13.1. Eliminating LE .Cf(G, T) with L/0 2 (L) not quasisimple
PROOF. By 13.1.4.5, 000 (L) = CL(Vc)· Hence (1) and the statement in (2)
that [Vc,L] =/= 1 follow from 13.1.4.3. If Ll E £1, then L* = Li by (1), so (2)
holds. D
We next establish an important technical result:LEMMA 13.1.9. {1) For each Ll E £1, Mc= !M(L1T).
{2) L = [L, J(T)].
PROOF. We will show in the first few paragraphs that (1) implies (2).
Set R := CT(L/0 00 (L)) and let Ll E £1. Observe by 13.1.4.3 that R is Sylow in
000 (LT), with R/02(LT) cyclic and D1(R/02(LT)) inverts 02 ,F(L)/0 2 ,ifl(F)(L).
Also for X E X, 02(X) S R, so that R E Sylz(XR). Set LR := NL(R)^00 ; by
a Frattini Argument, Ll = 000 (L1)NL 1 (R) = 000 (L1)NL 1 (R)^00 • As R :::l T,
Tacts on LR, so LR E C(G, T). As D1(R/02(LT)) inverts 02,F(L)/0 2 ,ifl(F)(L),
000 (LT) n LRT = R, so R = 02(LRT) and LR/02(LR) ~ L* ~ L 2 (p) for some
p > 3 by 13.1.8.1. As LE C1(G, T), LR E C1(G, T) by A.4.10.3. As NL 1 (R)^00 is
an R-invariant subgroup of LR and LR/02(LR) is simple, NL 1 (R)^00 =LR.
Now we assume that (1) holds, but (2) fails. We saw at the outset of the proof of
Theorem 13.1.7 that we may choose some HE 1-l*(T, Mc). We will appeal to 3.1.7
with Mo := LRT, so we begin to verify the hypotheses of that result: We've seen
that R = 02(M 0 ). As we are assuming that (2) fails, J(T) s 000 (LT) n LRT = R.
Thus it remains to verify Hypothesis 3.1.5.
Take V := Rz(Mo). As LR E C1(G,T), [V,Mo] =/= 1by1.2.10, so as Mo/R
is simple, R = CT(V). Finally we verify condition (I) of Hypothesis 3.1.5: Let
B := 02 (H n Mc)· As H 1:. Mc= !M(XT) by ( + ), X 1:. H, so as Tis irreducible
on X/0 2 ,ifl(X), B n X s 02,if!(X). As this holds for each XE X, B n 02,F(L) s
02 ,ifl(F)(L), so H n 000 (L) is 2-closed, and hence H n Mc acts on Rn L. Thus
H n Mc acts on LR = NL(R n L). This completes the verification of Hypothesis
3.1.5.
Applying our assumption that (1) holds to LR E £1, Mc= !M(LRT). Then
02 ((M 0 ,H)) = 1, which rules out conclusion (2) of 3.1.7. As Mc= !M(Ca(Z))
by (+), Z 1:. Z(H), which rules out the remaining conclusion (1) of 3.1.7. This
contradiction completes the proof that (1) implies (2).
So we may assume that L 1 E £1 and M E M(L1T) - {Mc}, and it remains
to derive a contradiction. By paragraph one, Ll = 000 (L1)LR, so LR E C(M, T).
Thus LR s LM E C(M) by 1.2.4, and as T normalizes LR, LM :::l M by 1.2.1.3.
We apply 13.1.6 to M and choose Y as in case (1) or (2) of that result. In
particular Y 1:. Mc, Y :::l M, and [Z, Y] =/= 1. We claim that [Y, LM] S 02(Y): In
case (2) ofl3.l.6, YE C(M), so by 1.2.1.2, either Y = LM or [Y, LM] S 02(Y). But
by 13.1.6, Y E Cj(G, T) with Y/0 2 (Y) quasisimple, so if Y = LM then Y =LR
by 13.1.2.5, contradicting Y 1:. Mc. Thus the claim holds in this case. Now assume
that case (1) of 13.1.6 holds and let M := M/02(M). In this case Y is of class
at most 2 and exponent d for an odd prime d, with ma(Y) S 2. Thus as both
Y and LM are normal in M, using 1.2.1.4, either [Y,LM] s 02(Y) as claimed, or
1 =/= D := [Od' (Y n LM),LM] :::l LM, with fJ ~ Ea2 or dl+^2 and LM irreducible
on D/if!(D). In the latter case Y = D by A.1.32.2 applied with D, Yin the roles
of "P, R". Then as [Z, Y] =/= 1, LM E C+(G, T), so Y S Ooo(LM) S Ca(Z) by
13.1.4.5, contrary to Y 1:. Mc = !M(Co(Z)). This contradiction completes the
proof of the claim.