12.1. A PRELIMINARY CASE: ELIMINATING L 0 (2) ON n EB n* 793~^9 with fixed axis B n ~^9 for i = 1 and 2. Thus as Vi is dual to Vi, m(U 1 ) :::; 1,
contradicting m(U1) = n - 2 = 2.
This contradiction shows that m(A) 2 n. Suppose n = 5. Then by lemma
H.9.2.3, we may take A :::; Ao, where A 0 is the centralizer in f' 1 of a 3-subspaceX of VJ.. Let W be a hyperplane of Vi containing X. Then m(A/CA(W)) :::;
m(Ao/CA 0 (W)) = 3, so W :S Cv(CA(W)):::; U by(*). As this holds for each such
hyperplane, we conclude Vi :::; U. But then 1 -/=- [V 1 , A] :::; V 1 n V9, contrary to
12.1.6.2.
Therefore n = 4. Then by lemma H.9.3.3, we may assume A is one of the
subgroups there denoted Ai for 0 :::; i :::; 2. Now. U < V as V 1:. Na(V9), soi-/=- 0 in view of(*) and lemma H.9.3.4. Therefore by parts (5) and (6) of lemma
H.9.3, m(U) 2 6, and Cu(A) is of rank 1 or 2 for i = 1 or 2, respectively. Next
as s(G, V) = 2 by 12.1.2.2, Cu(A) = Cu(V9) by E.3.6. Thus m(U/Cu(V9)) 2 5
or 4 in the respective cases; so as m 2 (M) = 4, we conclude that A = A 2 and
m(U/Cu(V^9 )) = 4. But as r(G, V) > m(V/U), Ca(U):::; Na(V); hence CA(V) =
CvB(U) since NA(U) is faithful on U by H.9.3.6. Therefore as m(A) = 4, CA(V) =Cv9 (U) is of rank 3. This contradicts parts ( 4)-(6) of lemma H.9.3, which say that
m(Cv9(U)) = 1, 2, or 4, since m(U/Cu(V9)) = 4. This completes the proof of
12.1.8. DHaving shown that w > 1, we turn to the other weak closure parameters of
section E.3; as usual we will obtain a contradiction from their interrelations.Recall by 3.3.2 that we may apply the results of section B.6 to any H E
1i*(T, M).
LEMMA 12.1.9. (1) If 1-/=-Xis of odd order in CM(V), then Na(X) SM.
(2) If HE 1i*(T, M), then n(H) = 2.
(3)r(G,V)2n+2.PROOF. Assume Xis as in (1); replacing X with Z(F(X)), we may assume
that X is abelian. Then [X, L] :::; 02(£). By Remark 4.4.2, Hypothesis 4.4.1 is
satisfied. As Vis not the sum of isomorphic natural modules for L/02(L) ~ Ln(2)
or nt(2), Na(X) SM by Theorem 4.4.3.Next suppose U :::; V with Gu := Ca(U) 1:. M. By 12.1.2.1, U is totally
singular. Conjugating in L, we may take Tu := Cr(U) Sylow in CM(U). Let H E
1i*(Tu, M) n Gu. By 12.1.8, Wi = Wi(T, V) :::; Q:::; Tu for i = 0, 1, so by E.3.15,
Wi = Wi(Tu, V) = Wi(Q, V), and Na(Tu):::; Na(Wi)· But M = !M(Na(Q)) by
1.4.1, so by E.3.34.2,Na(Wo(Tu, V)):::; M 2 Ca(Z(W1(Tu, V))) 2 Ca(C1(Tu, V)).
In particular Nau(Tu) :::; Mu, and hence Tu E Syb(Gu). Then as s(G, V) = 2
by 12.1.2.2, n(H) > 1 by E.3.19, so we may apply E.2.2 to conclude that a Cartan
subgroup B of the Borel subgroup H n M is nontrivial. For each odd prime p and
1-/=-X E Sylp(B), H = (H n M, NH(X)), so Na(X) f:. M as Hf:. M. If T =Tu
and p > 3, then as XT =TX, X:::; M = N 0 (V), while M =LT has no nontrival
p-subgroup permuting with f', we conclude [X, VJ = 1. This contradicts (1); thus
if HE 1i*(T, M) then p = 3 so that n(H) = 2, establishing (2).
Indeed this argument shows more generally that X -/=-1. As X is of odd order,