i3.4. THE TREATMENT OF THE 5-DIMENSIONAL MODULE FOR A 6 901As Z = Cz(Hi) x Cz(H2) with Cz(Hi) ~ Z2, there is v E Cz(H 2 ) - Cz(Hi);
set Vv := (vK^0 ).
Assume first that Ka/02(Ko) ~ L3(2); we will show (ii) holds. As v centralizes
H2 but not Hi, Vv is a quotient of the 7-dimensional permutation module for K(J
on the coset space Ko/H2, with m(Vv) > 1. Thus by H.5.3, [Vv,Ko] is either the
3-dimensional dual of I or the 6-dimensional core of the permutation module. Thus
as m(Z) = 2, dim(Vo) = 6, and so Vo= Vv EB I or Vv, respectively. This completes
the verification of (ii).
So to complete the treatment of the case H 0 not solvable, we may assume that
K 0 ~ A5; then to establish (iv), it remains to show that Ji (T) :::1 H 0. As above, Vvis a quotient of the 15-dimensional permutation module on H 0 / H 2 , so by G.5.3, Vv
has a K 0 -irreducible quotient W 2 isomorphic to the conjugate of Wi :=I /C 1 (K 0 )
under a graph automorphism. Therefore Ko has chief factors isomorphic to Wi
and W2 on Va. We may assume that Ji(T) 1:. 02(Ho) = CH 0 (Vo), so that there is
A E Ai(T) with A* I 1. By B.2.4.1, m(Vo/Cv 0 (A)) S m(A*) + 1, so either
(I) m(A*) = 1 and m(Wi/Cwi(A)) = 1fori=1 and 2, or(II) m(A*) 2:: 2, so that m(Wi/Cwi(A)) 2:: 2 for i = 1 and 2, and hence
m(A*) = 3 and A* is a strong FF*-offender on both Wiand W 2.
These cases are impossible since by B.3.4, no involution induces a transvectionon both Wi and W2, nor does there exists a subgroup which is a strong FF* -
offender on both Wi and W 2 • This contradiction completes the verification of (iv),
and establishes (1) when H 0 is nonsolvable.
Thus we have reduced to the case where H 0 is solvable, but [Ki, K2] I 1. Thistime let Ko := 02 (Ho) and let Qi := 02(Hi)·
Suppose first that Qi = Q 2. Then Qo = Qi, T = (t) is of order 2, and
Ki ~ 83 • Thus m(V 0 ) S 2m(Cv 0 (t)) = 2m(Z) = 4. Inspecting the solvable
subgroups H 0 of GL4(2) with 02(H 0 ) = 1 and generated by a paii of distinct
83 -subgroups with a common Sylow 2-subgroup, we conclude H 0 is Eg extended
by Z 2. But this contradicts the fact that [Ki, K2] I 1. This contradiction shows
that Qi! Q2.
Suppose next that Ki does not centralize Op(Ho/Qo), for some prime p > 3
and i = 1 or 2, say i = 1. Then by A.1.21 there is a supercritical subgroup P
of a Sylow p-subgroup of the preimage of Op(Ho/Qo). By a Frattini Argument,
Ho = NH 0 (P)Qo. Let Ho := Ho/CH 0 (P/iJJ(P))Qo. By A.1.25, Ho = (Hi, H2)
is a subgroup of GL 2 (p); of course H 0 is solvable and.Hi/02(Hi) ~ 83. Thus P
is noncyclic. Further if K2 I 1 then we conclude from Dickson's Theorem A.1.3
that Ko ~ 8L 2 (3) or Ko is cyclic, and in either case Qi = Q2, so that Qi = Q2,
contrary to the previous paragraph. Thus Kp := (K:^0 ) centralizes PQo/Qo. Let
1 =f. Po S Pi E 8ylp(Kp) with Pi acting on P; as mp(Ho) = 2 = mp(P), P contains
all elements of order pin P 0 , so AutKp (Po) is a p-group by A.1.21, and hence Kp
is p-nilpotent by the Frobenius Normal p-Complement Theorem 39.4 in [Asc86a].
As Kp is generated by 3-elements, Kp is a p'-group, so as Ki is a {2, 3} group, we
conclude Ko = (Ki, K2) = KiKp is a p'-group, contrary to PS Ko.Therefore Ki centralizes 03 (F(Ho/Qo)) for i = 1 and 2, so by F.6.9, Ho is
a {2, 3}-group. Then as CH 0 (Vo) is a 3'-group, we conclude that Qo = CH 0 (Vo).
Therefore H 0 = H 0 /Q 0 = Ht and Ho = KoT with Ko a {2, 3}-group. Further
(Hi, T*, H2) is a Goldschmidt amalgam by F.6.5.1. Since Qi I Q2 by an earlier
reduction, H 0 =Ht is described in Theorem F.6.18 by F.6.11.2. As His solvable