i4.7. FINISHING L 3 (2) WITH (vG1) ABELIAN 1045previous paragraph. Therefore as JTA : TiJ ::::; JT: TiJ = 2 = JTA : 02(LoTA)J, we
conclude TA E 8yh(G). As AnZ(TA) -=f. 1, LA E C 1 (G, TA), so LA E Cj(G, TA) by
14.3.4.2 and TinLA = 02(LoTA)nLA. In particular 02 (LA)::::; Ti::::; Na(K 0 ), so as
[LA, Ko] ::::; 02(LA), LA normalizes 02 (Ko02(LA)) = K 0 , as does Ti(TA n LA)=TA. Then as Na(LA) = !M(Na(LATA)) by 1.2.7.3, Ki ::::; Na(Ko) ::::; Na(LA)·
Therefore as [Vi, Ki] = 1, Ki acts on A = (V'iLA ), so as m(A) = 3, we conclude
Ki = 05 ' (Ki) centralizes A, whereas Ki/02(Ki) is fixed-point-free on Ui ;:::: A.
This contradiction shows that s = 1. Thus H = KiTLi, so Ki 1:. M. As
Li/0 2 (Li) is inverted in T, and involutions in GL(Ui) normalizing Ki centralize
Li, we conclude that T* ~ Z4. Thus fh (T) ::::; QHQ using parts (2) and (3) of
14.7.5. Then as all involutions in LT/Qare fused to involutions in T/Q inverting
Li/02(Li), fh(T)::::; Q. Therefore J(T)::::; Ji(T)::::; Q, sousingB.2.3.3weconclude
that Na(J(T)) and Na(Z(Ji(T))) lie in M = !M(LT). Therefore as Ki i. M,
Ki = [Ki, J(T)]. Then asp > 3, a standard result of Thompson (see 26.18.a in
[GLS96]) shows that Ki::::; Na(J(T))Na(Z(Ji(T)))::::; M, a contradiction. D
LEMMA 14.7.7. If Li :::] H, then 031(H) = 1.
PROOF. Suppose His a counterexample. Then 031(E(H*)) -=f. 1by14.7.6, so
there is K E C(H) with K* ~ 8z(2n) for some odd n;:::: 3. Let K 1 := (KT); by
14.5.19, KiL1T E Hz, so without loss H = KiLiT.
As Li :::] H*, 14.7.5 says that Li~ Z3 and UH= [UH,L1]. As 8z(2n) has
no FF-module by Theorem B.4.2, examining parts (4)-(6) of F.9.18 we conclude
that W :=[UH, K]/CUH,K is the natural module for K*. This is impossible as
[UH, Li]= UH and [Li,K*] = 1, whereas Endy 2 K•(W) ~ F2n has multiplicative
group of order coprime to 3 since n is odd. D
LEMMA 14.7.8. There is no HE Hz with 02 (H*) a cyclic 3-group.
PROOF. Assume 02 (H*) is a cyclic 3-group. Then as Li ::::; H, H =PT with
P ~ Z 3 n, and L 1 = 02 (0 1 (P)0 2 (H)) :::] H. Furthermore n > 1sinceH1:. M. But
then QH = 02(L1T) = Ri, so Uy::::; QH by 14.7.5, whereas u; -=f. 1by14.7.5.1. D
Observe that 14.7.8 eliminates case (2.iii) of 14.5.20, so we may strengthen
14.5.20 to read:
LEMMA 14.7.9. Assume Y = 02 (Y) :::] H with Y* a p-group of exponent p,
and 02 (Y) < Y n M. Then p = 3, and either
(1) Y = Li, or.
(2) Y ~ 31+^2 , Li= Z(Y), Tis irreducible on Y* /Li, and Li = 02 (Y n M).
LEMMA 14.7.10. Either
(1) L 1 has at most three noncentral 2-chief factors, or(2) Na(Baum(R1))'::s; M.
PROOF. Let 81 := Baum(R 1 ). We apply the Baumann Argument C.1.37 to
the action of LT on V. If (1) fails, then by C.1.37 there is a nontrivial characteristic
subgroup C of 8i normal in LT. Thus as M = !M(LT), Na(81) ::::; Na(C) ::::; M,
so (2) holds. · D
LEMMA 14.7.11. H* is not L3(2), A5, or 85.