15.1. INITIAL REDUCTIONS WHEN Cf(G, T) IS EMPTY 1087NOTATION 15.1.4. Set M := Mf. We choose V := V(M) in cases (1)-(5)
of 15.1.2, but in case (6) of 15.1.2 we choose V := [V(M), MJ], where MJ is the
preimage in M of }(M/CM(V(M)). Set M := M/CM(V) and Mo:= }(M, V).
Observe that except in case (6) of 15.1.2, Mo coincides with MJ. We review
some elementary but fundamental properties of V:
LEMMA 15.1.5. (1) V =((Zn V)M^0 ), so VE R2(M).
(2) Cc(V) :S: Mc.
PROOF. From the description of Vin 15.1.2, V = ((Zn V)M^0 ).completes the proof of (1). Part (2) follows since Mc= !M(Cc(Z)).
LEMMA 15.1.6. 02 (CM(Z)) :S: CM(V).
Then B.2.14
DPROOF. By 14.1.6.1, M^00 :S: CM(V). Let 8 := T n M^00 • By a Frattini Argu-
ment, CM(Z) = M^00 K, where K := CM(Z) nNM(8). Since K^00 ::;; NM=(8) and
NM= ( 8) is 2-closed, K is solvable. Thus K = XT, where X is a Hall 2' -subgroup
of K. Therefore it remains to show X :S: CM(V), so we may assume X =/=-1. From
the structure of M described in 15.1.2, M is 2-nilpotent, and hence so is X'i'.
Therefore X = O(XT) :::;! X'i'. Then as X =/=-1, Zn [V,X] = C[v,xi(T) =/=-1, and
C[v,xi(X) = 1 by Coprime Action, whereas X :S: K :S: CM(Z). This contradiction
completes the proof. DIn the next result we review the cases from 15.1.2 which can occur in our
counterexample, except that we reorder them according to the value of m(V):
LEMMA 15.1.7. One of the following holds:
(1) m(V) = 4, and M =Mo ~ 83.
(2) m(V) = 4, Mo~ 83, and M ~ 83 x Z3.
(3) m(V) = 4, and M =Mo= nt(V).
(4) m(V) = 4, Mo = P(f'> where P := 02 (M) ~ Eg and tis an involutioninverting P, and 'i' ~ Z4.
(5) m(V) = 4, M 0 ~ D10, 'i' ~ Z2 or Z4, and either F(M) = F(Mo) or
F(M) ~ Z15.
- t
(6) m(V) = 8, Mo = M1 x Mz where Mi ~ D2p with p = 3 or 5, M 1 = Mz for
some t ET, and V = V1 EB Vi, where Vi:= [V, Mi]·
(7) m(V) = 6, Mo = P(f'> where P := 02 (M) ~ 3i+^2 , tis an involution
inverting P / 'P ( P), and T acts irreducibly on P / 'P ( P).
Furthermore if V < V(M), then case (3) holds.
PROOF. Suppose first that V < V(M). Then by definition of Vin 15.1.4, case(6) of 15.1.2 holds and V = [V, MJ]; it follows that conclusion (3) holds. Thus in
the remainder of the proof we may assume that V = V ( M), and hence that one of
cases (1)-(5) of 15.1.2 holds.
Assume first that m(V) = 2. Then case (1) of 15.1.2 holds with (p, m) =. (3, 1)
and M ~ 83. Then as we observed at the start of section 14.2, 14.1.18 shows that
Hypothesis 14.2.1 is satisfied. Therefore we may apply Theorem 14.6.25 to conclude
that G is one of the groups listed in conclusion (1) of Theorem 15.1.3, contrary to
the choice of G as a counterexample.
Thus m(V) > 2. Also since G is a counterexample, conclusion (2) of Theorem
15.1.3 does not hold. Thus if case (3) of 15.1.2 holds, then m(Vi) = 4 for each i, so