790 i2. LARGER GROUPS OVER F2 IN .Cj(G, T)
A is faithful on each 1/i, and for u E Vi - Cv 1 (A), [u, Al = [Vi, Al is of rank 2.
Set A := V;^9. Now V :::; Na(A) by our assumption, and we saw that m(A) =
m(V/Cv(A)) = 4 < 5, so we have symmetry between V and A. Reversing the roles
of V and A, we conclude from that symmetry that V/Cv(A) is faithful on Ai, so
as [A, Vil is of rank 2, Vi induces transvections on Ai with center [A, Vil n Ai. This
is impossible as m(Vi/Cv 1 (A))= 2 > 1 and A2 is dual to Ai.
This contradiction establishes the claim that we may take m(A) 2: 5. Hence
by lemma H.9.2.3, we may take A :::; A 0 , where Ao is the centralizer in Ti of a
3-subspace of Vi. Then by H.9.2.5,
V = f'4,A(V) = f'4,A(V);so as r(G, V) 2: 5, V = f' 4 ,A(V) s Na(A) by E.3.32. Then [V, Al s VnA s Cv(A),
and applying lemma H.9.2.4 to the action of A on V, Cv(A) = [V,A] = V n A is
of rank 5. Thus m(V/Cv(A)) = 5, so we have symmetry between V and A, and
by that symmetry CA(V) = V n A is of rank 5. Hence m(A) = 5. We saw that
A:::; Ao, so A acts faithfully on each 1/i. In particular, lJ2 i. A, and for v E lJ2 -A,
[v,A] :::; An lJ2 =:=: Cv 2 (A), with m(Cv 2 (A)) = 2 by H.9.2.4. By symmetry, lJ2
normalizes but does not centralize V;^9 , so as m([V2,A]) = 2 and m(V2/Cv 2 (A)) > 1,
we have the same contradiction as in the previous paragraph. This completes the
proof of 12.1.3. DLEMMA 12.1.4. Assume n = 4 and letv generate Cv 1 (Ti). Then Ti E Syh(Gv)·
PROOF. Let Ti :::; To E Syh(Gv)· If the lemma fails, then as IT : Til = 2,
To E Syl2(G). But Ti E Syl2(Mv), so To i. M, and hence Na(Ti) i. M. If C is a
nontrivial characteristic subgroup of Ti normal in LvTi, then C :'.Si (T, Lv) =LT, soNa(Ti) :::; Na(C) :::; M = !M(LT), contrary to the previous sentence. Thus no such
C exists, so ( Lv, Ti) is an MS-pair in the sense of Definition C .1. 31. However Lv hasat least three noncentral 2-chief factors, and as v E Vi = [Vi, Lv] :::; Lv by H.9.1.3,
v E Z(Lv)· Hence Lv must satisfy case (4) of C.l.34. Therefore Zi := fh(Z(Ti))
is of rank at least 3, with m(Zi/Cz 1 (Lv)) = l. Now L = (Lv, Lt} fort ET - Ti,
so 1 =/= Cz 1 (Lv) nCz 1 (L~) S Cz 1 (L), and hence Cz(L) =/= l.
Next J(T) :::; Ti by B.l.5.4. Thus J(Ti) = J(T), so as Na(Ti) i. M,
Na(J(T)) i. M. In particular J(T) i. Q, and hence R2(LQ) = V EB Cz 1 (L) by
B.5.1.4. Now an FF*-offender inf' lies in P(T, V) by B.2.7, and by B.4.9.2iii theunique member J(T) of P(T, V) is the unipotent radical of the stabilizer in L of a 2-
subspace of Vi. Thus NLr(J(T)) = XT, where XE B(G,T) with XT/0 2 (XT) ~
83 wr Z2. As R2(LQ) = V EB Cz 1 (L) with Cv(X) = 1, Cz 1 (X) = Cz 1 (L). As
J(T) S Ti, To s Na(Ti) :<:::: Na(J(T)) =: GJ, and of course TX s GJ. Thus
H :=(To, TX):<:::: GJ, so HE 'H(XT). Suppose X :'.Si H. Then T 0 and Tact on Ti
and hence on Cz 1 (X) = Cz 1 (L), so To :::; Na(Cz 1 (L)) SM= !M(LT), contrary
to To i. M. Therefore X is not normal in H, so by 1.3.4, X < Ko := (KT) for
some KE C(H), and Ko is described in 1.3.4. Now KE C(G,T) and [Z,X] =/= 1,so K E Ct(G, T). By 1.3.9, K E Cj(G, T), so by 3.2.3 there is VK E R'2(KoT)
such that the pair K, VK satisfies the Fundamental Setup. Hence by Theorem 3.2.5,
this pair is listed in 3.2.5.3, 3.2.8, or 3.2.9. By Theorem 10.0.1, K = K 0 , so case
(1) of 1.3.4 does not hold. Theorem 11.0.1 eliminates case (3) of 1.3.4. Case (2),
and case (4) with K/0 2 (K) ~ M 11 , do not appear in the indicated lists for the
FSU. Thus KT/02(KT) ~ Ss or Aut(L5(2)). Let H* := H/CH(K/02(K)), so
that H* = K*T* since K*T* = Aut(K*). Then X* = 02 (P*), where P* is the