12.6. ELIMINATING As ON THE PERMUTATION MODULE 837This is a contradiction as T permutes with no nontrivial p-subgroups of L for p > 3
as L ~ £4(2).
Therefore K* ~ (8)£3(4) or L2(4). If K* ~ 8L3(4) and £ 1 i K, then as Mkcontains a Borel subgroup of H*, a Sylow 3-subgroup of MH is of order 27, whereas
by 12.6.1.5, a Sylow 3-subgroup of M is of order 9. Therefore if K* is (8)£ 3 (4),
then as T/02(L1T) projects on a noncyclic subgroup of Out(K*), (1) holds. So
suppose K* ~ L2(4). If K < Ko then as T/0 2 (L 1 T) is noncyclic, (2) holds. If
K = Ko our earlier analysis shows Li = L 0 x L'K with LiT* ~ 83 x 84 , so that
(3) holds. DLEMMA 12.6.32. K* ~ £2(4).
PROOF. Assume otherwise. Then case (1) of 12.6.31.1 holds, so Ka = K* ~
(8)£3(4), and T* K* /K* ~ E4. We pick i E Irr +(K, f~h, T), and by 12.6.30.3, we
may adopt the notation of F.9.18.4; in particular IH := (JH). As Tis nontrivialon the Dynkin diagram of K, it follows from B.5.1 and B.4.2.2 that KT* has no
FF-modules. Thus by F.9.18.7, [UH,K] = JH. If IH =I, then q(H*,i):::; 2 by
F.9.18.2; so as KT has no FF-modules, i/C1(K) must appear in B.4.5. As the
tensor-product module for £3(4) in B.4.5 has q > 2, we have a contradiction. Thus
I< IH, so case (iii) of F.9.18.4 occurs; that is, K* ~ 8L3(4) and JH = U 1 EE7 U 2 ,
where U1 = i is a natural K*-module, and U 2 = Uf, fort ET nontrivial on the
- -H - - -
Dynkin diagram of K. As V5 = [V5,Li.]:::; IH, UH= (V5 ) :::; IH, so UH =IH.
Next as Mk is the product of T with a Borel subgroup of K by 12.6.31.1,
MH = L1T, so T n K = 02(Mk) = Ri is Sylow in K. As V5 .is L1T-invariant
and centralized by Ri, we conclude i/5 = V5,1 EE7 i/5,2, with i/5,i = Cu, (T n K) an
- -H - - -
F 4-point in Ui. In particular V;,,i = Cv 5 (Xi) for some Xi of order 3 in £ 1 ; so
V5,i contains a nonsingular vector ui of V. Now CK* (ui) is a maximal parabolic
of K, so in particular CK ( Ui) does not lie in the Borel group MK: 0 • This is a
contradiction as Ca(ui) :::; M by Theorem 12.6.2. This contradiction completes the
proof of 12.6.32. DLEMMA 12.6.33. H* ~ 85 X 83.
PROOF. Assume otherwise. As 12.6.32 eliminates case (1) of 12.6.31, we mustbe in case (2), where H* of index at most 2 in 85 wr Z 2. Thus there is t E
T-Nr(K), and we let K 1 :=Kand K2 :=Kt. For XE 8yl 3 (L1), X = X 1 x X2
with Xi E 8yl3(Ki) and V5 = [V5,X]:::; [UH, Ko]= U1U2, where Ui :=[UH, Ki]·
Thus UH= U1U2 =[UH, Ko].
Pick i E Irr +(Ko, UH, T); by 12.6.30.3, we may adopt the notation of F.9.18.5.
We claim that either
(a) [UH/IH,K1,K2]:::; IH, and [IH,K1,K2] = 1, or
(b) UH is the n:t(4)-module for Ka·For notice by Theorems B.5.6 and B.4.2 that H* has no strong FF-modules. Thus it
follows from F.9.18.6 that either UH= JH, or both JH and UH/IH are FF-modules
for H*. Observe that only cases (i) and (iiia) of F.9.18.5 can arise. In case (i), JH
is not an FF-module for H*, so UH= JH and (b) holds. Suppose case (iiia) holds.
Then (a) holds if UH= JH, so assume otherwise. Thus UH/IH is an FF-module
for H*; then as UH= [UH, Ko], it follows from B.5.6 that [U1, K2] :::; IH, so again