13.1. ELIMINATING LE .Cr(G, T) WITH L/0 2 (L) NOT QUASISIMPLE 871K is nontrivial on [DL, R], we may choose A so that KA := [K, A] is nontrivial on
[ D L, R]. Also by A.1.17, k A is generated by the fixed points of hyperplanes of A;
so we may choose a hyperplane B of A and a subgroup kB= [kB, A] of CkA (B)
of order d, such that kB is nontrivial on [DL, R]. By the Thompson Ax B-Lemma,
kB is nontrivial on C[D,LRJ(B). In this case as m(A)::::; 2, E = 0 or 1.
Let I be an irreducible KEA-submodule of W. Then as H 1 = LR x KB.A,Clifford's Theorem says that W is the direct sum of r copies of I for some r, and
CaL(W)(KBA) ~ GLr(F), where F := EndkB(I). As LR::::; CaL(W)(KB) with LR
nonabelian, r > 1.
As A E A1(R), by B.2.4.1,
m(D/CD(A))::::; j + m(A).
Thenr ·m(I/Cr(A)) = m(W/Cw(A))::::; m(D/CD(A))::::; j +m(A) = j + E+m(A.). (*)
Thus as r > 1 ~ E, it follows from (*) that m(I/Cr(A.)) ::::; m(A.), so that I is
an FF-module for KB.A. Therefore by Theorem B.5.6, KB is not ad-group for a
primed> 3. Therefore KB = K, E = 0, and K ~ Z3, A5, or L3(2). Further if
I is the natural L 2 (4)-module, then m(A.) ::::; m2(Aut(L 2 (4)) = 2 ::::; m(I/Cr(A)),
contrary to (*). Thus if k ~ A5, then I is the A5-module by B.4.2. It follows
from B.4.2 that F = F 2 for each of the possible irreducible FF-modules I for each
K, so LR ::::; GLr(Fz). Since LR ~ Lz(p) for p ~ 5, we conclude r ~ 3, with
LR ~ Lz(7) ~ L 3 (2) and each IR E Irr +(LR, W) of rank 3 in case of equality.
Now m(A) ::::; m 2 (Aut(K)) ::::; 2 and m(W/Cw(A)) ~ r ~ 3, so all inequalities
in (*) must be an equalities, and in particular LR ~ L 3 (2), m(IR) = r = 3,
m(I/Cr(A)) = 1, j = 1, and m(A.) = 2. As r = 3, K ~ L 3 (2) and m(J) = 3, so
NaL(W)(LR) ~ Ls(2) x Ls(2); hence A::::; K, LRK ~ L3(2) x L3(2), and Wis the
tensor product of natural modules for the two factors. Furthermore as m(A.) = 2
and m(I/Cr(A.)) = 1, A is the group of transvections in K with a common axis
on I. In particular A is the unique such subgroup of T n K, so A = J ~ E4, and
hence J = J 1 (0 2 (H1)A). Then NK(A) ::::; Na(J) ::::; Mc by an earlier observation.
Now k is simple, so A is faithful on each nontrivial k-section of D. Since (*) is an
equality, we conclude that [D, k] =WE Irr +(LRKT). Now ZnW is a 1-subspace
of W centralized by the parabolic of k stabilizing the group of transvections on
I with a common center, and Ca(Z n W) ::::; Mc = !M(Cc(Z)) by (+). Thus
K = (CK(ZnW), NK(A.))::::; Mc, contrary to(!). This contradiction completes the
proof of (1), and hence of 13.1.9. D
LEMMA 13.1.10. One of the following holds:
(1) L* ~ L 2 (4) and U/Cu(L) is the Lz(4)-module.
(2) LT ~ 85 and U is the 85 -module.
(3) L* ~ L 3 (2) and U is the sum of at most two isomorphic natural modules.
(4) L* ~ L3(2), m(U) = 4, and [Z, L] = 1.
PROOF. By 13.1.9.2, Vc is an FF-module for M; with L* ::::; J(M;, Ve)· By
13.1.8.1, L ~ L 2 (p) for a prime p ~ 5. Thus L is isomorphic to L 2 (5) or Lz(7) by
B.5.5.liv. Then cases (2) and ( 4) of Theorem B.5.6 do not hold; and in cases (3)