il.5. THE FINAL CONTRADICTION 779(1) or (2) of 11.1.2. In case (1) of 11.1.2, 02,21(K) i. M, so we may choose Ri :::;
Hi :S Ri02,2^1 (K), and then n(Hi) = 1 by E.1.13. As Ri E Syh(Ca(Li/0 2 (Li)),
Ri E Syb(H1), completing the proof in these two cases by paragraph one.
In case (2) of 11.1.2, K/02(K) ~ L2(P) for a prime p 2:: 11, and Li has nonontrivial 2-signalizer in Aut(K/02(K)), so Ri = 02(KT). Thus K:::; Na(Ri) :::;
Na(Wo(Ri, V)):::; M, a contradiction. D
As Gz i. M, 11.4.3.1 says Ca.(K/0 2 (K)) or K is not contained in H. Thus
the following choice is possible:
From now on we choose H so that either [K, 02 (H)]:::; 02 (K), or0^2 (H):::; K.
In particular notice that if K =Li, then as Hi. M, [K, 02 (H)]:::; 02 (K). By
11.4.4, n(H) = 1. Recall Gi = Na(Vi) and set Gi := Gi/Vi.
LEMMA 11.5.2. (1) If z E V n V^9 then VB E VG•. That is Gz is transitive on
{V9: z E VB}.
(2) K = K(VB,z) for each g E Gz.
PROOF. By 11.4.1.1, Lis transitive on V#, so (1) holds using A.1.7.1.
As K :SI Gz, for g E Gz
K = K^9 = K(V,z)^9 = K(V^9 ,z),so (2) holds. D
LEMMA 11.5.3. Assume K =Li and set m := 2n, 3n; for L ~ SL3(q), Sp4(q),
respectively. Then
(1) For each g E G - Na(V), V n V^9 :::; Vf for some y EL.(2) r(G, V) 2:: m.
(3) (VG^1 ) is nonabelian.PROOF. Let g E G - Na(V). Our first goal is to prove (1), so we may suppose
1 i= U := V n VB. By transitivity of L on V#, we may assume Un Vi i= 1, and we
may suppose that U i. Vi. For u EU#, VB= V9u for some 9u E Ca(u) by 11.5.2.1.
Now u EV{ for some x EL, and K(u) := K(V,u) = K(V9u,u) = K(VB,u)
by 11.5.2.2, while K(u) = Lf by our hypothesis that K = Li. Since U i. Vi,
L = (K(u) : u E U#); so as UB-
1
.'.S V, LB-
1
= L, and hence g E M; indeed
g E Mv since Vis a TI-set under M by 11.0.3.4. This contradicts the choice of g,
so (1) is established.
If U :::; V with m(V/U) < m, then U i. Viy for any y E L, so Ca(U) :::; Mv by
(1). Thus (2) holds.
Suppose that (VG^1 ) is abelian. By (2), Ca(V 2 ):::; M. Hence W 0 := Wo(T, V) i.
02 (H) by 11.2.5. Then since H is a minimal parabolic with H n M the unique
maximal overgroup of T, NH(Wo):::; H n M. As m(Mv, V) > 1, s(G, V) > 1 by
(2). Hence as n(H) = 1by11.4.4, it will suffice to show Ca(Ci(T,V)):::; M, since
then E.3.19 supplies a contradiction. Indeed as Ca(Vi) :::; M, it suffices to show
Vi :S Ci(T, V).
So suppose A := VB n T is of corank at most 1 in VB, but [Vi, A] i= 1. Let
B := CA(Vi). Then
m(V^9 / B) :S m2(AutM(Vi)) + 1=n+1 < 2n :Sm, ·