952 13. MID-SIZE GROUPS OVER F2
By 13.8.13, H is nonsolvable, so there exists K E C(H). By 13.8.14, F(H*) =
Z(0^2 (H*)), so K* is quasisimple. Then by 13.8.12.1:LEMMA 13.8.15. Kt;_ M, so (KT)L 1 T E Hz. Further K* is quasisimple.
LEMMA 13.8.16. (1) K ::::) H, so KL1T E Hz. In particular, F.9.18.4 applies.
(2) K/02(K) is not. Sz(2n).
PROOF. Assume Ko = (KT) > K. Then Ko = KKt fort ET - Nr(K) by
1.2.1.3. Let K 1 := K and K 2 := Kt. By 13.8.15, we may take H = KoL1T.
By F.9.18.5, K ~ L 2 (2n), Sz(2n), or L3(2). Further unless K ~ Sz(2n), Ko =
0
31(H) by 1.2.2.a so L1 :::; Ko.
Suppose first that K* ~ L3(2). Then 'L1 :::; H1 :::; Fi where Hi/02(H1) ~
83 wr Z 2 , so L 1 = B(H n M) = 02 (H1) using 13.7.3.9. As m3(0^2 (H1)) = 2,
L/0 2 (L) ~ A.6, so that AutM(Li/02(L1)) ~ E4, whereas we have seen just above
that AutHnM(Li/02(L1)) ~ Ds..
Therefore K ~ L 2 (2n) or Sz(2n). Let B 0 be a Borel subgroup of K 0 containing
T 0 := T n K 0 , and set B := 02 (B 0 ). As L 1 T = TL 1 , L 1 acts on B 0 • Therefore
Bo :::; M by 13.8.13.
Let W denote an H-submodule of UH maximal subject to [UH, Ko] 1. W; thus
[UH,Ko]W/W is an irreducible K 0 -module. As K 0 T* has no strong FF-modules
by B.4.2, it follows from parts (5) and (6) of F.9.18 that either
(a) UH/Wand Ware FF-modules for K 0 T*, or
(b) [UH, Ko]= fH = (fH) for some f E Irr+(Ko,UH,T), and [W,Ko] = 0.
Let U := UH/W or iH in case (a) or (b), respectively, and let Vu denote the
projection of Vs on U.Suppose for the moment that case (a) holds. Then by Theorems B.5.1 and
B.5.6, K* ~ L2(2n), and U = U1 EBU2, where Ui is the natural module or orthogonal
module for Ki, and [Ki, U3-i] = 0. Further as UH = (Vl), Vii 1. W, so as L1 is
irreducible on V 3 , Vu is isomorphic to V 3.Now suppose for the moment that case (b) holds. Then by F.9.18.5, either
(bl) U = U1 + U2 with Ui := [U, Ki] and Ui/CuJKi) the natural or A5-module
for K*, or
(b2) U is the n~tural orthogonal module for K 0 ~ nt(2n).
Here if Vii :::; U, then U = (Vl) = UH. In particular this sub case holds when K* ~
L 2 (2n), since there we saw that L 1 :::; K 0 , so that Vii:::; [UH,L1]:::; [UH, Ko]= U.
We first eliminate the case K* ~ L 2 (2n). Since L1:::; Ko, L1:::; NK 0 (Bo) =Bo,
and hence n is even. Then m 3 (B 0 ) = 2, so as Bo:::; M, L/0 2 (L) ~ A 6 by 13.7.3.9.
Ast ET - Nr(K) acts on Lo and L 1 ,+, these groups are diagonally embedded in
Ko. Let B := 02 (B 0 ). As L 1 ,+/0 2 (L 1 ,+) is inverted bys ET n L, and [B, s]:::; L,
[B, s] is a {2, 3}-group. We conclude that n = 2 and L 1 = B.Assume that case (a) or (bl) holds. Then Vu 1. Ui as Vu is T-invariant.
Thus the projections VJ of Vu on Ui are nontrivial. As Vu = [Vu, L 1 ,+], also
VJ= [VJ, L1,+l· Similarly Lo centralizes VJ. This is impossible, as L 0 K 2 = L 1 ,+K 2
since Lo and L1,+ are diagonally embedded in K 0 , and [U 1 , K 2 ] = 0.
Therefore case (b2) holds, so U = ti H is the orthogonal module. In par-
ticular H* contains no F2-transvections, so case (2) of 13.8.8 holds. Hence the