14.7. FINISHING Ls(2) WITH (vG1) ABELIAN 1049PROOF. Recall E = [U, Q] by definition, while [E, Q] = V from the action of
Q* on the module U. Then (1) and (2) follow from the corresponding parts of
14.7.13. Furthermore [E, U 0 J = 1by14.7.18, and [U, Ua] = V by 14.7.16.3; then in
view of 14.7.15.5 and 14.7.16.5, (3) and (4) follow from the corresponding parts of
14.7.13. DRecall that K = 02 (H), P = 02(K), and CH(U) = CqH(U) as QH = CH(U).
LEMMA 14.7.20. (1) QH/CH(U) is isomorphic to P/Cp(U) and to the dual of
U as an H -module.
(2) Either(i) [CH(U), K]::::; U, or
(ii) H has a unique noncentral chief factor W on CH (U) /U, W is of rank6, and H* is faithful on W.
(3) 02(L1) = 02(K) = P.
(4) IP: PnQI = 4 and (PnQ)/Cp(U) = [P/Cp(U),Q].
PROOF. By 14.7.4.1, [U, QH] #-1, so as H is irreducible on U by 14.7.16.1,
Cu(QH) =Vi. Next QH/CH(U) is dual to U as an H-module by 14.5.21.1, so as
U = [U,K], also QH/CH(U) = [QH/CH(U),K]. Thus QH = PCH(U), so that (1)
holds. As m(V) = 2, the duality shows that Cp(V) = P n Q is of corank 2 in P,
and also that (4) holds, since V = Cu(Q) by 14.7.16.3.
By 14.7.18, Ea= Cu°' (U) is of rank 5, and V::::; Ua n U::::; Ea using 14.7.16.3,so m(EaU/U)::::; 2 as m(V) = 3. By 14.7.16.4, Ua =UY for some y EL; thus V{::::;
V::::; U. Then CH(U) ::::; CH(V{)::::; NH(Ua) since H = Ca(V1), so [CH(U), Ua] ::::;Cu°' (U) = Ea. Hence if W := Wi/W2 with U ::::; W1 ::::; W2 ::::; CH(U) is a noncentral
chief factor for Hon CH(U)/U, then m([W, Ua]) ::::; 2 as we saw m(EaU/U) ::::; 2.
Therefore as U~ has rank 1by14.7.16.2, Q(H*, W) is nonempty. Then by D.2.17,
Wis a 6-dimensional faithful module, and [CH(U), Ua] ::::; W2, so Wis the unique
noncentral chief factor for K* on CH(U)/U. Therefore conclusion (ii) of (2) holds
in this case, while conclusion (i) holds if no such chief factor exists; hence (2) is
established.
By (1) and (2), all noncentral chief factors X for Kon P satisfy X = [X, L1],
so 02 (£1) = 02(K) = P, establishing (3). DLEMMA 14.7.21. (1) H = Ca(z) EM.
(2) Z(P)::::; Z(K).
·PROOF. By 14.7.15.4, HK:= Na(K) = !M(H). By 14.7.15.1, L1 :::l HK. Set
CK:= CHK(K/02(K)), and YK := CHK(Li/02(L1)), so that YK is of index 2 in
HK= YKT. Then as Ri = 02(L1T), Riis Sylow in YK, and hence in CKR1. As
CAut(K*))(Li))/AutK(K*) ~ SL2(3) is 2-closed,CKR1 :::l HK. Let BE Sy[s(HK), BK:= BnK, and Bo:= BnCK. Asm3(HK)::::;
2 = m3(K), Bo is cyclic with Bi := fh(Bo) =BK n Bo Sylow in L1. Then as
Ri = 02(L1T), [B1,R1]::::; 02(L1) n CK::::; 02(CK); so as Riis Sylow· in CKR1,
Bo is not inverted in its normalizer in CKR1. Therefore by Burnside's Normal p-
complement Theorem 39.1 in [Asc86a], CKR 1 has a normal 3-complement. Then
by a Frattini Argument, we may take B = BKBM, where Bo ::::; BM := N3(R1),
and BM ::::; M by 14.7.16.7. Therefore as M = LCM(L/02(L)), BM = Bi x B 0