6 JOHN BAMBERG AND NICOLA DURANTE
then be a triangle in
a contradiction. So all the four points R, S,
T , U of a quadrangle must be aﬃne.
Recall that u is the polarity defining
to SU and that SU ∩
is a Baer subline with a point on O. Indeed
spans a Baer subplane fully contained in
and it meets O in a
Baer subline and since
is a Baer subline contained in
then SU ∩
has a point in O. Likewise
equal to RT and
is a Baer subline with a point in O. So SU and RT are polar
to each other under u, but then each point of
on SU is collinear
with each point of
on RT , while the points of O are pairwise non-
collinear, a contradiction. Hence Γ has no quadrangles.
Suppose we have a pentagon R, S, T , U, W of Γ. Now points of type
(b), which are aﬃne points, are collinear in Γ if they are incident with a
common element of Ω. Since each element of Ω spans a generator, points
of type (b) are also collinear in
is a generalised
quadrangle, we see immediately that each point of our pentagon is an aﬃne
point. Suppose, by way of contradiction, that our pentagon has a point of
type (a), that is, a generator of
Then we would have four gener-
forming a quadrangle and we obtain a similar “forbidden”
quadrangle of aﬃne points (i.e., RSTU) from the above argument. So there
are no pentagons in Γ.
A trivial counting argument shows that L has size
− 1)/(q − 1),
which is equal to the sum of the number of aﬃne points and the number of
and so it follows that Γ is a generalised hexagon (of
order (q, q)).
2.2. Exhibiting a suitable set of Baer subgenerators. In this sec-
tion, we describe a natural candidate for a set Ω of Baer subgenerators satis-
fying the hypotheses of Theorem 1.2. Consider the stabiliser GO in PGU4(q)
of the Hermitian curve O = π ∩
where π consists of the elements
whose last coordinate is zero. Then the elements of GO can be thought of
(projectively) as matrices MA of the form
0 0 0 1
, A ∈ GU3(q).
Lemma 2.1. The group GO acts transitively on the set of Baer subgen-
erators which have a point in O.
Proof. Inside the group PGU4(q), the stabiliser J of a generator in-
acting 3-transitively on the points of . So the stabiliser
in J of a point P of acts transitively on the Baer sublines within which
contain P . Now J meets GO in the stabiliser of a point of , and so GO, acts
transitively on Baer subgenerators contained in . Since GO acts transitively
on O, the result follows.