15.1. INITIAL REDUCTIONS WHEN .Cf(G, T) IS EMPTY 1089Since Mis maximal in M(T) under::-.,, we may now apply 14.1.4 to conclude that
(2) holds; and apply 15.1.5.1, (2), and 14.1.2 to complete the proof of (1). Observe
that NM(R) E 'He by 1.1.3.2. Using case (b) of the hypothesis of A.5.7.2 rather
than case (a), the proof of 15.1.1.1 shows that (5) and (6) hold, and case (II) of
Hypothesis 3.1.5 is satisfied with NM(R) in the role of "Mo" for any HE 'H*(T, M).
Further M = !M(NM(R)) by (5), so (3) follows from 3.1.7. Then (4) follows from
(3) and 15.1.6. Finally Na(T) ::; M by (1), so (7) follows from 3.1.3.2. D
LEMMA 15.1.10. If case (6) of 15.1. 7 holds with p = 3, then M ~ 83 wr Z 2.
PROOF. Since Coa(M)(f' n Mi) is cyclic by A.1.31.1, 02 (Mo) = 02 (M). Then
2 - - - -as 0 (M) acts on Mi and Vi for i = 1, 2, CaL(V;)(Mi) ~ £ 2 (2), and 02 (M) = 1 by
15.1.5.1, the result follows.. DLEMMA 15.1.11. For HE 'H*(T, M):
(1) V::; 02(H).
(2) UH := (VH) is elementary abelian.
PROOF. Set R := Cr(V). By 15.1.9.5, 02((NM(R),H)) = 1, so Hypothe-
sis F.7.1 is satisfied with NM(R), H in the roles of "G 1 , G 2 ". Further as R =02(NM(R)) by 15.1.9.6, R = 02(CNM(R)(V)), so that Hypothesis F.7.6 is also sat-
isfied. Now Vis not an FF-module for AutNM(R)(V) by 15.1.8, so if (1) holds, we
may apply F.7.11.8 to obtain (2).
So we may assume that Vi 02 (H), and it remains to derive a contradiction.
By 3.1.3.1, H n M is the unique maximal subgroup of H containing T, and by
15.1.9.7, H is described in B.6.8. Then our assumption V i 02(H) implies V i
kerHnM(H) by B.6.8.5. Thus Hypothesis E.2.8 is satisfied with HnM in the role of
"M". Then by E.2.15, r := q(M, V) < 2, so that by 15.1.8, m(V) = 4, M = 0!(2),
and r = 3/2. Also by 15.1.9.4, 02 (H n M) ::; CH(V). Hence by E.2.17, Y = (VH)
is isomorphic to 83 /Q~, L3(2)/D~, or (Z 2 x L3(2))/D~. However in the last two
cases, IAutr(V)I :::'.:'. 8 by E.2.17, contrary to IMl2 = 4. Therefore Y ~ 83/Q~. Set
P := 02(Y), Xo := 02 (NM(R)), and X := 02 ([Xo, P]). As Autp(V) ~ E4 ~ T
and R = Cr(V), T =PR, so X = Xo ~ Eg. Next
[R, P] ::; Cp(V) = V n P,
so P centralizes R/V, and hence X ::; [X 0 , P] centralizes R/V. Then V = [R, X]
so as F*(M) = 02 (M) ::; R, Cx(V) is a 2-group by Coprime Action. Then as
X ~ E 9 and X = 02 (X), it follows that X ~ A4 x A4 and R = V x CR(X). By
(*), [CR(X), P]::; Cv(X) = 1, so TX= PRX =PX x CR(X) with PX~ 84 x 84.
But now [V, J(T)] #-1, contrary to 15.1.9.1. D
LEMMA 15.1.12. Let HE 'H*(T,M) and UH:= (VH). Then
(1) H has exactly two noncentral chief factors U1 and U2 on UH·
(2) There exists A E A(T) - A(0 2 (H)), and for each such A chosen with
A02(H)/02(H) minimal, A is quadratic on UH, and setting B :=An 02(H), we
have:
2m(A/B) = m(UH/CuH(A)) = 2m(B/CB(UH));