810 12. LARGER GROUPS OVER F2 IN .Cj(a, T)Ko = [Ko, J(T)] does not centralize Vo, but 000 (Ko) centralizes Vo by 3.2.14, we
conclude that K 0 /Cx 0 (V 0 ) ~ L 2 (p) for p = 5 or p 2 11. But p 2 11 is ruled
out by Theorem B.4.2, so Ko = Cx 0 (Vo)Ke. However as e, e(6) E Vo, Cx 0 (Vo) ::::;
Ge(G) n Ge ::::; Me by 12.3.6, and then Ko acts on Ke, a contradiction.
Thus K 0 /0 2 (K 0 ) is quasisimple, so by Remark 12.2.4, Hypothesis 12.2.3 is sat-
isfied with Ko in the role of "L". Thus Ko and its action on any IE Irr +(Ko, Vo, T)are described in Theorem 12.2.2.3. Then comparing that list with the possible em-
beddings in A.3.14, we conclude that either Ke= Ko or Ko/Cx 0 (Vo) ~ A1.
Suppose first that Ke< K. Then as usual Ki Mand K <Ko is eliminated,
since by A.3.14, there is no KE £(G,T) with Ke< K <Ko when Ko/Cx 0 (Vo) ~
A 7. Then K =Ko E £j(G,T), so that K satisfies our hypothesis in this sectionthat K/02,z(K) ~ A1. Hence we may apply the results in this section to K. In
particular by 12.3.3 and 12.3.7, I := [Vo, Ko] is the A1-module, Vo =I EB Cz(K),
and
[V,Ke] = [!11(Z(02(KeT))),Ke] = [Vo,Kel·Pick v E [V, Ke] of weight 4. Then ev is of weight 6 in V, so Ca ( ev) ::::; M by 12.3.6.
But Cx(v) = Cx(ev) as K::::; Ge, so K = (Ke,Cx(v))::::; M, contrary to Ki M.
This contradiction shows that Ke= K :SJ Ge. Then as Out(K/0 2 (K)) is a 2-
group, Ge= KTY, where Y :=Ca. (K/0 2 (K)), so it remains to show that Y::::; M. ·Set U := (za.) and G; := Ge/Ca.(U). Then U E R 2 (Ge) by B.2.14. As K =
[K, J(T)], Theorems B.5.1 and B.5.6 imply [U, K] = [V, K]; so as Endx· ([V, K]) =
F2, Y::::; Ca.([V,K])::::; Ca.(ev)::::; M, completing the proof of 12.3.8. D
Recall the weak closure parameters r := r(G, V) and w := w(G, V) from Defi-
nitions E.3.3 and E.3.23.LEMMA 12.3.9. (1) If g E G - Na(V), then V# n Vg ~ 04.
(2) r(G, V) 2 4.
PROOF. By 12.2.6, V is a TI-set in M; so by 12.3.5 and A.1.7.3, if u E V#
with Gu ::::; M, then u is in a unique conjugate of V. Thus (1) follows from 12.3.6
and 12.3.8. Up to conjugation, (e 1 , 2 , 3 ,4, e(4)) is the unique maximal subspace U ofV with U# ~ 04, so (1) implies (2) since m(V) = 6. D
LEMMA 12.3.10. W1(T, V)::::; CT(V), so w(G, V) > 1.
PROOF. Assume the lemma fails. Then we may choose A:= vg n M::::; T::::;
Na(V) to be a w-offender in the sense of subsection: E.3.3. Thus A "I- 1 and
w := m(Vg /A) ::::; 1. Now from the action of 87 on V, for each a E A#, [V, a]# i 04.
But if V ::::; Na(Vg), then [V, a] ::::; V n Vg, contrary to 12.3.9.1, so we conclude
Vi Na(Vg). Therefore m(Vg /CA(V)) 2 r(G, V) 2 4 by 12.3.9.2, so that m(A) 23 = m2(LT). Thus these inequalities must be equalities, so m(A) = 3, w = 1, and
r(G, V) = 4. Hence A is fused under L to
A1 := ((1, 2), (3, 4), (5, 6)) or A2 := ((1, 2), (3, 4)(5, 6), (3, 5)(4, 6)).
Now the Fundamental Weak Closure Inequality of Remark E.3.29 is an equality, so
by E.3.31.1:VA := (Cv(a) : a EA#) ::::; Na(Vg).
Therefore [A, VA]::::; VnVg, and hence [A, VA]#~ 04by12.3.9.1. We compute that
this does not hold if A= A 1. Similarly [VA, A] ::::; Vg ::::; Ca(A), so that [VA, A, A] =