1162 15. THE CASE .Cr(G, T) = r/JPROOF. Pick YE Y n M, and apply Theorem 3.1.1 to S, Na(S), YT in the
roles of "R, M 0 , H". Note that by 15.4.12.4, S E Syl2(YS), and by 15.4.12.6,
Y S is a minimal parabolic in the sense of Definition B.6.1, so the hypotheses of
3.1.1 are indeed satisfied. Therefore by Theorem 3.1.1, 02( (Na(S), YT)) -I 1, soNa(S) :::;; M by 15.4.12.2. D
LEMMA 15.4.17. Assume BCa(B) :::;; H:::;; Na(B) for some 1 -I B:::;; T such
that TH := T n H E Sylz(H). Assume TH :::;; I :::;; H, and for z E z#, letMz E M(Ca(z)). Then
(1) The hypotheses of 1.1.5 are satisfied with I, Mz, z in the roles of "H, M,
z".
(2) If Zn 02(I) -=f. 1, then F*(I) = 02(I).
PROOF. As TH E Sylz(H), and Z :S; CT(B) :S; TH by hypothesis, 1.1.6 says
that the hypotheses of 1.1.5 are satisfied with H, Mz, z in the roles of "H, M, z".
Then as TH is Sylow in Hand I, 02 (HnCa(z)):::;; 02 (JnCa(z)) by A.1.6, so that
(1) holds. Assume Zn 02 (J) -=f. 1. Then N := Na(0 2 (J)) E 'He by 1.1.4.3, so as
BCN(B) :S; H n N :S; NN(B), H n NE 'He by 1.1.3.2. Hence as TH :S; I :S; H n N,
and TH is Sylow in H, we conclude IE 'He from 1.1.4.4. D
LEMMA 15.4.18. s(Y) = 2 for each YE Y.
PROOF. Assume Y E Y with s(Y) = 1, and let M1 := Na(Y); then M1 =
!M(YT) by 15.4.12.2. Pick M 2 E M(T) - {M 1 }; by 15.4.12, we may choose XE
Y n M 2 , and again M 2 = N 0 (X) = !M(XT). By 15.4.15, s(X) = 2, so by 15.4.12,
X = Y2Yi where Y2 = 02 (Y 2 ) = [Y2, J(T)] is S-invariant with Y2/02(Y2) ~ Z3 and
t ET - NT(Y2). Set T1 := NT(Y2), Yi := Y, Gi := YiT1, and I:= (G1, G2)· By
15.4.12.4, SE Syl2(YiS), so as S:::;; T1, T1 E Syl2(Gi)· Notice IT: T1I = 2.As S E Syl2(YiS), E = D1(Z(J(T))) = Ei x Fi, where Ei := CE(Yi), and
Fi := [E, Yi] n E is of order 2. In particular Ei is a hyperplane of E. Similarly
E =Ex x Fx, where Ex:= CE(X) and Fx := [E,X] n E ~ E 4. As Tacts on
Ex n Ei, if Ex n Ei # 1, then Cz(X) n Cz(Y) -=f. 1, then M1 = M2 by 15.4.12.2,
contrary to the choice of M2. Thus Ex n Ei = 1, so as E 1 is a hyperplane of E,
m(Ex) :::;; 1. By 15.4.14, 1 -I Cz(X) :::;; Ex, so Cz(X) = Ex is of rank 1, and
m(E) = 3. Thus as Ei is a hyperplane of E, E 1 n E 2 =: Ea -I 1; and Gi centralizes
Eo, so I:::;; Ca(Eo). In particular, IE 0 E 'H, so that I is an SQTK-group.
Next S = Baum(T):::;; T1, so that S = Baum(T1) by B.2.3.5. Thus Na(T 1 ) :::;;Na(S) :::;; Mi by 15.4.16, so as T1 = NT(Y2) E Syl2(CM 2 (Eo)), we conclude that
T1 is Sylow in GE := Ca(Eo), and hence also T1 E Sylz(I). Thus (I, Gi, G2) is
a Goldschmidt triple. Let I* := I/Q 3 ,(I). As T :::;; M 1 , Y 2 1:. M 1 since M 2 =
!M(XT). Then as M1 = !M(YT) and 02(G1) ::::1 YT, we conclude 02(G1) -=f.
02(G2). Thus a:= (GJ:, Tj, G'.2) is a Goldschmidt amalgam by F.6.11.2, so a and
I* are described in Theorem F.6.18.As Z:::;; T1 E Syl2(GE), we may apply 15.4.17 with I, GE, E 0 in the roles of
"I, H, B". We conclude that for z E z# and Mz E M(Ca(z)), the hypotheses of
1.1.5 are satisfied with I, Mz, z in the roles of "H, M, z", and F*(J) = 02 (!) if
Zn 02(I) # 1. By 1.1.5.1, F*(J n Mz) = 02(I n Mz), so by 1.1.3.2, F*(C1(z)) =
02(C1(z)). As [Z, Yi] :::;; C 1 (0(J)) by A.1.26.1, and 1 -=f. Zn [Z, Y 1 ], O(I) = 1 by
1.1.5.2.Suppose first that F*(J) -=f. 02 (!). Then there is a component L of I, and Z is
faithful on L by 1.1.5.3. Now Lis described in one of cases (3)-(13) of F.6.18, so as