1092 15. THE CASE .C.r(G, T) =^0Until the proof of Theorem 15.1.15 is complete, assume G is a counterexample.
Thus we are in case (4), (6), or (7) of 15.1.7. As in 15.1.13, let Ei := Di(Z(Ji(T))).By 15.1.13.1, Cc(Ei) "f:. M.
As case (3) of 15.1.7 does not hold, Ji(T) = f' n Mo and J1(M) = Mo by
15.1.13.4. Also V = V(M) by 15.1.7.We begin by determining VE := Cv(Ji(T)) in each of our three cases, and
defining some notation:
NOTATION 15.1.16. (a) In case (6) of 15.1.7, V =Vi E9 V2 for Vi defined there,and VE = Z1 E9 Z2, where Zi := Cv;. (T n Mo) 9'! E4·
(b) In case (4) of 15.1.7, V =Vi EEiVz, where Vi and V2 are the two 4-subgroups
of V such that Mi := N.M(Vi) is not a 2-group; in this case VE = Zi E9 Z2 where
zi := VE n Vi is of order 2.
(c) Finally in case (7) of 15.1.7, VE 9'! Em. In this last case, P := Oa(M) 9'!
31+^2 • Let Pz := Z(P), pick Pi of order 3 in Mo inverted by T n Mo for i = 1, 2
with P = PzP 1 P 2 , and set Zi := Cv(Pi) and V 2 := [V, Pi], so that V = Z1 E9 Vz,
zi <:;::! E4, and Vz <:;::!Em. In this case VE = Z1 E9 Z2.
In each case, set S := c;(Zi), Gi := Cc(Z1), Mz := Gin Mc, and Qi :=
02(Mz).
Observe that in each of the cases in Notation 15.1.16, Case I holds by construc-
tion: Namely Z 1 :::; VE= Cv(Ji(T)), and M = (CM(Zi), T). Also:
LEMMA 15.1.17. SE Syb(Gi n M), J(S) = J(T), Baum(S) = Baum(T), and
C(G, Baum(S)):::; M.
PROOF. By construction in Notation 15.1.16, Sis Sylow in GinM and Cr(V) :::;
Cr(Zi) = S. So as J(T) :::; Cr(V) by 15.1.9.1, J(S) = J(T) and Baum(S) =
Baum(T) by (3) and (5) of B.2.3. Then 15.1.9.1 completes the proof. DLEMMA 15.1.18. (1) Z 1 :::; VE, and 02 (Cc(Z)):::; Mz.
(2) 02 (M n Mc n G1) = 02 (CM(V)) and CM(Zi) "f:. Mz.
(3) S, Mz, and Qi are T-invariant.
(4) SE Syl2(Gi).
(5) C(Gi, Qi)= Mz = Nc 1 (Qi), so Hypothesis C.2.3 is satisfied with Gi, Qi,
Mz in the roles of ''H, R, MH"·(6) Hypothesis 1.1.5 is satisfied with G 1 , Mc in the roles of ''H, M", for any
lf=.zEZ.
(7) Mc= !M(MzT) and C(G, Qi):::; Mc.
PROOF. We observed earlier that Case I holds, so in particular, Z 1 :::; VE
and M = (CM(Zi), T). Then as 02 (Cc(Z)) :::; Cc(VE) by 15.1.14.1, and Mc =
!M(Cc(Z)), (1) follows; and as M "f:. Mc, CM(Zi) i Mz. By 15.1.17, S E
Syl2(Gi n M) and Nc(S) :::; M, so (4) holds. Hence Sis also Sylow in Mz and in
Mn Mz ·= G1 n Mn Mc. Since Zi :::; V, 15.1.5.2 says that CM(V) :::; G 1 n Mn Mc.
By construction in 15.1.16, 02 (CM(Zi)) is of prime order, so as CM(Zi) i Mz
and SE Syb(Mz), it follows that 02 (MnMcnGi) = 02 (CM(V)), completing the
proof of (2). We check in each case in 15.1.16 that S = C't(VE), so that S :':'.] T.