KY/0 2 (KY), so as G 1 = KYM 1 by 11.4.3.2 and M 1 acts on S, we conclude that
S::::; 02 (G1), completing the proof of the claim.
As S ::::; 02 (G 1 ), [U, SJ = 1 by 11.5.52. Consequently r = 0 = s, so that
S = 82 ::::; U and Lis an SL3(q)-block. Since H^1 (L, V) = 0 by l.1.6, C.1.13.b says
that S = Cs(L)V. As 82 ::::; Un E and qi(E) = 1, Cs(L) = Cu(L) is abelian. As
tJ = R 1 = [R 1 , L 1 ], U = [U, L 1 ]Cu(L) = (Un L)Cu(L) and (Un L)/CunL(L) is
special of order q^5. Further Cu(L 1 ) = Cz(U)(L)V 1 = Z(U), so that U/Z(U) is of
rank 4n. As L 1 centralizes Z(U), also [K, L 1 ] = K centralizes Z(U).
Assume that L 1 < K. Then by 11.1.2, n = 2 and q = 4, so that U/Z(U) is
of rank 4n = 8 by an earlier observation. As we are assuming that L ~ SL3(4),
case (2) of 11.1.2 does not arise, so by 11.4.1.2, K is described in case (1) of 11.1.2.
As L 1 :::l M 1 and V ::::; U, CK(U) acts on L1, so CK(U) ::::; 02,z(K). Then asK centralizes Z(U), CK(U/Z(U)) ::::; 02,z(K), impossible as K/02,z(K) is not a
section of GLs(2).Therefore L 1 = K. As L is an SL 3 (q)-block, L 1 has two noncentral 2-chief
factors, so 11.5.5.3 says that U is a sum of exactly two copies of the natural module
for K / 02 ( K) ~ L2 ( q). In particular I UI = q^5. Therefore as U = C Z(U) (L) (Un L)
and l(U n L)/CunL(L)I = q^5 , we conclude that U = Un L = T n L = 02(L1) =02(K), and CunL(L) = 1 so that V = 02(L).
As K = L 1 ::::; M, KH := 02 (H) ::::; Y by our choice of H. Let X 1 :=
Cx(K/02(K) and C := (R1,KH,X1). Further since C ::::; Ca 1 (K/02(K)) and
R1 = Cr(K/02(K), R1 E Syl2(C). Set 6 := C/Cc(U). As U is a sum of two
absolutely irreducible modules for K/02(K) over Fq, 6::::; CGL(u)(K/02(K)) ~
GL 2 (q). Since Lis an SL 3 (q)-block with V = 02 (L), as before C.1.13.b says
In particular as U::::; Land tJ = R1, R1 = UCR 1 (L) by(*) and CR 1 (L) centralizes
U. Thus 6 is a subgroup of GL 2 (q) of odd order. Next CH(U)::::; NH(V)::::; HnM,
so CH(U) ::::; kerHnM(H). Therefore by B.6.8, KH = 02(KH)D for some p-group
D with D n M = qi(D). By 11.5.6.1, (p, q - 1) = 1, so as 6 is a subgroup of
GL 2 (q) of odd order, fJ is cyclic of order dividing q + 1 and [D,X 1 ] = 1. If n iseven then X 0 := Cx(L/V) is a subgroup of X 1 of order 3. Then as fJ centralizes
X1, H = DT acts on [U,X 0 ] = V, contradicting KH 1:. M. Therefore n is odd so
T =TL. Then using (*) and our earlier observations that 02 (K) = U = T n L,
T = (T n L)Cr(L) = UCr(L) = UCr(U).
Further [Cr(U),KH] ::::; CKH(U) ::::; 02,w(KH), so H acts on Cr(U) and hence by
(**), H::::; Na(T) ::::; Musing Theorem 3.3.1, a contradiction. This completes the
treatment of the case L ~ SL 3 (q).
Therefore by 11.5.1 it remains to treat the case L ~ Sp 4 (q). At several places
we use the fact that:
(!) L1.X is indecomposable on 02(L 1 ) with chief series 1 < Z(L 1 ) < 02 (£ 1 ),
and Z(L1) = C.Mv(V3).We first observe that V::::; 02 (G 1 ): For V = [V, X], so by parts (4) and (5) of
11.5.6, V centralizes KY/0 2 (KY). Then recalling that G 1 = KYM 1 and V :::l M 1