3.3. NORMALIZERS OF UNIQUENESS GROUPS CONTAIN No(T) 599Finally if m = 1, let Pi, i = 1, 2, denote the maximal parabolics of LT over T.
Then Pi has just two noncentral 2-chief factors, so D acts on 02 (Pi) by 3.3.20.4.
Thus D acts on To := 02(P1) n 02(P2), completing the proof of the lemma. D
In the proof of the next lemma, we encounter the shadow of the non-maximal
parabolic in Z3/0t(2), and we eliminate this shadow using 3.3.20.
LEMMA 3.3.24. Lis a block of type A6, A6, G2(2), or L3(2).
PROOF. We observed earlier that either L is a block of type A 6 , A 6 , or G 2 (2),
or L ~ L 3 (2). Thus appealing to 3.3.23, we only need to eliminate the cases arising
in 3.3.23.2, where L has k := 2 or 3 noncentral 2-chief factors.
Let Q := [02(LT), L]. When k = 2, 0.1.34.2 says that Q is the direct sum
of two isomorphic natural modules for L/02(L); then LT acts on at least one of
the three natural submodules Vo of Q, and we set Zo := Zn V 0 • When k = 3,
Vo := Z(Q) is a natural L/0 2 (L) module, and Q/Vo is the direct sum of two copies
of the dual of V 0. In this case we again set Zo :=Zn Vo. Thus in either case Zo is
of rank 1 and V 0 = (Z{j) = [Z 0 , L] is an LT-invariant natural L/0 2 (L)-module.
Set R := 02(CL(Z)T), X := 02 (CL(Z)), and Y := (XD). Then X has
k + 1 ::; 4 noncentral 2-chief factors. By 3.3.23.2, D does not act on R, so we can
apply 3.3.20.5.to conclude that Y has a unique noncentral 2-chief factor W, and that
Z 0 ::; (02(X)) :::; Z(Y). Set YT:= YT/Zo, Ry := 02(YT) and U := (Vl). As
Xis irreducible on V 0 , we may apply G.2.2.1 with Vo, Zo, YT in the roles of "V, Vi,
H", to conclude that U::; 01(Z(Ry)), so (U)::; Zo. As Vo= [Vo,X], U = [U, Y],
so by uniqueness of W, W = U/Uo where Uo := Cu(Y). By 3.3.20.3, 02(X)::; Ry,
so as X = 02 (X), 02 (X) = [0 2 (X), X] ::; [Ry, Y] = U. Then (02(X)) ::; Zo,
eliminating the case k = 2, for there (0 2 (X)) = Cq(L) is of rank 2. Thus k = 3,
and here we compute that Q/(02(X) n Q) ~ E4 and [02(X), a] 1:. Zo for each
a E Q-0 2 (X). Therefore setting (YT):= YT/Ry, Q ~ E4. This is impossible,
since by 3.3.20.3, XT ~ D12, whereas Q ::::] XT*. D
LEMMA 3.3.25. (1) Lis a block of type A5, G2(2), or L3(2).
(2) Assume Cr(L) of. 1 and L is not L3(2), and let X := 02 (CL(Z)) andR := 02 (XT). Then D acts on X and R, but does not act on any nontrivial
D-invariant subgroup of R normal in LT.
(3) If Cr(L) = 1, then either V = 02(LT), or L is an A5-block.
PROOF. Let X := 02 (CL(Z)) and R := 02 (XT). Inspecting the cases listed
in 3.3.24, XT/R ~ 83.
We first prove (2), so suppose Cr(L) of. 1 and Lis not L 3 (2). Then Cz(L) of. 1,
so as usual Ca(Z)::; Ca(Cz(L))::; M = !M(LT), and then Ca(Z) = CM(Z). As
Lis not L 3 (2), m3(L) = 2 and so by A.3.18, Lis the subgroup B(M) generated by
all elements of M of order 3. Therefore X = B(Ca(Z)), so D acts on X and hence
also on R. Then the final statement of (2) follows from 3.3.6.b.
In view of 3.3.24, to prove (1) we may assume L ~ A6, and it remains to derive
a contradiction. By B.4.2, J(R) ::; Cr(V) = 02 (LT), so that J(R) = J(02(LT))
by B.2.3.3. Therefore Cr(L) = 1 by (2). Then as the A5-module has trivial 1-
cohomology by I.1.6, V = 02 (LT) by 0.1.13.b. But again using B.4.2, there is a
unique member A of P(T, V), m(A) = 2, and Cv(A) = Cv(a) for each a EA#.