1070 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY
LEMMA 14.7.65. VUo = Cu(R1) and V = [Cu(R1),L1].
In Notation 14.7.l we chose h EH with 'Yo= 72h, a:= 7h, and Ua :::; R1. Let
Za :=A}.
LEMMA 14.7.66. U~ :::; Q* E 8yl2(K*).
PROOF. By 14.7.4.4, Q = Ri, so Q E 8ylz(K*), and the lemma follows as
~:::;~. D
F.
LEMMA 14.7.67. (1) G2:::; M.
(2) If F is a hyperplane of V, then V is the unique member of VG containing
(3) KE C*(G, T).
(4) Nc(K) = H EM.
(5) LT= Nc(V).
•
PROOF. First as H = Cc(z), Cc(V 2 ) = CH(Vz) :::; T from the action of H
on U, so (1) holds since L 2 T induces GL(V 2 ) on Vz. Then as L is transitive on
hyperplanes of V, (1) and A.1.7.1 imply (2). Similarly AutLr(V) = GL(V), so
Nc(V) = LTCc(V) with Cc(V) = CH(V) :::; CH(Vz) :::; T, so (5) holds.
Suppose K <IE C(G, T). As K = 02 (Cc(z)), [z, I] f=. 1, so IE C1(G, T), and
hence I /0 2 (I) is A 5 or L 3 (2) by 14.3.4.1. But then A.3.14 supplies a contradiction,
establishing (3).
Let MK := Nc(K); by (3) and 1.2.7.3, MK = !M(H). It remains only to
prove (4), so we may assume H < MK, and we must derive a contradiction. Let
D := CMK(K/02(K)); then MK= KDT so 02 (D) f=. 1.
Set D1 := 02 (D n M). Then KT normalizes 02 (D 102 (K)) = D1, and D1
normalizes 02 (L102(K)) = L 1. Thus D 1 centralizes Li/02(L1), and D n L1 :::;
02(L1) as L 1 :::; K, so as D 1 is T-invariant and L1 = [L1, T n L], we conclude that
D 1 centralizes L/0 2 (L). Thus LT normalizes 02 (D 102 (L)) = D 1 , so if D 1 f=. 1
then I[.:::; Nc(D 1 ) :::; M = !M(LT), a contradiction.
Therefore D1 = 1, so that D n M :::; T. Also D n H :::; CH(K/0 2 (K)) :::; T
as H =KT. As [D, L1] :::; 02 (L1), D n T:::; R1, and hence R1 E 8yl 2 (DR1). Let
81 := Baum(R 1 ). Now L1 has two noncentral chief factors on U, and hence also
two on QH /CH(U) by the duality in 14.5.21.1. Thus L 1 has at least four noncentral
2-chief factors, so Nc(8 1 ):::; M by 14.7.10.
Let E := (Vp); then EE R 2 (DR 1 ) by B.2.14, since DE He by 1.1.3.1. Further
CD(E):::; DnH:::; T, so CDR 1 (E) = 02(DR1) = CR 1 (E). Thus if J(R1) centralizes
E, then 81 = Baum(0 2 (DR 1 )) by: B.2.3.5, and then 1 f=. 02 (D) :::; Nc(8 1 ) :::; M,
contrary to D n M:::; T. Therefore J(R1) does not centralize E, so by Thompson
Factorization B.2.15, Eis an FF-module for (DR1)+ := DRi/0 2 (DR 1 ).
Suppose there exists KDE C(J(DR1)). Then as [E, KD] f=. 1, KDE C1(G, T)
by 1.2.10, so we conclude from 14.3.4.1 that KD/02(KD) ~ Ls(2) or A 5 , KD :::1 MK,
and for each VK E Irr +(KD, E, T), VK is the L 3 (2)-module or A 5 -module and is
T-invariant. As KD = [KD, J(R 1 )], we conclude using Theorem B.5.1 and B.2.14
that E = [E, KD] EB CE(KDR1), and [E, KD] is the A5-module or the sum of
at most two isomorphic L 3 (2)-modules. Thus 02 (CKn(V 1 )) f=. 1, impossible as
02 (CKn (V1)):::; D n H:::; T.
Thus J := J(DR 1 ) is solvable by 1.2.1.1. As D centralizes K/0 2 (K) and
ms(MK) :::; 2, m 3 (J) = 1 and hence Jj0 2 (J) ~ 8s by Solvable Thompson Fac-