726 8. ELIMINATING SHADOWS AND CHARACTERIZING THE J4 EXAMPLE
PROOF. We first prove (1). Assume the first statement in (1) fails. We claimthen that n(H) = 1 for each H E 1{,(T,M) with H ~ KzT· We examine the
groups listed in 8.3.7. The claim follows in case (i) of 8.3.7 from E.1.14.6 when
p 2 7, and in case (ii) from E.1.14.1. Thus the claim is established, and of course
it contradicts 8.3.9. Thus the first part of (1) holds, and as the Borel subgroup
X(KznT) of Kz is the unique T-invariant maximal subgroup of Kz, the remaining
statements of (1) hold.
Next by 8.3.7, Gz = KzMz. Assume that Gz > KzT. Then Y :=
02 (CaJKz/0 2 (Kz).)-:/= l, and Y ~ Mz. But Y is a 3^1 -subgroup of Mz by 1.2.2.a,
so as Mz is a {2, 3}-group, Y centralizes V. Then [Lo, Y] ~ CL 0 (V) = 02(Lo), so
that L 0 T normalizes 02 (YLo) = Y, and hence Gz ~ Na(Y) ~ M = !M(LoT),
contradicting Kz i M. Thus Gz = KzT, so Mz = XT, and hence CM(V) is a
2-group. Therefore M = L 0 T, completing the proof of (2). DLEMMA 8.3.11. r(G, V) > 3.
PROOF. Recall r(G, V) 2 3 by 7.3.2. Assume r(G, V) = 3. Then there is
U ~ V with m(V/U) = 3 and Ca(U) i M. By E.6.12, Q < CM(U), and CM(U) =
CLoT(U) by 8.3.10.2. Therefore by H.4.12.3 and H.4.10, U = Cv(I) for some
involutiof!-IE Lo'f'. By H.4.12.3, CfVJ(U) is a 2-group, so by E.6.27, U is centralized
by an (F - 1)-offender. Thus IE 10 by H.4.10.3. Consequently as m(V/Cv(I)) =
3, we may assume I E R 1 , so that U = Cv(R 1 ). But of course Ri < R andCa(Cv(R)) ~ M by 8.3.8. This contradiction establishes 8.3.11. D
LEMMA 8.3.12. V ~ 02(Gz)·
PROOF. Let Qz := 02(KzT). If V ~ Qz, then the lemma holds, since Gz =
KzMz by 8.3.7 and V :::1 M. So we assume V i Qz. Let Gz := Gz/ (z) and
K;T* := KzT/Qz. By H.4.9.2, XT is irreducible on i/5 and V/V5, where 115
denotes the 5-dimensional space in H.4.9.2.
We claim that V 5 = V n Qz: to see this, we apply G.2.2, which is designed for
such situations. Note that Hypothesis G.2.1 is satisfied with (z), 115, L 0 , X, KzTin the roles of "V 1 , V, L, L 1 , H". We conclude from G.2.2 that
U := (V5K") ~ Z(Qz),
and U is a 2-reduced module for K;. Further as Vi Qz and XT is irreducible on
V/115, 115 = V n Qz as claimed.
Notice as U ~ Qz ~ T that [U, VJ~ VnQz = V5; so as m(i/5) = 4 = m(V/115),U is a dual FF-module for K;T, with dual F F-offender V. Now V is a normal
E 16 -subgroup of the Borel subgroup (Mn KzT)* in 8.3.10, so V* E Sylz(K;).
In particular there is h E Kz with K; = (V*, V*h). Observe U = [U,K;], since
V5 = [i/5, X] and U = (V 5 Kz). Then as [i/5, V*] ~ V5 and K; = (V*, V*h), we
conclude
U -= [U, - Kz] * = V5 - + V-h
5 ,so that Uhas dimension at most 8, and hence_ is itself an FF-module, with quadratic
FF-offender V.,By Theorems B.5.6 and B.5.1, the only possibility is U =UK EB
Uk for UK a natural module for K*. But now P of order 3 in X diagonally
embedded in KK^8 is fixed-point-free on U, and hence on i/5 of rank 4. Also X is
fixed-point-free on V*, so Cv(X) = (z), contradicting H.4.12.1. This completes the
proof of 8.3.12. D