15.2. FINISHING THE REDUCTION TO Mr/CMf(V(Mf)) ~ Oj"(2) 1115and [K1, K2] S Kin K2· Now S acts on X, F*(I) = 02 (J) by hypothesis, and SE
Syl2(I) by 15.2.18.1; thus F(XS) = 02 (XS) by 1.1.4.4. Then F(X) = 02 (X) by
1.1.3.1, so that 02(X) =/-1. It follows that 02 (XXt) =/-1. Then as Tacts on xxt,
XXtT E 1i(T). This is a contradiction, as Y 1 YlT S XXtT and M = !M(Y 1 YlT)
by 15.2.2, so that K = KiK2 ::::; M, contrary to Hi. M. DRecall CMc(Zs) = Ca(Zs) by 15.2.16. During the remainder of the proof of
Theorem 15.2.15, take I minimal subject to IE 1i(Ca(Zs)) and Ii. Mc. Recall
Mr <I, so as M1 =In Mc, Mr is a maximal subgroup of I.
For X::::; G, let B(X) be the subgroup generated by all elements of X of order
3.
LEMMA 15.2.23. F*(I) =/-02(!).
PROOF. We assume F*(I) = 02 (J) and derive a contradiction. We begin with
some preliminary reductions.
Suppose first that there is X 0 = 031 (X 0 ) = X 0 ::;! I with X 0 nontriv-
ial on Zo := n1(Z(02(Xo))) and Xo S Mc. As Xo = 0
31(Xo), Xo S Kc by
15.2.12.4. Since S E Syb(I) by 15.2.18.1, Sn 02(Kc) = In 02(Kc), so Xo
acts on Sn 02(Kc)· Therefore as IT : Bl = 2 by 15.2.17.3, I02(Kc) : Sn
02(Kc)I s 2, so [02(Kc), Xo] s Sn 02(Kc) s Na(Xo), Thus Xo = (Xo02(Kc))^00
is 0 2 (Kc)-invariant. Hence [Xo, 02 (Kc)] :::; 02(Xo) :::; Ca(Zo), so by the Thompson
Ax B-lemma, Xo is nontrivial on Cz 0 (02(Kc)). But since Kc E He by 1.1.3.1,
Cz 0 (0 2 (Kc) ::::; n 1 (Z(02(Kc))) =: Zc. Hence Kc is nontrivial on Zc, so that
Kc E Ct(G, T) by 1.2.10, contradicting part (1) of Hypothesis 14.1.5. Thus no
such X 0 exists.
Next assume that either X is a x-block of I, or X E C(J) with X/0 2 (X) ~
L 3 (2) and X is described in C.1.34. Set Xo := (XS) and Uo := (Zff^0 ). Observe
that X 0 :::) I either by 1.2.1.3, or when Xis an A 3 -block since IXrl ::::; m3(J) ::::; 2.
If Xis an A 7 -block, then X = 031 (I) by A.3.18, so K::::; X. In the remaining cases,
m 3 (X) = 1 and K acts on X, so as m3(KX):::; 2, Ko:= 02 (K n X) =/-1.
Consider first the subcase where X is an A 3 -block or an L 2 (2n)-block. By
15.2.11.2, Zs centralizes K, so that Zs n Uo centralizes (K6) = Xo; this is impos-
sible, as Zs n Uo i. Cu 0 (Xo) in these blocks (cf I.2.3.1).
This leaves the subcases where either X is an A5-block or an A1-block, or
X/0 2 (X) ~ L3(2). Then Xo i. Mc by paragraph two, so Mo := Xo n Mr < Xo.
Notice Cx 0 (Zs) S Mo by 15.2.12.3. If Xis an A5-block, then Cx 0 (Zs) is a Borel
subgroup of X 0 , so Mo is that Borel subgroup. If Xis an A1-block, then we saw K S
X, so K ::::; Mo by 15.2.11.2, and hence Mo = K(X n S) is the maximal subgroup
stabilizing the partition { {1, 2, 3, 4}, {5, 6, 7} }, using the notation of section B.3.
Finally if X/02(X) ~ L3(2), then since Cx 0 (Zs) ::::; Mo< Xo, case (5) of C.1.34
is eliminated by B.4.8.2, as in that case Zs centralizes Xo. Thus Cx 0 (Zs) is a
maximal parabolic of X 0 , so Mo is that maximal parabolic.
Let Qr:= 02(KS); in each case Mo :::) KS, so Qr= 02(M0Qr). Further Mo
contains a Sylow 2-group of Xo, so 02(X0Qr)::::; Qr by A.1.6. Next Qr :::) Has IT:
SI= 2, so C(I,Qr)::::; Mr by 15.2.12.3. Then as Mo< Xo, J(Qr) i. 02(X0Qr)),
so there is an FF*-offender in AutQI(Uo) by B.2.10. Hence by B.3.2.4, X is not
an A 5 -block or an A 7 -block. Further cases (2)-(4) of C.1.34 are eliminated since
Cx 0 (Z$)::::; M 0 , leaving case (1) of C.1.34 where Xis an L3(2)-block. Thus we have