667. Let X := (DL, H). Then XE 'H(T) by 5.1.7.2.iii, as VL is not the S 5 -module.
Set U := (zx), Qx := 02(X) and X* := X/Cx(U). As VL is the natural module
and Z S V, for d E Df we have Cz(d) = Cz(L) < Z, so that DL is faithful on
U. Thus CvLT(U) = CT(U). Also CH(U) S CH(UH) = 02(H) from an earlier
reduction. Thus CT(U) = CH(U), so CT(U) is normal in X = (DL, H). Finally
Qx s CT(U) as u E n2(X), so Qx = CT(U) is Sylow in Cx(U).
We next show that D£ does not act on K, so we assume that D£ S Nx• (K),
and derive a contradiction during the next few paragraphs. First DL acts onthe preimage KCx(U) of K*. Recall DL acts on S, so that DL normalizes
[Cu(S),K] = [Cu(S),K] =: Ux. We saw that SE Syl2(SK), so that Ux E
n2(SK) by B.2.14. As K = [K, J(T)], we may apply E.2.3.2 to Ux to conclude
KS = Hi x · · · x H; and Ux = Ui E9 · · · E9 Us with s S 2, Hi ~ S3, and
ui := [Ux, Hi] ~ E4. As s s 2, DL normalizes Hi and ui. Therefore DL acts
on Cui (S) ~ Z2, so D L centralizes K S* and UK. Then as T normalizes K and
Cz(DL) = Cz(L),
1 <Zn Ux S Cz(DL) = Cz(L),
so that Cx(U) S Cx(Z n Ux) SM= !M(LT). Thus Cx(U) S Oz S Na(TL) n
Na(VL) using paragraph three. Set Xo := 02 (0x(U)) and C := Cx 0 (VL).
Suppose for the moment that there exists an odd prime divisor p of IXo I co-
prime to 2n - 1. Then as 02 (Cz/TL) is a subgroup of Z 2 n_i by paragraph three,
OP' (Xo) S 0. In this case set Xi := OP' (X 0 ); then Xi char X 0 ::::] X, so that
Xi ::::] X. Now suppose instead that q is any prime divisor of 2n - 1. Then
mq(M) S 2 as Mis an SQTK-group, so as DL is faithful on U, mq(Xo) S 1. Thus
if all odd prime divisors of IXol divide 2n -1, and C is not a 2-group, then for some
odd prime p, Xi :=OP' (0 2 ,p(O)) of. 1, and X 0 has cyclic Sylow p-groups, so again
Xi char Xo, and Xi ::::] X.
We have shown that if C is not a 2-group, then there is 1 #Xi= 02 (Xi) SC
with Xi ::::] X. Thus [L,Xi] S CL([V,L]) = 02(L), so that LT normalizes
02 (0 2 (L)Xi) = Xi. But then X S Na(Xi) S M = !M(LT), contradicting
Hf;. M. We conclude that C is a 2-group, and so Cx 0 T(VL) = CT(VL)C = Q from
paragraph two. Then as we saw that Cx(U) normalizes VL and TL, Xo normalizes
Baum(TLQ) = S. Therefore as DL acts on Sand KX 0 , DL acts on (SKXo) =(SK),
and hence on 02 ((SK)) = K.
Let K 1 := 02 (K n Hi). We saw that H appears in case (2) of E.2.3, so S acts
on Ki with S Sylow in SKi and SK1/02(SKi) ~ S3. As DL normalizes Hi, DL
normalizes KiS. Thus parts (a)-(d) of Hypothesis F.1.1 hold with LS, KiS in the
roles of "Li", "L 2 ". By Theorem 4.3.2, M = !M(LS), so 02( (LS, Ki)) = 1, giving
part (e). Finally as LS ::::] LT, LS E 'He by 1.1.3.1, and similarly KiS E 'He,
giving part (f). Thus a := (LS, SDL, KiDLS) is a weak BN-pair of rank 2 by
F.1.9. Indeed as NL 2 (S) SS, a is described in F.1.12. Then a is not of type L3(q)
since n(Ki) = 1 < n(L). In all other cases of F.1.12, one of LS or KiS centralizes
Z ( S) 2: Z, which is not the case. This contradiction shows that D£ does not act
on K*.
Recall that H = J(H)T, and UH is an FF-module for H/0 2 (H) ~ S3 or
S 3 wr Z 2. Thus U is also an FF-module for X. By Theorem B.5.6, J(X) =
L1 x · · · x L; is a direct product of s S 2 subgroups Li permuted by H, with
either Li~ L 2 (2) or F*(Li) quasisimple. In particular ass S 2, 02 (X) normalizes