596 3. DETERMINING THE OASES FOR LE .Cj(G, T)Next by 3.3.9, and appealing to B.5.1.6 in case (al):
(b) In case (al), CL(Z) ~ 83 /2^1 +4.
(c) In case (a2), CL(Z) ~ (83 x 83)/E15.
Let R := 02 (CL(Z)T). Then R is the unipotent radical of the parabolic CL(Z)
of L, so NM(R) :::; NM(0^2 (CL(Z))). By B.5.1.6 and B.4.2.10, J(R) :::; Cr(V) =
02 (LT), so that J(R) = J(0 2 (LT)) by B.2.3.3, and hence Na(R) :::; Na(J(R)) :::;
M as M = !M(LT). Therefore as we just showed that 02 (CL(Z)) is normal in
NM(R):
(d) J(R) = J(0 2 (LT)), and 02 (CL(Z)) :S) Na(R):::; M. Thus D does not act on
R, and hence does not act on 02 (CL(Z)).
Next we show:
(e) Cz(L) = 1, so Z:::; [V,L] = V. Further when (a2) holds, Lis irreducible on V.For if Cz(L) =f 1, then Ca(Z) :::; Ca(Cz(L)) :::; M = !M(LT), so 0
31
(Ca(Z)) =03 ' (CM(Z)) = 02 (CL(Z)) is D-invariant, contrary to (d). Then since V =
[V,L]Cz(L) by 3.3.7.4, V = [V,L].
Our final technical result requires a lengthier proof:
(f) Tis nontrivial on the Dynkin diagram of L.
Assume that T is trivial on the Dynkin diagram of L. Then f' :::; Pi :::; L,for i = 1, 2, with Pi ~ L 3 (2)/ Es. Let Yi := P[>^0 , so that Yi E .C(L, T). Then
LT= (Y1, Y2, T).
We now repeat some of the proof of 3.3.18: By 3.3.15 we may assume there
is no nontrivial characteristic subgroup of T normal in YT for Y := Y 1 , so the
MS-pair (YT, T) is described in C.1.34. As T is Sylow in G, case (5) of C.1.34
does not hold. By (a) and (e), m(Cz(Y)) :::; 1, so case (4) does not hold. NextY has a nontrivial 2-chief factor on 02 (Y) and two on [V, L] from (al) and (a2),
eliminating cases (1) and (2) of C.1.34 where there are at most two such factors.
Therefore case (3) of C.1.34 holds. Set Q := [0 2 (YT), Y] and U := Z(Q); then U
is a natural module for Y/0 2 (Y) and Q/U the sum of two copies of the dual of
U. In particular, Y has exactly three noncentral 2-chief factors. Then CQ(Y) = 1,
eliminating case (al) where C[v,Yj(Y) =f 1 and [V, Y] :::; Q. Thus case (a2) holds
and L is an As-block.
As T :::; L, LT = 02(LT)L. By (e), Cr(L) = 1, so by C.1.13.b and B.3.3,
either V = 02 (LT) or 02 (LT) is the 7-dimensional quotient of the permutation
module for L. But in the latter case, as T = 02 (LT)(L n T), J(T) :::; Cr(V) by
B.3.2.4, contradicting 3.3. 7.2.
Thus 02 (LT) = V, so T:::; Land ITI = 212. Let Li, i = 1, 2, be the rank-1
parabolics of CL(Z) over T, and set Xi := 02 (Li), and Ri := 02 (Li)· By (d), D
does not act on R, so as R = R 1 n R 2 , D does not act on Ri for some i, say i = 1.
We now apply 3.3.20 to X1 in the role of "X": Let Y := (Xf), and observe that
the number'"'( of noncentral 2-chief factors of X 1 is four, and Z:::; Z(YT). Thus as
ITI = 212 , part (c) of 3.3.20.5 supplies a contradiction, which establishes (f).
We now complete the proof of lemma 3.3.21.
Let P be the parabolic of L with P/0 2 (P) ~ 83 x 83 , and set H :=PT. Thenby (f), H/02(H) ~ 83 wr Z2, so by 3.3.13, D :::; Na(H). However in case (a2),
J(02(H)) :::; Cr(V) by B.3.2, so that J(02(H)) = J(02(LT)) by B.2.3.3; hence
D normalizes J(0 2 (LT)), contradicting 3.3.6.b. Therefore case (al) must hold.