11.3. ELIMINATING THE SHADOW OF L4(q) 771PROOF. Assume that Wo(T, V) :::; 02(H). We will show that 11.3.2 holds,
contrary to our choice of G as a counterexample. When L ~ SL 3 (q), we have
Ca(V2) :::; Ca(V1) :::; M by Hypothesis 11.3.1.2, so we may apply 11.2.5 to conclude
that (VNa(Vi)) is nonabelian. As V is a TI-set under M by 11.0.3.4, this forces
Nc(Vi) i M, so conclusion (2) of 11.3.2 follows from 11.2.3.2. Next Cv(L) :::;
V1 ::::1 T, so if Cv(L)-=/= 1 then Cv 1 (LT)-=/= 1, and hence Ca(V1):::; Ca(Cv 1 (LT)):::;
M = !M(LT), whereas we just saw 11.3.2.2 holds; this contradiction establishes
conclusion (3) of 11.3.2. Hence Na(V 1 ) = Cc(V 1 )NM(V 1 ) by 11.2.3.1, so (VCa(Vi))
is nonabelian as NM(Vi):::; Mv, establishing conclusion (4) of 11.3.2.
By Hypothesis 11.3.1.2 and as 11.3.2.2 holds, L is not SL 3 (q). Since H i
M = !M(Nc(02(LT)), while H normalizes W 0 (T, V) by E.3.15 since Wo(T, V):::;
02(H), Lis not G 2 (q) by 11.2.2.5. Thus L ~ Sp4(q), so that conclusion (1) of
11.3.2 holds. But now the choice of Gas a counterexample is contradicted. DSet Wo := W(T, V). By 11.3.3, Wo i 02(H), so part (4) of Proposition 11.3.2
is vacously satisfied. Thus we only need to establish parts (1)-(3).Let VR :=(ZR), H* := H/CR(VR), m := s(G, V), and k := n(H). By B.2.14,
VR E R 2 (H). As Wo i 02 (H) while CR(VR) is 2-closed by B.6.8, there exists
A:= Vg:::; T with A*-=/= 1.LEMMA 11.3..4. (1) A E Am(H, VR)·
(2) Lis SL3(q) or Sp4(q).
(3) Either k = n, or k > n and conclusion (2) of 11.2.1 holds.(4) Nc(V1):::; Mv.
PROOF. Part (1) follows from E.3.6. By E.3.20, k ~ m. By 11.2.2, m ~ n, and
m = 2n if L is G 2 (2n). Finally by 11.2.1, either k :::; n, or conclusion (2) of 11.2.1
holds. Thus (2) and (3) hold.
Suppose Ca(V 1 ) i M. Then conclusion (2) of Proposition 11.3.2 holds, and
hence (as we saw during the proof of 11.3.3) also conclusion (3) of 11.3.2 holds.
Then by Hypothesis 11.3.1.2, Lis not SL 3 (q), so that conclusion (1) of Proposition
11.3.2 holds, contrary to our choice of Gas a counterexample. Thus Ca(Vi) :::; M,
and hence Nc(V 1 ) :::; Mv by 11.2.3.2 and 11.0.3.4. This establishes conclusion (4),
and completes the proof. DLEMMA 11.3.5. (1) 02 (H) =(KR), with KE C(H) and K/0 2 (K) ~ L 2 (2k).
(2) If k > n assume k = 2n. Then K = 02 (H).
PROOF. By 11.3.4.3, k ~ n > 1. Therefore by E.2.2, 02 (H) = (KR), for
K E C(H) described in E.2.2; in particular K/0 2 (K) is of Lie type over F 2 k.
As [Z, HJ -=/= 1 by Hypothesis 11.3.1.1, K E Ct(G, T), so K :::; K+ E Cj(G, T)
by 1.2.9.2. Now the possibilities for the embedding of K in K+ are described in
the list of A.3.12. In particular if K/0 2 (K) is not L 2 (2k), then we conclude by
comparing that list with those of Theorems B.5.1 and B.5.6, that K+T has no
FF-module-contrary to Theorem 7.0.1.
Thus (1) is established, so we assume the hypotheses of (2) with K < 02 (H).
By 11.3.4.3 and the hypotheses of (2), k = n or 2n.
Let D be a Hall 2'-subgroup of H n M, pa prime divisor of q - 1, and Dv :=
fh(Ov(D)). Ask= nor 2n, Dv-=/= 1. As K < 02 (H), D = D1 xDt for D1 := DnK