956 i3. MID-SIZE GROUPS OVER F2PROOF. We begin with the proof of (2); as usual, we may take H = KLiT.
By 13.8.17 and 13.8.20, K* ~ L 4 (2), L 5 (2), A5, A7, G2(2)', A5, M22, or M22-Thus
to establish (2) we may assume K/0 2 (K) ~ A5, M22, or M22, and it remains to
derive a contradiction.
By A.3.18, Li:=:: ()(H) = K. Then Li is solvable and normal in J := KnM. It
follows when K/0 2 (K) ~ A 6 that J/0 2 ,z(K) is a maximal parabolic subgroup of
K/0 2 ,z(K), and when K/02,z(K) ~ M22 that J/02,z(K) is a maximal parabolic
of the subgroup Ki/02,z(K) ~ A 6 / E 24 of K/02,z(K).Assume K/02(K) ~ M22· By the previous paragraph, IL1l3 = 3, so L/02(L) ~
A 6 rather than A5. Further case (i) ofF.9.18.4 holds with f E Irr(K, UH), and f is
the code module in view of F.9.18.2 and B.4.5. As M 22 has no FF-modules by B.4.2,
f = [UH, K] by F.9.18.7, so that V3 = [f'3, Li] :=:: J. By the previous paragraph,C1(02(LiT)) :::: C1(02(KiT)), while m(C1(02(KiT))) = 1 by H.16.2.1. This is
a contradiction, since LiT induces GL(V3) on V 3 of rank 2 inf, so that 02(LiT)
centralizes f'3. ·
Thus we may assume that K/0 2 (K) ~ A5 or M22. Then Y := 02 (02,z(K)) -=f.
1; by 13.8.13, Y :=:: M, so Y :=:: ()(H n M) =Li by 13.7.3.9. Then if Y =Li, eachsolvable overgroup of YT in H is contained in M by 13.8.13. However there is
Ki E £(KT, T) with Kif 02,z(Ki) ~ A5, so either Ki E C(H n M) or K = Ki
and T is nontrivial on the Dynkin diagram of K*. In the former case Ki = L,
contradicting M = !M(LT). As q(H*, UH) :=:: 2, the latter is impossible by B.4.5.
Thus Y <Li, so L/02(L) ~ A5. Then as Na(Lo) =Musing 13.2.2.9, Y-=/:-Lo, so
Y =Li,+· Further if K/0 2 (K) ~ M 22 , replacing K by Ki, we reduce to the case
K/02(K) ~ A5.
Now Li = ()(H n M) by 13.7.3.9, and H n Mis a maximal parabolic of H. As
V3 = [V, Li,+] and Y =Li,+ :::] H, UH= [UH, Y].
Since K* is A5, case (2) of 13.8.8 holds, so that u; E Q(H*' UH)· Let j E
Irr +(K, UH, T); by B.4.5, f is a 6-dimensional module for H*. Further as H* hasno faithful strong FF-modules by B.4.2.8, F.9.18.6 says that either f =UH or UH/f
is 6-dimensional. Set W :=UH or UH/i in the respective cases. Now LiT acts on
V3 and hence also on its image in W, so LiT* is the stabilizer of an F 4 -point in
W. Choose a as in case (2) of 13.8.8; then VO: :::: 02 (LiT*) ~ E 4 , so by 13.8.18.4,VO: = 02(LiT*). This is a contradiction as VO: is quadratic on UH by 13.8.4.6.
Thus ( 2) is established.. We next prove (1); we may continue to assume H = KLiT but Li f:_ K.
Therefore m3(K) = 1 by (2) and A.3.18, so K* ~ L 3 (2). Let X := Li+ if
L/02(L) ~ A5, and X :=Li if L/0 2 (L) ~ A5. As X = [X, T], either X :=::Kor
[X, K] :::: 02(K).
Assume first that L/0 2 (L) ~ A 6. Then Li= X centralizes K by the previous
paragraph, so that F(H) = K x X. As V 3 = [f3,X] and X :::] H*, UH=
[UH,X]. If H*/CH·(K*) is not Aut(L3(2)), then KX is generated by a pair of
solvable overgroups of X, so that KX :=:: M by 13.8.13, contrary to 13.8.12.1.
On the other hand if H*/CH·(K*) ~ Aut(L 3 (2)), then since UH= [UH,X], foreach chief factor W for H on UH with [ K, W] -=f. 1, W consists of either a pair
of Steinberg modules, or a pair of natural modules and a pair of duals for K*,
contradicting q(H*, UH):=:: 2 by B.4.5.