13.5. THE TREATMENT OF A5 AND Aa WHEN (v;^1 ) IS NONABELIAN 917
LEMMA 13.5.6. Assume n = 6; then(1) V::; 02(G2).
(2JVPnV=Vl.
(3) If V2 ::; V n V^9 then either [V, V^9 ] = 1 or [V, V^9 ] = V 2 , and in the lattercase V9 ::; Mv and V^9 = 02(L2).
PROOF. As L has two orbits on 4-subgroups of V, either (2) holds or for some
g E G, Vi is a nondegenerate 2-subspace of V.
Assume the latter holds. Then Qo := Cr(V:f) E Syb(CM(Vi)), and either T::;
Land Qo = Q := 02(LT) = Cr(V), or Qo is the preimage in LT of the subgroupgenerated by a transposition. Now if Na(Qo) ::; M, then Q 0 E Syb(Ca(Vi)),
contradicting JCa(Vi)l2 = JTJ/2 > JQol·
Thus Na(Qo) 1:. M, so Q < Q 0 , and as M = !M(LT), no nontrivial charac-
teristic subgroup of Qo is normal in LT. Therefore Lis an A 6 -block by case (c) ofC.l.24. Now Q = 02(CM(V/)), and 031 (NM(Vi)) = X1 x X2 is the product of
A 3 -blocks, with V = 021 (X 1 X 2 ). Let X := 02 (! 2 ), in the notation of 13.5.4. Now
X 1 X 2 acts on X^9 by 13.5.4.1, so [X 1 X 2 , X9] ::; 02 (X9) by 13.5.4.2. Therefore asm 3 (G 2 ) ::; 2, X9 ::; X 1 X 2 , impossible as 021 (X) is nonabelian by 13.5.4.6, while
V = 021 (X1X2) is abelian.
This contradiction establishes (2); thus it remains to prove (1) and (3). Observethat Hypothesis G.2.1 is satisfied with Vi, V in the roles of "V 1 , V" as L 2 is
irreducible on V/Vi. Let U := (VG^2 ). By G.2.2, (1) holds and V 2 ;::: <I>(U).
Finally suppose V 2 ::; V n V^9. By (2) and A.l. 7.1 we may take g E G2, soV9 ::; U ::; Mv and [V, V9] ::; V n V9 as V ::::! U. Further if U is abelian, then
[V, V9] = 1 and (3) holds. Thus we may take U nonabelian. As X ::::! G 2 by 13.5.4.1
while V = [V, X], U = [U, X] ::; 02 (X). Therefore using 13.5.4.6, Cr = 02(L2) is
of rank 2. Thus V9 ::; 02 (£ 2 ), so [V, V9] = V 2 and m(V/Cv(V^9 )) = 2, so by
symmetry, m(V9 /Cvg(V)) = 2 and hence V9 = 02 (£ 2 ), completing the proof of
W- D
LEMMA 13.5.7. For each HE Hz, Hypothesis F.9.1 is satisfied with Vi in the
role of "V+ ".
PROOF. Most of this proof is exactly parallel to that of 13.3.18.1: namely
part (c) of F.9.1 follows, this time using 13.5.5 rather than 13.3.17.1 to obtain
G 1 n G 3 ::; Mv; parts (b) and (d) follow just as before; and part (a) is proved as
before. Thus it remains to verify F.9.1.e.
Assume 1 -f. [V, V9] ::; V n V9; then V9 is quadratic on V. To verify hypothesis
F.9.1.3, we may assume that g E G 1 with [V9, V3] = 1 = [V, Vf]. Then V^9 ::;
02 (L 1 T) = R 1 , so as V9 is quadratic on V, either m(V9) = 1; or n = 6, m(V9) = 2,
and conjugating in L 1 , we may assume that V 2 = [V, V^9 ]. But in the latter case
as [V, V9] ::; V n V9, V 2 ::; V n V9; then by 13.5.6.3, V^9 = 02 (£ 2 ), contradicting
V^9 ::; R1.
Therefore m(V9 /Cvg(V)) = 1, and hence also 1 = m(V/Cv(V^9 )) by symmetry.
Suppose [Vi, V9] = l. Then as [V, V9] /=-1, n /=-5, since in that case CM(V3) =
CM(V). Thus n = 6 and V9 is generated by a transvection with center Vi, so
[V, V9] = V 1. Thus V induces a transvection on V^9 with center Vi, so Cvg (V) =
Vf = Vf; hence [Vf, V] = 1, and F.9.1.e holds.