1044 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cf(G, T) IS EMPTY
PROOF. As Li ::::I H, H normalizes 02 (LiQH) =Li. Then 14.5.15.3 says
that Li 3;! Z 3 and (5) holds. As LiT / Ri 3;! 83, it follows that JT : CT· (Li)J = 2.
Further Li ::::I G 10 ,-y 1 , so as "( 2 is conjugate to 'Yo in H ::::; Gi, Li ::::I G, 1 , 12.
Then as Li 3;! Z 3 , Li centralizes 02 (G~ 1 m). Thus (4) follows as U 1 ::::; 02(G 1 i, 12 )
by F.9.13.2, and similarly [UH,L 1 ] ::::; 02 (G 1 ), where L 1 := Lfb. Therefore as
U 1 /Ai = [U 1 /Ai,L 1 ] by (5) where L 1 has action of 9rder 3 commuting with that
of UH, m([U 1 /Ai, u]) is even for each u E UH, sou does not induce a transvection
on U 1 /Ai. Thus D 1 < U 1 by 14.5.18.1, establishing (1).
Part (2) is contained in 14.7.4.4. Then by (2), Q* = CT·(Li), completing the
proof of (3). D
LEMMA 14.7.6. F(H*) is a 3-group.
PROOF. Suppose H is a minimal counterexample, let p > 3 be prime with
Hi := fh(Z(Op(H*)) -=/= 1, and pick P E 8ylp(Hi) where Hi is the preimage
of Hi. Since p > 3, Hi n M = QH by 14.5.20, so H = P LiT by minimality of
H. Similarly H is irreducible on P. By 14.5.18.3, q(H*, UH) ::::; 2, so by D.2.17,
p = 5 and P::::; K ::::1 H with K* =Ki x · · · x K;, Kt 3;! D10, and Ui := [UH,Ki]
of rank 4. As usual s = ms(H*) ::::; 2 as H is an SQTK-group, so Li = 02 (Li)
normalizes Kt. Then [Kt,Li] = 1, so that Li ::::I KLiT = H. Hence 14.7.5
says we may adopt Notation 14.7.1, Q* = CT·(Li), u; centralizes Li of order 3,
and UH = [UH,Li]. In particular, u; is faithful on P since F(H) = LiP.
Then since u; E Q(H*, QH), either Z2 3;! u; ::::; Kt for some i, or s = 2 and
E4 ~ u; ::::; K'j_K2. In either case 2m(U;) = m(UH/Cu-H(U;)), so by 14.5.18.2,
u; induces a faithful group of transvections with center A 1 on a subspace DH of
UH of codimension m(U;). But if u; is of rank 2, this is not the case, so we may
choose notation so that Z 2 ~ u; ::::; Ki- Therefore A := [UH, U 1 ] ::::; Ui. Since Li
centralizes u;' Li normalizes A.
Now by the choice of 'Yin Notation 14.7.1, 1 = m(U;) ;:::: m(UH/DH) ;:::: 1,
so m(U;) = 1 = m(UH/DH), and hence as discussed in Remark F.9.17, our hy-
potheses are symmetric between 'Yi and 'Y· As U 1 centralizes no hyperplane of UH,
Ai= [DH, U 1 ]::::; A by F.9.13.6. Thus by the symmetry, Vi::::; A, so that m(A) = 3
as m(A) = 2, and Li acts on A as Li acts on A.
Assume first thats= 2, and let Ti := NT(Ki). Then Ti is of index 2 in a Sylow
2-subgroup of G, Ti E 8yb(NH(A)), and by 14.5.21.1, LiTi induces A4 or 84 on A
and centralizes Vi. Again by the symmetry between 'Y and 'Yr, Na"! (A) induces A 4
or 84 on A and centralizes Ai-=/= Vi, so we conclude that Na(A) induces GL(A) ~
L3(2) on A. Therefore by 1.2.1.1, Na(A) = LACa(A) for some LA E C(Na(A))
with LA/CLA (A) 3;! L3(2). By 1.2.1.4, LA/02(LA) ~ L3(2) or 8L2(7)/E49, and
in either case Aut(LA/0 2 (LA)) is a 5^1 -group. Thus as Ko := 05 ' (K 2 ) acts on
A, [LA, Ko] ::::; 02(LA)· Also Li is nontrivial on A, so either Li ::::; LA or Li is
diagonally embedded in LACa(A). As [Li, K2] = 1, Li acts on K 0 •
Next [Ko, Tin LA] ::::; Kon 02(LA) ::::; 02(Ko) ::::; QH, so (Tin LA)* centralizes
K 0. Therefore as CaL(u 2 )(K2) 3;! Z 15 , Tin LA centralizes U 2 , and hence also
centralizes Li and Li/02(Li).
Let Lo be the preimage in LA of AutL 1 (A). As AutL 1 (A) ~ A4, Na(Lo) n
Na(A) contains a Sylow 2-group ofNa(A). Thus Ti::::; TA E 8yb(Na(A)) with TA
acting on Lo. Further each t ETA - 02(LoTA) inverts Lo/02(Lo), sot¢:. Ti by the