734 9. ELIMINATING nt(2n) ON ITS ORTHOGONAL MODULE
PROOF. By 9.2.7, B = CD(Vi) SK is diagonally embedded in LLt, so D =
B(DnL) s (K,L) s K 0. As Mo= (LDT1,K), 02 (Mo) = (L,K) s Ko, so
Mo= K 0 Ti and Ko= 02 (M 0 ). Next using parts (2) and (3) of 9.2.2, Oi(Z(Ti)) S
Cv(Ti) s Vi. Hence as 02(Mo) n Oi(Z(Ti)) =f. 1, and as Dis irreducible on Vi,
Vi s 02(Mo). Therefore Na(02(Mo)) E He by 1.1.4.l. Next Vi= (VP) S 02(Mo),
and then
Cr(02(Mo)) s Cr(Vi) s Nr(L) S Tis Mo,
so Mo E He by 1.1.4.4 with Na(0 2 (Mo)) in the role of "M". Also Z(Mo) = 1 as
Oi (Z(Ti)) s Vi and Cv 1 (D) = l. D
We now proceed as in the last paragraph of the proof of 9.3.l. Let U := (ViM^0 ).
As Vi = (ZD), U = (ZM^0 ), so by B.2.14, U E R'2(Mo). Set M 0 := Mo/CM 0 (U).
By 9.3.5.1, K* =f. l. Now as in the proof of 9.3.1, K = [K, J(T)] and hence
[U, J(T)] =f. 1, so U is an FF-module for K 0. Then we obtain the same four
possiblities for K 0 as in the proof of 9.3.1, and eliminate the fourth case K 0 ~ A7
as in that proof, to conclude:
LEMMA 9.3.7. K 0 ~ SL 3 (4), Sp4(4), or G2(4), and U is an FF-module for
Mo.
LEMMA 9.3.8. K 0 is not SL 3 (4).
PROOF. Otherwise Z(K 0 ) = CD(L) = (D n Lt), as each is of order 3. But
then K/0 2 (K) ~ L 2 (4) is centralized by ((D n Lt)NT(D)) = D, a contradiction
since B s D n K. D
LEMMA 9.3.9. K 0 ~ Sp4(2n).
PROOF. If not, by 9.3.7 and 9.3.8, K 0 ~ G2(4). Now Land Kare normalized
by T, so L = Pf' and K = P2, where Pi and P2 are the maximal parabolics of K 0
containing (TnK 0 )*. By 9.3.7, U is an FF-module for K 0 , and by 9.3.6, Z(M 0 ) =
1-so U is the natural G 2 (4)-module by Theorems B.5.1 and B.4.2.4. Therefore by
B.4.6.14, D n L centralizes K/0 2 (K). We again use the action of Nr(D) to obtain
the same contradiction obtained at the end of the proof of 9.3.8. D
LEMMA 9.3.10. Ko is an Sp4(4)-block.PROOF. Recall Ti is of index 2 in T. If 1 =f. C char Ti with C :::;! M 0 , then
Nc(C) contains Mo :i. Mand (L, T) = L 0 T, contradicting M = !M(L 0 T). Thus
no such characteristic subgroup exists, giving the condition (MS3) of Definition
C.l.31. We obtain (MSl) and (MS2) using 9.3.9. Then the lemma follows from
C.l.32.3. D
We are now in a position to obtain a contradiction, eliminating the case n(H) =
2. For by 9.3.6, Z(Mo) = 1, so U is the natural module for the Sp4(4)-block Ko by
9.3.10. Now V/V2 is the natural module for L/0 2 (L). However L = P^00 for some
maximal parabolic P* of K 0 , so 02 (L)/U is an indecomposable of F-dimension 3
with no natural submodule. Therefore V s U, so V = U as both are of order 28.
Then Kos Nc(U) = Nc(V) = M, a contradiction.