11.2. WEAK-CLOSURE PARAMETER VALUES, AND (vNa(V1)) 769Assume Cv(L) = 1. Then X is regular on vt, so by A.1.12, X is normal
in any overgroup of odd order in GL(V 1 ). Hence O(Y) ::; NaL(Vi)(X), where
the latter group consists of X extended by Zn, and contains T. Then XT ::;I
O(Y)T = Y, so by a Frattini Argument, Y = X Ny(T) ::; NM(ili)*, since
Na(T)::; M by Theorem 3.3.1. Thus (1) holds.Assume G1 := Ca(ili) ::; M. In view of (1), we may also assume that Cv(L) =f.
1. Hence Na(R1) ::; M by 11.1.5. As G1 ::; M, Li ::;I G1, so as T acts on Li,
Li ::;I Na(V1) by 1.2.1.3. Then as R1 E Syh(Ca 1 (Li/02(L1)), by a Frattini
Argument Y = Ca 1 (L1/02(L1))Ny(R1) ::; M, so that (2) holds. D
LEMMA 11.2.4. Assume (VNa(Vi)) is abelian, and [V, W 0 (T, V)J =f. 1. If L ~
SL3(q), assume further that Ca(V 2 ) ::; M. Then
{1) Wo(T, V)::; R2.
(2) If Vg::; T with [V, VgJ =/=-1, then Vi Na(Vg).
{3) r(G, V)::; 2n.
PROOF. Byhypothesisw(G, V) = 0, so by 11.2.2.5, Lis not G 2 (q) and s(G, V) =
n by 11.2.2.3. Furthermore there is A := Vg ::; T with A =/=-1. As s( G, V) = n,
A E An(T, V) by E.3.10. Let A:= A/CA(Lg) and j,g :=Lg /CLY(A).
Our hypothesis that Ca(V2) ::; M when L ~ SL 3 (q), together with 11.2.2.3,says that r(G, V) > n. Thus if m(A/ B) ::; n, then Ca(B) ::; Na(A). Also by
hypothesis (VNa(Yi)) is abelian, so g ~ Na(V 1 ) as [V, AJ =/=-1.
We next claim there is no W::; V with [W,AJ = V 1 and m(A/CA(W)) = n =
m(W/Cw(A)). For if so, W::; Ca(CA(W)) ::; Na(A) by the previous paragraph,
and then W induces transvections on the F q-space A with axis c:;(W). If L is
SL3(q) then Cv(L) = 1 and by hypothesis V1 = [A, WJ, so V 1 is a 1-dimensional
F q-subspace of A. If Lis Sp 4 (q) then as m(W/Cw(A)) = n, Autw(A) is a root sub-
group of j,g inducing transvections on A, so [A, WJ is a 1-dimensional F q-subspace
of A, and CA(Lg) ::; [A, WJ by I.2.3.1.ii.b. Thus as [W,AJ = V 1 , [W,AJ = V 1.
Now in either case Lg is transitive on 1-subspaces of A with representative Vf,
so conjugating in Na(A) we may assume g E Na(ili), contrary to the previous
paragraph. ·
Next assume that Ai R2. Then AutA(V2) E An(AutT(V2), V2), so as R2 cen-
tralizes V2 and V 2 is the natural module for L 2 /02(L2), AutA(V2) E Syl2(AutL 2 (V2)).
Hence [V2,AJ = V1, and m(A/CA(V2)) = n = m(V2/Cv 2 (A)), contary to the claim
applied to V 2 in the role of W. Thus (1) is established.
By (1), A ::; R 2 , and hence [V, AJ ::; V 2. Suppose that [V, AJ < V2. Then
m([A, VJ) = n and A is contained in the root subgroup of a transvection in R2.
In particular, m(A) = m(A/CA(V)) = n and conjugating in L 2 , we may assume
[V, AJ = Vi, contrary to the claim applied to V in the role of W. Therefore
[V, AJ = V2, so Cv(A) = V2 and C.y-(A) = V2 since A::; R2.
We next reduce (3) to (2). Namely as A::; R2,
V = (Cv(B): m(A/B)::; 2n),so if r(G, V) > 2n then V::; Na(A), contrary to (2).
Thus it remains to prove (2), so we may assume that V::; Na(A).
Assume first that V2 =[A, VJ. Then as V::; Na(A), V2 =[A, VJ ::; V n A. By
symmetry between A and V, since V2 = C.y-(A), \'.>2 = c;(V") is an Lg-conjugate of