14.7. FINISHING Ls(2) WITH (vG1) ABELIAN 1069Pick Fo E :F. As Vz and Fo are distinct hyperplanes of Fo V 1 ~ Es, Vz n Fo = Vf
for a suitable l E L2T interchanging Vi and Vf. Set (3 := "(ol. Then Z13 ::::; F 0 ::::;
[U, Ua] ::::; Ua, and as V1 i:. Ua:
(b) For each FE :F, Z13 = Vz n Ua = Vz n Fis a complement to Vi in Vz.
Set U13 := U13/Z13 and consider the generalized hexagon Q(U13). Since V 2 = Z13
is a singular point of U, conjugating by l it follows that V 1 is a singular point of
U13.
Next Fis the set of lines in Q(U) through V 2 = Z13, while by 14.7.56.2, £ 2 fixes
V(Vz) = { FVi : F E :F} pointwise. Therefore conjugating by l, we conclude:
(c) {FV1 : FE :F} is the set of lines through V1 in Q(U13).
Now by 14.7.55.8,
F = {[w, Ua] : 1 -1-w E W}.
Therefore as [Ua, w]::::; Ua and F = Ua n FV 1 for each FE :F,
(d) {[Ua,w]: w E W} =:F= {[U,b]: b EB}.Applying symmetry to (a), and using (d) to conclude that :Fis invariant when
interchanging 'Yl and a, it follows that
(a') V(Za, Vz,a) = {FZa: FE :F},
and then from (a') and ( c) that:
(c') {FZa: FE :F} is the set of lines through Zain Q(U13).But now choosing F 1 and F 2 to be distinct members of :F, it follows from ( c)
. and ( c') that Za, F1, V1, F2, Za is a 4-cycle in the collinearity graph of g (U 13 ),
contrary to 14.7.55.7.This contradiction completes the proof of Theorem 14. 7.53.
14.7.5. Identifying Ru when 02 (H*) = A 5. We summarize the major re-
ductions achieved so far in this section:THEOREM 14.7.63. H = Ca(z) is the unique member of 1-lz, H =KT where
K := 02 (H) E C(H), H/02(H) ~ Ss, U is an indecomposable K-module, and
U /Cu(K) is the L2(4)-module for K/02(K).
PROOF. By Theorem 14.7.52.1, Ca(z) =His the unique member Hof Hz. By
14.7.48.1, H ~KT for some KE C(H); thus K = 02 (H). By Theorem 14.7.52.2and Theorem 14.7.53, K/02(K) is As. Then Theorem 14.7.40 says H* ~ (^85)
and U/Uo is the L2(4)-module. Thus U is indecomposable as UH = [UH,K] by
14.7.48.2. D
REMARK 14. 7.64. We will be working with the following special case of I.1.6.1:
Let Ube the largest F 2 H*-module such that U = [U, H*] and U /Co(H*) ~ N :=
U /Cu(K*). (cf. 17.12 in [Asc86a]) As N is the natural module for K* ~ £ 2 (4), Uhas the structure of an F4K-module, and as dimF 4 (H^1 (K,N)) = 1, dimF 4 (U) =
- Set U 0 := Co(K). There exists a 4-dimensional orthogonal space U 1 over F 4
with H ::::; ro(U1) such that Uo is a nonsingular point of U1 and fj = ut. This
facilitates later calculations in the image U of U.
Observe that by Theorem 14.7.63 and 14.3.3.6, M1 = HnM = NH(V) = L 1 T,
and Ri = 02(Li) E Syl2(K*). Let Uo be the preimage in U of Cu(K). As
V'= [V, £ 1 ] ~ E4 and U is a quotient of the module U in Remark 14.7.64: