669hence Cz(DoT) n [U, L1] is a vector of weight 6, so that Ox* (Cz(D 0 T)) ~Sf,. Now
Cz(DoT) = Cz(DL) = Cz(L), so Cx(Cz(DoT)) ::::; M = !M(LT). But this is
impossible as Do :::l X n M, completing the proof of 6.1.5. D
LEMMA 6.1.6. (1) Cz(L) = 1, and hence Cr(L) = 1.
(2) VL is the natural module for L, and V = VL if L = [L, J(T)].
(3) VL = [R2(LT), L].PROOF. If Cz(L)-/= 1, then Ca(Z) ::::; Ca(Cz(L))::::; M = !M(LT). But then
for H E 'H*(T, M), H ::::; M by 6.1.5, contrary to H i M. This contradictionestablishes (1). Then 6.1.4 and (1) imply VL is the natural module for L. The final
statement of (2) follows as V = Cv(L)[V, L] by E.2.3.2. Finally V::::; R 2 (LT), soVL ::::; [R2(LT), L]. On the other hand, applying (2) to R2(LT) in the role of "V",
L is irreducible on [R2 (LT), L], so (3) .holds. DNow replacing V by VL if necessary, we assume throughout the rest of this
section that
V=V£.
Thus by 6.1.6.2, Vis the natural module for L ~ L 2 (2n). Since L :::l M, and Lis
irreducible on V, using 6.1.3.1 we have:LEMMA 6.1.7. (1) Vis a TI-set under M. Thus if 1-/= U::::; V, then NM(U)::::;
NM(V) = Mv.
(2) Hypothesis E.6.1 holds, so we may apply results from section E.6.Using 3.1.4.1, 6.1.7, and 6.1.5 we have:LEMMA 6.1.8. If H::::; Na(U) for 1 -/= U ::::; V, then H n M = NH(V). In
particular H n M = NH (V) for each H E 'H* (T, M).Let Zs := Cv(TL), so that Zs is a 1-dimensional F2n-subspace of the natural
module V. Let S := Cr(Zs).
LEMMA 6.1.9. (1) S = TL0 2 (LT) and 8 E Syl2(Ca(Zs)):
(2) Na(S) ::::; M.
(3) F*(Na(Zs)) = 02(Na(Zs)).
(4) V::::; 02(Ca(Zs)) and V/Zs::::; Z(S/Zs).
(5) Na(Zs) = Ca(Zs)NM(Zs) = Ca(Zs)NMv(Zs).
(6) J(T) = J(S) and Baum(T) = Baum(S).
PROOF. As T ::::; Na(Zs), (3) holds by 1.1.4.6, and also Cr(Zs) = S E
Syl 2 (Ca(Zs)). As V is the natural module for L, the remaining assertion of (1)
holds, and also V/Zs::::; Z(S/Zs). Then an application of G.2.2.1, with Na(Zs) in
the role of "H", establishes the remaining assertion of (4).
Now using (1), we may apply a Frattini Argument to conclude that Na(Zs) =
Ca(Zs)(Na(Zs) n Na(S)). Thus (5) will follow from (2) since Vis a TI-set in M
by 6.1.7; so it remains to prove (2) and (6). ·
If J(T) ::::; Cr(V), then in particular J(T) ::::; S. On the other hand, if J(T)
does not centralize V, then as Vis the natural module for L, J(T) ::::; S by B.4.2.1.