natural. Therefore by B.4.14, Wis the adjoint module for K/02(K) and Cw(Xi)
is indecomposable of F 4 -dimension 4 for L+/0 2 (L+)· In particular Cw(Xi) does
not split over [Cw(Xi), L+], a contradiction as CR 1 (X1)/CR 1 (L+X1) is the natural
module for L+ / 02 ( L+). This contradiction finally completes the elimination of the
case where (*)holds and L/0 2 (L) ~ L ~ SL3(4).
Thus in view of 11.0.2.2, we have shown that if(*) holds then L/02(L) ~ L ~
Sp 4 (4) or G 2 (4), and to complete our treatment of the case i = 1 it remains to show
that() implies the remaining statements in (3). Recall I/02(I) ~ SL3(4), Sp4(4),
or G 2 (4), so IE £(G,T) by 1.2.8.4, and then as [Z,L2] i= 1, even IE £j(G,T).
This begins to establish some symmetry between L and I; in particular applying
11.0.2 to I, we conclude there is Vr E R 2 (IT) with VrfCVi(I) the natural module.
Assume [Z, K] i= 1. Then K::::; K+ E £j(G, T) by 1.2.9. Now since K/02(K) ~
L 3 (4), by A.3.12, either K = K+ or K+/02(K+) ~ M23. By B.4.2, neither L3(4)
nor M 23 has an FF-module, so Theorem 7.0.1 supplies a contradiction.
Therefore [Z,K] = 1, so if Cv 1 (I) i= 1 then Zr:= Cz(I) i= 1. But then by
1.2.7.3, Na(I) = !M(IT) = !M(Co(Zr)), so Li ::::; K::::; Na(I) 2: L2, and hence
K::::; M = !M(LT), for our usual contradiction. Thus Cv 1 (I) = 1. As [Z,K] = 1,
Ko stabilizes the 1-dimensional F4-subspace Vr,1 of Vr stabilized by T. Thus Ko
plays the same role in I that Li plays in L. As XT = TX and we saw X ::::; I , X
is also a Cartan subgroup of I and Vr, 1 = [Zn Vr,1, Xx] ::::; Ca(K). Therefore K is
the member of C(No(Vr, 1 )) containing Ko, so K plays the role of "K" for I as wellas for L. In particular, (*) is also satisfied by I. Therefore applying our previous
reduction to I, I/02(I) is not SL3(4). Notice also that L 2 plays the same role in
both Land I: L 2 is the derived group of the stabilizer of a line of V and T/j.
Suppose L/0 2 (L) ~ G 2 (4). Then Xi ::::; L 2 by B.4.6.14. From the previous
paragraph, Ko centralizes Vr,i, so if I/0 2 (!) ~ G2(4), then by the same argument,
Cx(Ko/02(Ko)) = X n L2 = X1.
But from(*), [Ko,X1] 1:. 02(Ko), a contradiction. Therefore if L/0 2 (L) ~ G2(4),
then I/02(I) ~ Sp4(4) and so (3) holds.
This leaves the case L/02(L) ~ Sp4(4). Interchanging the roles of Land Iif necessary, and appealing to the previous paragraph, we may assume that also
I/02(I) ~ Sp4(4). As L/02(L) ~ Sp4(4), Xx = X n Li and Xi are the two
diagonally-embedded subgroups of order 3 with respect to the decomposition
X = X2 x (X n L2)·
Therefore as Ko centralizes Vr,i, and Xx= XnKo, from the structure of I /02(I) ~
Sp4(4), the second diagonal subgroup Xi centralizes Ko/0 2 (Ko). But again this
does not hold in (*), a contradiction completing the treatment of the case i = 1.Now we turn to the easier case i = 2. We assume that L 2 < K 2 = K, and
it remains to derive a contradiction. Here Vi/Cv 2 (L) is the natural module for
L2/02(L2) ~ L2(q), and by 11.0.3.3, either Cv 2 (L) = 1, or Lis Sp4(q) or G2(q)
with m(Cv 2 (L))::::; n. Thus m(Vi)::::; 3n .. Examining the possibilities in (2) and (3)
of 11.1.1 for cases where K possesses a nontrivial module of rank at most 3n, we
conclude that one of the following holds:
(a) K/02(K) ~ SL3(q) and Vi is the natural module.
(b) q = 4, K/02(K) ~ Ay, and Vi is the natural module.