862 12. LARGER GROUPS OVER F2 IN .Cj(a, T)
establishing (3). D
LEMMA 12.9.7. Gi:::; Mv for 1 < i < n.
PROOF. Recall Mi :::; Mv as V is a TI-set in M by 12.2.6, so it suffices to show
Gi :::; M. Let 1 < i < n. As GL(Vi) = AutM(Vi), Gi = MiCa(Vi), so it suffices
to show Ca(Vi) :::; M. As Ca(Vi) :::; Ca(V2), it remains to show Ca(V2) :::; M.
Set (K 1 T)* := K 1 T/0 2 (K 1 T). As Out(Ki) is a 2-group, G1 = DK1T, where
D := Ca 1 (Ki) and D :::; M1 by 12.9.5.1. As [D, L]'] = 1, [D, L1] :::; 02(L1),
so D centralizes i/2. Thus Ca 1 (i/2) = DCK 1 T(V2), so it suffices to show that
Y := CK 1 r(V2)T = CK 1 r(i/2) ::; LlT·
Now Y is an overgroup of Tin K 1 T with I := 02 (L 1 n G 2 ) :::; Y, and IT is a
parabolic of LT of Lie rank n - 3 contained in LlT. By 12.9.5.2, Ki ~ A1, L4(2),
or L5(2).
We assume that Y 1:. L 1 T and derive a contradiction. Then IT < Y. If
1q ~ L 4 (2) or L 5 (2), then Y is a parabolic in KlT of rank at least n - 2 ~ 2,
so (Y n Kl)02(Y)/02(Y) ~ 83 x 83, L3(2), 83 x L3(2), or L4(2). If Ki ~ A1,
then examining overgroups of (TnK 1 ), we conclude that (YnK1)02(Y)/02(Y) is
L 3 (2), A 6 , or a subgroup of index 2 in 84 x 83. However by 12.9.6.3, m3(Y) ::; 1, so
(YnK 1 )0 2 (Y)/0 2 (Y) ~ L 3 (2). Then as Y has Lie rank at least n-2, we conclude
that n = 4, so that Li/0 2 (L 1 ) ~ L 3 (2) and IT/02(IT) ~ 83. As T induces inner
automorphisms on Li/02(L1), T:::; Ki, so Y*:::; Ki and Y/02(Y) ~ L3(2).
Set H := (Y, L 1 ), so that H E Hz. Now Y and LlT are of Lie rank 2 and
intersect in IT of Lie rank 1, so we conclude from the lattice of overgroups of T
in KiT that H/02(H) is A1 or L4(2). In either case as L]' does not centralize V2,
CH(V2) = Y; so as Y/0 2 (Y) ~ L3(2), we conclude from B.4.12 that UH= (V 2 H) is
a 4-dimensional module for H/0 2 (H) = A 7 or L 4 (2). Thus 02 (Y) is irreducible on
UH/112, so UH= (V^02 CY)). Define I2 as in 12.9.6.2. By that result, I2 normalizes
V, Iz <I G2, and I2/02(I2) ~ 83; therefore as Y :::; G2 with Y/02(Y) ~ L3(2),
02 (Y) centralizes I 2 /0 2 (I 2 ), and hence I 2 normalizes 02 (0^2 (Y)0 2 (I)) = 02 (Y).
Hence Iz acts on (V
02
CY)) =UH. But then LT= (Jz,L 1 ) ::; Na(UH), so thatNa(UH):::; M = !M(LT), contrary to H 1:. M. This contradiction completes the
proof of 12.9.7. D
LEMMA 12.9.8. (1) m(V n Vg) ::::: 1 for g E G - M.
(2) If V n Vg =f:. 1, then [V, Vg] = 1.
PROOF. We may assume V n Vg < V as Na(V) :::; M. Then if V n Vg =f:. 1,
by 12.8.3.2 we may assume V n Vg = Vi for some 1 :::; i < n, and take g E Gi by
12.8.3.3. If i > 1, we have Gi :::; M by 12.9.7, proving (1). Part (2) follows from
12.9.4 and 12.8.6. D
LEMMA 12.9.9. (1) Wo := Wo(T, V) centralizes V, so Na(Wo) :::; M.
( 2) If A := Vg n M is a hyperplane of Vg contained in T, then CA (V) = 1.
PROOF. Suppose A:= Vg n M:::; T with [A, VJ =f:. 1 and m(Vg /A) :::; 1. Let
I := Nv(A). By 12.9.8.2, An V = 1, so [A, I] :::; An V = 1 and hence I < V.
By 12.9.7, for each noncyclic subgroup B of A, Cv(B):::; Nv(Vg):::; Nv(A) =I=
Cv(A) :::; Cv(B), so Cv(B) =I. Thus m(CA(W)) :::; 1 for each A-submodule W
of V not contained in I; in particular as I< V, m(CA(V)) ::; 1.