15.4. COMPLETING THE PROOF OF THE MAIN THEOREM 1157{1) 3!M(XT), so that XE C(G, T).
{2) X:::; 02,p(M) for ME M(XT). In particular, 02 (X):::; 02 (M).
{3) [Z,X] #-1, so XE ~t(G,T).
(4) p = 5.PROOF. Let M E M(XT), and set M* := M/0 2 (M). If q #-p is an odd
prime such that [Oq(M), X] "I-1, then Autx(Oq(M*)) is embedded in GL 2 (q) by
A.1.25.2, sop divides q - E for E := ±1. This is impossible asp;:::: q by maximality
of p, with p and q odd. Therefore X :::; CM·(OP(F(M))) =: H*, and as M
is solvable by 15.4.2.2, F(H) = Op(H) x OP(Z(H)). Asp > 3, the solvable
subgroup AutH(Op(H*)) of GL2(p) is p-closed using Dickson's Theorem A.1.3, so
that X =OP' (X) :::; Op(H) :::; Op(M), establishing (2) for each M E M(XT).
Let Nc(X) :::; Mx E M. We will show that M = Mx; then since Mis· an
· arbitrary member of M(XT), (1) will be established. Let K := 02 ,F(M) and
Y := 02 (NK(X)); by (2), X :::; Y :::; Mn Mx =: I. Let Xo be the preimage
in M of X; as X is T-invariant, X = 02 (X 0 ), so NM(X) = NM(X). Thus
Y = 02 (NK (X)). Next CK (Y) :::; CK (X) :::; Y as K* is of odd order. Thus
the hypotheses of case (b) of A.4.4 are satisfied with M, Mx, Y0 2 (M), in the roles
of "H, K, X". Therefore R := 02(1) = 02(M) by A.4.4.1, and C(Mx, R):::; I by
A.4.4.2, so Hypothesis C.2.3 is satisfied by Mx, I in the roles of "H, MH". Further
C.2.6.2 says that either 02,F(Mx) :::; I, or Mx has a:ri. A 3 -block L with L i I.
In the first case M = Mx by A.4.4.3, as desired. In the second [L, Y] :::; 02 (L),so taking Yz to be the preimage in M of Z(Op(M)) :::; Y and Yo := 02 (Yz), L
normalizes 02 (Yo0 2 (L)) = Y 0. Then L:::; Nc(Y 0 ) = M as M EM, contrary to
Li I. This contradiction completes the proof of (1).
If [Z, X] = 1 then XT :::; Cc(Z) .and M = !M(XT) by (1), so that alsoM = !M(Cc(Z)), contrary to Hypothesis 15.4.1.2. Thus (3) holds.
Next V := (Zx) E R 2 (XT) and V:::; 02(M) by B.2.14, and as [Z,X] #-1 ~nd
X/0 2 (X) has prime order, CxT(V) = 02(CxT(V)) = CT(V). Then by (1) we may
apply 15.4.2.4 with XT in the role of "Mo", to conclude q(XT /CxT(V), V) :::; 2.
Then.asp> 3 by hypothesis, D.2.13.1 shows p = 5, so (4) holds. DLEMMA 15.4.4. Each member of M(T) is a {2, 3, 5}-group.
PROOF. .Suppose some ME M(T) has order divisible by p > 5, and choose p
maximal subject to this constraint. As M is solvable by 15.4.2.2, there is a Hall
{2,p}-subgroup H of M containing T. Let P denote a Sylow p-subgroup of the
preimage in Hof D 1 (Z(H/0 2 (H))); then TP E H(T) with P elementary abelian,
and mp(P) :::; 2 since His an SQTK-group. If mp(P) = 2 and Tis irreducible on P,
then HE 3(G, T), contrary to 15.4.2.3. Thus there is X:::; TP with XE ~(G, T),
contrary to 15.4.3.4. DLEMMA 15.4.5. Cc(Z) is a {2, 3}-group.
PROOF. If not, arguing as in the proof of 15.4.4, there is a nontrivial elementary
abelian 5-subgroup P of Cc(Z) with PT E H(T), and there is X :::; ,PT withX E 3(G, T) or ~(G, T). Since X :::; Cc(Z), the former is impossible by 15.4.2.3,
and the latter by 15.4.3.3. D