- FINISHING THE TREATMENT OF A 6 WHEN (VG1) IS NONABELIAN 941
PROOF. Observe (1) follows from parts (4) and (6) of 13.7.3, and (2) followsfrom part (8) of 13.7.3. Finally (3) follows from (2) and 13.7.4.1. D
Pick g E L with !P = 1 and V.f! not orthogonal to Vj_. Set I:= (VH, V_k) and
Z1 := VH n V_k. Observe Q normalizes VH, and also V_k since g E Na(Q), so Qnormalizes I.
Recall that we can appeal to results in section F.9. In particular, as in F.9.6
define DH :=UH n Q'Jr, DH9 := Ufr n QH, EH := VH n Q'Jr and EH9 := V_k n QH.
Since we chose g^2 = 1, F.9.6.2 says that
(DH)^9 = DH9 and (EH)^9 = EH9·Let A:= VJ} n Q, Uo := CuH(QH), U"}; := UH/Uo, and recall He= CH(UH)·
By 13.7.9.3, H* is faithful on U"}f, and by 13.7.4.2, QH/He is dual to U"Ji as an
H-module.
Let UL := (UJ;). By 13.7.3.3, UL s Q. In particular Ufr s Q, so that
Ufr s Vj}nQ=A.LEMMA 13.7.10. (1) V{ i. UH.
(2) 02(I) = (VH n Q)(Vj} n Q) and I/02(I) ~ 83.
( 3) 02 (I)/ Z 1 is elementary abelian and the sum of natural modules for I/ 02 (I),
and Z1 /V1 V.f! is centralized by I.(4) (Ufr)Z1 = UHUfIZ1 and UfIZ1 = {x E VJ}: [VH,x] S UHZ1}.
(5) (DJt) = DHDH9 =Vi V.f!(DH n DH9) S Z1.
(6) [DH,A] = 1 and [DHB, VH] S V1.
(7) EH= EH9 = Z1 s Q n He, so [EH9, VH] s V1.
(8) L 1 has m(A*) + 2 noncentral 2-chief factors.
(9) Ufr n Vi is a complement to V1 in V3, and Vi s Z1.
(10) An QH = EH9·PROOF. If V.f! S UH, then V = Vi V{ S UH, so V H S UH is abelian, contrary
to the choice of Gas a counterexample to Theorem 13.7.8. Thus (1) holds.
By 13.7.9.1 and the choice of g, l ~ 83; e.g., if g = (4, 5) then l = ((5, 6), (4, 6)).
Let p := (VH n Q)(Vj} n Q). By 13.7.9.1, <P(VH) = V1, so <P(VH) s V1 V.f! :::! I;
e.g., Vi V.f! = (e 5 , 6 , e 4 , 6 ). Arguing as in G.2.3 with I, Vi V.f! in the roles of "L, V",
(2) and (3) hold. In particular Z1 s 02(I) s Q, so that Z1 s A.
Let P := P/Z1. For v E VHnQ-Z1, Pv :=(fl)~ E4 as Pis the sum of natural
modules for I/P. Thusifv E UH, thenPv S f)Hf)k and hence (Ufr)Zr = UHUJrZ1,
proving (4).
By F.9.6.3, [DH, Ufr] s VJ.^9 n UH = 1 using (1). Then by symmetry, DH9 s
He, so by 13.7.3.7, [DHB, VH] S V1 and [DH,A] S V.f! n DH = 1. Hence (6) is
established.
By 13.7.9.1 and (6), [I,DHDH 9 ] S V1V.f!, so
(Dk)= DHV./! = DH9 Vi
and hence (5) holds.
By 13.7.3.6, [EHB, VH] S UH, so for v E VH - Q, [EHB,v] SUH, and hence
EH9 S UfIZ1 by (4). Thus
EH9 = EH9 n UfIZ1 = (EH9 n UH9)Z1 = DH9ZJ