gog 12. LARGER GROUPS OVER F 2 IN .C:j(G, T)Thus we may assume that V is a 4-dimensional irreducible for A1, and it
remains to derive a contradiction. Then L is transitive on V#. As V is not invariant
under 87 , L = Mv ~ A 7. Since the groups in conclusions (2)-(4) of Theorem
12.2.13 do not have a member LE Cj(G, T) of this form, conclusion (1) of Theorem
12.2.13 must hold:^2 that is, Gv i M for each v E V#. Now Zn V = (z) is of
order 2, so that Gz i M. Recall from Notation 12.2.5 that Lz = 02 (CMv(z));set Kz := L';°. From the structure of V as an £-module, Lz = Kz ~ L3(2)
and V = [V, Lz] is the indecomposable Lz-module of B.4.8.2 with V/ (z) a natural
module. Thus Q = 02(KzT) E Syh(Ccz(Kz/02(Kz)) and Hypothesis C.2.8 issatisfied with Gz, Mz, Kz, Q in the roles of "H, MH, LH, R" by 12.2.12. Now
Kz E C1(G,T), so by 1.2.4, Kz :S; KE C(Gz), and then KE C1(G,T) by 1.2.9.l.^3
We claim that Kz = K. Suppose that Kz < K. Then K i Mz by 12.2.5.3a.
If K/02(K) is not quasisimple, then K/02(K) ~ SL2(7)/E49 by A.3.12. On
the other hand if K/0 2 (K) is quasisimple, then K/02(K) ~ L4(2) or L5(2) by
Theorem C.4.1. In either case V ::;; VK := [fh(Z(02(KT))), K] by 1.2.9.l. But
if K/0 2 (K) ~ SL2(7)/E 49 , then by 3.2.14, B1(K) :S; Cc(VK) :S; Cc(V) :S; M, so
K = B1(K)Lz :S; Mz, contrary to K i Mz. Thus K/02(K) is Ln(2) for n = 4or 5. As our tuple satisfies Hypothesis C.2.8, it also satisfies Hypothesis C.2.3.
Hence by C.2.7.2, J(Q) i 02 (KQ) and VK is an FF-module for KT/02(KT).
Then VK is described in Theorem B.5.1. As z E V ::;; VK, CvK(K) -/= 0, so by
Theorem B.5.1.2, n = 4, VK E Irr+(K, VK), and VK/CvK(K) is the 6-dimensional
irreducible for K/0 2 (K) ~Ag. Indeed as the 1-cohomology of that module is 1-
dimensional by I.l.6.1, VK is the 7-dimensional core of the permutation module for
Ag. But then from the structure of that module, 02 (NK(V)) induces the full group
of transvections with center (z) on V, contrary to 02 (NK(V)) ::;; 02 (KzT) = Q::;;
Cc(V).
Therefore Kz = K ~ Gz, so V = [V, Kz] :S; Kz = K. Let Y := Ccz (K/02(K))
and recall that Q E Syh(Y). Then by a Frattini Argument, Gz = YNc,,(Q) =
YMz, and hence Y i Mas Gz i M. Turther m3(Gz) ::;; 2 as Gz is an SQTK-
group, and Lz contains a subgroup of order 3 intersecting Y trivially, so m 3 (Y) ::;; .1.
Notice Y E He by 1.1.3.l. Then as Q E Syh(Y), while C(G, Q) ::;; M by 1.4.1.1,
Hypothesis C.2.3 is satisfied now with Y, Y n M, Q in the roles of "H, MH, R".
Therefore as m3(Y) ::;; 1, we conclude from C.2.5 that Y = (Y n M)X, where
X i Mis a block of type A3, A5, or L2(2n) for some n, and Xis normal in Gz.
As [K,X] ::;; 02(K), we conclude from C.l.10 that K centralizes X. Hence X::;;
Cc(K) ::;; Cc(V) ::;; M, a contradiction which completes the proof of 12.3.3. D
By 12.3.3, V is the 6-dimensional irreducible module for L ~ A 7 , so we now
adopt the notation of section B.3 in discussing the action of Mv on V. In particular:
LEMMA 12.3.4. (1) L has three orbits Om, m = 2,4,6, on V#, where Om is
the set of vectors in V of weight m.
(2) (Zn V)# = {e(m) : m = 2,4,6}, with e(m) := ee= of weight m, where
fh := {1, 2}, ()4 := {3, 4, 5, 6}, and ()6 := n -{7}.(^2) This application of 12.2.13 eliminates the "shadow" of M22 in Theorem 12.3.1.
(^3) Notice this eliminates the shadow of G = McL, in which K ~As; thus 02 (K) = (z), and
hence K It Ct(G, T).