1130 15. THE CASE .Cr(G, T) = 0holds, since in the other cases in C.1.34 there are at most two such factors. Case( 4) is eliminated as m(0 1 (Z(S))) ~ 3, contrary to 15.3.8. Suppose case (3) holds,
set Qp := [02(P),P], and let WL = W1 EB W2 with Wi E Irr+(L, WL); we may
choose notation so that [W1, P] is of rank 3 and W2 = [W2, P]. Then Z ( Q p) is a
natural module for P/Qp and Qp/Z(Qp) is the sum of two copies of the dual of
Z(Qp), impossible as [W1,P]Cw 2 (P):::; Z(Qp).
So n = 5. Let P be a maximal parabolic of L over SL containing Y+. Since
Y+S/R ~ 83 x 83, Lis generated by such parabolics, so we may assume P i
M. Thus PS E H+, and we obtain a contradiction from earlier reductions as
P/02(P) ~ 83 x L3(2) or L4(2). DObserve that 15.3.22, 15.3.23, and 15.3.24 establish Theorem 15.3.13.
We now complete the elimination of the case F*(I) = 02 (I) under Hypothesis
15.3.10, by treating the remaining subcase where m 3 (Y+) = 1 in the following
result:
THEOREM 15.3.25. If IE H+ then F*(I) =/:-02(I).
Until the proof of Theorem 15.3.25 is complete, assume I is a counterexample.The proof will be largely parallel to that of Theorem 15.3.13, except this time
our list of possibilities for L* will come from Theorem B.5.6 rather than C.2.7.3,
and the elimination of those cases will be somewhat simpler. As F*(I) = 02(I),
U1 = (Z^1 ) E R2(I) by :B.2.14.
LEMMA 15.3.26. (1) Case (2) of Hypothesis 15.3.10 holds, Y+S/R ~ 83, Y+ =
0
31(M 1 ), and R = Cs(V2).
(2) There is L E C(J(I)) with L i M1, L = [L, Y+J, and L* and L/02(L)
quasisimple.(3) Each solvable Y+S-invariant subgroup of I is contained in M1.
PROOF. By Theorem 15.3.13, m 3 (Y+) = 1. Then (1) follows from 15.3.11.5.
The proofs of (2) and (3) are the same as those in 15.3.15. D
During the remainder of the proof of Theorem 15.3.25, pick Las in 15.3.26.2.As L is a component of J(I), L = [L, J(S)*] and UL := [U1, L] is an FF-module
for AutLJ(S)(UL) by B.2.7, so L is described in Theorem B.5.6.
Recall H+, denotes the set of members of of H+ minimal under inclusion.
LEMMA 15.3.27. (1) L :SJ I.
(2) Y+:::; L.
(3) L* is not SL 3 (2n), n even, or A 6.
(4) If I E H+,*' then I = LS, and M1 is the unique maximal subgroup of I
containing Y+S.PROOF. Suppose first that L is not normal in I, and let Lo := (L^8 ). By
1.2.1.3 and Theorem B.5.6, L* ~ L 2 (2n) or L 3 (2). By 1.2.2, Lo = 031 (I), so
Y+ :S Lo. Then as Y+S/ R ~ 83 by 15.3.26.1, and S E Syh(I) by 15.3.11.1,
L ~ L 2 (2n)-since when L ~ L 3 (2), there is no S-invariant subgroup of Lo with
Sylow 3-group of order 3. Let B 0 be the Borel subgroup of Lo containing Sn L 0 ;
then Mo :=Mn Lo :::; Bo, and Bo is M1-invariant and solvable, so Mo =Bo by
15.3.26.3. Then as Y+ :S Bo, n is even. But now m3(M1);::::: m 3 (B 0 ) > 1, contrary
to 15.3.26.1. This contradiction establishes (1).