PROOF. Let HE H*(T, M). If Cv(L) -/= 1, then Lis not SL3(q) by 11.0.3.3,
and [Z, H] -/= 1 as Hi_ M = !M(LT), contrary to 11.3.2.3. Then 11.0.2.2 completes
the proof of (1).
Suppose 11.1.2.3 holds. Observe I satisfies the hypotheses for L, so applying
11.0.2.2 and (1) to V 1 E Irr +(I, R 2 (IT)) in the Fundamental Setup (3.2.1), we
conclude Vi is the natural module for I* := I/C1(Vi) = I/02(I) ~ Sp4(4). From
the proof of 11.1.2.3, L 2 stabilizes a line of VJ. In 11.1.2.3, we also have L/02(L) ~
L ~ G 2 (4). By 11.2.2.5, Wo := Wo(T, V) :::; Cr(V), and hence Wo :::; CL 2 r(V2) =
02(L 2 T); furthermore Na(Wo) :::; M, so as I i M, Wo i 02(IT) = CIT(Vi ).
Thus 1 -/= W 0 :::; 02 (L2T*) = 02(L2) = R2, and R2 is of rank 6. Let A := Vg :::; T
with A -/= 1. Let R be a root subgroup of 02 (L2), and AR the preimage in A of
A n R. Then m(A /A'R):::; m(02(L2)/R) = 4. By 11.2.2.3 r(G, V) > 2n = 4,
so Ca(AR) :::; Na(A). Hence from the action of R2 on Vi,
Vi = (Cv 1 (AR) : R* :::; 02(L2) ) :::; Na(A).
On the other hand by 11.2.2.3, s(G, V) = 4, so m(A*) :?:: 4 by E.3.10. It follows that
[Vi, A]= Cv 1 (A) is of rank 4 so m(Vi/CVi(A)) = 4; thus as A is the natural module
for Lg, AutVi(A) is contained in a long root group of AutLY(A) ~ G2(4) of rank
2 (e.g. see (6) and (13) of B.4.6), forcing m(Vi/Cv 1 (A)) :::; 2. This contradiction
establishes (2).
If L1 = K, then (3) is immediate. Otherwise by (2), K is described in case
(1) or (2) of 11.1.2. In those cases, observe that L 1 has no nontrivial 2-signalizers
in Aut(K/O=(K)), so that R 1 :::; O=(KT). Since 02 ,F(KT) is of index 1 or 2 in
O=(KT), [R1, X] :::; 02(KT), so that (3) holds in these cases also. D
We can now return to our study of the embedding of L1 in Ca(Vo) for 1-/= Vo :::;
V1, begun in the initial section of the chapter. For 1-/= Vo :::; Vi with T:::; Na(Vo),
define K(V, Vo):= (Lfa(Vo)); and for z E Vt, let K(V, z) = K(V, (z)).
LEMMA 11.4.2. Let z E Cv(T)#. Then K(V,z) = K(V, Vi).
PROOF. Let K := K(V,z) and K1 := K(V, Vi). By 11.1.1, KE C(Na((z)))
and K1 E C(Na(Vi)). Then K1:::; Ca(V1):::; Ca(z), so K1 = (Lf^1 ):::; K.
We assume that K1 < K and derive a contradiction. As K 1 :SJ Ca(Vi),
K i_ Ca(Vi). Further L 1 < K, so K is described in case (2) or (3) of 11.1.1. As
KE L(G,T), K:::; K+ E L*(G,T).
Assume first that K < K+. Then the embedding of Kin K+ is described
in A.3.12. The pairs K/0 2 (K), K+/0 2 (K+) appearing there with K described in
case (2) or (3) of 11.1.1 (cf. also A.3.13) are: L3(4), M2 3 ; A7, M23; L 2 (p), (S)LHp),
for a prime p:?:: 11; SL2(p)/Epo, (S)L 3 (p) for a prime p:?:: 5; M22, M23; M22, J4;
and SL2(5)/Po, SL2(5)/P1, where Po and P1 are suitable nilpotent groups of odd
order. Moreover as K < K+, [z, K+] -/= 1, so in the last case [z, O=(K+)] -/= 1,
contrary to 3.2.14. Therefore K+/02(K+) is quasisimple. Further as [z,K+]-/= l,
K+ E Lj(G, T). But this contradicts Theorem 7.0.1, since K+ has no FF-module
by Theorem B.4.2.
Therefore K = K+ E L*(G,T). As K 1 < K, [V1,K] -/= 1. But by A.1.6,
02(KT):::; Ri, and Vi:::; Z(R1), so
Vi:::; D1(Z(02(KT))) =: VK·
In particular [VK,K]-/= l, so KE Lj(G,T) by A.4.9.