14.7. FINISHING L 3 (2) WITH (Val) ABELIAN 1067(2) QQH = Rl. Thus Q* = Ri = 02(Mi) is the unipotent radical of M{.
(3) If H* ~ G2(2) then M1 is transitive on the three conjugates in Ri ofA := A(Vi, Vi) in 14. 7.55.4.
PROOF. Recall M1 = NH (V), so Mi is the minimal parabolic NH CV) = LiT
of H*, and hence (1) holds. Next R 1 = QQH by 14.7.4, so (2) follows from (1).
Finally. (3) follows from (1) and B.4.6.13. D
Recall from Notation 14.7.1 that h EH with /o = 12h, a:= 1h, and Ua ::; Rl.
Set Za :=A~, and let Uo denote the preimage in U of Co(H). Let D :=Un Qa.
LEMMA 14.7.58. Assume Za::; V. Then
(1) There exists g E G interchanging /l and a.
(2) Vi::; Ua and m(U:X) = m(U/D).
PROOF. Part (1) follows as Lis 2-transitive on V#. Then (1) implies (2). DSet Ha := CH(Za) and U_ := U(Vi, Vi)Uo. As H = G1 by Theorem 14.7.52,
Ha acts on Ua and hence on u:;,, so that:
LEMMA 14.7.59. 02(H:;,) -=/= 1.LEMMA 14.7.60. Assume Za::; U. Then
(1) Replacing a by a suitable conjugate under M 1 , we may assume ZaUo =
V2Uo.
(2) H:;, = CH·CV2).
(3) Either
(a) u:;, =A := A(V1, Vi), D::; U""", and H ~ G2(2)), or
(b) u:;, = B := B*(V1, Vi), and either D::; u or fJ[J =Vi-.
PROOF. As Za ::; U, Ha is a subgroup of index at most 2 in CH(Za), so that
02 (CH·(Za)) i= 1 by 14.7.59. Therefore as 02(H*) = 1, Za i. Uo. It follows
that there is a E H with ZaUo = V 2 aUo. Indeed by 14.5.21.2, [QH, Za] = Vi, so
H~ = CH·(Za) is the parabolic subgroup of H* centralizing Za. Thus u:;, :::! H~
with <P(U~) = 1, so it follows from the structure of the parabolic H~ that U:X is
one of the two subgroups B(V 1 , V 2 a) or A(Vi, V 2 a) described in 14.7.55. In either
case, 14.7.55.4 says that Co(U:X) = U(V1, v;nuo =: U<:. But u:;,::; Ri = CH·CV)
using 14,7.57, so the doubly singular line Vis contained in U'!:. Therefore v 2 a ::; V
by 14.7.56.1 and the fact that the generalized hexagon Q(U) contains no cycle of
length 3. Then as M 1 = NH(V) is transitive on fr#, conjugating in M1, we may
take v 2 a = V 2 , and maintain the constraint Ua ::; R 1. Hence (1) holds. We saw
[QH, Za] = V1, so (2) holds. Further Ha acts on Un Qa = D, so from the action
of H~ on U, - DU -- is U, V-_l - - -
2 , or U. As [D, Ua] :::; Za by F.9.13.6, the third case
is impossible as H* induces no transvections on the module UH in 14.7.54. In the
second u:;, = B*(Vi, V 2 ) by 14.7.55.4. Thus (3) is established. D
LEMMA 14.7.61. (1) H ~ G2(2), and replacing a by a suitable M1 -conjugate,
we may assume u:;, =A:= A*(V1, Vi).
(2) [U, Ua] = unua = Co(A*) = U_.
(3) D = U_,