15.2. FINISHING THE REDUCTION TO Mf/CMf(V(Mf)) '.:::: o;t(2) 1105{ii} Case {3} of 15.2.1 holds, Re ~ Z2 inverts 02 (M 0 ) = 02 (M) = Y,
Y* ~ 3i+^2 , and 02 (0 2 ,z(Y)) is the subgroup B(CM(V)) generated by all elementsof CM(V) of order 3.
{iii} Case {3} of 15.2.1 holds, Re~ Z 2 inverts Y ~ Y* ~ Z 3 , and a Sylow
3-subgroup of M is isomorphic to Z3 x Z3n for some n ;:::: 1.
(2) Y :s1 M.
{3} IfY = 02 (Mo), then M = !M(YT).
(4) M =(Mn Mc)0^2 (Mo).
( 5) R~ Y* centralizes CM (V) *.PROOF. Part (3) follows from 15.2.2. Set Yo := 02 (M 0 ). To establish the
remaining parts, we apply case (b) of 14.1.17 with Mc in the role of "M 1 ". By
14.1.17: Re f.1, Y = [Y 0 ,Rc], and R~Y centralizes CM(V), so that (5) holds. As
Re f. 1 = 02(M) using 15.1.5.1, Y f. l.
Assume for the moment that Y is cyclic. Then Y is of prime order from 15.2.1,
and Y is inverted in Re. Hence as R~ centralizes Cy(V)* by (5), we conclude
Cy(V) = 1, so that Y ~ Y. ·
We next prove (1). If case (3) of 15.2.1 does not hold, then Y 0 is of prime
order, so Y = Yo, and then conclusion (i) of (1) holds by the previous para-
graph. Thus we may assume that case (3) of 15.2.1 holds. Since R~ is faithful
on F(M) :::; Y 0 CM(V), but R~ centralizes CM(V), R~ is faithful on 03(M),
so that m3(CM(V)):::; 1by14.1.17.4.
Suppose first that Y = Y 0 , so that Y = 02 (M). Since Cy(V) :::; Z(Y) by (5),
we conclude from A.l.21 and A.l.24 that either Y ~ Y ~ E 9 or Y ~ 3i+^2. In the
former case, conclusion (i) of (1) holds: for CM(V) is centralized by Y ~ Eg by
(5), so that CM(V) is a 3'-group since m3(M) = 2. In the latter case as f' ~ E4,
we conclude from (5) that Re is the subgroup of order 2 in f' which centralizes
Cy(V)*, and hence inverts Y; then since m 3 (CM(V)) :::; 1, conclusion (ii) of (1)
holds.
Thus we may suppose that Y < Yo. Therefore Y is of order 3, so Re is of order
2, and Y ~ Y by the second paragraph of the proof. Since Y centralizes CM(V)*
and m 3 (CM(V)):::; 1, conclusion (iii) of (1) holds, completing the proof of (1).
We next prove (4). First assume the subcase of case (4) of 15.2.l where f' ~ Z4
and O(M) ~ Z 15 does not hold. In the remaining cases, M = YoN1VJ(T), and
N1VJ(T) = NM(T) by a Frattini Argument, so as NM(T) :::; NM(Z) :::::; Mc =
!M(Ca(Z)), (4) holds. Now consider the excluded subcase. By 15.1.13.4, J1(T) =
f' n M 0 , so J 1 (T) centralizes 03 (M). Then 031 (M) acts on Ve := Cv(J1(T)), so
that 031 (M) :::; Mc by 15.1.14.2. Hence M = YT0^31 (M) = Yo(MnMc), completing
the proof of (4). ,
As Mn Mc acts on Re and Yo, Mn Mc and Yo act on [Yo, Re] = Y. Then as
M =(Mn Mc) Yo by (4), (2) holds. D
Recall that 15.1.12 describes the possible structures for H E 1i*(T, M). We
next eliminate one subcase of 15.1.12.3:
LEMMA 15.2.4. If HE 1i*(T,M), then H/0 2 (H) is not S5 wr Z2.
PROOF. Assume otherwise, define Re and Yasin 15.2.3, set M* := M/02(M)