1106 i5. THE CASE .Cf(G, T) = 0does not hold, as in that case m 3 (CM(V)) S 1by15.2.3.1. Thus Y = 02 (Mo) ~ Y*
by 15.2.3.1, so Y* X* = Y* x X* as Y* centralizes CM(V)* by 15.2.3.5. Then as
Mis an SQTK-group, 02 (M 0 ) ~ Y* is a 3^1 -group, so case (4) of 15.2.l holds.
Next K := .. ,0^2 (H) = KiKf for t E T - NT(Ki) and Ki E C(H) with
Ki/02(Ki) ~ A5 ap_d Ki i M. Observe that F*(Ki) = 02(Ki) by 1.1.3.l. Let
Xi:= XnKi and S := NT(Ki), and observe that 02 (XT) SS, while J(T) SS by15.1.12.3. By 15.2.3.2, 02(Y) S 02(M) S Re, and hence Re E Syl2(YRc)· Then
as XS Mn Mc, Re S 02(XT) SS, so SE Syb(Gi), where Gi := YXiS. Also
SE Syl2(G2), where G2 := KiS. Let Go:= (Gi, G2)·Suppose first that 02 (G 0 ) = 1. This assumption gives part (e) of Hypothesis
F.1.1 with YRc, Ki in the roles of "Li, L 2 "; most other parts are straightfor-
ward, but we mention: As X* centralizes Y* R~, NK 1 (Sn Ki) = Xi(S n Ki) S
XS S Na(YRc)· Recall that Y = [Y, Re] by construction in 15.2.3, so that
YRc/0 2 (YRc) ~ D 10 or Sz(2). Thus the amalgam a:= (Gi,XiS,G2) is a weakEN-pair of rank 2 by F.1.9. Furthermore as S = Nys(Rc), a is described in F.1.12.
This is a contradiction, as YRc/02(YRc) ~ D10 or Sz(2) while Ki/02(Ki)) ~ A5,
and no such pair appears in F.1.12.Therefore 02 (Go) =/= 1. Let S S To E Syb(Go). We saw J(T) S S, so that
J(T) = J(S) by B.2.3.3. As IT : SI = 2, ITo : SI S 2, so that To normalizes S.
We conclude from.15.1.9.1 that To S Na(J(T)) S M, so that either To = S or
T 0 E Syb(M). But in the latter case, M = !M(YTo) by 15.2.3.3, so Ki S Go S
M, whereas we saw Ki i M. Thus S E Syb(Go). Let Go := Go/02(Go) and
Mi :=Go n M. If F*(Gi) = 02(Gi) for i = 1 and 2, then as above (Gi, XiS, G2)is a weak EN-pair described in F.1.12, for the same contradiction as before. Thus
either Y ~ Z5 or Ki~ A5.
As Ki E £(Go, S) and SE Syb(Go), Ki S Li E C(Go) by 1.2.4. As Ki i M,
Li i M. As S normalizes Ki, Li ::::1 Go by 1.2.1.3. Indeed since Ki S Li and
Gi S Mi, Go= (Gi, G2) = LiMi. As IT: SI= 2 and SS Mi, F*(Mi) = 02(Mi)
by 1.1.4.7.
We claim that Go E He: If Y ~ Z5, then V = [V, Y] S 02(Y) S 02(Go),
so that G 0 E He by 1.1.4.3. Suppose on the other hand that k ~ A 5. We haveseen that Go = LiMi and F*(Mi) = 02(Mi). Further Na(02(Y)) = M since
Y ::::1 M and M E M. So it suffices by A.1.10 to show that F*(Li) = 02 (Li).
_Now as Ki S Li and Ki~ A5, 02(Ki) SS 02(Li). Therefore Li~ Li/02(Li) is
quasisimple by 1.2.1.4. Then as F*(Ki) = 02(Ki), Li does not centralize 02(Li),
so that F*(Li) = 02 (Li). This completes the proof of the claim that G 0 E He.
Let R := 02 (YS). As Y and Sare T-invariant, so is R; so as M = !M(YT) by
15.2.3.3, C(G, R) SM, and hence C(Go, R) S Mi. Further as Y ::::1 Mi, C.1.2.4
says RE B2(Mi) and RE Syl2((RM^1 ), so RE B 2 (G 0 ). Thus Hypothesis C.2.3 issatisfied with Go, Mi in the roles of "H, MH".
Suppose first that RE Syb(RLi). Then Li is a xo-block by C.2.5, so as Ki E
£(Li, S), we conclude from A.3.14 that Li= Ki. As Y = 051 (Y) normalizes R, it
centralizes the Sylow group Rn Ki of Ki= Li, and hence centralizes Li. Therefore
Ki normalizes 02 (Y0 2 (G 0 )) = Y, and hence Ki S Na(Y) S M = !M(YT), a·
contradiction. Thus R is not Sylow in RL1. However if Y i Li, then Y normalizes
YS n Li =Sn Li, so Sn Li S 02(YS) = R, contradicting R,~ Syl2(RLi).