i5.l. INITIAL REDUCTIONS WHEN .Cr(G, T) IS EMPTY 1103(4) UL is a sum of at most n - 1 isomorphic natural modules for L* ~ Ln(2),
where n = 4 or 5.
By B.2.14, U = ULCu(L). Let Zu be the projection of Z 2 on UL with respect
to this decomposition.LEMMA 15.1.28. UL is a natural module and ML = CL(Z) is the stabilizer of
the point Zu in UL.
PROOF. First by 15.1.18.5, C(Gi, Qi)= Mz, and Hypothesis C.2.3 is satisfied
with Gi, Qi, Mz in the roles of "H, R, MH". Thus by C.2.1.2, 02 (LS) :::; Qi.
Further Lis normal in Gi by 15.1.25.1, so we may apply C.2.7.2 to conclude that
Qi contains an FF-offender on U.
As Ca 1 (Z) = Ca 1 (Z2), CLs(Zu) = CLs(Z) :::; Mz. That is, ML is an S-
invariant proper parabolic containing CL(Zu ).
Suppose case (1), (2), or (4) of 15.1.27 holds. Then CL(Zu) is a maximal
parabolic, acting irreducibly on 02 (CL(Zu)*), so by the previous paragraph ML=
CL(Zu) and Qi= 02(M£). Therefore as Qi contains an FF*-offender, we concludefrom B.3.2 or B.4.2 that case (1) holds with UL the natural module for L, so that
the lemma holds in this case.Thus we may assume case that (3) of 15.1.27 holds. Therefore U = Ui EB U 2 ,
where Ui is a natural module for L*, and U2 is its dual. Let Zo := CuL(S), so
that Zu :::; Zo. Then either S acts on Ui and U2 with Zo = Zu,i EB Zu, 2 , where
Zu,i is the point of ui fixed by SL, or else s is nontrivial on the Dynkin diagram
of L*, with Zo = (ziz2) where Zu,i := (zi)· In either case, CL(Zo) contains the
parabolic P determined by the interior node(s) of the diagram for L*. Thus as
CL(Zo) :::; CL(Zu) :::; ML, Qi :::; 02(P*). But then by B.4.9.2, Qi contains no
FF* -offenders, contrary to an earlier remark. D
LEMMA 15.1.29. UL is not a natural module.PROOF. By 15.1.24.1, we are in case (4) or (6) of 15.1.7. Adopt the notationof the proof of 15.1.28. By 15.1.28, the projection Zu of Z 2 on UL is of order 2.
We saw U = ULCu(L) and Z is a full diagonal subgroup of ZiZ2 with [Zi,L] = 1,
L <£.Mc, and Ca(z) :::; Mc for each z E z#. Thus Z projects faithfully on UL, soIZI = 2. Therefore case (4) of 15.1.7 holds, rather than case (6) with p = 3, so V'2
is of rank 2. As S acts on V2, and V'2 = [V2, Y] :::; [U, L] :::; UL, it follows that V2
is the line in UL stabilized by S. Thus NL(V2) contains the minimal parabolic Po
of LS over S which is not contained in the maximal parabolic MLS = CL(Z)S.By 15.1.9.1, J(T) :::; CT(V2), and by 15.1.17, J(T) = J(S) and Na(J(S)) :::; M.
Hence J(S) = J(Cs(V2)) = J(02(Po)) using B.2.3.3, so that Po :::; M, and hence
Yo := 02 (P 0 ) :::; Y with V'2 = [V'2, Yo]. Let P2 be the minimal parabolic adjacent to
Po with respect to the Dynkin diagram of L, let Y2 := 02 (P2), and let K := (Y 0 , Y 2 ).
Thus KS/02(KS) ~ L3(2) with KS the rank-2 parabolic corresponding to an end
node and its neighbor, and KSnMLS = P2. Let Q := CT(V) and t E T-S. Let P 3be the remaining end-node minimal parabolic of L, and set Lo := 031 (ML)· Thus
LoS/02(LoS) ~ Ln-i(2), and P2 and P3 are the end-node minimal parabolics of
LoS. By 15.1.25.1 and 15.1.28, Lo = 031 (Ca 1 (Z)), so as 02 (Ca(Z)) :::; Ca(VE)
by 15.1.14.1, and Zi :::; VE by 15.1.18.1, Lo = 031 (Ca(Z)). Thus T acts on L 0.
Hence as P2 and P3 are the end-node minimal parabolics of L 0 , Yi is either Y 2 or