ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 21 (2021) #P2.10 / 309–317
https://doi.org/10.26493/1855-3974.2049.3db
(Also available at http://amc-journal.eu)
Point-primitive generalised hexagons and
octagons and projective linear groups*
Stephen P. Glasby † , Emilio Pierro ‡ , Cheryl E. Praeger
Centre for the Mathematics of Symmetry and Computation, Department of Mathematics
and Statistics, The University of Western Australia
Received 16 July 2019, accepted 8 April 2021, published online 11 November 2021
Abstract
We discuss recent progress on the problem of classifying point-primitive generalised
polygons. In the case of generalised hexagons and generalised octagons, this has reduced
the problem to primitive actions of almost simple groups of Lie type. To illustrate how
the natural geometry of these groups may be used in this study, we show that if S is a
finite thick generalised hexagon or octagon with G ⩽ Aut(S) acting point-primitively and
the socle of G isomorphic to PSLn(q) where n ⩾ 2, then the stabiliser of a point acts
irreducibly on the natural module. We describe a strategy to prove that such a generalised
hexagon or octagon S does not exist.
Keywords: Generalised hexagon, generalised octagon, generalised polygon, primitive permutation
group.
Math. Subj. Class. (2020): 51E12, 20B15, 05B25
1 Introduction
We show in this paper that the Aschbacher–Dynkin [2] classification of maximal subgroups
of classical groups is a potentially useful tool to investigate whether or not a finite thick
*This work was supported by the Australian Research Council. Discovery Project Grants DP140100416
and DP190100450. The research reported in the paper forms part of the Australian Research Council (ARC)
Discovery Project Grant DP140100416 of the first and third authors. The authors gratefully acknowledge sup-
port from the Australian Research Council Discovery Project Grants DP140100416 and DP190100450; and also
advice from an anonymous referee on exposition. This research was conducted at the University of Western Aus-
tralia whose campus we acknowledge is located on the traditional lands of the Whadjuk people of the Noongar
Nation. We pay our respects to Noongar Elders past, present and emerging.
†Corresponding author.
‡Author gratefully acknowledges the support of the ARC Discovery Project Grant DP140100416.
E-mail addresses: stephen.glasby@uwa.edu.au (Stephen P. Glasby), e.pierro@mail.bbk.ac.uk (Emilio
Pierro), cheryl.praeger@uwa.edu.au (Cheryl E. Praeger)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
310 Ars Math. Contemp. 21 (2021) #P2.10 / 309–317
generalised hexagon or octagon admits a large rank classical group as an automorphism
group with a point-primitive action.
The notion of a generalised polygon arose from the investigations of Tits [12] and is
connected with the groups of Lie type having twisted Lie rank 2. They belong to a wider
class of geometric objects known as buildings, which were also introduced by Tits, whose
motivation was to find natural geometric objects on which the finite groups of Lie type
act, in order to work towards a proof of the classification of finite simple groups. Indeed,
all families of simple groups of Lie type having twisted Lie rank 2 arise as automorphism
groups of generalised polygons.
An incidence geometry S = (P,L, I) of rank 2 consists of a point set P , a line set L
and an incidence relation I ⊆ P×L such that P and L are disjoint non-empty sets. We say
that S is finite if |P∪L| is finite. The dual of S is SD = (L,P, ID), where (p, ℓ) ∈ I if and
only if (ℓ, p) ∈ ID. We say that S is thick if each point is incident with at least three lines
and each line is incident with at least three points. A flag of S is a set {p, ℓ} with p ∈ P ,
ℓ ∈ L and (p, ℓ) ∈ I. The incidence graph of S is the bipartite graph whose vertices are
P ∪ L and whose edges are the flags of S. A generalised n-gon is, then, a thick incidence
geometry of rank 2 whose incidence graph is connected of diameter n and girth 2n such that
each vertex lies on at least three edges [13, Lemma 1.3.6]. It is not immediate, but if S is a
thick generalised n-gon, then there exist constants s, t ⩾ 2 such that each point is incident
with t + 1 lines and each line is incident with s + 1 points [13, Corollary 1.5.3]. We then
say that the order of S is (s, t). A collineation of S is a pair (α, β) ∈ Sym(P)× Sym(L)
that preserves the subset I ⊆ P ×L. The subset of all collineations of Sym(P)× Sym(L)
is a subgroup denoted Aut(S). A celebrated result of Feit and Higman [8] states that if
S is a finite thick generalised n-gon, then n ∈ {2, 3, 4, 6, 8}. We refer the reader to Van
Maldeghem’s book [13] both for further details about classical generalised polygons, and
for a full introduction to the theory of generalised polygons.
In this paper we shall only be concerned in the case that S is a finite thick generalised
hexagon or octagon. At present the only known examples of these are the split Cayley
hexagon H(q), the twisted triality hexagon T (q, q3), the Ree–Tits octagon O(22m+1) and
their duals. These correspond to the groups G2(q), 3D4(q) and 2F4(22m+1) and complete
descriptions of these can be found in [13].
The point graph of a generalised polygon S is the graph with points as vertices and
with two points adjacent if they are collinear. The classification of (not necessarily thick)
generalised polygons admitting an automorphism group which acts distance-transitively
on the point graph of S is due to Buekenhout and Van Maldeghem [6]. In addition, they
show that distance-transitivity implies that G acts primitively on P . If S is also thick, then
Buekenhout and Van Maldeghem show that the socle of G is a finite simple group of Lie
type having twisted Lie rank 2. The assumption of distance-transitivity for this graph is
strong, and in recent years there has been work by a number of authors to show that the
assumption of distance-transitivity can be relaxed.
Schneider and Van Maldeghem [11, Theorem 2.1] showed that if G ⩽ Aut(S) acts
flag-transitively, point-primitively and line-primitively, then G is an almost simple group
of Lie type. The following theorem, which significantly strengthened this result, provided
motivation for the present paper.
Theorem 1.1 ([3, Theorem 1.2]). Let S be a finite thick generalised hexagon or octagon.
If a subgroup G of Aut(S) acts point-primitively, then G is an almost simple group of Lie
type.
S. P. Glasby et al.: Point-primitive generalised hexagons and octagons and projective . . . 311
The proof of Theorem 1.1 relies on the classification of finite simple groups. In order to
rule out certain possibilities for soc(G), it is sufficient to consider the primitive actions of
the almost simple groups of Lie type, or equivalently, their maximal subgroups. For an ex-
ceptional Lie type group that has a faithful projective representation in defining characteris-
tic of degree at most 12, a complete classification of its maximal subgroups is summarised
in [4, Chapter 7]. Using this classification it was proved by Morgan and Popiel in [9] that
under the hypothesis of the above theorem, if in addition it is assumed that the socle of G is
isomorphic to one of the Suzuki–Ree groups, 2B2(22m+1)′, 2G2(32m+1)′ or 2F4(22m+1)′,
where m ⩾ 0, then up to point-line duality, S is the Ree–Tits octagon O(22m+1). For a
general classical group G, however, we appeal to the Aschbacher–Dynkin classification [2]
of its maximal subgroups. The maximal subgroups of G fall into eight families of “geo-
metric” subgroups, those which preserve a natural geometric structure, and a ninth class of
exceptions. These classes are denoted Ci for 1 ⩽ i ⩽ 9, and some authors denote C9 as
S . The class C1 consists of stabilisers of subspaces and includes the maximal parabolic
subgroups of G. Our main result is as follows.
Theorem 1.2. Let S be a finite thick generalised hexagon or octagon. If G ⩽ Aut(S) acts
point-primitively on S and the socle of G is isomorphic to PSLn(q) where n ⩾ 2, then
the stabiliser of a point of S is not the stabiliser in G of a subspace of the natural module
V = (Fq)n.
The subspace stabilisers considered in Theorem 1.2 are all maximal parabolic sub-
groups. Given this result, and in the light of the result of Morgan and Popiel [9] mentioned
above, it would in the first instance be good to handle all primitive coset actions of Lie type
groups on maximal parabolic subgroups.
Problem 1.3. Extend Theorem 1.2 to show that, if S is a finite thick generalised hexagon
or octagon and G ⩽ Aut(S) is an almost simple group of Lie type such that the stabiliser
Gx of a point x is a maximal parabolic subgroup, then (S,G) is one of the known classical
examples.
Problem 1.3 has been solved for the Suzuki–Ree groups in [9], and it has also been
solved by Popiel and the second author [10] for the groups G2(q)′. It would be especially
interesting to have a solution to Problem 1.3 for the groups of (twisted or untwisted) Lie
rank 2, and in particular for the family 3D4(q)′ which is the only untreated case where
the groups are known to act on a generalised hexagon or octagon. Moreover, it would be
even more interesting to have a characterisation of all point-primitive actions of groups
with socle G2(q)′ or 3D4(q)′ on a thick generalised hexagon or octagon (not just the coset
actions on maximal parabolic subgroups). This, however, seems to be a substantially harder
problem.
Maximal parabolic subgroups mentioned in Problem 1.3 are examples of large sub-
groups, a notion introduced by Alavi and Burness in [1], namely a subgroup H of a finite
group G is large if |H|3 > |G|. In [1] all large subgroups of all finite simple groups are
determined. In our view the next level of attack on the general classification problem would
be to handle actions on cosets of large subgroups.
Problem 1.4. Extend Theorem 1.2 to show that, if S is a finite thick generalised hexagon
or octagon and G ⩽ Aut(S) is an almost simple group of Lie type such that the stabiliser Gx
of a point is a large maximal subgroup, then (S,G) is one of the known classical examples.
312 Ars Math. Contemp. 21 (2021) #P2.10 / 309–317
Popiel and the second author [10] have almost solved Problem 1.4 for groups G with
socle G2(q)′. The only unresolved point-primitive action is on a generalised hexagon with
stabiliser satisfying Gx ∩ soc(G) ∼= G2(q1/2).
The result of Alavi and Burness [1, Theorem 4] for groups G ∼= PSLn(q), taking into
account Theorem 1.2 for parabolic actions and using properties of the parameters of a gen-
eralised n-gon, shows that a solution to Problem 1.4 for these groups involves consideration
of just four kinds of point actions. We follow Alavi and Burness in using type to denote a
rough approximation of the structure of a subgroup.
Proposition 1.5. Let S be a finite thick generalised hexagon or octagon of order (s, t).
Suppose that G ⩽ Aut(S) with G ∼= PSLn(q), and G acts point-primitively on S such that
the stabiliser Gx of a point x is a large subgroup. Then one of the following holds:
(a) Gx is a C2-subgroup of type GLn/k(q) ≀ Sk, where k = 2 or k = 3;
(b) Gx is a C3-subgroup of type GLn/k(qk), where k = 2 or k = 3;
(c) Gx is a C5-subgroup of type GLn(q0) with q = qk0 , and either k = 2 or k = 3, or;
(d) Gx ∈ C8 of type Spn(q) (n even), SUn(q0) (q = q20), SOn(q) (nq odd), or SO
ϵ
n(q)
(n even, ϵ = ±).
Proof of Proposition 1.5. In addition to the classes asserted in the statement of the propo-
sition, Alavi and Burness show that either Gx ∈ C1, which is excluded by Theorem 1.2, or
Gx is one of finitely many cases belonging to classes C6 or C9 [1, Proposition 4.7 and The-
orem 4(ii)]. Of these, the cases where G is a group appearing in the Atlas [7] are excluded
by [5, Theorems 1.1 and 1.2]. The remaining possibilities for (G,Gx) are:
G PSL5(3) PSL4(5) PSL4(7) PSL2(q) q ∈ {41, 49, 59, 61, 71}
Gx M11 24. A6 PSU4(2) A5
The number |P| of points is the polynomial f(s, t) = (s + 1)(s2t2 + st + 1) if S is
a generalised hexagon, and f(s, t) = (s + 1)(s3t3 + s2t2 + st + 1) if S is a generalised
octagon. Running through the possibilities for |P| = |G : Gx| from the table above, we find
that there are no solutions to the equation |P| = f(s, t) with s, t ⩾ 2. This completes the
proof.
Extending Theorem 1.2 to include the large subgroups in class C2 has also proven to
be unexpectedly challenging to the authors.
2 The proof of Theorem 1.2
To prove Theorem 1.2, we assume for a contradiction that S is a thick generalised hexagon
or octagon and G ⩽ Aut(S), with soc(G) = PSLn(q), is such that a point stabiliser is
maximal in G and is the stabiliser of a k-subspace of the natural module V = (Fq)n, where
0 < k < n. Hence we may identify the point set P of S with the set of k-subspaces of
V , which we denote by
(
V
k
)
. If G contains a graph automorphism then k = n/2 and, for
its index 2 subgroup H = G ∩ PΓLn(q), the stabiliser HU is maximal in H. Thus we
may assume that PSLn(q) P G ⩽ PΓLn(q). It is convenient in the proofs to work with a
S. P. Glasby et al.: Point-primitive generalised hexagons and octagons and projective . . . 313
group G such that SLn(q) P G ⩽ ΓLn(q) acting linearly on V , with the scalar matrices
acting trivially on
(
V
k
)
, so G = G/Z where Z is the subgroup of scalars. Since a graph
automorphism of G maps
(
V
k
)
to
(
V
n−k
)
, and hence maps S to an isomorphic generalised
polygon with point set identified with
(
V
n−k
)
, we may assume further that 1 ⩽ k ⩽ n/2,
and so the following hypotheses hold.
Hypothesis 2.1. Let S = (P,L) be a finite thick generalised hexagon or octagon of order
(s, t), such that P is identified with the set
(
V
k
)
of k-subspaces of V = (Fq)n, where
1 ⩽ k ⩽ n/2. Suppose that SLn(q) P G ⩽ ΓLn(q) and that G induces a group of
automorphisms of S acting naturally on P , (so that a point stabiliser belongs to class C1).
Our proof of Theorem 1.2 uses the following three lemmas. The first is from [3].
Lemma 2.2 ([3, Lemma 2.1(iv)]). Let S be a finite thick generalised hexagon or octagon
of order (s, t), and let P denote the set of points of S. Let x, y1, y2 ∈ P such that x ∼ y1
and x ∼ y2, and let g ∈ Aut(S) such that xg ̸= x. If g fixes y1 and y2, then x, y1, y2, xg
all lie on a common line.
The second lemma is not difficult to prove, and its proof is left to the reader.
Lemma 2.3. Suppose SLn(q) P G ⩽ ΓLn(q), V = (Fq)n and k ⩽ n/2. Then, if
dim(V ) = n and k ⩽ n/2, then the orbits of G on
(
V
k
)
×
(
V
k
)
are
Γi =
{
(x, y) ∈
(
V
k
)
×
(
V
k
)
| dim(x ∩ y) = i
}
where 0 ⩽ i ⩽ k. (2.1)
Moreover, for x ∈
(
V
k
)
the orbits of Gx, are
Γi(x) =
{
y ∈
(
V
k
)
| dim(x ∩ y) = i
}
where 0 ⩽ i ⩽ k.
The third lemma allows us to characterise adjacency in S.
Lemma 2.4. Assume Hypothesis 2.1 and let x, y ∈ P . Then the following properties hold.
(F1) For every i ∈ {0, . . . , k}, if x, y are collinear and dim(x ∩ y) = i, then any x′, y′ ∈
P with dim(x′ ∩ y′) = i are also collinear.
(F2) For every i ∈ {0, . . . , k − 1}, if x, y are collinear and dim(x ∩ y) = i, then there
exists y′ ∈ P such that dim(x ∩ y′) = i and y′ ̸∼ y.
Proof. Property (F1) follows from Lemma 2.3. For (F2), suppose towards a contradiction
that every point y′ with dim(x ∩ y′) = i is collinear with y. By (F1), every such point y′
is also collinear with x, and hence lies on the line ℓ through x and y (because otherwise
{x, y, y′} would form a triangle and S contains no triangles). Let J = J(n, k)i denote
the generalised Johnson graph with vertex set V (J) =
(
V
k
)
and two vertices adjacent if
and only if they intersect in an i-subspace. Since G acts primitively on
(
V
k
)
, and since the
connected components are G-invariant, J is a connected graph. Note that Property (F1)
implies that adjacency in J implies collinearity, but the converse is not necessarily true. By
definition of J , y, y′ ∈ J1(x), the set of vertices adjacent to x in J . By the above argument,
{x} ∪ J1(x) is contained in the line ℓ. Since G acts transitively on J and since adjacency
314 Ars Math. Contemp. 21 (2021) #P2.10 / 309–317
is preserved by this action, it is true for all u ∈ P that {u} ∪ J1(u) is contained in a line of
S. Since S has more than one line, the diameter of J is at least 2.
We now prove by induction on the distance d, where 2 ⩽ d ⩽ diam(J), that, for any
vertices u, v of J , if the distance d = δ(u, v) and (u0, u1, . . . , ud) is a path of length d
in J from u = u0 to v = ud, then {u0, . . . , ud} is contained in the line ℓ containing
{u} ∪ J1(u). First we prove this for d = 2. Suppose that δ(u, v) = 2 and let (u,w, v)
be a path of length 2 in J from u to v. Note that w ∈ J1(u) ⊆ ℓ. Also u, v both lie in
{w} ∪ J1(w) which, as we have shown, is contained in some line ℓ′ of S. Then u,w are
contained in both ℓ and ℓ′, and since two points lie in at most one line of S it follows that
ℓ′ = ℓ, and so u,w, v all lie in ℓ and the inductive assertion is proved for d = 2. Now
suppose inductively that 3 ⩽ d ⩽ diam(J) and that the assertion is true for all integers
from 2 to d − 1. Suppose that δ(u, v) = d and that (u0, u1, . . . , ud) is a path in J from
u = u0 to v = ud. Then δ(u, ud−1) = d − 1, so by induction {u0, . . . , ud−1} ⊆ ℓ. Also
ud−2, v ∈ {ud−1} ∪ J(ud−1), which we have shown to be contained in some line ℓ′; since
ud−2, ud−1 are contained in both ℓ and ℓ′, it follows that ℓ′ = ℓ, and the inductive assertion
is proved for d. Hence by induction the assertion holds for all d ⩽ diam(J). However, this
is a contradiction because the points of S do not all lie on a single line.
We are now in a position to prove Theorem 1.2.
Proof of Theorem 1.2. As discussed at the beginning of this section we may assume that
Hypothesis 2.1 holds. Thus P =
(
V
k
)
and k ⩽ n/2.
CLAIM 1: k ⩾ 4. Consider the action of G on P × P . For each i with 0 ⩽ i ⩽
k − 1, G acts transitively on the set Γi defined in (2.1) by Lemma 2.3. It is a standard
result in the theory of permutation groups that the orbits of G on P × P are in one-to-one
correspondence with the orbits of Gx on P , and there must be at least one Gx-orbit for
each possible distance from x in the point graph of S. If k < 3, then the number of orbits
of Gx is less than four, so no point of P \{x} is at distance 3 from x in S , contradicting the
assumption that S is either a generalised hexagon or a generalised octagon. If k = 3, then
for the same reason S is not a generalised octagon, and so S is a generalised hexagon and
G acts distance transitively on the point graph. By the main result of Buekenhout and Van
Maldeghem in [6], S is a classical generalised hexagon and its distance transitive group has
socle G2(rf ) for some prime power rf , which is a contradiction. Hence k ⩾ 4 as claimed.
Now let {e1, . . . , en} be a basis of V and take x = ⟨e1, . . . , ek⟩. Let k1 < k be
maximal such that there exists a point y ∼ x with (x, y) ∈ Γk1 (as defined in (2.1)). Note
that, by Claim 1, n ⩾ 2k ⩾ 8.
CLAIM 2: k1 < k − 1. For a contradiction, assume that k1 = k − 1 and without loss
of generality that y = ⟨e1, . . . , ek−1, ek+1⟩. By (F2) there exists a point y′ ∈ P such that
(x, y′) ∈ Γk−1 and y ≁ y′ and by (F1) we have x ∼ y′ and so dim(x ∩ y′) = k − 1
and dim(y ∩ y′) ⩽ k − 2. Now dim(x ∩ y) = dim(x ∩ y′) = k − 1 implies that
dim(x ∩ y ∩ y′) ⩾ k − 2, and hence dim(y ∩ y′) = dim(x ∩ y ∩ y′) = k − 2. We may
assume without loss of generality that y′ = ⟨e2, . . . , ek, ek+2⟩. But now the permutation
matrix corresponding to (1, k+1)(k, k+2) leaves y and y′ fixed, but not x. By Lemma 2.2,
this implies that y ∼ y′, a contradiction. Hence k1 < k − 1, as required.
CLAIM 3: k1 = 0. Assume to the contrary that k1 > 0. Recall that k1 < k is maximal
such that there exists a point y ∼ x with y ∈ Γk1 . Thus we may assume that
y = ⟨e1, . . . , ek1 , ek+1, . . . , e2k−k1⟩.
S. P. Glasby et al.: Point-primitive generalised hexagons and octagons and projective . . . 315
If 2k − k1 + 1 ⩽ n, let
z = ⟨e1, . . . , ek1 , ek+2, . . . , e2k−k1+1⟩
so that dim(x∩z) = k1 and dim(y∩z) = k−1 > k1. It then follows from (F1) that x ∼ z
and from Claim 2 and the maximality of k1 that y ≁ z. Since 1 ⩽ k1 ⩽ k − 2, we have
k + 2 ⩽ 2k − k1 and hence the permutation matrix corresponding to (1, k + 2)(k, k − 1)
fixes y and z but not x. But once again Lemma 2.2 implies that y ∼ z, a contradiction.
Therefore k1 = 0 as claimed.
An immediate corollary of Claim 3 and (F1) is that G acts flag-transitively on S.
CLAIM 4: n = 2k or 2k + 1. For a contradiction, suppose that 2k + 1 < n and recall
k ⩽ n/2. Let y = ⟨ek+1, . . . , e2k⟩ and z = ⟨ek+2, . . . , e2k+1⟩. Observe that x ∼ y, x ∼ z
by (F1); furthermore dim(y ∩ z) = k − 2 > 0, so y ≁ z by the maximality of k1. Since
k ⩾ 4 by Claim 1, the permutation matrix corresponding to (1, 2k + 2)(2, 3) fixes y and z
but not x, contradicting Lemma 2.2.
CLAIM 5: n = 2k. Assume n = 2k + 1. Let y be as in Claim 4 and let z =
⟨ek+1, . . . , e2k−1, e1+ e2k+1⟩. Then x ∼ y and x ∼ z by Claim 3 and since dim(y∩ z) =
k − 1 > 0, we see that y ≁ z by Claim 3. Once again we apply Lemma 2.2 by noting that
since k ⩾ 4, the permutation matrix for (1, 2k + 1)(k + 1, k + 2) leaves y and z fixed but
not x, contradicting Lemma 2.2. Hence n = 2k and Claim 5 is true.
To complete the proof let
x = ⟨e1, . . . , ek⟩, y = ⟨ek+1, . . . , e2k⟩,
z = ⟨e1 + ek+1, . . . , ei + ei+k, . . . , ek + e2k⟩.
(2.2)
Then dim(x ∩ y) = dim(x ∩ z) = dim(y ∩ z) = 0, and so x, y and z are pairwise
collinear by Claim 3. Then, since S does not contain any triangles, x, y and z lie on a
line of S, say ℓ. Consider the stabiliser Gℓ. Note that ℓ is the unique line containing any
pair of the elements x, y or z and so in particular, Gℓ ⩾ ⟨Gxy, Gxz, Gyz⟩. Writing vectors
in V as n-dimensional row vectors over Fq relative to the basis e1, . . . , en, and writing
matrices relative to this basis, we see that x consists of all vectors of the form (X, 0),
where X, 0 denote k-dimensional row vectors, and the stabiliser Gx consists of all matrices
M =
(
A B
C D
)
∈ G for which (I | 0)M has the form (X | 0) where X ∈ GLk(Fq). Our aim
is to show that ⟨Gxy, Gxz, Gyz⟩ contains SLn(q). Let H = SLn(q) and let Hx = H ∩Gx
and define Hy , Hz , Hxy , Hxz , Hyz and Hℓ analogously. Let Mk(q) denote the ring of all
k × k matrices over Fq , and recall that k = n/2. Then
Hx = ⟨
(
A 0
C D
)
∈ SLn(q) | A,D ∈ GLk(q), C ∈ Mk(q)⟩.
Similarly,
Hy = ⟨
(
A B
0 D
)
∈ SLn(q) | A,D ∈ GLk(q), B ∈ Mk(q)⟩
and
Hz = ⟨
(
A B
C D
)
∈ SLn(q) | A+ C = B +D⟩.
From this we see that
Hxy = ⟨
(
A 0
0 D
)
∈ GLn(q) | A,D ∈ GLk(q),det(AD) = 1⟩.
Similarly,
Hxz = ⟨
(
A 0
D−A D
)
∈ GLn(q) | A,D ∈ GLk(q),det(AD) = 1⟩
316 Ars Math. Contemp. 21 (2021) #P2.10 / 309–317
and
Hyz = ⟨
(
A A−D
0 D
)
∈ GLn(q) | A,D ∈ GLk(q),det(AD) = 1⟩.
Our aim is now to show that Hℓ := ⟨Hxy, Hxz, Hyz⟩ is equal to H . We interrupt our proof
of Theorem 1.2 to prove this in the following lemma.
Lemma 2.5. The group generated by all matrices of the form
(
A 0
0 D
)
,
(
A A−D
0 D
)
and(
A 0
D−A D
)
where A,D ∈ GLk(q) and det(AD) = 1 equals H := SL2k(q).
Proof. Let L = ⟨
(
A 0
0 D
)
,
(
A A−D
0 D
)
,
(
A 0
D−A D
)
| A,D ∈ GLk(q),det(AD) = 1⟩ and
let x, y, z ∈
(
V
k
)
be as in (2.2). Then, L contains the matrix
(
A 0
D−A D
)(
A−1 0
0 D−1
)
=(
I 0
DA−1−I I
)
. In particular, choosing A = I and D = I+E1,2 we have DA−1 = I+E1,2
so that L contains the matrix M =
(
I 0
E1,2 I
)
. An element h = hA,D =
(
A 0
0 D
)
conjugates
M =
(
I 0
E1,2 I
)
to
( I 0
D−1E1,2A I
)
. Since L contains
(
A 0
0 A
)
for each permutation matrix
A, it follows that L contains
(
I 0
Ei,j I
)
for each i, j. Hence L contains Hx. Similarly, L
contains Hy . Since Hx is maximal in H and Hx ̸= Hy , we conclude that L = H .
Resuming our proof: Lemma 2.5 implies that Gℓ contains SLn(q) and since G is primi-
tive on points it follows that SLn(q) and hence also Gℓ, is transitive on points. This implies
that ℓ is incident with all points, which is a contradiction. This completes the proof of
Theorem 1.2.
ORCID iDs
Stephen P. Glasby https://orcid.org/0000-0002-0326-1455
Emilio Pierro https://orcid.org/0000-0003-1300-7984
Cheryl E. Praeger https://orcid.org/0000-0002-0881-7336
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