14.4. FINISHING THE TREATMENT OF (VG1) ·NONABELIAN 1011hyperplane W of U, so Vx 2 S [U1, W9] SE. Thus X 2 E Y, and hence (4) holds.
As X2 E Y, X2T/02(X2T) 2:! 83 by 14.4.8.2, so (5) holds, completing the proof of
the lemma for the L 2 (4)-module.Thus as conclusion (5) of 14.4.2 holds, we may assume instead that U 1 is the
A5-module, and it remains to derive a contradiction.
As U1 is the A5-module, X2 centralizes V2; so that X 2 rf. Y. Hence we conclude
from (1) and 14.4.8.1 that L1 = X1 and V = Vx 1 • Then X 2 centralizes (V2x^1 ) = V.
ThusX2 S Ca(V) S MusingCoprimeAction, and then [L,X 2 ] s CL(V) = 02 (L),so that X2 acts on L2 and hence on (UL^2 ) = I 2. Let G 0 := (h, K 0 ), V 0 := (z^00 ),
and Gt:= Go/Ca 0 (Vo).
Suppose 02(Go) = 1. Then Hypothesis F.1.1 is satisfied with Ko, h Tin
the roles of "L1, L2, 8"; for example we just saw that B1 := NK 0 (T n Ko) =
X2(T n Ko) normalizes h Thus a := (KoT, T X2, I2X2) is a weak BN-pair by
F.1.9. Further B2 := N1 2 (K 0 ) = Tnh, so T :SJ TB2, and hence the hypotheses of
F.1.12 are satisfied. Therefore a is described in F.1.12. This is a contradiction as
U = 02(Ko) 2:! Q~ and Ko/U 2:! L2(4), a configuration not appearing in F.1.12.
Thus 02(Go) =f. 1, so Go E 'H(T), and Vo E R2(Go) by B.2.14. By 1.2.4,
Ko SJ E C(Go). Then 1 =f. [V2, Ko] S [Vo, J], so that J E Ct(G, T) by 1.2.10. Then
J E Cj(G, T) by 14.3.4, so that Ko= J by 13.1.2.5. Now I 2 = 02 (h) normalizes
Ko by 1.2.1.3, and hence acts on Z(02(Ko)) = V1, contradicting hi. G1. DLEMMA 14.4.11. Assume case (4) of 14.4.2 holds. Then
(1) We can represent H* 2:! 81 on n := {1, ... , 7} so that T preserves the
partition { {1, 2, 3, 4}, {5, 6}, {7}} of n..
(2) Y = {X1, X 2 }, where X1 := 02 (H1,2,3,4) and X2 = 02 (H5,6,7 ). In particu-
lar, X1X2T/02(X1X2T) 2:! 83 x 83.PROOF. Part (1) is trivial; cf. the convention in section B.3. Further X =
{X 1 , X2, X 3 }, where X1 and X2 are defined in (2), X3 := 02 (P) for P the stabilizer
of the partition { {1, 2}, {3, 4}, {5, 6}, {7} }, and X1X2T /02(X1X2T) 2:! 83 x 83.
Next tJ = U1 EB U 2 , where U1 is a 4-dimensional irreducible for K, and U2 = Uf
for x* E W9* - K* is dual to U1. Now Co.(NT(U1)) = (ui) for suitable ui, so
s = u 1 u 2 , with CH· (s) = P* = X3T* from the structure of the sum of U1 and
its dual. In particular, X 3 rf. Y. Recall W9* :SJ CH· (\/ 2 ) = P* and m(W9*) = 3
by 14.4.2, so W^9 * = 02(P*) = (xi : 1 S i s 3), where xi := (2i - 1, 2i) on n.
Let [U, xi] =: Di. Then Di S W and bi is of rank 4, so bi is the L2(4)-modulefor Yi :=CK· (xi) 2:! 85 since elements of order 3 in Yi are fixed-point-free on U.
As such elements lie in Xi for i = 1, 2, Vxi is of rank 3. Further X2 s Y3 with
Vx 2 S [D3, xix2] s [D3, W^9 ] S E by 12.8.11.1, so X2 E Y. Similarly a Sylow .3-
group B of Xi is contained in Yi* with Vx 1 = [\/2, B] = [\/2, x2x3] S [b1, W^9 ] S E,
so X 1 E Y, completing the proof of (2). D
LEMMA 14.4.12. Assume case (6) of 14.4.2 holds. Then Y = {X1, X2} where
X1 := 02 (02,3(H)) and X1X2T/02(X1X2T) ~ 83 x 83.
PROOF. First (cf. H.12.1.5) Cu(T) = V2 and CH.(V2) ~ 85/E32. We haveseen that W9 :SJ CH.(V2), and by 12.8.1, m(W9) = 5, so W^9 * = 02 (CH•(V 2 )).
Next we calculate that X = {X1,X2,X3}, where X1 := 02 (0 2 ,3(H)), X 3 :=