13.7. FINISHING THE TREATMENT OF Aa WHEN (vG1) IS NONABELIAN 935G1 =KT, or G1 = 0
31(G1)T with 031 (G1) ~ PGL 3 (4). Hence the proof of the
lemma is complete. DLet H := G1 and K := G'f'. Now H E 1-lz, so by 13.5.7, Hypothesis F.9.1
is satisfied with V3 in the role of "V+". Then by 13.6.24.2, Hypothesis F.9.8.f
is satisfied, while case (i) of Hypothesis F.9.8.g holds in view of 13.2.3.2. We now
adopt the standard conventions from section F.9 given in Notation 13.5.8, including
H* := H/QH, UH := (V 3 H), and fI := H/Vi. By 13.6.27, F*(H*) = K* ~L3 ( 4), M23, or J2, and in the first case T* is nontrivial on the Dynkin diagram
of K. Therefore q(H, UH) > 2 by B.4.2 and B.4.5, contrary to F.9.16.3. This
contradiction completes the proof of Theorem 13.6.1.
- Finishing the treatment of A 6 when (VG^1 ) is nonabelian
In this section, and also in the final section 13.8 of the chapter, we adopt a
hypothesis excluding the groups identified in previous sections:
HYPOTHESIS 13.7.1. Hypothesis 13.3.1 holds, L/CL(V) ~ A 6 , and G is not
Sp5(2) or U4(3).
Thus since Hypothesis 13.7.1 includes Hypothesis 13.3.1 and Hypothesis 13.5.1,
we may appeal to results in sections 13.4 and 13.5.
Set Q := 02(LT). We continue with the notation established in section 13.5:
Namely we adopt the notational conventions of section B.3 and Notations 12.2.5
and 13.2.1.
By 13.5.2.3, Cv(L) = 1, so that V is the core of permutation module for
L ~ A 6 , given by the vectors es for subsets S of even order inn:= {1, 2, 3, 4, 5, 6},
modulo en. In particular Vi= Zn Vis generated by z := ei,2,3,4 = e5,6·
By 13.5.7, Hypothesis F.9.1 is satisfied with V 3 in the role of "V+", so we may
use results from section F.9. We also adopt the conventions from that section given
in Notation 13.5.8, including G\ := Gi/V 1. As usual define
By 13.3.6, G 1 E 1-lz, and so 1-lz is nonempty.
In the remainder of the section, H denotes a member of 1-lz.
From Notation 13.5.8 UH:= (Vl), VH := (VH), QH := 02(H), and H := H/QH
so that 02(H) = 1. By F.9.2.3, QH = CH(UH)· Set He .-CH(UH)i then
Hc:::;QH.
By Theorem 13.5.12:
LEMMA 13.7.2. (V 3 °^1 ) is abelian, so UH is abelian.
There are no quasithin examples satisfying 13.7.2, so in the remainder of this
section we will be working toward a contradiction. As far as we can tell, there are
not even any shadows.
13. 7.1. Preliminary results. We begin with several consequences of Hy-
pothesis 13.7.1 and 13.7.2, which we can apply both in the next subsection where
(V^01 ) is nonabelian, and in the final section 13.8 where (V^01 ) is abelian.