i5.l. INITIAL REDUCTIONS WHEN .Cr(G, T) IS EMPTY 1095Now by Remark 0.1.19, we may take C2(SH) = C2(T) and Ci (T) :::; Ci (SH). Then
Na(C2(SH)) :::; M by 15.1.17, so G2(SH) is not normal in H by (1). ThereforeL :::; Ca(Ci(SH)) :::; Ca(C1(T)) :::; Mc since Ci(T) :::; Z and Mc = !M(Ca(Z)).
However this contradicts our hypothesis that L 1:. Mz.
Suppose next that L* ~ SL3(2n), Sp4(2n), or G 2 (2n) with n > 1. Let Pi,
i = 1,2, be the maximal parabolics of Hover SH, and Li := 02 (Pi)· Then
Li E L(Gi,S) with Li/02(Li) ~ L2(2n), but Li is not a block since 02 (L;) i= 1
and [U, Li] =f. 1, so by the previous paragraph, Li:::; Mz. Thus L = (Li, L 2 ) :::; Mz,
contrary to hypothesis.
If L ~ L4(2) or L5(2), then as SL :::; ML < L and SL E Syh(L), ML is
a proper parabolic subgroup, so conclusion (c) of (2) holds. If L is one of the
remaining possibilities, then L/0 2 ,z(L) is listed in conclusion (b) of (2). Thus to
complete the proof of (2), it remains to assume that L E C(G 1 ) with L* ~ L 3 (2),
A5, A1, A5, or G 2 (2)', and to verify the final two sentences of (2b).
As Lis not a xo-block, by 15.1.18.5 we may apply 0;2.4 to conclude that Qi acts
on L. So since L EC( Gi), the hypotheses of 0.2.7 are satisfied, and hence Lis listed
in 0.2.7.3. Set Zs:= Cu(SH) and Zu := Gru,LJ(SH); then Z:::; Un Z(SH) =Zs.
By B.2.14, U = Cz 5 (L)[U,L], so Zs= Cz 8 (L)Zu. Since Mc= !M(Ca(Z)),
CL(Zu) = CL(Zs):::; CL(Z):::; ML. (*)
Suppose either that L/02,z(L) is of rank 2 over F2, or that Lis an exceptional A1-
block. The module [U, Ll/Cu,L(L) is described in case (i) or (ii) of Theorem B.5.1.1,
and in each module U, CL(Zu) is a maximal subgroup of L. Therefore as ML< L,
the inequalities in(*) are equalities, so that ML= CL(Zu) = CL(Z(SL)). Suppose
instead that L is an ordinary A1-block; then Zu contains vectors Zw of weights
w = 2, 4, 6, and there is z E ZnCz 8 (L)zw for some w. Then CL(zw) :::; CL(z) :::; ML.
But unless w = 2, Auto 2 (cLsH (.zw)) ([U, L]) contains no FF* -offenders by B.3.2.4,
contrary to 0.2.7.2. Thus w = 2 and as CL(z 2 ) is maximal in L, CL(z 2 ) =ML.
We have shown that if L is one of the four blocks listed in (2b), then (2b)
holds, so we may assume L is not one of these blocks. If L is A 6 or G 2 (2)',
then by 0.2.7.3, Lis an A 6 -block or G 2 (2)-block, contrary to this assumption. If
L ~ L 3 (2), then by 0.2.7.3, Lis described in 0.1.34. By the previous paragraph,
Mz is the parabolic centralizing Zs, so case (1) or (5) of 0.1.34 holds as the other
cases exclude Qi normal in that parabolic; thus Lis an L 3 (2)-block, again contrary
to assumption. If L/02,z(L) ~ A1, then by 0.2.7.3, Lis either an A1-block or an
exceptional A 7 -block. Again the first case contradicts our assumption, and in the
second case, we showed in the previous paragraph that M£ is the maximal subgroup
of index 15 fixing Zu, rather than the subgroup of index 35 in L * appearing in case
( d) of 0.2. 7.3.
Thus it only remains to eliminate case ( c) of 0.2. 7.3, where L is a block of
type A 6 : Here by B.4.2, the only parabolic P of L such that 02 (P) contains an
FF-offender is not the parabolic CL(Z) =ML, contrary to 0.2.7.2. This completes
the proof of (2).
Finally we prove (3), so assume L is a component of Gi. By 15.1.18.6 we
can apply 1.1.5, and in particular Lis described in 1.1.5.3. Since O(G1) = 1 by
15.1.19.1, Z(L) = 02(L) is a 2-group. By 1.1.5.1, Mz E He, so ML E He by 1.1.3.1.
By 1.1.5.3, Z is faithfully represented on L with Autz(L) :::; Z(Auts(L)). Thus if