898 13. MID-SIZE GROUPS OVER F2
(3) Irr +(K, R 2 (KT)) ~Irr +(KT, R2(H)), there is VK E Irr +(K, R2(KT), T),
and for each such VK, VK :::] T, the pair K, VK satisifies the FSU, and VK =
((Zn VK)K) is either a natural module for K/CK(VK) ~ A5, L3(2), or A5, or a5-dimensional module for K/CK(VK) ~ A5.
(4) For VK as in {3), 02 z(K) = CK(VK) = CK(R2(H)). In particular, R2(H)
contains no faithful 6-dime~sional modules when K / 02 ( K) ~ A6.
PROOF. As H is nonsolvable, there exists K E C(H) by 1.2.1.1. If K :::::; M
then we obtain a contradiction by applying 13.3.8.2 with L, M, (KT) in the roles
of "K, MK, Y". Thus K i M, so [Zv, K] =f. 1 by 13.4.2.1. -Thus as Zv :::::; Z,
[R2(H), K] =f. 1, so KE Ct(G, T) by 1.2.10. Therefore Na(K) = !M(KT), and (1)
holds by parts (1) and (2) of 13.3.2, since Theorem 13.3.16 rules out K/02(K) ~
U3(3). As Ki Mand M = !M(LT), Li Na(K) so (2) holds.
By 13.3.2.3, there is VK E Irr+(K,R 2 (KT),T) and VK :::] T. As R 2 (KT):::::;
R2(H) by A.1.11, VK E Irr +(K, R2(H)). Further the action of Kon VK described
in 13.3.2.3, and VK =((Zn VK)K) by 13.3.4.2, completing the proof of (3).
By (3), either CK(VK) = 02(K), or K/02(K) ~ A.6 with CK(VK) = 02,z(K).
Therefore as 02(K) :::::; CK(R2(H)) :::::; CK(VK) = 02,z(K), either (4) holds, or
K/02(K) ~ A.6 with CK(R2(H)) = 02(K). However in the latter case by A.1.42,
there is IE Irr +(K, R2(H), T) with K/CK(I) ~ A5, and IE Irr +(K, R 2 (KT), T)
by A.1.41, contrary to (3). Thus (4) holds. D
When G is Sp5(2), Gz is solvable; thus we must eventually eliminate thecase where Gz is nonsolvable. In that case by 13.4.5 there is K E C(Gz) with
KE Cj(G, T), so that we can use our knowledge of groups in Cj(G, T) to restrict
the structure of Gz. We begin with 13.4.6; notice in particular the very strong
restrictions in part ( 4).
LEMMA 13.4.6. Assume H E 'Hz is nonsolvable. Then
{1) There exists KE C(H), and for each such K, KE Cj(G, T) and K :::] H.
Further Ki M, Li Na(K), and [Zv,K] =f.1.{2) K:::::; Gz:::::; Na(K).
· {3) K = [K, J(T)].
(4) Either M =LT, or L/02(L) ~ A.6 and M = LXT, where X is a cyclicSylow 3-subgroup of CM(L/02(L)) = CM(V).
PROOF. Part (1) is a restatement of parts (1) and (2) of 13.4.5. As KT:::::; H:::::;
Gz and N := Na(K) = !M(KT) by (1), Gz :::::; N, so (2) holds.
Let L := L2 if L/02(L) ~ A5, and L := L2,+ if L/0 2 (L) ~ A5. Then
L = (L1,L-), and by (1), Ll :::::; N but L 1,. N; thus LT E 'H.*(T,N). Now
[Z, L] =f. 1 as L_ does not centralize V 1 =Zn V, so (3) follows by applying 3.1.8.3
with Na(K), R2(KT) in the roles of "M, V".
Let Y := 02 (CM(V)). As M = Lf', M = LTY and Y :::] M. Further
Y :::::; Gz :::::; N by (2), so the hypotheses of 13.3.8 are satisfied with M, Nin the
roles of "H, MK". Therefore Y is a {2, 3}-group by 13.3.8.2. In particular if
m3(CM(V)) = 0, then CM(V) is a 2-group, so that M = LT and (4) holds. So
assume m3(GM(V)) ::'.": 1. Then by 13.3.7, L/02(L) ~ A.6 and m 3 (CM(V)) = 1 with
Lo= e(CM(V)). Thus CM(V) = CT(V)X, where x E Sy[s(CM(V)) is cyclic, and