6 JOHN BAMBERG AND NICOLA DURANTE
then be a triangle in
H(3,q2);
a contradiction. So all the four points R, S,
T , U of a quadrangle must be affine.
Recall that u is the polarity defining
H(3,q2).
Note that
Ru
∩T
u
is equal
to SU and that SU
H(3,q2)
is a Baer subline with a point on O. Indeed
R∗
spans a Baer subplane fully contained in
H(3,q2)
and it meets O in a
Baer subline and since
Ru
T
u

H(3,q2)
is a Baer subline contained in
R∗
then SU
H(3,q2)
has a point in O. Likewise
Su
U
u
equal to RT and
RT
H(3,q2)
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
H(3,q2)
on SU is collinear
with each point of
H(3,q2)
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
H(3,q2).
So since
H(3,q2)
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
H(3,q2).
Then we would have four gener-
ators of
H(3,q2)
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
(q6
1)/(q 1),
which is equal to the sum of the number of affine points and the number of
generators of
H(3,q2),
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
H(3,q2),
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
MA :=
0
A 0
0
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-
duces a
PGL2(q2)
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.
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