1090 15. THE CASE .C.f(G, T) = 0
for each h E H with [B, Vh] -=f. l; m(A/B) = m(Ui/Cui(A)); and CuH(A) =
CuH(B).
(3) H/CH(Ui) 9'1. 83, 85, 83 wr Z2, or 85 wr Z2, with Ui the direct sum of the
natural modules [Ui, F], as F varies over the 83-factors or 85-factors of H/CH(Ui).
Further J(H)CH(Ui)/CH(Ui) S=! 83, 85, 83 x 83, or 85 x 85, respectively.
(4) [!11(Z(J1(T))), 02 (H)] ~ 1.
PROOF. Let R := CT(V). By 15.1.9.6, M = NM(R). We check that the
hypothesis of 3.1.9 holds, with N M(R) in the role of "Mo": First case (II) of
Hypothesis 3.1.5 is satisfied by 15.1.9.6. By 15.1.11, V ~ 02(H), giving (c). By
15.1.7 and B.1.8, Vis not a dual FF-module for M = NM(R), giving (d). By 15.1.8,
q(M, V) = 2, giving (a). By 15.1.9.5, M = !M(NM(R)), giving (b). Finally by
15.1.9.4, 02 (H n M) ~ Ca(V), so the hypotheses of part (5) of 3.1.9 are satisfied.
Therefore by 3.1.9, (1)-(3) hold.
As A is an FF-offender on Ui, it follows from (3) that there is a subgroup X
of Y with A E 8yl 2 (X), 02(H) ~ X, X/0 2 (X) S=! 83, and H = (0^2 (X), T). Now
we chose A E A(T), and UH is elementary abelian by 15.1.11.2, with An UH ~
An 02(H) = B, so CuH(A) =An UH= B n UH~ CB(UH)· Next by (2),
m(A/CB(UH)) = m(A/B) + m(B/CB(UH)) = 2m(A/B) = m(UH/CuH(A)),
so m(UHCB(UH)) 2': m(A). Hence UHCB(UH) E A(T), so as UHCB(UH) ~
02(H) ~ 02(X), also UHCB(UH) E A(02(X)). Therefore by B.2.3.7, !11(Z(J(T)))
and !11(Z(J(02(X)))) are contained in UHCB(UH ), so by B.2.3.2, !11(Z(J1(T))) =:
E and !11(Z(J1(02(X)))) =: D are also contained in UHCB(UH)· In particular,
E ~ 02(X), so E ~ D.
If[E,0^2 (X)] = 1, thenH = (0^2 (X),T) ~ Na(E), andhenceK ~ (0^2 (X)H) ~
Ca(E), so that (4) holds. Thus we may assume that [E, 02 (X)] -=f. 1, and it re-
mains to derive a contradiction. We saw E ~ D, so also [D, 02 (X)] -=f. 1. Then as
02 (X) = [0^2 (X), A] by construction, [D, A] -=f. 1, so in particular Di. An02(X) =:
Bx. Observe that Bx 2': An02(H) = B. On the other hand, Bx E Ai(02(X)) as
X/02(X) S=! Ss, so D centralizes Bx, and then as Di. Bx, m(DBx) > m(Bx) =
m(A) - 1, so DBx E A(T). Then as D ~ UHCB(UH) ~ 02 (H), by minimal-
ity of A02(H)/02(H), Bx ~ 02(H) n A = B, so that Bx = B. But by (2),
CuH (B) = CuH (A), so
D ~ CB(UH)UH n Ca(B) = CB(UH)CuH(B) = CB(UH)CuH(A) ~ Ca(A),
contrary to an earlier observation. This contradiction completes the proof of
15.1.12.. DLEMMA 15.1.13. Let E1 := !11(Z(Ji(T))). Then
(1) Ca(E1) i. M.
(2) [V, Ji(T)] -=f. 1.(3) Either.
(i) for all A E Ai(T) with A# 1, I.Al= 2 and A E Q*(M, V), or
(ii) case (3) of 15.1. 7 holds.
(4) Either
(a) Ji(T) = T n Mo and Ji(M) =Mo, or