520 2. CLASSIFYING THE GROUPS WITH [M(T)[ = 1
groups in 1.1.5.3, Lis a Bender group with Autu(L) = S11(S n L). In particular,
U acts on each component of H.
Let UL and Uc be the projections of U on L and CH(L), respectively. As
z EU:::; ULUc, Na(ULUc) SM by (*). As L = [L, z], the projection of z on L
is nontrivial, while as L is a Bender group, NL(UL) is irreducible on S11(S n £).
Therefore UL= [z, NL(UL)] :=:; U and hence U =Uc x UL. In particular m(U) =
m(UL) + m(Uc).
Now pick u E .:J so that Lis maximal among components of Gj for j E .:J. Let
v E utf. Since Gv contains Li M, v E .:J, so by earlier remarks, U acts on each
component of Gv and O(Gv) = 1. Then as u EU, u acts on each component of Gv,
so Lis contained in a component Lv of Gv by I.3.2. Hence L = Lv by maximality
of L.
Suppose g EM with U!; n Uc i= l; we claim that L = £9, so that Uc= U!;
as M = Nc(U). Assume the c!aim fails and let 1 i= v E Uc n U!; =: V. By the
previous paragraph, L and £9 are components of Gv, and we may assume L i= L^9 ,
so that [£, £9] = 1. It will suffice to show that M acts on {L, £9}, since then M
permutes {Uc, U!;}, and hence M acts on 1 i= V =Uc n U!;:::; Uc, contradicting
the irreducible action of Mon U. Now
m(UL) + m(Uc) ==' m(U) = 2m(UL) + m(V),
so m(Uc) = m(UL) + m(V) > m(U)/2 since m(V) > 0. Then for each x E M,
1 i= UcnU 0. HenceifLx ~ {L,£9}, bysymmetrybetweenxandg, also [L,Lx] = 1.
Then UL :::; U!; n U 0 , so also [ Lx, £9] = 1. But now for p an odd prime divisor of
JNL(UL)J, m 2 ,p(££9£x) > 2, contradicting G quasithin. This completes the proof
of the claim.
The claim shows that Uc is a TI-set under M. Further AutL(U) is cyclic and
regular on uf, and is invariant under NAutM(u)(Uc). Hence (AutM(U), U) is a
Goldschmidt-O'Nan pair in the sense of Definition 14.1 of [GLS96]. So by O'Nan's
lemma, Proposition 14.2 in [GLS96], one of the four conclusions of that result holds.
Neither conclusion (i) nor (iii) holds, as M is irreducible on U. As Gz :=:; M but
.:Ji= 0, Mis not transitive on U#, so conclusion (iv) does not hold. Thus conclusion
(ii) holds, so that Nc(Uc) is of index 2 in M. However as Mis the unique point fixedby z in G / M, by 7.4 in [ Asc94], M controls G-fusion of 2-elements of M. Therefore
by Generalized Thompson Transfer A.1.37.2, 02 (G) n M:::; NM(Uc), contrary to
the simplicity of G. This contradiction completes the proof of Proposition 2.2.2. D
Recall that S2 ( G) is the set of nonidentity 2-subgroups of G, and (cf. chapter
- that 8HG) consists of those SE S 2 (G) such that Na(S) E He. We next verify:
LEMMA 2.2.3. The Alperin-Goldschmidt conjugation family lies in 8HG).
PROOF. By (b) and ( c) of the definition of the Alperin-Goldschmidt conjuga-
tion family D for Tin G, 0
21
(Ca(D)) :=:; D for each D E 'D. Thus as D :::; T,
Z(T) SD. Therefore DE S2(G) using 1.1.4.3. D
NOTATION 2.2.4. Define O = OM to consist of those D E S2(G) such that
D:::; M, but Na(D) i M. Leto*= OM denote the maximal members of o under
inclusion.THEOREM 2.2.5. If o = 0, then G is a Bender group.