ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 23 (2023) #P2.04
https://doi.org/10.26493/1855-3974.2507.a1d
(Also available at http://amc-journal.eu)
A parametrisation for symmetric designs
admitting a flag-transitive, point-primitive
automorphism group with a product action*
Eugenia O’Reilly-Regueiro †
Instituto de Matemáticas, Universidad Nacional Autónoma de México (UNAM), Área de
la Investigación Cientı́fica, Circuito Exterior, Ciudad Universitaria, Coyoacán, 04510,
Mexico City, Mexico
José Emanuel Rodrı́guez-Fitta
Facultad de Estudios Superiores Acatlán, Universidad Nacional Autónoma de México
(UNAM), Av. Alcanfores y San Juan Totoltepec s/n, Santa Cruz Acatlán, Naucalpan,
53150, Edo. de México, Mexico
Received 17 December 2020, accepted 2 May 2022, published online 11 November 2022
Abstract
We study (v, k, λ)-symmetric designs having a flag-transitive, point-primitive automor-
phism group, with v = m2 and (k, λ) = t > 1, and prove that if D is such a design with
m even admitting a flag-transitive, point-primitive automorphism group G, then either:
(1) D is a design with parameters
(
(2t+ s− 1)2, 2t
2−(2−s)t
s ,
t2−t
s2
)
with s ≥ 1 odd, or
(2) G does not have a non-trivial product action.
We observe that the parameters in (1), when s = 1, correspond to Menon designs.
We also prove that if D is a (v, k, λ)-symmetric design with a flag-transitive, point-
primitive automorphism group of product action type with v = ml and l ≥ 2 then the
complement of D does not admit a flag-transitive automorphism group.
Keywords: Symmetric-designs, flag-transitivity, primitive groups, automorphism groups of designs.
Math. Subj. Class. (2020): 05B05, 51E05, 20B15, 20B25
*The authors would like to express their gratitude to the referee who made very helpful comments and sugges-
tions that improved our paper.
†Corresponding author.
E-mail addresses: eugenia@im.unam.mx (Eugenia O’Reilly-Regueiro), 887191@pcpuma.acatlan.unam.mx
(José Emanuel Rodrı́guez-Fitta)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 23 (2023) #P2.04
1 Introduction
If D = (P,B) is a (v, k, λ)-symmetric design, a flag of D is an ordered pair (p,B) such
that p ∈ P is a point of D, B ∈ B is a block of D, and p ∈ B. The order of D is n = k−λ.
There are some symmetric designs in which the parameters are related in some special
way, such as Hadamard designs in which v = 4n + 3, k = 2n + 1, and λ = n (n ∈ Z+),
and Menon designs, in which v = 4t2, k = 2t2 − t, and λ = t2 − t for some positive
integer t. These last ones will be relevant in the present work.
If G = Aut(D), then G is point-transitive if it is transitive on P (the set of points of
D), and it is flag-transitive if it is transitive on the set of flags of D. If G is point-transitive,
it can either be point-primitive, that is, there is no G-invariant non-trivial partition of P , or
point-imprimitive, which is when there is a non-trivial partition of the points of D invariant
under the action of G.
Primitive groups are classified by the O’Nan-Scott Theorem, we will use the classifi-
cation in [4] by Liebeck, Praeger, and Saxl, with five types, namely affine, almost simple,
product, simple diagonal, and twisted wreath.
Buekenhout, Delandtsheer, and Doyen proved in [1] that if a 2-design with λ = 1
(linear space) admits a point-primitive, flag-transitive automorphism group G, then it must
be of affine or almost simple type. O’Reilly-Regueiro proved the same result for symmetric
2-designs with 2 ≤ λ ≤ 4 in [5, 6]. All designs in this paper will be 2-designs.
In [7], Tian and Zhou extended this result to λ ≤ 100, and conjectured that it holds
for all values of λ. Having an upper bound on λ, in [7] they ruled out the simple diagonal,
product, and twisted wreath action by finding possible groups and/or sets of parameters
of designs and then ruling them out by arithmetic constraints and the use of GAP [2].
Additionally, in [3, 8, 9, 10], Zhou et al. have tackled this issue from different perspectives,
and have ruled out the product action for flag-transitive (v, k, λ) symmetric designs in
which λ ≥ (k, λ)2, as well as, for those cases in which λ is prime.
We have tried to prove that if D is a (v, k, λ)-symmetric design with v = m2 even and
any λ admitting a point-primitve, flag-transitive automorphism group G, then G does not
have a product action. In this paper we present our results, namely, a parametrisation for
such designs which in some cases correspond to Menon designs.
In 1998, Zieschang proved in [11] that if a (not necessarily symmetric) 2-design in
which (r, λ) = 1 (where r is the number of blocks incident with any given point) admits a
flag-transitive group G, then G is of affine or almost simple type. Given this result, in our
work we will assume (k, λ) = t > 1.
2 Product action
We start with a result from [5], which will be useful later.
Corollary 2.1. If G is a flag-transitive automorphism group of a (v, k, λ)-symmetric design
D = (P,B), then k divides λ(v − 1, |Gx|) for every point-stabiliser Gx.
The next lemma gives us an arithmetic condition that will be used throughout this work.
Suppose that the group G has a product action on the set of points P . Then there is a
finite set Γ with |Γ| ≥ 5 and a group H acting primitively on Γ, with an almost simple or
simple diagonal action, such that
P = Γl and G ≤ H l ⋊ Sl = HwrSl, with l ≥ 2.
E. O’Reilly-Regueiro et al.: A parametrisation for symmetric designs . . . 3
Lemma 2.2. If G is a point-primitive group acting flag-transitively on a (v, k, λ)-symmetric
design D = (P,B), with a product action on P , then k divides λl(|Γ| − 1) and v = |Γ|l ≤
λl2(|Γ| − 1)2.
Proof. Take x ∈ P = Γl. If x = (γ1, ..., γl), define for 1 ≥ j ≥ l the Cartesian line of the
jth parallel class through x to be the set:
Gx,j = {(γ1, ..., γj−1, γ, γj+1, ..., γl)|γ ∈ Γ},
(So there are l Cartesian lines through x).
Denote |Γ| = m.
Since G is primitive, Gx is transitive on the l Cartesian lines through x. Denote by ∆
the union of those lines (excluding x). Then ∆ is a union of orbits of Gx, and so every
block through x intersects it in the same number of points. Hence k divides λl(|Γ| − 1).
Also, k2 > λ(ml − 1), so (ml − 1) < λl2(m− 1)2.
Hence
v = ml ≤ λl2(m− 1)2. (2.1)
3 Results
In this section we will only consider l = 2, further work may be done for greater values
of l.
When l = 2, m = r
2+2r+4λ
4λ−r2 so
(m+ 1)r2 + 2r − 4λ(m− 1) = 0 (3.1)
solving for r we have
r =
−2±
√
4 + 16λ(m− 1)(m+ 1)
2(m+ 1)
=
−1±
√
1 + 4λ(m2 − 1)
m+ 1
therefore
r =
2(k − 1)
m+ 1
. (3.2)
Suppose that (k, λ) = t > 1 (the case where (k, λ) = 1 was done by Paul-Hermann
Zieschang [11]), so there exist positive integers a and b such that
k = at, λ = bt. (3.3)
Then, by Lemma 2.2 we have
k =
2λ(m− 1)
r
, (3.4)
and substituting (3.3) in the last one and also in k(k − 1) = λ(v − 1) we obtain
a =
2b(m− 1)
r
, (3.5)
a(at− 1) = b(m2 − 1). (3.6)
4 Ars Math. Contemp. 23 (2023) #P2.04
From (3.5) we can see that a divides b(m − 1). But (k, λ) = t so t = (at, bt) implies
(a, b) = 1. Therefore a divides m − 1, that is, there exists a positive integer s such that
m− 1 = as and substituting in (3.5), we obtain r = 2bs. Then since (a, b) = 1, this forces
s = (m− 1, r2 ).
We have te following results with respect to the new parameters a and s.
Lemma 3.1. Let D be a (v, k, λ)-symmetric design with v = m2 admitting a flag-transitive,
point-primitive automorphism group with product action. If k = at and λ = bt with
t = (k, λ), then a ̸= 1.
Proof. If a = 1 then k = t and λ = kb with b ≥ 1. This is a contradiction because k > λ,
therefore a ̸= 1.
Lemma 3.2. Let D be a (v, k, λ)-symmetric design with v = m2 admitting a flag-transitive,
point-primitive automorphism group with product action. If k = at and λ = bt with
t = (k, λ), then (a, s) = 1 where s is a positive integer such that m− 1 = as and r = 2bs.
Proof. Note (3.2) can be rewritten as:
r + 1 = k − (m− 1)r
2
.
Using the expressions k = at, λ = bt, m − 1 = as and r = 2bs we obtain 1 =
a(t− bs2)− 2bs and here we can see that (a, s) = 1 .
The fact that the parameter s = 1 is a necessary and sufficient condition for Menon
designs is seen in the following result:
Lemma 3.3. Let D be a (v, k, λ)-symmetric design with v = m2 admitting a flag-transitive,
point-primitive automorphism group with product action. If t = (k, λ) and s ∈ Z+ is such
that m − 1 = as and r = 2bs, then s = 1 if and only if v = 4t2, k = 2t2 − t, and
λ = t2 − t.
Proof. Suppose first that s = 1, so m− 1 = a which implies k = (m− 1)t. We also have
r
2 = b, so λ =
r
2 t. Now from m =
r2+2r+4λ
4λ−r2 we obtain
m =
b+ t+ 1
t− b
(3.7)
then
a = m− 1 = 2b+ 1
t− b
. (3.8)
Now if t = b then λ = tb = b2 =
r2
4
, and substituting in (3.1) we obtain
r2(m+ 1) + 2r − r2(m− 1) = 0, so r(r + 1) = 0.
This forces r = 0 or r = −1, which is a contradiction and so t ̸= b. From (3.8) we
have t− b ≥ 1.
E. O’Reilly-Regueiro et al.: A parametrisation for symmetric designs . . . 5
Suppose that t− b > 1, and let x > 1 be an integer such that t = b+ x. We will prove
that x is an odd number. If x = 2y for some y ∈ Z, then t = b + 2y, and substituting in
(3.8) we obtain
a =
2b+ 1
2y
,
which is a contradiction since a ∈ Z, so x is odd. Therefore there exists a positive integer
y such that x = 2y + 1 > 1 and with this we obtain t = b + 2y + 1, substituting in (3.8)
results in
a =
2b+ 1
2y + 1
.
Using the last expression for a together with a > b we obtain 2b+1 > b(2y+1) which
results in
1 > b(2y − 1). (3.9)
But we assumed t− b > 1 so x = 2y+1 > 1, that is, 2y− 1 > −1. This together with
the expression (3.9) implies that the equation 2y − 1 = 0 should hold. But that implies
y = 12 , which is a contradiction since we assumed y ∈ Z.
From the above we can conclude that b = t − 1 and this implies λ = t(t − 1). Then
substituting this expression in (3.8) we have
a = 2(t− 1) + 1 = 2t− 1,
so k = t(2t− 1) and m = a+ 1 = 2t, therefore v = m2 = 4t2.
Now, suppose that we have a symmetric design with parameters v = 4t2, k = 2t2 − t
and λ = t2 − t. Then a = 2t − 1 and b = t − 1. In addition, we have m = 2t and all of
these combined imply m − 1 = 2t − 1 = a. But m − 1 = as, and so s = 1. Hence the
result.
Remark 3.4. When we fix m − 1 and we vary r2 we get many possible values for λ that
satisfy the equation m = r
2+2r+4λ
4λ−r2 , at this point we observe that if m− 1 is a power of an
odd prime, then the parameters satisfy the conditions of Menon designs.
Lemma 3.5. Let D be a (v, k, λ)-symmetric design with v = m2 admitting a flag-transitive,
point-primitive automorphism group with product action such that t = (k, λ) and s ∈ Z+
such that m − 1 = as and r = 2bs. If m − 1 = pd with p an odd prime and d ∈ N, then
v = 4t2, k = 2t2 − t, and λ = t2 − t.
Proof. From m− 1 = as = pd then we have the following possible cases:
1. s = pi and a = pd−i for some natural number i < d.
This case is not possible because this would imply that (a, s) = pj for some natural
number j, and this contradicts Lemma 3.2.
2. a = pi and s = pd−i for some natural number i < d.
This case is not possible because this would imply that (a, s) = pj for some natural
number j, contradicting Lemma 3.2.
3. s = pd and a = 1.
This is not possible because it contradicts Lemma 3.1.
6 Ars Math. Contemp. 23 (2023) #P2.04
4. a = pd and s = 1.
Recall that s = 1 (Lemma 3.3), so in this case v = 4t2, k = 2t2 − t and λ =
t2 − t (these are the conditions for Menon designs). With this we have proved the
lemma.
Remark 3.6. We cannot claim the previous result for any odd m− 1 because the parame-
ters (4900, 3267, 2178), (16900, 2752, 448) and (44100, 8019, 1458) are counterexamples
to that possible generalisation. However we have neither confirmed nor discarded the exis-
tence of designs with these parameters. These (and Menon designs) are the only admissible
parameters for v ≤ (210)2.
Recall the definition of the Cartesian lines from Lemma 2.2. In the case we are study-
ing, when l = 2, there are two Cartesian lines through any point in the design, so we have
two possibilities. Either:
(i) there exists a point x and a block that contains it such that it intersects only one
Cartesian line through x, or
(ii) for any point x in the design, every block that contains it intersects each one of the
Cartesian lines through x.
We now study these cases separately. Although there are similarities between both
proofs, due to their length and enough differences we present two theorems for clarity.
Theorem 3.7 (Case (i)). Let D be a (v, k, λ)-symmetric design with v = m2 admitting a
flag-transitive, point-primitive automorphism group with product action. If there exists a
flag (x,A) in the design such that A intersects only one Cartesian line through x then r+1
divides k.
Proof. Let (x,A) be the flag such that A intersects a Cartesian line through x := (a0, b0).
Suppose that A intersects the second Cartesian line through x.
First, let us prove that for any element of the block A, A intersects only the second
Cartesian line through that point. We have, two subcases: either a point y ∈ A is in the sec-
ond Cartesian line through x, or a point y ∈ A is not in the second Cartesian line through x.
First subcase: we can see that if we take a point y ∈ A so that it is also in the second
Cartesian line through x, then y = (a0, ν) for some ν ∈ Γ. In this way, the set of elements
in the second Cartesian line through y which are also in A is the same as the intersection
of A with the second Cartesian line through x. Also, since by Lemma 2.2 the size of the
intersection of A with the second Cartesian line through x is r + 1, the size of the set of
elements in the second Cartesian line through y which are also in A is r + 1, and since the
size of the intersection of A with the Cartesian lines through any point is r + 1, there are
no more elements of any of the two Cartesian lines through y in A. In particular, there are
no elements of the first Cartesian line through y in A and so the statement is proved for this
subcase.
Second subcase: Now we are going to take y ∈ A such that it is not in the second Cartesian
line through x, in particular y ̸= x, so if y := (a1, b1) then a1 ̸= a0. Let us consider the flag
(y,A). Since the group G is flag-transitive, there is a g ∈ G such that g(x,A) = (y,A),
that is, g(x) = y, so
g(a0, b0) = (a1, b1), (3.10)
E. O’Reilly-Regueiro et al.: A parametrisation for symmetric designs . . . 7
this implies g|Γ(a0) = a1. So, for any µ ∈ Γ such that (a0, µ) ∈ A we have g(a0, µ) =
(a1, ν) for some ν ∈ Γ. Thus the element g ∈ G sends every element of the second
Cartesian line through x which is also in A to an element of the second Cartesian line
through y which is also in A. In this way A intersects only the second Cartesian line
through y. This is true for any y which is not in the second Cartesian line through x and
by Lemma 2.2 the size of this intersection is r + 1 and with this the statement the second
subcase is proved.
Let A0 be the set of points in the second Cartesian line through x which are also in A,
including x, the size of this set is r + 1. Now let us take an element x1 ∈ A \ A0. By
previous arguments A intersects only the second Cartesian line through x1, therefore, if A1
is the set of points in the second Cartesian line through x1 that are in A including x1 then
the size of this set is also r + 1. In the same way as before, we take x2 ∈ A \ (A0 ∪ A1)
and define the set A2 as the set of points in the second Cartesian line through x2 that are in
A including x2 and again its size is r + 1.
The process is continued in this way until no more points can be taken in A, thus we
get a set of points x0, x1, ..., xi ∈ A along with a collection of sets A0, A1, ..., Ai for some
natural number i, such that Aj is the intersection of the second Cartesian line through xj
with A. So, the size of Aj is r + 1 for all j. Also, A =
⋃j=i
j=0 Aj and by construction if
xg ̸= xh then Ag ̸= Ah with 1 ≤ g, h ≤ i.
It remains to prove that each pair of sets in this collection is disjoint, that is, if Ae ̸= Af
are two sets in the collection that was previously constructed, we must prove that Ae∩Af =
∅ with 1 ≤ e, f ≤ i, e ̸= f . Suppose that there exists an element p ∈ Ae ∩ Af , with
xe := (ae, be) and xf := (af , bf ). Then p = (ae, µ) = (af , ν) for some µ, ν ∈ Γ. We can
see that ae = af , which implies that xe is in the second Cartesian line through xf . This is
a contradiction since Ae ̸= Af .
From all of the above we can conclude that we obtain a partition of the block A. We
know that the size of A is k, and on the other hand A =
⋃j=i
j=0 Aj .
They are all disjoint and the size of each Aj is r+1, so k = i(r+1), and r+1 divides
k.
Let (x,A) be a flag such as in Theorem 3.7, that is, A intersects only the second Carte-
sian line through x. We count the number of flags (y, C) such that x ∈ C and y ̸= x is in
the second Cartesian line through x.
The number of these flags is the same as the number of blocks that contain x as well as
elements of the second Cartesian line through x, (we denote this number by z), multiplied
by the number of elements of the second Cartesian line through x (excluding x) which are
in these blocks, that is, r, therefore the number of such flags (y, C) is zr.
On the other hand, x and y are together in λ blocks and there are m − 1 points of the
second Cartesian line through x, so when we count these flags (y, C) we obtain λ(m− 1).
The above implies the equation zr = λ(m − 1), but the equation kr = 2λ(m − 1)
also holds, hence z = k2 and since z ∈ N, k is even. This means that half of the blocks
that contain x intersect with the second Cartesian line through x and the other half intersect
with the first Cartesian line through x. This is possible since the previous argument is also
valid for the first Cartesian line through x.
In the following theorem we examine Case (ii).
Theorem 3.8 (Case (ii)). Let D be a (v, k, λ)-symmetric design with v = m2 admitting a
flag-transitive, point-primitive automorphism group with product action. If for every point
8 Ars Math. Contemp. 23 (2023) #P2.04
x in the design, every block that contains it intersects with the two Cartesian lines through
x, then r2 + 1 divides k.
Proof. Let x = (a0, b0) be an arbitrary point in the design, and let A be a block containing
x, then there are r1 elements of the first Cartesian line through x (excluding x) in A and
there are r2 elements of the second Cartesian line through x (excluding x) in A. The
numbers r1 and r2 satisfy the equation r = r1 + r2, by Lemma 2.2.
If C is another block containing x, then it must intersect the two Cartesian lines through
x. Since G acts transitively on the flags, there is an element g ∈ G such that g(x,A) =
(x,C) and from this we can see that g(x) = x, that is, g|Γ fixes a0 and b0.
First, we will prove that g sends the elements of the first Cartesian line through x
which are also in A to elements of the first Cartesian line through x which are also in
C. Let (µ, b0) be an element of the first Cartesian line through x which is also in A.
Then g(µ, b0) = (ν, b0) ∈ C for some ν ∈ Γ since g|Γ fixes b0. Similarly g sends the
elements of the second Cartesian line through x which are also in A to elements of the
second Cartesian line through x which are also in C. Let(a0, µ) be an element of the first
Cartesian line through x which is also in A, then g(a0, µ) = (a0, ν) ∈ C for some ν ∈ Γ
since g|Γ fixes a0. Therefore the block C has as many elements of the first Cartesian line
through x as A, and as many elements of the second Cartesian line through x as A. The
above is true for every block that contains x.
Now let us count the number of flags (y, C) of the design such that y ̸= x is an element
of the first Cartesian line through x and C is a block containing x. Every block contains r1
elements of the first Cartesian line through x, when we exclude x, and there are k blocks
containing x. All of them intersect the first Cartesian line through x, therefore there are kr1
flags of this type. On the other hand y and x are together in λ blocks and there are m − 1
elements of the first Cartesian line through x (excluding x), hence there are λ(m − 1)
flags of this type. This yields the equation kr1 = λ(m − 1), but from Lemma 2.2 the
equation kr = 2λ(m− 1) also holds and we conclude that r1 = r2 . However r1 + r2 = r,
so the intersection of every block containing x with the second Cartesian line through x
(excluding x) has r2 = r2 elements.
The above is true for every x, that is, for every point in the design, every block that
contains it intersects the first Cartesian line through that point in r2 other points and the
same holds for the second Cartesian line through that point (excluding the point itself).
In what follows we will consider A0 to be the set of points of the second Cartesian
line through x which are also in A including x itself. The number of elements in that set
is r2 + 1. Let us consider x1 ∈ A \ A0 so from the previous paragraphs A intersects the
second Cartesian line through x1 in r2 elements, thus if A1 is the set of points of the second
Cartesian line through x1 which are also in A including x1 itself, the number of elements
in A1 is r2 + 1. Now we take an element x2 ∈ A \ (A0 ∪ A1) in the same way as before,
and let A2 be the set of points of the second Cartesian line through x2 which are also in A
including x2 itself. The number of elements in A2 is r2 + 1.
We can continue this process in this way until there are no more elements in A, (ev-
erything is finite), so we obtain a collection of points x0, x1, ..., xi ∈ A and a collection
of sets A0, A1, ..., Ai for some natural number i such that for all j = 0, . . . , i Aj is the
intersection of the second Cartesian line through xj with A, and Aj has r2 +1 elements. By
construction, A =
⋃j=i
j=0 Aj and the construction implies that if xg ̸= xh then Ag ̸= Ah
with 1 ≤ g, h ≤ i.
E. O’Reilly-Regueiro et al.: A parametrisation for symmetric designs . . . 9
It remains to prove that every two sets in this collection are disjoint, that is, we must
prove that if Ae ̸= Af then Ae∩Af = ∅ (with 1 ≤ e, f ≤ i and e ̸= f ). Suppose there is an
element p ∈ Ae ∩Af , with xe := (ae, be) and xf := (af , bf ). Then p = (ae, µ) = (af , ν)
for some µ, ν ∈ Γ. We can see that ae = af , which implies that xe is in the second
Cartesian line through xf , a contradiction since Ae ̸= Af .
Therefore we have a partition of the block A =
⋃j=i
j=0 Aj . The size of
⋃j=i
j=0 Aj is
i( r2 +1) since they are all disjoint, and the size of A is k, therefore k = i(
r
2 +1) and
r
2 +1
divides k.
Now we will present some consequences of Theorem 3.8.
Corollary 3.9. With the same hypotheses of Theorem 3.8, r2 + 1 divides m.
Proof. From (3.2) we have
k =
r
2
m+
r
2
+ 1,
and there is an integer p such that k = p( r2 + 1). Substituting in the previous equation we
obtain (p− 1)
(
r
2 + 1
)
= r2m. Since
(
r
2 + 1,
r
2
)
= 1, r2 + 1 necessarily divides m.
Corollary 3.10. With the same hypotheses of Theorem 3.8, r2 + 1 divides λ.
Proof. There is an integer p such that k = p( r2 + 1), and substituting this and (3.2) in
k(k − 1) = λ(m− 1)(m+ 1), we obtain
p
r
2
(r
2
+ 1
)
= λ(m− 1).
By Corollary 3.9, r2 + 1 divides m, so
(
r
2 + 1,m− 1
)
= 1 and r2 + 1 divides λ.
Since t is the greatest common divisor of k and λ, the following holds:
Corollary 3.11. With the same hypotheses of Theorem 3.8, r2 + 1 divides t.
Proof. Since r2 + 1 divides k and λ, and also (k, λ) = t we conclude
r
2 + 1 divides t.
In the next results, we will introduce a particular case in which we have obtained the
parameters of a Menon design, as a consequence of Corollary 3.11. Since r2 + 1 divides
t > 1, we will first consider the case in which t is a prime number. The following result is
a first approach to our main result.
Lemma 3.12. Let D be a (v, k, λ)-symmetric design with (k, λ) = t > 1 a prime number
and v = m2, admitting a flag-transitive, point-primitive automorphism group G. If for
every point x in the design, every block that contains it intersects the two Cartesian lines
through x, then either G does not have a product action or D is a Menon design.
Proof. From Corollary 3.11, we have r2 + 1 divides t. Since t is a prime we obtain
r
2 = 0
or r2 + 1 = t.
If r2 = 0 then from (3.2) we have k − 1 = 0 and this is impossible.
If on the other hand r2 + 1 = t, then m =
r2+2r+4λ
4λ−r2 so
m =
bs2 + s+ t
t− bs2
(3.11)
10 Ars Math. Contemp. 23 (2023) #P2.04
which implies t ≥ bs2. If t = bs2 then λ = b2s2 = r
2
4 . Substituting this in (3.1) we
obtain
r2(m+ 1) + 2r − r2(m− 1) = 0, therefore r(r + 1) = 0,
so r = 0 or r = −1 which is a contradiction, so t > bs2.
This forces t = r2 + 1 = bs+ 1 > bs
2, so 1 > bs(s− 1) and s = 1. From Lemma 3.3,
v = 4t2, k = 2t2 − t and λ = t2 − t, which are the parameters of a Menon design.
Remark 3.13. The triples of parameters (4900, 3267, 2178), (16900, 2752, 448), and
(44100, 8019, 1458), do not correspond to Menon designs but they satisfy all known nec-
essary arithmetic conditions on the existence of a symmetric design with v even, so we do
not prove the conjeture for v ≤ (210)2 (we have not tried computational methods).
Lemma 3.14. Let D be a (v, k, λ)-symmetric design with (k, λ) = t > 1 and v = m2 ≤
(210)2 with m even admitting a flag-transitive point-primitive automorphism group G,
then either G does not have a non-trivial product action or one of the following conditions
holds:
1. D is a Menon design with parameters (4t2, 2t2 − t, t2 − t), where t > 1, or
2. D has parameters (16900, 2752, 448).
Proof. For m ≤ 210 the admissible parameters that do not satisfy the conditions of Menon
designs and in which k − λ is a square are (4900, 3267, 2178), (16900, 2752, 448), and
(44100, 8019, 1458). For these, r2 + 1 is 47, 22 and 39 respectively, so they do not satisfy
Theorem 3.8. Now for those parameters r + 1 is 93, 43 and 78 respectively. The first and
third of them do not satisfy Theorem 3.7, but the parameters (16900, 2752, 448) do. Thus,
these are the only possible parameters for m2 ≤ (210)2.
In this case, k is even, which is consistent with Theorem 3.7. We also have s = 3 so
from Lemma 3.3 we know that these parameters cannot correspond to a Menon design.
Remark 3.15. The triple (16900, 2752, 448) does not correspond to a Menon design since
s = 3, although it satisfies all the arithmetic conditions for a symmetric design. We make
no claim as to whether such a design exists, but perhaps it is not the case that when l = 2
only Menon designs are possible (if at all).
The following is our main result, the proof follows Cases (i) and (ii) from
Theorems 3.7 and 3.8, that is, either: there is a flag (x,A), such that the block A inter-
sects only one Cartesian line through x (Case (i)), or for every point x, every block that
contains it intersects both of the Cartesian lines through x (Case (ii)). The proof based on
Case (i) is similar to the proof of the case in which m − 1 is the power of an odd prime.
In this sense it is a generalisation of this proof, but because of the existence of the param-
eter s this generalisation was not obtained in an obvious way. For this reason, we need an
additional arithmetic condition, which is found in Corollary 3.11.
In the proof based on Case (ii) we also obtain an arithmetic condition for a and so
also for k. We believe we do not necessarily obtain parameters for Menon designs for an
arbitrary λ when we study symmetric designs admitting a flag-transitive point-primitive
E. O’Reilly-Regueiro et al.: A parametrisation for symmetric designs . . . 11
automorphism group with product action when l = 2. This case also gives us a parametri-
sation of (v, k, λ) in terms of t and s, and if s = 1 then the parameters correspond to Menon
designs, that is, our parameterisation is a generalisation of the parameterisation of Menon
designs.
Theorem 3.16. Let D a (v, k, λ)-symmetric design admitting a flag-transitive, point-
primitive, automorphism group G with (k, λ) = t > 1 and v = m2 with m even. Then
either:
(i) G does not have a non-trivial product action, or
(ii) D is a design with parameters
(
(2t+ s− 1)2, 2t
2−(2−s)t
s ,
t2 − t
s2
)
with s ≥ 1 odd.
When s = 1 D is a Menon design and if s > 1 then t is even.
Proof. From the hypotheses and from Lemma 2.2 there are two possible cases. For any
given point, either each block that contains it intersects the two Cartesian lines through it,
or there is a point such that a block containing it only intersects one Cartesian line through
it. As we have seen, the latter implies that every block intersects only one Cartesian line
through each point it contains.
First we will study this last case. Here, Theorem 3.7 is satisfied, so r+1 divides k with
r > 1 an integer such that kr = 2λ(m − 1). This implies there is an integer p such that
k = p(r + 1). Also k − 1 = r2 (m+ 1) holds, so k =
r
2m+
r
2 + 1, and therefore
m− 1 = (r + 1)(m+ 1− 2p).
Then r + 1 divides m − 1, but m − 1 = as = x(r + 1) where x := m + 1 − 2p and
since r = 2bs we have (r+1, s) = 1 so r+1 divides a. Also a divides r+1 which forces
r+1 = a, and this implies k = (r+1)t. This all implies t− bs2 = 1, so b = t−1s2 , and we
obtain the parameters λ = t−1s2 t, k =
2t+s−2
s t, v = (2t+ s− 1)
2.
The proof of Theorem 3.7 states that k should be an even number and since r + 1 is an
odd number then t should be an even number. Since m = 2t + s − 1 and m is an even
number then s is an odd number.
The triple (16900, 2752, 448) satisfies the conditions we obtained, with t = 64 and
s = 3, so this is not a Menon design.
When s = 1 we obtain the parametrisation for Menon designs with t an even number.
Now suppose that for every point, every block that contains it intersects both Cartesian
lines through it. Here the hypotheses of Theorem 3.8 hold and from Corollary 3.11, there
exists x ≥ 1 such that
t =
(r
2
+ 1
)
x = (bs+ 1)x. (3.12)
From (3.11) we obtain m = s(bs+1)+tt−bs2 =
s(bs+1)+(bs+1)x
t−bs2 , that is
m =
(
s+ x
t− bs2
)(r
2
+ 1
)
. (3.13)
Using (3.12) we obtain
t− bs2 = x+ bs(x− s), (3.14)
which we divide into the following cases:
12 Ars Math. Contemp. 23 (2023) #P2.04
1. x < s
From Lemma 3.12, we have t−bs2 > 0, and from (3.14) x > bs(s−x) > bx(s−x).
Since x < s then 1 > b(s−x) > 0, but this cannot be the case since b(s−x) should
be an integer.
2. s < x
From (3.13), s+x ≥ t−bs2. If s+x = t−bs2 we have from (3.13) that m = r
2
+1
so as = m − 1 = r
2
= bs and therefore a = b, but (a, b) = 1 so a = 1 this is
impossible by Lemma 3.1. Therefore s + x > t − bs2, and from (3.14) we have
s+x > x+ bs(x− s) so 1 > b(x− s) > 0 and this is also imposisble since b(x− s)
is an integer.
3. s = x
From (3.12), s divides t, and from (3.2) at − bs(as + 2) = 1, so (t, s) = 1 and
therefore s = 1. Also from Lemma 3.3, v = 4t2, k = 2t2 − t, and λ = t2 − t
This concludes the proof.
The parameters of Menon designs are not the only ones we can obtain when we assume
that the automorphism group of the design has a product action on the points of the design,
and the parameters (16900, 2752, 448) are an example of this. However we note that a
design with the possible parameters which arise and do not correspond to Menon designs
must satisfy that each block only intersects one Cartesian line through each point in that
block.
It is not the case that the way in which we consider product action to obtain possi-
ble Menon designs does not work because here is a potential counterexample, but rather
that with this theorem we give explicit expressions for the parameters v, k, λ, in terms of
parameters s, t, and when s = 1 they do correspond to Menon designs.
4 One further result
Here we present an additional result for any l ≥ 2.
Theorem 4.1. Let D a (v, k, λ)-symmetric design with v = ml admitting a flag-transitive,
point-primitive, automorphism group G with a non-trivial product action. Then the com-
plement of the design is not flag-transitive.
Proof. Suppose that D′ is the complement of the design D, so its parameters are
(v′, k′, λ′) = (v, v − k, v − 2k + λ). If we also assume D′ is flag-transitive, then the
following equation holds:
(v − k)(v − k − 1) = (v − 2k + λ)(m− 1)(ml−1 +ml−2 + ...+ 1). (4.1)
If D has a point-primitive automorphisms group G, then D′ has the same point-primitive
automorphisms group G and we can consider the Cartesian lines through a point, since G
is transitive on the points of D′. Thus k′ divides λ′l(m − 1), so there is an integer p such
that
(v − k)p = l(v − 2k + λ)(m− 1). (4.2)
E. O’Reilly-Regueiro et al.: A parametrisation for symmetric designs . . . 13
Substituting this in (4.1) we obtain
l(v − k)(v − 1− k) = (v − k)p(ml−1 +ml−2 + ...+ 1)
so l((m− 1)(ml−1 +ml−2 + ...+ 1)− k) = p(ml−1 +ml−2 + ...+ 1),
hence
lk = q(ml−1 +ml−2 + ...+ 1) (4.3)
with q = l(m− 1)− p > 0.
But for D we know that k =
lλ(m− 1)
r
and k(k − 1) = λ(v − 1) so
k(k − 1) = kr
l(m− 1)
(ml − 1)
and we obtain a generalisation of (3.2):
l(k − 1) = r(ml−1 +ml−2 + ...+ 1). (4.4)
If we substitute (4.3) in (4.4) then
q(ml−1 +ml−2 + ...+ 1)− l = r(ml−1 +ml−2 + ...+ 1)
so (q − r)(ml−1 +ml−2 + ...+ 1) = l,
and therefore ml−1 +ml−2 + ... + 1 ≤ l if m > 1 and l ≥ ml−1 +ml−2 + ... + 1 > l,
which is impossible.
We conclude m ≤ 1, but this is a contradiction since m ≥ 5.
ORCID iDs
Eugenia O’Reilly-Regueiro https://orcid.org/0000-0001-5867-7258
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