15.2. FINISHING THE REDUCTION TO Mr/CMf(V(M£)) ~ Of(2) 1113LEMMA 15.2.18. (1) SE Syb(I).
(2) B := Baum(T) = Baum(S) and O(I, B):::;: Nr.
(3) Either
(i) Nr:::;: Mr, or
(ii) Nr i Mr, case (ii) of 15.2.13.1 holds, with M = nt(V), and Nr =OM(Z1) is of index 6 in M, for some complement Z1 to Z in Zs.
PROOF. Recall Zs is of order 4 by 15.2.11.3. Let S ::S: Tr E Syb(I). By
15.2.14.3 and B.2.3.5, B = Baum(S) = Baum(Tr), and O(G,B):::;: M by 15.1.9.l.
Thus (2) holds and also Tr :::;: Nr(B) :::;: M, so as Nrii.I(S) = f' and IT : SI = 2 by
15.2.17, Tr is either T or S. But if Tr = T, then I :::;: M by 15.2.5 since I i Mc,
and we saw K:::;: Mr; but this is contrary to Hi M. Thus (1) is established.
Next using 15.2.16, OM(V) :::;: OM(Zs) :::;: Mr, so if 02 (Nr) :::;: OM(V), then
conclusion (i) of (3) holds. Thus we may assume X := 02 (Nr) i OM(V).
Suppose X :::! M. Then T acts on SXOM(V) and K, so T acts on G 0 :=(SXOM(V),K). Now 02(I) :::;: S:::;: Go :::;: I, so TGo E 1-l(T). But by 15.2.14.5,
Mn Mc = OM(V)T, so as X i OM(V), TGo i Mc. Hence M = !M(TG 0 ) by
15.2.5, so K :::;: G 0 :::;: M, contrary to H i M.
Therefore Xis not normal in M, so case (ii) of 15.2.13.1 holds, and X is one
of the two subgroups of O(M) of order 3 not normal in M. Thus conclusion (ii) of
(3) holds, completing the proof. DRecall 01 (8) from Definition C.l.18.LEMMA 15.2.19. Define B := Baum(S).
(1) If conclusion (i) of 15.2.18.3 holds, then O(I, B):::;: Mr 2: Or(01(S)).
(2) If conclusion (ii) of 15.2.18.3 holds, then Hypothesis C.2.3 is satisfied with
I, 02(Nr), Nr in the roles of "H, R, MH"·
PROOF. Assume first that conclusion (i) of 15.2.18.3 holds. Then Nr :::;: Mr,
so O(I, B) :::;: Nr :::;: Mr by 15.2.18.2. Further 01 (S) :::! T as S :::! T by 15.2.17.3,
so 1 =f. Zn 01(8) and hence Or(01(S)) :::;: Oa(Z n 01(8)) :::;: Mc = !M(Oa(Z)),
completing the proof of (1).Next assume conclusion (ii) of 15.2.18.3 holds, and let R := 02 (Nr ). Then Nr ~
83, so R:::;: OM(V). But OM(V) :::;: OM(Zs) :::;: Nr, so R = 02(0M(V)) = 02(M),
and hence O(I, R) :::;: Nr. Then as R = 02(Nr ), the remaining two conditions of
Hypothesis C.2.3 are trivially satisfied, so (2) holds. DLEMMA 15.2.20. (1) The hypotheses of 1.1.5 are satisfied with I, Mc in the
roles of "H, M" for each z E z#..
(2) F*(Mr) = 02(Mr).
(3) O(I) = l.
PROOF. Let Io E M(I); then part (1) holds for Io in the role of "I" by 1.1.6.
Then by 15.2.18.1, Sis Sylow in.I and Io. In particular, 02 (I 0 nMc):::;: 02 (InMc) :::;:
S by A.1.6. Hence as Io satisfies the hypotheses of 1.1.5,.Oo 2 (M 0 )(02(J n Mc)) ::S: Oo 2 (M 0 )(02(Io n Mc)) ::S: T n Io= S ::S: I,
and so (1) holds. Then (2) follows from (1) and 1.1.5.l. As usual 1 =f. UK:= [UH, K]
centralizes O(I) by A.l.26.1, and Z :::;: UK since Z = fh(Z(T)) has order 2 by
15.2.17.2, so (3) follows from 1.1.5.2. D