1010 14. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTYAs the preimage W1 of W1 satisfies W1 = Cu(V), Wf = Cwg•(V) = W^9 n K
since CH·(V) ::; K ~ L 2 (4) x L 2 (4) as case (3) of 14.4.2 holds. Therefore as
sk E W - W 1 , for some x E G there is i := z"' E W9 with i tf K.
Then K = K 1 Kf, where K 1 E C(H) and Ki ~ L2(4). As case (3) of 14.4.2
holds, m([U,i]) = 4. Thus by Exercise 2.8 in [Asc94], CH•(i) = CH·(i). Let
Ko := 02 (CH(i)). Then K 0 ~ L 2 (4) is diagonally embedded in K, and Ko
centralizes (i, z), so Ko= 02 (CH(i)). Of course Ko acts on [U, i], and since diagonal
subgroups of K of order 3 centralize a subspace of U of rank exactly 4, it follows
that [U, i] is the A5-module for K 0.
Let D := [U, i](i, z). Then D ~ E 64 since Ko is irreducible on [U, i] of rank 4.
Further as U is extraspecial, K 0 U acts on D with CD(u) ::; [U,i]Vi for each u E
U-[U,i]Vi, and U/[U,i]Vi induces the full group oftransvections on [U,i]V1 with
center V 1. In particular CD(U) = (z), D = CuD(D), and U/[U,i]Vi is also the A5-
module for KoU/U. Further as Q = U by 14.4.7, D = 02(CH(i)) = 02(Ca((z,i)))
and UD = 02(KoUD).
Next Ko= 02 (CH.,(z)), so we conclude that z interchanges the two membersof C(H"'). Thus we have symmetry between i and z, and so U"' acts on D with
CD(U"') = (i). Therefore as Dis an indecomposable K 0 U-module with chief series
1 <Vi < [U, i]Vi < D, it follows that Y := (K 0 U, U"') is irreducible on D.Now let TD:= Nr(D) E 8yb(NH(D)), and GD:= No(D). As Y is irreducible
on D, D ::; Z(02(GD)), so as D = CuD(D), D = UD n 02(GD). As CD(TD) ::;
CD(U) = (z), Na(TD)::; H so that TD E 8yb(GD)·
Next Ko E .C(GD, TD), so Ko ::; K+ E C(GD) by 1.2.4, and as D = DU n
02(GD), K < K+· However A.3.14 contains no "B" with 02 (B) the A 5 -module
UD/D for K 0 D/UD. This contradiction completes the proof of 14.4.9. D
The elimination of the A5-module in part (3) of the next lemma 14.4.10 rulesout the shadow of the non-quasithin group 08(2). Again we obtain a contradiction
working in the 2-local corresponding to the local ff6 (2) / E 26 in the shadow.
LEMMA 14.4.10. Assume case {5) of 14.4.2 holds. Then
{1) X = {X1,X2} where X1 := 02 (02,3(H)) and X2 := 02 (B) for B a T-
invariant Borel subgroup of Ko := H^00 •
0-1.{2) There is a unique T-invariant chief factor U1 for Ko, and V2 ::; Co(T) ::;
(3) U1 is the L2(4)-module for K 0.
(4)X=Y.(5) H* ~ 83 x 85 and X1X2T/02(X1X2T) ~ 83 x 83.
PROOF. Assume conclusion (5) of 14.4.2 holds. Let Ko := H^00 • It is easy to
check that (1) and (2) hold, with U 1 := [U, x] for x E Cwg• (K 0 ); such an x exists
since W^9 * is of rank 3 by 14.4.2. Also [U, X 1 ] = U, so Vx 1 is of rank 3, and there
is a Ko-complement U2 to U1. Recall that [W9, W] ::; Eby 12.8.11.1, and that
W = Vl-. Then computing in either module for A 5 in case (5) of 14.4.2, we obtain
- -J_ - -
Vxl n U2 ::; [V2 n U2, W^9 * n K~] ::; E.
So as Vx 1 = (112, Vx 1 n U2), X1 E Y.
Assume first that U1 is the L2(4)-module for K 0. Then (3) holds, and Vx 2 =
[l/2, X2] is the F 4-point in U1 containing V 2. Now [U, x] ::; W as x acts on the
- -J_ - -