670 6. REDUCING L2(2n) TO n =^2 AND V ORTHOGONALLEMMA 6.1.10. (1) r(G, V) ~ n.(2) s(G, V) = m(AutM(V), V) = n.
(3) Suppose that VB normalizes but does not centralize V for some g E G. Thenm(VB / Cvg (V)) = n.
PROOF. As Vis the natural module for L, m(AutM(V), V) = n. By 6.1.7.2,
V satisfies Hypothesis E.6.1. Thus if n > 2, (1) and (2) hold by Theorem E.6.3. So
assume n = 2, and let U :<:::; V with m(V/U) = 1. As V is the natural module, Lis transitive on Frhyperplanes of V, so we may choose U :::] T. Then E.6.13 says
Ca(U) :<:::; M. Thus in any case, r(G, V) ~ n = m(AutM(V), V), so that (1) and
(2) are established.Assume the hypotheses of (3), and set U := Cvg(V). As VB:<:::; Na(V),
m(VB /U) :<:::; m2(LT/CLT(V)) = n.
On the other hand as V i. Ca(VB), m(VB /U) ~ s(G, V) = n by E.3.7 and (2),
establishing (3). D
LEMMA 6.1.11. Suppose VB :<:::; T with 1 =f. [V, VB] :<:::; V n VB. Then Zs =
[V, VB] = v n VB and VB E vca(Zs).PROOF. Let A:= VB. By 6.1.10.3, m(A/CA(V)) = m(V/Cv(A)) = n, so that
A is an FF*-offender on V. Therefore by B.4.2.1, A E Syl 2 (L) and Zs= [A, VJ=Cv(A). As V normalizes A by hypothesis, we have symmetry between A and V,
so Zs = CA(V). Therefore zf
1= Cv(VB-
1
) is~ 1-dimensional F2n subspace of
-1V, and hence Z~ = Z~ for some h E L by transitivity of L on such subspaces.
Thus VB= VhB with hg E Na(Zs), so VB E vca(Zs) by 6.1.9.5. D
LEMMA 6.1.12. (1) Either Na(Wo(T, V)) i. M or Ca(C1(T, V)) "f:. M.
(2) If n > 2, then Zs:<:::; C1(T, V).(3) Wo(T, V) :<:::; S.
PROOF. By Hypotheses 6.1.1, n(H) = 1 for each H E 1-l*(T, M). Hence as
H i. M, part (1) follows from 6.1.10.2 and E.3.19. Assume A :<:::; VB n T, with
w := m(VB /A) satisfying n - w ~ 2. By 6.1.10, n = s(G, V), so by E.3.10, either
A = 1 or A E A2(T, V). In either case, A :<:::; TL, so that A :<:::; S by 6.1.9.1. Since
n ~ 2, (3) follows from this observation in the case w = 0. If n > 2, (2) follows
from the observation in the case w = 1. DLEMMA 6.1.13. Let U::::; V with m(V/U) = n. Then one of the following holds:
(1) Ca(U) :<:::; Na(V).
(2) U E Z~.
(3) n is even, and U = Cv(t) for some t E M inducing an involutory field
automorphism on L.
PROOF. If U does not satisfy either (2) or (3), then CM(U) = CM(V). Then
as r(G, V) ~ n > 1 by 6.1.10.l, (1) holds by'E.6.12. DLEMMA 6.1.14. Assume n is even and U = Cv(t) for some t E T inducing
·an involutory field automorphism on L. Choose notation so that Tu := NT(U) E
Syb(NM(U)). Then
(1) R := Q(t) E Syb(Ca(U)), Na(J(R)) :<:::; M, and Tu E Syl2(Na(U)).
(2) Na(U) and Ca(U) are in 1-le.