1122 15. THE CASE .Cr(G, T) = 0
Thus M = (M1 x M2)(f';, where Mi:= CM(V3-i), Vi= [V, Mi], Mi~ o;-(Vi) ~ 83,
and tis an involution with Mi = M2. Set S := NT(V1) and Zs := Cv(S).
LEMMA 15.3.5. Let E := fh(Z(J(T))) and B := Baum(T). Then either
(1) [V, J(T)] = 1, B ='== Baum(CT(V)), V:::; E, and C(G, B):::; M, or
(2) B = J(T) = S ~ E 4 and En V =Zs ~ E4.
PROOF. If J(T) :::; CT(V) then V :::; E and B = Baum(CT(V)) by B.2.3.5;thus M = CM(V)NM(B) by a Frattini Argument, and hence C(G, B) :::; M by
15.3.2.4. Otherwise J(T) -=f. 1, and then (2) follows from B.1.8. D
LEMMA 15.3.6. (1) Baum(T) = Baum(S).(2) Na(S):::; Na(Baum(S)):::; M.
(3) SE Syl2(Ca(Cv 1 (S))).
(4) SE Syb(Na(V1)).PROOF. By 15.3.5, Baum(T) :::; S, so that (1) and (2) follow from 15.3.3.2. As
SE Syb(CM(CvJS))), (2) implies (3), and similarly (2) implies (4). D
LEMMA 15.3.7. LetRc := 02 (MnMc) andY := 02 ((Rf;'1)). ThenY = 02 (M),
Mn Mc= CM(V)T,.
M = Na(Y) = !M(YT),
02 (YT) = CT(V), and either
(1) 02 (Y) = c;(V) with Y/0 2 (Y) ~ E 9 , and Y = 0
31(M); or
(2)Y/0 2 (Y) ~ 3i+^2 , 02 ,z(Y) = Cy(V), Re is cyclic, Y = fJ(M), and MnM~
has cyclic Sylow 3-subgroups.PROOF. We apply case (b) of 14.1.17 with Mc in the role of "M1", and Yo:=
02 (M). By 14.1.17.1, Re -=f. 1. As Re :::l '1', Re contains Z(T) =: (r), and r
inverts Yo by 15.3.2.1, so Yo = [Y 0 , Re] and Mn Mc= CM(V)T. Further applying
parts (2) and (3) of 14.1.17, we conclude Y =Yo and Y*R~ centralizes CM(V)*,
where M* := M/0 2 (M). In particular, M = !M(YT) by 15.3.2.4, and of course
M = Na(Y) as Y :::l ME M. Also V = (ZY), so that VE R2(YT) by B.2.14.
As r inverts Yo = 02 (M) and [r*, Cy(V)*] = 1, r inverts y of order 3 in eachcoset of Cy(V) in Y. Therefore since Y centralizes CM(V), Y = fh(Y) is a
3-group. As <.P(Y) :::; Cy(V) :::; Cy.(r), it follows that <.P(Y) :::; Z(Y*), and
hence Y* ~ Y/0 2 (Y) ~ E 9 or 3i+^2 by A.1.24. Further 02 (YT) = CT(V).
If Y* ~ E 9 , then as M = YCM(Y/0 2 (Y))T and m 3 (M) :::; 2, Y = 031 (M),
so that (1) holds. If Y* ~ 3i+^2 , m 3 (CM(V)) = 1 by 14.1.17.4, so that Y =B(M), and CM(V)T has cyclic Sylow 3-subgroups. Further Re is embedded in
CAut(Y•)(Z(Y*)) ~ Qs, while T ~ Ds, so T contains no Qs-subgroup. Thus Re is
cyclic, completing the proof of (2). D
In the remainder of this section, define Yasin lemma 15.3.7.
LEMMA 15.3.8. Zs= fh(Z(S)) ~ E4.
PROOF. Let Z 0 := fh(Z(S)). As T/S is of order 2, Z = Cz 0 (T) is of rank atleast m(Z 0 )/2, so m(Zo) :::; 2 as Z is of order 2. As Zs = Cv(S) is of rank 2, the
lemma follows. DFinally we eliminate a configuration which appears at various points later, the
case V = 02(Y):