836 12. LARGER GROUPS OVER F2 IN .Cj(G, T)PROOF. By 12.6.30.3 we may apply F.9.18 to conclude that K /Z(K) is a
Bender group, L 3 (2n), 8p4(2n)', G2(2n)', L4(2), L5(2), or A1, or K 0 = K* ~ M22
or M 22. By 12.6.29, n(H) > 1 as H 1:. M, so by E.1.14 either(i) K/02,z(K) is a Bender group over F2,,,, L3(2n), 8p4(2n), or G2(2n), with
n := n(H) > 1, or
(ii) K 0 ~ M 22 or M22, and n(H) = 2.
Pick i E Irr +(K 0 , f) H, T), and adopt the notation of F.9.18; in particular IH :=
(JH). Let Le := 02 (CL 1 (K 0 )) and LK := 02 (L1 n Ko). In case (i), a Borel
subgroup of Ko over T n K 0 is contained in M by 12.6.29.
Suppose first that m 3 (K 0 ) = 0. Then K ~ 8z(2n). In particular, LK = 1 as
L 1 = 031 (L 1 ). By F.9.18.3, q(AutKaT(J), J) ::::; 2, so J is described in B.4.2 or B.4.5.
Hence f is the natural module for K, and either F.9.18.4i holds with K =Ko and
I= IH, or F.9.18.5iiia holds, with K <Ko and fH = f EB ft fort ET - NT(K).
Now 8z(2n) has no FF-modules by B.4.2, so by F.9.18.7, IH = [UH, Ko]. As
L 1 = [L 1 , T], while Out(K) is cyclic, either Li =Le; or K <Ko, Le is of order
3, and an element of order 3 in Li - Le acts as a nontrivial field automorphism on
each component of K 0. As Le is L 1 T-invariant and nontrivial, V 5 = [l/5,Lc] since
LiT is irreducible on l/5. Then as Le '.S! Hand UH= (V 5 H), UH= [UH, Le] and
Le is a 3-group, so CuH(Le) = 1 by Coprime Action. However Le stabilizes K
and I, and EndK(f) ~ F 2 ,,, with n odd so that 3 does not divide 2n -1. Therefore
[IH,Le] = 1, contradicting CuH(Le) = 1. Therefore m3(Ko) > 0.
Suppose next that m 3 (K 0 ) > 1. Then comparing A.3.18 to the list of groups in
(i) and (ii), either Ko= 8(K 0 ), so that Li::::; Ko, or LiKo/02z(Ko) ~ PGL'3(2n)
or L~'^0 (2n).
Suppose first that K ~ M 2 2 or M2 2. Then there is Ho E 'H(T) n H with
02 (H 0 /02,z(H 0 )) ~ A5. Therefore n(Ho) = 1 by E.1.11, E.1.13, and E.1.14.1, so
Ho ::::; M by 12.6.29. But then 031 (Ho) ::; 031 (M) = L by 12.6.1.5. impossible as
L has no T-invariant A5-section.
Therefore as m3(Ko) = 2, K 0 is one of the Lie-type groups L 2 (2n) x L 2 (2n),
(8)L3(2n), 8p4(2n), or G2(2n) determined earlier. As L]'::::; 02,3(MH), and MHT
is the normalizer of a parabolic subgroup while n > 1, it follows that MKo = MnK 0
is a Borel subgroup of K 0 , with n even; hence K is not (8)U 3 (2n) as m 3 (K) = 2.
Recall LiT/02(LiT) ~ 83 x 83 or 83 wr Z2, so T/02(LiT) is noncyclic. Thus as
TnKo::; 02(LiT), T/02(LiT) projects on a noncyclic 2-subgroup of Out(K 0 ), so
K 0 ~ (8)L3(2n) or L2(2n) x L 2 (2n). We return to these cases in a moment.
We now consider the case m 3 (Ko) = 1. Here K = K 0 , and K ~ L 2 (2n),
L3(2m), m odd, or U3(2k), k even. As Li = [L 1 , T], and Out(K) is abelian, L 1
induces inner automorphisms on. K, so that L]' = Le x L'K. Then Le -=/=-1 as
m3(K) = 1, while Le < L]' as L]' has 3-rank 2. Now T normalizes K and L 1 , and
hence normalizes LK and Le; hence for X E {K, C}, LxT/0 2 (LxT) ~ 83. As
TLK = LKT, K is not L 3 (2m) since mis odd, and if K ~ L 2 (2n), then n is even.
Wenowhandletogethertheremainingcases: K 0 ~ (8)L 3 (2n), L 2 (2n)xL 2 (2n),
L2(2n), and U3(2n), with n even. Let TL := T n L; then Li = [L1,TL] and TL
acts on LK, so LK = [LK, TL]· Moreover from the structure of Aut(K 0 ), unless
K* ~ (8)L3(4) or L 2 (4), there exists a prime p > 3 and a nontrivial p-subgroup X
of the Borel subgroup MKo with XT = TX and X = [X, TL], so that
X = [X,TL]::::; [X,LJ::::; L.