832 i2. LARGER GROUPS OVER F2 IN .Cj(G,T)and Vtf, so from the structure of 6 := Out(U1) ~ ot(2), i::::: Ca(Li) ~ 83 x 83.
But now m 3 (H) > 2, whereas His an SQTK-group. This contradiction shows that
V :<:::: 02(CH(L1/02(Li))).
We next show that V* centralizes F(H*). Assume otherwise. Let PE 8yl3(Li)
and set C* := Cos(H*)(P*). If [0 3 (H*), V*] -/=-1, then as V* centralizes P*, V*inverts a subgroup X of C of order 3 by the Thompson A x B Lemma; but
then m 3 (P X) = 3, contrary to H an SQTK-group. Therefore V* centralizes
03 (H), so we may assume that [V, Op(H*)] -/=- 1 for some prime p > 3. We
saw that elements of odd order in H inverted by V are of order 3 or 5, so
p = 5. Let Y be a supercritical subgroup of 05 (H). By the previous paragraph
V :<:::: 02(CH(Li/0 2 (Li))), so 05 (H) f:. CH(Li). Therefore we conclude from
A.1.25 that Y is E 25 or 51+^2 and P is irreducible on Y /(Y). Thus as V
centralizes P*, V* inverts Y* /<I>(Y*). If Y* ~ 51+^2 , then a faithful chief section
W for Y* V* on UH is of dimension 20 over F 2 , and m([W, V*]) 2:: 8. If Y* ~ E25,then a faithful YV -chief section W for PY is of dimension 12 over F2 and
m([W, V*]) = 6. So in any case m([UH, V*]) ;::: 6, whereas we saw [UH, V*] = V5 isof rank 4. This contradiction establishes the claim that V centralizes F(H).
So as 02 (H) = 1, [K, V] -/=- 1 for some K E C(H) such that K is a
component of H*. Then K f:. Mas V :<:::: 02 (M). As V* :<:::: Z(T*), V* normalizes
K*, so that K* = [K*, V*]. Further Li= 02 (Li) normalizes K by 1.2.1.3. As V :<::::
02(CH(Li/02(L1))), K* f:. CH* (Li), so that K* = [K*, Li]. Set X := KLi V and
X := X/Cx(K*). By F.9.5.3, CH*(V*) = NH(V)*. Further NH(V) :<:::: HnM :<::::NH(Li) and Cx(V) = CK(V)Li V. Now CK(V) centralizes V Z(K)/Z(K*), so
as 02 (K*) = 1, C.K(V) = C~) :<:::: KnM, and then
Li:<:::: 02,3(C_x(V)) with V of order 2 in Z(T) and 3 E 7r(L1). (*)
Since all elements of odd order in k inverted by V are of order 3 or 5, we conclude
k is not 8z(2n). Hence m 3 (K) = 1 or 2.
Suppose first that m3(K) = 2. Then as m3(P) = 2 = m3(PK), either Pisfaithful on K, or 1 -/=-P n 0 2 ,z(K), so that by l.2.l.4b, K ~ 8LHq), A.6, A1, or
M22- Further in the latter case, K/02(K) ~ A7 or M22 by (*).
Suppose that K* ~ A7. Then Ri = 02(LiT) ~ E4, while NK(Ri) :<:::: M
by 12.6.1.4, so 0
31(NK(Ri)) :<:::: 0
31(M) = L by 12.6.1.5. Thus 0
31
(NK(Ri)) :<::::02 (0.r,(z)) = Li. This is impossible, as NK(Ri) has Sylow 3-group 31+^2 , while
Eg ~ P E 8y[z(Li).Suppose next that K* ~ M 22. As H :<:::: Gz, H n M = NH(V) by 12.2.6, so
(MnK) = NK(V) = CK(V), and hence (MnK) ~ (8 3 x Z2)/(Z 3 x Q~). We
saw V5 = [UH, V*] is of rank 4, so we conclude from H.12.1.3 that UK := [UH, K]
is the 12-dimensional irreducible for K*. Choose v 2 of weight 2 in V 5 ; by parts (5)
and (7) of H.12.1, C.K(v2) ~ 85/E 16 or A5/E 16. In particular CK(v2) involves A5,
so CK(v 2 ) f:. M, as we saw earlier that (Mn K)* is solvable. But this contradicts
Theorem 12.6.2.
We have shown that if m 3 (K) = 2 then Pis faithful on K*, and hence also on
k. Then m3(P) = 2, so that m 3 (0 2 , 3 (C_x(V)) = 2 by (*). But no simple group
k of 3-rank 2 in the list of Theorem C satisfies this restriction. This contradiction