672 6. REDUCING L2(2°) TO n = 2 AND V ORTHOGONALsolvable. This contradiction establishes (5), and so completes the proof of the
lemma. D
PROPOSITION 6.1.15. Either
(1) Cc(Zs) i. M, or
(2) n = 2, and either Nc(Wo(T, V)) i. Mor W1(T, V) i. S.
PROOF. Set W 0 := W 0 (T, V). Suppose first that Nc(Wo) i. M. Then as
M = !M(LT), W 0 i. 02 (LT) = CT(V) by E.3.16.1, so there is Vg::::; T::::; Nc(V)which does not centralize V. Set U := Cvg (V); then m(Vg /U) = n by 6.1.10.3,
so that 6.1.13 applies to U with Vg is in the role of "V". If V acts on Vg, then
by 6.1.11, Vg E vca(Zs), while VM ::::; 02 (L) ::::; Cc(V), so (1) holds. Therefore
we may assume V i. Nc(Vg). In particular Cc(U) i. Na(Vg), so that case (1) of
6.1.13 does not hold. If case (2) of 6.1.13 holds, then again (1) holds. If case (3) of
6.1.13 holds with n > 2, then v EV - Cv(Vg) induces a field automorphism on Vg
with U = Cvg(v) and Vis Vg-invariant with 1 =/= [V, Vg], so by 6.1.14.5, (1) holds
yet again. Finally if n = 2, then (2) holds as we are assuming that Nc(Wo) i. M.
Thus we may instead assume that Nc(Wo) ::::; M. Therefore by 6.1.12.1,
Cc(C 1 (T, V)) i. M. Thus if Zs::::; C1(T, V), then (1) holds. On the other hand if
Zs i. C1(T, V) then n = 2 by 6.1.12.2, and also W1(T, V) i. S, so (2) holds. D
6.1.2. Reducing to Cc(Zs) ::::; Mand n = 2. In this subsection, we consider
the first case of 6.1.15, where Cc(Zs) i. M. Our object is to establish a contra-
diction and so eliminate that case; this is accomplished in Theorem 6.1.27. In the
following chapter, we show that in the second case, G is isomorphic to M 22.
Hence in this subsection, we assume:HYPOTHESIS 6.1.16. Cc(Zs) i. M, where Zs:= Cv(TL)·
Let I:= Cc(Zs) and
1is :={HE 1i(T): Hi.Mand 02 (H)::::; I}.In particular IT E 1is, so that 1is is nonempty.
Let H denote some arbitrary member of 1is. As 02 (H) ::::; I, H = 02 (H)T::::;
IT::::; Nc(Zs). Set UH:= (VH), Hs := CH(Zs), QH := 02(Hs), and fI := H/Zs.
Notice that UIT = (V^1 ), (IT)s =I, and QIT = 02 (I). Also a Hall 2'-subgroup
DL of NL(TL) normalizes Zs and hence I, but DL n I= 1. Then as Nc(Zs) is an
SQTK-group,
mp(DLI) ::::; 2 for each odd prime p.
LEMMA 6.1.17. (1) v::::; QH, s E Sylz(Hs), F*(Hs) = 02(Hs) = QH, and
Hs SJ H =HsT.
(2) UH E 'R2(Hs), so UH::::; Z(QH)·
(3) QH = CH 8 (UH)·
(4) Fors E S-Cs(V) and Zs::::; Y::::; V, [V, s] =Zs and m([Y, s]) = m(Y/Zs).
(5) If Zs::::; Y::::; V with IV: Y[ = 2, and So is a noncyclic subgroup of S, then
Zs= [Y,So].
PROOF. As H E 1i(T), F*(H) = 02(H) by 1.1.4.6. We saw H ::::; Nc(Zs),
so that Hs = CH(Zs) SJ H; then F*(Hs) = 02 (Hs) = QH by 1.1.3.1, and S =
CT( Zs) E Sylz(Hs). Recall also T::::; H::::; IT, so that H = T(H n I)= THs. As