossible as V is quadratic on
u.