1112 15. THE CASE .Cf(G, T) =^0
holds. Then we compute that X acts faithfully on [V, BJ =: Zs, so X :::; Na (Zs) :S
Mc by 15.2.12.3. Hence X :::; Ca(Z) by 15.2.12.4, ifupossible as X is nontrivial
on Zs, and 1 =!= Zn Zs. Therefore the latter case holds for each B E :B'(V), and
hence conclusion (ii) of (1) holds. This completes the proof of (1). Then (1) implies
(2). D
For the remainder of the section, we define Bo and Zs := [V, Bo] as in 15.2.13.2.
Thus Zs= [V, B] for each BE :B'(V) by 15.2.13.2. Let S := Cr(Zs).
LEMMA 15.2.14. (1) Mc= Ca(Z).
(2) Z:::; Zs, and either
(a)S=TandZ=Zs~E4, or 1
(b) M ~ Sz(2) or Dt(V), Z is of order 2, and IT: Bl= 2.(3) Baum(T) :::; S.
(4) Zs= D1(Z(S)).
(5) Mn Mc= CM(Z) = CM(V)T.(6) m(M, V) = 2 and a(M, V) = 1.
PROOF. As Mis solvable, a(M, V) = 1 by E.4.1. Then by inspection of the
cases in 15.2.13.1, and recalling V = V(M) by 15.2.12.5 so that Z:::; V, (2) and (6)
hold, and CM(Z) = f'. Recall CM(V) :::; Mc by 15.1.5.2. In case (i) of 15.2.13.1,
f' is maximal in M, so CM(V)T = Mn Mc as Mc 1:. M. In case (ii) of 15.2.13.1,
this holds as 031 (Mn Mc):::; CM(V) by 15.2.12.4. Thus (5) is established. Further
Mc = (Mn Mc)CMJV(Mc)) since Mc.:!:., M by 15.1.9.2, so Mc centralizes Z by
(5), and hence (1) holds.
If Zs = Z, then S = Tso (3) and (4) are trivial; thus we may assume that
M ~ Sz(2) or Dt(2) with Z of order 2. Then Baum(T) :::; Cr(V) :::; S by 15.1.9.1,
completing the proof of (3). Finally Zs:::; D1(Z(S)) =: Z 0 and Z 2 ~ Z = Cz 0 (T);
so as T/S is of order 2, m(Z 0 ) :::; 2m(Z) = 2 = m(Zs) using 15.2.11.3, and hence
Zo =Zs, establishing (4). D
15.2.2. A uniqueness theorem. This subsection is devoted to establishing
the following uniqueness theorem:
THEOREM 15.2.15. Mc= !M(CM 0 (Zs)).
The proof of Theorem 15.2.15 involves a series of reductions. Until it is com-
plete, we assume IE Ji(CMJZs)) with I 1:. Mc, and work toward a contradiction.
Set M1 := Mc n I and N1 := M n I. In particular
M1 <I.
Since Z :::; Zs by 15.2.14.2, and Mc = Ca(Z) by 15.2.14.1, while we chose
CMJZs) :::; I:
LEMMA 15.2.16. Ca(Zs) = CM 0 (Zs) :S M1.
Recall H:::; Mc by Notation 15.2.9; so as Zs:::; Z(K) by 15.2.11.2, KS:::; M 1.
As CM 0 (Zs) :S I 1:. Mc and Mc = Ca(Z), Z < Zs. Hence case (b) of 15.2.14.2
holds, so by that result:
LEMMA 15.2.17. (1) M ~ Sz(2) or nt(2).
(2) IZI = 2.
(3) IT: SI= 2.