i4.7. FINISHING La(2) WITH (vGi) ABELIAN 104714.7.2. Eliminating solvable members of Hz· As was the case in Theorem
14.6.18 where LT/02(LT) ~ L 2 (2), in Theorem 14.7.29 of this subsection we will be
able to show that no member of Hz is solvable. The most complicated configuration
we must treat is that of case (2) of 14.7.9. We eliminate that case in the following
result:
THEOREM 14.7.14. There exists no HE Hz such that 02 (H*) ~ 31+^2.
Until the proof of Theorem 14.7.14 is complete, assume His a counterexample.
Let K := 02 (H) and P := 02 (K). By hypothesis, K* ~ 31+^2.
LEMMA 14.7.15. (1) Li :::;I H with Li= Z(K*).
{2) Ri = Q =CT (Li)~ Z4 or Qs.
(3) H = Gi = KT is the unique member of Hz.
(4) Na(K) = !M(KT).
(5) D'Y < u'Y, so that u; -::f. 1.
PROOF. Let Kz be the preimage of Z(K) in K, and Ko := 02 (Kz). Then
L 1 acts on Ko, so if Ko = [Ko, T] then Ko S M by 14.5.3.2. If Ko > [K 0 , T], then
Kos Nc(T) s M by Theorem 3.3.1.
Thus in any case, Ko SM, so we may apply 14.7.9 to Kin the role of "Y" to
conclude that L 1 = K 0 , T is irreducible on K /Li, and L 1 = 02 (K n M). Thus
(1) holds, and by (1) we can apply 14.7.5. By 14.7.5.1, (5) holds. By 14.7.5.3,
R]' = Q = CT (Li) is of index 2 in T, while as T is irreducible on K /Li, the
remainder of (2) follows from the structure of Out(K) ~ GL2(3); and we also
conclude that KE 3(G, T) in the sense of chapter 1. Then by 1.3.6, KE 3(G, T),
so (4) follows from 1.3.7. In particular K :::;I Gi, so also L 1 :::;I 01.
Let (Ji := Gi/02(G1), Ci := Ca 1 (K), and Yi E Syl3(Ci). As m3(G1) s 2,
Y1 is cyclic. Thus Oi(Yi) = L1 S Z(6i), and hence Yi s Z(N 01 (Yi)), so 61 is
3-nilpotent by Burnside's Normal p-complement Theorem 39.1 in [Asc86a]. As
Li :::;I Gi, we may apply 14.7.7 with Gi in the role of "H" to conclude that
031(G1) = 1, so that 61 = Y1 is a cyclic 3-group. Thus Y1 SM by 14.7.8. Then
as M = LCM(L/02(L)), Y1 = (Y1 n Ll) x Cy 1 (L/02(L)), so we conciude IY1I = 3
as Y 1 is cyclic. Then as Y 1 = 61 , k = F*(G 1 ). Therefore as 02 (G1) S GL3(4),
either G 1 =KT, or 02 (G 1 ) is the split extension of 31+^2 by SL 2 (3) in view of (2).
In the latter case, m 3 (G 1 ) = 3, contradicting G1 an SQTK-group, so the former
case holds with H =KT= 01 , completing the proof of (3). D
By 14.7.15.3, G 1 = H is the unique member of Hz, so UH = (VG^1 ) = U.
Similarly set D :=DH. Also in view of 14.7.15.5 and 14.7.5.1:
During the remainder of the proof of Theorem 14. 7.14, we adopt Notation
14.7.1.
LEMMA 14.7.16. {1) fJ = [fJ, Li] is a 6-dimensional faithful irreducible module
for K.
(2) u; = Z(T) is of order 2.
(3) [U, Ua] = V and V = Co(Q*).
(4) UL= {U} U U[:_^1 T.
{5)b=3.
(6) m(U/D) = 1 = m(U;).
(7) Na(Baum(R1)) SM.