15.3. THE ELIMINATION OF Mf/CMf(V(Mf)) = Ss wr Z215.3.1. Preliminary results. Recall by Hypothesis 14.1.5.2 that
Mc= !M(Ca(Z)).
1121Throughout section 15.3 we assume G is a counterexample to Theorem 15.3.1.
Let M := M1 be the unique maximal member of M(T) :_{Mc} under the partial
order ;S of Definition A.5.2, supplied by 14.1.12. Recall in particular that M E
M(Na(C2)), where C2 := C2(Baum(T)) is the characteristic subgroup ofBaum(T)
from C.1.18. Let V := V(M).
We summarize what has been established in this chapter ·so far:LEMMA 15.3.2. (1) m(V) = 4 and M = ot(V).
(2) Z = Cv(T) is of order 2.
(3) M(T) = {M,Mc}·
(4) If Ts X s M, then either(i) 02 (X):::; CM(V), or
(ii) X =Mand M = !M(X).
(5) M is the unique maximal member of M(T) under :S.(6) Na(T):::; M. In particular, members of 1-l*(T,M) are minimal parabolics
described in B.6.8, and in E.2.2 when nonsolvable.
PROOF. As G is a counterexample to Theorem 15.3.1, and the groups in 15.3.1
appear as conclusions in Theorem 15.1.3, conclusion (2) of 15.1.3 holds, giving (1).
Then (1) implies (2) as V = V(M) = (ZM), and (2) and 14.1.12.1 imply (5).
Since Na(T) preserves the relation ;Son M(T), Na(T) :::; M by (5); then 3.1.3.2
completes the proof of (6).
Assume the hypotheses of (4). By (1), Tis maximal in M, so either 02 (X) :::;
CM(V) so that (i) holds, or X = M. In the latter case, (ii) holds by A.5.7.1. Thus
( 4) is established.
Finally suppose Ml E M(T). By (5),-Ml = (Mn Ml)CM 1 (V(M1)). As usual
CM 1 (V(M1)):::; Mc since Mc= !M(Ca(Z)). We apply (4) to Mn Ml in the role
of "X": .in case (i) of (4), Mn Ml:::; CM(V):::; Ca(Z):::; Mc, so that Ml S Mc; in
case (ii) of (4), M = !M(M n M 1 ) so that M 1 = M. Thus (3) holds. D
LEMMA 15.3.3. (1) M = !M(NM(C2(Baum(T)))).
(2) If Baum(T) :::; S:::; T, then Baum(T) = Baum(S), and further Na(S) sNa(Baum(S)):::; M.
PROOF. Set 02 := C 2 (Baum(T)), and recall M E M(Na(C2)). By 15.3.2.2
and 14.1.11, M = CM(V)NM(C 2 ), so that (1) follows from 15.3.2.4. Choose Sas
in (2). Then Baum(T) = Baum(S) by B.2.3.4, and Na(S) :::; Na(Baum(S)) :::;
Na(C2), so that (2) follows from (1). D
LEMMA 15.3.4. Mc= Ca(Z).
PROOF. Let U := V(Mc), and assume the lemma fails; then U > Z. Next
<.
Mc '"'"'M by 15.3.2.5, so Mc = CM(U)X where X :=Mn Mc, and hence U =
(Zx). Thus as Z < U, 02 (X) fo Ca(V), so M = !M(X) by 15.3.2.4, contrary to
~#~ D
By 15.3.2.1, M = Ot(V) preserves an orthogonal-space structure on V, so