673Vis the natural module for L, [S, VJ= Zs; therefore Vis central in SE Syl 2 (Hs),
and hence UH~ (VH) E R2(Hs) by B.2.14. This establishes (2).
Next CH(UH) :S NH(V) :::; M, and further X := 02 (CH 8 (UH)) :::; CM(V) :::;
CM(L/02(L)), so that LT normalizes 02 (X0 2 (L)) = X. Hence if X i-1, then
H :::; Nc(X) :::; M = lM(LT), contradicting H i_ M. Therefore X = 1, soCH 8 (UH) :::; 02(Hs) = QH; then (3) follows from (2). Parts (4) and (5) follow
from the fact that V is the natural module for L. D
Let G1 := LT, G2 := H, and Go := (G1, G2). Notice Hypothesis F.7.6 is
satisfied: in particular 02(Go) = 1 as G2 i_ M = lM(G1). Form the coset geometryr := r(Go; G 1 , G 2 ) as in Definition F.7.2, and adopt the notation in section F.7. In
particular for i = 1, 2 write 'Yi-l for Gi regarded as a vertex of r, let b := b(r, V),
and pick ')' E r with d( ')'o, ')') = b and V i_ av). Without loss, ')'1 is on the geodesic')'o,')'1, · · · ,')'b := 'Y·
Observe in particular that UH plays the role played by "V.Yl" in section F. 7. For
a:= ')'oX E I'o let Va := vx. For (3 := ')'1Y E I'1 let Zf3 := Z~ and Uf3 = U'Jr.
Notice that by 6.1.17.1 and F.7.7.2, V:::; QH:::; G~~l, so that by F.7.9.3:
LEMMA 6.1.18. b > 1.
LEMMA 6.1.19. Suppose there exists H E Hs n H*(T, M) with b odd. Then
n = 2 and (V^1 ) is nonabelian.
PROOF. Assume bis odd. By 6.1.18, b > 1, sob 2: 3. Then UH is elementary
abelian by F.7.11.4.
Further by F.7.11.5, UH :::; G'Y and U'Y :SH, so applying 6.1.12.3 to suitable
Sylow 2-subgroups of G"I and H, we obtain:
UH :S Cc'Y(Z"I), and U'Y:::; Cc'Y 1 (Zs)= Hs. (!)
Observe that the hypotheses of F. 7.13 are satisfied: We just verified hypothesis
(a) of F.7.13, and hypothesis (c) holds by 6.1.1.2 as H E H*(T,M). Also as
H E H* (T, M), H n M is the unique maximal subgroup of H containing T by
3.3.2.4. Finally H n M = NH(V) by 6.1.8, so hypothesis (b) of F.7.13 holds.
Applying F.7.13 to A:= UH, we conclude there is a Er(')') with B := NA(Va) of
index 2 in A. Write E := Va. If [E, BJ = 1, then as s(G, V) > 1 by 6.1.10.2, for
eachhEH
E :S Cc(B) :::; Ca(B n Vh) :::; Ca(Vh).
But then [E, AJ = 1, contrary to B <A. Therefore [E, BJ i-1. So as A::; Ca'Y (Z"I)
by (l), [E,BJ = Z"I by 6.1.17.4.
Suppose that E :::; QH. Then [A, EJ :::; Zs by 6.1.17.2, so that Z'Y = [B, EJ ::;
[A, EJ ::; Zs. Hence Zs= Z"I, as these groups are conjugate and so have the same
order. This is impossible, as V:::; 02(Cc(Zs)) by 6.1.17.2, while Vi_ 02(G'Y) by