602 3. DETERMINING THE CASES FOR L E .Cj (G, T)on the members of Af normal in T, so that A1 is the only such member. Thus
{A, V} = {B E A.(T) : B ::::;! T} is D-invariant, so as usual D acts on V. Then
D S Na(V) SM, contrary to 3.3.6.a. D
By 3.3.30, LT ~ A 6 or 86 , so we can represent LT on n := {1, ... , 6} so that
T has orbits {1, 2, 3, 4} and {5, 6}, and permutes the set of pairs { {1, 2}, {3, 4}}.
Further we adopt the notation of section B.3.
LEMMA 3.3.31. To ~ 1.
PROOF. Assume otherwise; then in particular, Cz(L) -:/=-1. By 3.3.25.2, Dacts on Y := 02 (CL(Z)) and on R := 02 (YT). Then by 3.3.26, there is d E
D - M with Vd i. Q£. As Vd ::::;! YT, either Vd = ((5, 6)), or Vd contains
((1, 2)(3, 4), (1, 3)(2, 4)). The latter is impossible, since as Vd ::::;! T, Vd acts
quadratically on V. Thus Vd = ((5,6)), and in particular LT/QL ~ 85 rather
than A5.
By Sylow's Theorem, D acts on some B of order 3 in Y, and so D acts on
CR(B). Now for v E Cv(B) - Z, V = (vT), so v tj. Q'1, since Vi. Q'1,. Therefore
by symmetry, vd tj. QL, and thus vd = (5, 6).
Next ICR(B): CQL(B)I = 2andCR(B) = (vd)CQL(B) sinceYT/QL ~ 84XZ2.
As QL = CR(V), CQL(B) = CQL(VB) = CR(VB). Conjugating by d, ICR(B) :
CR(Vd B) I = 2, so as CR(B) = (vd)CQL (Vd), ICQL (B) : CQL (Vd) I = 2. Then as
[Cv(B)/Cv(L),vd]-:/=-1, CQL(B) = Cv(B)CQL(BVd), so as Tc S CQL(B) Tc S
CQL (BVd) and hence Vd centralizes Tc. Thus CR(B) = (vd)Cv(B)CQL (BVd).
Finally by Ooprime Action, QL = VCQL(B), so QL = VCQL(BVd).
Set s := QLVd. As Cr(B) = vd = s, CT(B) = Cs(B) = CR(B) and
[Vd, B] s [ QL, B] = [V, B]. So by symmetry, [V, B] s [Vd, BJ = [V, B]d, and hence
[V, B] = [Vd, B] = [V, B]d as these groups have the same order. Thus d acts on
CT(B)[V,B] = (vd)Cv(B)CQL(BVd)[V,B] = (vd)VCQL(BVd) = VdQL = S.
By 3.3.7.4, v = [V, L]Cz(L), so that z = ([V, L] n Z)Cz(L). Therefore IZ :
Cz(L)I = l(Z n [V, L]) : C[v,LJ(L)I = 2. We saw Cz(L)-:/=-1, so as Tc n T8 = 1 by
3.3.27.1, Z ~ E4 and Cz(L) ~ Z2.
Suppose (QL) = 1. Then as d normalizes 8 and S = Vd is of order 2,
A.(8) = {QL, Q'i}, so as dis of odd order, d E Na(QL) s M = !M(LT), contrary
to our choice of d E D - M. Thus (QL) -:/=-1. So as (Tc) = 1 by 3.3.27.2,
TcV < QL. As we saw QL = VCQL(VdB), we may choose u E CQL(VdB)-TcV.
Now IQL : TcVI s 2 by 0.1.13.b and B.3.1, so QL = (u)TcV and T =
(u)(TnL)TcVd. Also (Tc) = 1, Tc commutes with L by definition, and we saw Vd
centralizes Tc. Therefore as T = (u)(TnL)TcVd, 1-:/=-CT 0 (u) =Zn Tc s Cz(L),
so as Cz(L) is of order 2, CT 0 (u) is of order 2. As u^2 E VTc s Ca(Tc) and Tc is
elementary abelian, it follows that m(Tc) s 2.
Assume first that To ~ Z2. Then as ( Q L) -:/=- 1 while ( Q L) S Tc by
0.1.13.a, u^2 generates Tc. Recall we chose u to centralize Vd and Vd centralizes Tc.
Therefore Z(S) = Cv(Vd)Tc(u), with Cv(Vd)Tc elementary, so that (Z(S)) =
Tc is d-invariant, contradicting 3.3.27.1.
Thus Tc ~ E4, so (u)Tc ~ Ds. Hence 8 = 81 x 82 x E, where Si ~ D 8 and
E ~ E4. But then as dis of odd order, the Krull-Schmidt Theorem A.1.15, says d
acts on Z(8)Si for i = 1 and 2, so d centralizes (Z(S)Si) of order 2, and hence
also centralizes (S). This contradicts 3.3.27.1, since Tc n (8) -:/=-1. D