730 9. ELIMINATING nt(2n) ON ITS ORTHOGONAL MODULE9.2. Reducing to n =
Our first goal is to show that n :::; 2. We cannot use the uniform approach
of chapters 7 and 8, but we can still use some of the underlying techniques. For
example we will not be able to bound r as in 7.4.l using E.6.28 (which relies onE.6.27), but we can instead use extended Thompson factorization to achieve the
hypotheses of E.6.26, which we use in place of E.6.27:LEMMA 9.2.1. (1) [V, Jn-2(T)] = 1.
(2) Either [V, J 1 (T)] = 1, or n = 2 and rr E f'.
PROOF. This follows from H.1.1.2 and B.2.4.1.Recall Z := D1(Z(T)).
LEMMA 9.2.2. (1) V = fl1(Z(Q)).(2) If Q:::; S:::; T, then D1(Z(S)):::; Cv(S).
(3) Z:::; V 1 = Cv(TnL).
D
PROOF. By 3.2.10.9, Cz(Lo) = 1. Assume (1) fails. Now H^1 (Lo, V) = 0
(e.g., using Exercise 6.4 in [Asc86a]). So we obtain [D 1 (Z(Q)), Lo] i V. But
q(L 0 f', V) > 1 since V is not an FF-module, and q(AutLoT(W), W) ;:::: 1 for any
nontrivial LoT-chief factor W on D1(Z(Q)) by B.6.9.1, so q(AutL 0 T(Di(Z(Q)),
D 1 (Z(Q))) > 2, contrary to 3.1.8.1. Thus (1) is established. Further for Q :::; S:::; T,Z(S):::; Q since L 0 T E 1-{e, so (1) implies (2) and (3). D
We can now prove the analogue of 7.3.2 in the case L 0 ~ nt(2n), using 9.2.2
as an alternative to E.6.3 when n = 2:LEMMA 9.2.3. r(G, V);:::: n.PROOF. As m(M, V) = n, this follows from Theorem E.6.3 when n > 2. Thus
we may assume that n = 2 and r = 1; that is Ca(U) i M for some U of corank
1 in V-and without loss, NT(U) E Syl2(NM(U)). Now U contains a unique F-
hyperplane U 0 , and from 9.1.2.3, there are two M-orbits on F-hyperplanes, each ofthe form W-1 for an F-point W of V. Next To:= NTnL(Uo):::; NT(U), so that
Cv(NT(U)):::; Cv(To):::; Uo:::; U. (*)
But Un Z =/= 1, so by E.6.10.4, D1(Z(NT(U))) i U. On the other hand by 9.2.2.2,
D1(Z(NT(U))):::; Cv(NT(U)), so Cv(NT(U)) i U, contradicting(*). DFrom now on, let H E 1-{* (T, M). Recall that H is a minimal parabolic in
the sense of Definition B.6.1 by 3.3.2.4. Further by 3.1.8, H centralizes Z. Set
K := 02 (H). If n(H) > 1, let B be a Cartan subgroup of H n M.
LEMMA 9.2.4. (1) n(H) ;:::: n - 1.
(2) If n(H) = 1, then [V, J1(T)] =/= 1 and n = 2.
PROOF. To prove (1), we may assume n;:::: 3; we will apply E.6.26 with j :=
n - 2;:::: 1. By 9.2.3, r > j, and by 9.2.1.1, Jj(T) :::; CT(V); therefore (1) follows