816 12. LARGER GROUPS OVER F2 IN .Cj(G, T)z inverts Ou 0 := O(Gu 0 ). Similarly z inverts Ou, as we may also apply 1.1.6 and
1.1.5 to U. Now Ou is a nontrivial Q-invariant subgroup of O(Ccu 0 (U)).
Suppose first that Ou acts nontrivially on Ko, for some component Ko of Gu 0 •
Then 1 # Autou(Ko) :::; O(CAut(Ko)(U)) is Q-invariant. Inspecting the list of
1.1.5.3 for such a centralizer, we conclude K 0 /Z(K 0 ) t;::! A1, U induces a group of
inner automorphisms of order 2 on Ko, and Autou (Ko) t;::! Z3. But by l.l.5.3d, z
induces an involution of cycle type 23 , so that V = ZvU is not normal in CKoZv (z),
contradicting G z = Na (V).
Therefore Ou centralizes E(Gu 0 ). As z inverts Ou and Ou 0 , Ou centralizes Ou 0 •
By 31.14.1 in [Asc86a], Ou centralizes 02(Gu 0 ). Thus Ou :::; Cau 0 (F*(Gu 0 )) :::;
F(Gu 0 ), so in particular Ou:::; Ou 0 • Further Ou 0 abelian since it is inverted by z.
Now given any l EM -NM(U), we may choose UnU^1 as our hyperplane Uo of
U. Then (Ou, Ofr) is contained in the abelian group Ounuz , and in particular, Ou
and Ofr commute. Therefore -1 # P := (Off) is abelian of odd order. Thus LT:::;
Na(P) <Gas G is simple; and Na(P) is quasithin. As m2(Na(P)) 2: m(V) 2: 4,
V cannot act faithfully on Op(P) for any odd prime p by A.1.5, so mp(P) :::; 2 for
each odd prime p. Therefore V = [V, L] centralizes P by A.1.26, which is impossibieas z inverts Ou and so acts nontrivially on P. This contradiction finally completes
the proof of 12.4.8. D
LEMMA 12.4.9. If A:= VY:::; Na(V) with [A, VJ# 1, then Vi:. Na(A).PROOF. Assume otherwise. Interchanging the roles of A and V if necessary,
we may assume m(A) 2: m(V/Cv(A)), so that A contains a member of P(M, V)
by B.1.4.4. Then by B.4.6.13 or B.4.8.2, A is determined up to conjugacy in M,
. and m(A) = m(V/Cv(A)). In particular, we have symmetry between A and V.
Suppose L t;::! L 3 (2). Then E = [A, V], so by symmetry E = EY. Then as
Zv is weakly closed in Eby 12.4.7, g E Ca(Zv) = M = Na(V), contradicting[V, VY]# 1.
Therefore Lt;::! G2(2)' and AL= Mt;::! G2(2). Thus by B.4.6.3, [V, A] = Cv(A),
and again Zv :::; Cv(A) and by symmetry [V, A] = [V, A]Y. Then again Zv is weakly
closed in [V,A] by 12.4.7, and we obtain the same contradiction. This completes
the proof of 12.4.9. D
We are now in a position to complete the proof of Theorem 12.4.2. Recallm(M, V) = 2, so by 12.4.8, s(G, V) > 1. Then by 12.4.3.1, we may apply 12.4.1 to
conclude that there is g E G with 1 # [V, VY] :::; V n VY. In particular, VY :::; Na (V)
and V :::; Nc(VY), contrary to 12.4.9. This contradiction completes the proof of
Theorem 12.4.2.12.5. Eliminating L 5 (2) on the 10-dimensional module
In this section we eliminate the exterior-square module in case (3c) of Theorem
12.2.2, hence reducing the treatment of L 5 (2) to the natural module in case (3a).
This is analogous to the reduction for L 4 (2) in Theorem 12.6.34 of the next section.
Specifically we prove:
THEOREM 12.5.1. Assume Hypothesis 12.2.3 with L/0 2 (L) t;::! L5(2). Then V
is the natural module for L/0 2 (L).