13.1. ELIMINATING LE .c;(G,T) WITH L/0 2 (L) NOT QUASISIMPLE 867LEMMA 13.1.4. Assume LE £+(G, T). Then
(1) Either
(i) LE £*(G,T), or(ii) L/02,F(L) ~ 8£2(5), and L+ E £+(G, T) for each L+ E £(G, T) with
L<L+.
. (2) There is Mc E M(T) with Mc = !M(LT). If L E £*(G, T), then Mc =
Na(L).
(3) For some prime p > 3, X := 'Bp(L) f. 1, and for each such X, XE B(G, T)
and either
(i) X/02(X) ~ Ep2 and L/X ~ SL 2 (p), or(ii) L/02,F(L) ~ SL2(5).
(4) XE s+(G,T), and Mc= !M(XT) = Nc(X).
(5) CL(R2(Mc)) = 000 (£). In particular, X centralizes R2(Mc)·(6) Mc= !M(Cc(Z)).
PROOF. Assume L ~ L+ E £(G,T). As L E £+(G,T), L E £1(G,T), so
L+ E £1(G,T) by 1.2.9.1. Recall Tacts on L, so Tacts on L+ by 1.2.4.
As L E £+(G, T), X := 'Bp(L) f. 1 for some prime p > 3 by 1.2.1.4, and
XE B(G, T) by 1.3.3. Indeed (3) holds by 1.2.1.4.Suppose that X is not normal in L+. Then by 1.3.4, L+ appears on the list
of 1.3.4; in particular L+/02(L+) is quasisimple in each case. As T acts on L+,
conclusion (1) of 1.3.4 does not hold, and asp > 3, conclusion ( 4) does not hold.
Thus L+/0 2 (L+) ~ (S)L 3 (p) or Sp 4 (2n), with n even. Furthermore L+ E £j(G, T)
using 1.3.9.1. But this is contrary to the list of possibilities in 13.1.2.3.This contradiction shows that X :':::I L+, so L+/02(L+) is not quasisimple and
hence L+ E £+(G, T) by definition. Further taking L+ maximal, L+ E £*(G, T).
Therefore XE B*(G, T) by 1.3.8. If L = L+, then (Ii) holds. Otherwise by 1.2.4,the inclusion L < L+ is described in A.3.12 (see A.3.13 for further detail in this
case); so 1 f. 000 (£) ~ 000 (L+) and (lii) holds. This completes the proof of (1).
As XE B*(G,T), Mc:= Na(X) = !M(XT) by 1.3.7. Asp> 3 and AutL(X)
is not a 2-group, XE B+(G,T); thus XE s+(G,T), completing the proof of (4).
Further as X ~ L, it follows that also Mc = !M (LT). If L E £* ( G, T), then
LE C(Mc) by 1.2.7.1, and then L :':::I Mc by 1.2.1.3, completing the proof of (2).Recall£+ E £*(G,T), soL+ :':::I Mc by (2). AsL+ E £1(G,T), CL+(R2(Mc)) <
L+ by A.4.11. We also saw earlier that 000 (£) ~ 000 (L+)· Let Y := 02 (02,F(L+)).
Then Y centralizes R 2 (L+T) by 3.2.14, and Y :':::I Mc, so Y centralizes R2(Mc) by
A.4.11. Then as L+ :':::I Mc and R2(Mc) is 2-reduced, 02,F(L+) ~ CL+(R2(Mc)),
and hence 000 (L+) = 02,F,2(L+) = CL+(R2(Mc)). So as 000 (£) = 000 (L+) nL,we conclude that (5) holds.
Finally as Mc E 1-ie by 1.1.4.6, Z ~ R2(Mc) by B.2.14, so (4) and (5) imply
(~. DLet £+(G, T) denote the maximal members of £+(G, T); thus £+(G, T) is
nonempty whenever £+(G, T) is nonempty. By 13.1.4.1,
c+(a, T) <;;;; £j(G, T).
LEMMA 13.1.5. Assume B+(G, T) f. r/J. Then
(1) There is Mc E M(T) with Mc= !M(Cc(Z)).
(2) s+(G,T) <;;;;Mc, so Mc= Nc(X) = !M(XT) for each XE s+(G,T).