1022 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN Lf(G, T) IS EMPTYCQ 1 (u) i CQ 1 (V). Since Qi ::::; QH, CQH(u) i CQH(V), and since L = 02 (L)
induces Z 3 on V, also
L= [L,CQ 1 (u)] = [L,CQH(u)],
completing the proof of (3).Next Na 1 (Tu) normalizes TuQ1 =To using (1), so Na 1 (Tu) :S T by (2); hence
again using (1), Na 1 (Tu) = NT(Tu) =Ta.
We now show that to prove (4) it will suffice to establish that I:= Na(Tu) ::::;
G 1 : For in that case I = T 0 by the previous paragraph, establishing the first
assertion of (4). Next let Tu::::; 8 E 8yh(Ca(u)). Then Ns(Tu) ::::; I= To::::; H, so
as Tu E 8yl2(CH(u)) by (Ul), 8 =Tu. In particular u tf. zG as ITul < ITI. This
completes the proof that (4) holds if I::::; G 1 , so we may assume that Ii G1, andit remains to establish a contradiction. We saw earlier that To E 8yh(I), so in
particular T 0 < I as To ::::; G1.
We claim that N1(C) = T 0 for each.1 -=/=-C::::; T 0 with C :SI T, so we assume
that T 0 < N1(C) and derive a contradiction. We saw that To = NaJTu), so
N 1 (C) i G 1. Hence as M(T) = {M, G 1 } by 14.2.2.5, we must have Na(C)::::; M.
Therefore as IM: MnG 1 / = 3by14.2.2.1, and N1(C) i G1, M = (MnG1)N1(C);
hence as I normalizes Z(Tu),
V = (ZM) = (ZNI(C)) :S Z(Tu)·
But then CQH(u) = TunQH::::; CQH(V), contrary to (3), so the claim is established.In particular C(I, To)= To as To :SI T.
We have seen that To E 8yh(I), with /To: Tu/= 2, so that I/Tu and hence also
I is solvable by Cyclic Sylow 2-Subgroups A.1.38. Also F*(I) = 02 (!) by 1.1.4.3
as Z::::; Tu. So since C(I,T 0 ) =To< I, we may apply the Local C(G,T)-Theorem
C.1.29 to conclude that I= T 0 B, where Bis the product of s := 1 or 2 blocks oftype A3 which are not contained in G1. Further N1(J(To)) =To as C(I,To) =To,
so Solvable Thompson Factorization B.2.16 says that I/0 2 (J) contains the direct
product of s copies of 83. Therefore if s = 2, then I /0 2 (!) contains 83 x 83 ,
contradicting Tu :SJ I and /To : Tu/ = 2. Thus s = 1, so B ~ A4 by C.1.13.c.Now the hypotheses of Theorem C.6.1 are satisfied with I, T, T 0 in the roles
of "H, A, TH"; for example, part (iv) of that hypothesis follows from the claim
and the facts that To <I and /T: T 0 / = 2. Therefore case (a) or (b) of Theorem
C.6.1.6 holds since s = 1; thus I~ 84 or Z 2 x 84, and in particular Tu = 02 (!).
By C.6.1.1, To = J(To) = 02(I)0 2 (J)x for each x ET -T 0 , and hence T 0 = TuT:(},.
However by (3), Tu is nontrivial on V, so that T = ToCT(V) since /T: CT(V)/ = 2;
thus we may take x E CT(V). Next as T 0 = 02 (J)0 2 (J)x, Ct 0 (x) = Z 0 (bbx), where
Zo := Z(To) and 02 (0^2 (1)) =: (b, z). Further if I~ Z2 x 84 , then Zo ~ E4, and
hence [Z 0 , x] = Z as Z has order 2. However in either case, bbx is of order 4, so
that fh(CT 0 (x)) = Z; this is a contradiction, as x E CT(V) and V::::; Tu::::; T 0. This
contradiction completes the proof of (4), and hence of 14.6.3. DFor the remainder of this subsection, u denotes a member of U(H).
Define I := I(T, u) to be the set of I E 1i(Tu) such that I is contained in
neither G1 nor M. We will see later (cf. 14.6.17.5 and 14.6.24.4) that for suitable
u E U(H), Ca(u) EI, so that I is nonempty.
Let I* consist of those I E I such that T n I is not properly contained in T n J
for any J E I. Finally let I be the minimal members of I under inclusion.