13.5 .. THE TREATMENT OF A5 AND Aa WHEN (v;^1 ) IS NONABELIAN 915
We will now obtain a contradiction to our assumption that H is not solvable.
This contradiction will complete the proof of Theorem 13.4.1.
Pick g EH as in 13.4.17.2. Then V9 = ((5, 6)), so there isl EL with V9^1 ==
((3,4)). Let y := gl. Then A:= VY:::; T with L 1 = [L 1 ,A]. Let K := 02 (H),so that K E C(H) by 13.4.10 and our assumption that H is not solvable. As
L1 = [L1,AJ, K = [K,A], and hence [VH,A] # 1 as [VH,K] # 1. Let U := VHnMY
so that [U,A]:::; Un A, and set (KT)*:= KT/CKr(VH)·
Suppose first that [A, U] # 1. By 13.4.17.4, H is transitive on Vj/ and Zv :::;
VH, so Z{;;-:::; [A, U] ~An U for some h EH. Then as Vh is the unique member
of v^0 containing zt by 13.4.2.3, vh =A= VY, and hence zt = Z{;r as Na(V) =
M = Ca(Zv ). Indeed this argument shows Z{;r is the unique point of VH n A, andhence of [A, U]; thus [A, U] = Z{;r, and hence U induces transvections on A with
center Z{;r, whereas M contains no such transvection, since CM(V) = CM(V/Zv)
by 13.4.2.4.This contradiction shows that [A, U] = 1. In particular VH 1:. MY, as [VH, A] -=f.
1; hence as r(G, V) 2: 3 by 13.4.17.5, m 2 (KT) 2: m(A*) > 2, and then examining
the cases listed in 13.4.13, we conclude that H ~ 85 and m(A) = 3. Hence
for 1 # b E A, (b) = B for some B :::; A with m(A/ B) :::; 2, so CvH (b*) =
CvH(B):::; MY as r(G, V) 2: 3, and therefore
(CvH(b*): 1 # b* EA*):::; U:::; CvH(A*),
so that A* E A 3 (T*, VH ). However H* has no such rank-3 subgroup, since each such
subgroup is the radical of some minimal parabolic and hence contains a transvectionwhose axis is centralized only by that transvection.
This contradiction establishes Theorem 13.4.1.13.5. The treatment of A 5 and A 6 when (v;^1 ) is nonabelian
In this section, we continue our treatment of the remaining alternating groups
A 5 and A5, postponing treatment of the final group L3(2) of Frtype until the
following chapter. More specifically, this section begins the treatment of the case
where (V^01 ) is nonabelian, by handling in Theorem 13.5.12 the subcase (V 3 °^1 )
nonabelian. In fact if L/0 2 (L) is A 5 and (V 3 °^1 ) is abelian, we will see that (V^01 )is also abelian; thus in this section we also deal with the case where L/02(L) ~ A 5
and (V^01 ) is nonabelian.In this section, with Theorem 13.4.l now established, we assume the following
hypothesis:HYPOTHESIS 13.5.1. Hypothesis 13.3.1 holds and G is not Sp5(2).
In addition we continue the notation established earlier in the chapter, and the
notational conventions of section B.3. In particular we adopt Notations 12.2.5 and
13.2.1.
LEMMA 13.5.2. Assume Hypothesis 13.5.1. If KE .C1(G, T), then
(1) K/02(K) ~ A5, L3(2), A5, or A5.
(2) K :'.9 KT and KE .C*(G,T).