Symmetric Point Sets with Few Intersection Numbers in PG(r,q)

Abstract

In 1965, H. Retkin and E. Stein defined a symmetric point set as a set of points with the same intersection numbers. In this paper, we perform a detailed analysis of symmetric point sets of the finite projective space which shows that the class of symmetric sets is very broad including caps, two-character sets and transitive sets. We derive necessary conditions for the existence of such sets. Since the most studied sets are caps and two-character sets and not much seems to be known in the general case of sets, which are different from caps, with more than two intersection numbers, by using incidence-preserving group actions, symmetric point sets with few intersection numbers are provided. The results indicate that any finite projective space contains symmetric sets with few intersection numbers.

1. Introduction

Let 𝔽_q be the finite field of order q, with q = p^h, a prime power. In PG(r,q), the projective space of dimension r and order q, let K denote a k–set, i.e., a set of k points. A set K has class [m₁,m₂,…,m_s] if every hyperplane meets K in m_i points for some i = 1,…,s, cf. [1]. If each value m_i occurs as an intersection number, then we say K has type (m₁, m₂,…,m_s), and it is called an s-intersection set, cf. [2,3]. A cap is a set of points such that at most two of them are on a hyperplane. Caps and sets with few intersection numbers have been extensively studied due to their connections to coding theory, cf. [4,5,6,7,8]. A point P has type $\left(v_{m_1}\left(P\right),v_{m_2}\left(P\right),\dots ,v_{m_s}\left(P\right)\right)$ if P is contained in $v_{m_i}\left(P\right)$ hyperplanes meeting K in m_i points. Following [9] p. 393, a k-set K is said to be symmetric if for each value of m_i the value of $v_{m_i}\left(P\right)$ is independent of P, in which case we use the symbol $v_{m_i}$. The fundamental objective of the research developed in this paper is to initiate the investigation of symmetric sets and discuss whether such sets exist in finite projective spaces. We derive necessary conditions for the existence of such sets and construct a class of symmetric sets by using the cyclic structure of a finite projective space. The paper is organized as follows. In Section 2, we observe that sets of class [0,m,n] are a broad class of symmetric point sets including caps and two-character sets. As they have been extensively studied, in Section 3, after showing that transitive point sets represent another large class of symmetric point sets, we generalize the construction of Brower [10] to get transitive few-intersection sets. In Section 4, we explicitly construct transitive sets with at most three intersection numbers in the smallest two projective planes PG(2,q) with q≡1 mod 3. In Section 5, we derive transitive sets with at most four intersection numbers in the smallest projective space PG(3,q) with q≡1 mod 4.

2. Sets of Class [0,m,n] are Symmetric

In this section, we show that symmetric sets are far from being trivial. Such sets include caps, i.e., sets of type (0,1,2), and two-character sets and these have been subject to extensive study since the middle of the last century, c.f. [11]. Let K denote a k-set of class [0,m,n] in PG(r,q). We prove the following:

Lemma 1.

A set of class [0,m,n] is symmetric.

Proof. 

Double counting the number of hyperplanes through a point P∈K, the number of pairs (P,H), where P∈K and H is a hyperplane through P, we get the following equations on the integers v_i(P):

$$\left\{\begin{array}{c}{v}_m\left(P\right)+v_n\left(P\right)=q^{r-1}+\cdots +q+1\\ {\left(m-1\right)v}_m\left(P\right)+\left(n-1\right)v_n\left(P\right)=\left(k-1\right)\left(q^{r-2}+\cdots +q+1\right)\end{array}\right.$$

By Cramer’s rule, the system of linear equations has one solution.

Therefore, the integers v_i = v_i(P) do not depend on the point P and we have that

$$\left\{\begin{array}{c}{v}_m={\displaystyle \frac{n-k}{n-m}}\left(q^{r-2}+\cdots +q+1\right)+{\displaystyle \frac{n-1}{n-m}}{q}^{r-1}\\ v_n={\displaystyle \frac{k-m}{n-m}}\left(q^{r-2}+\cdots +q+1\right)-{\displaystyle \frac{m-1}{n-m}}{q}^{r-1}\end{array}\right.$$

and the set K is symmetric. □

The existence of a set of class [0,m,n] requires particular admissible parameters. Here, we provide a necessary condition for the existence of a set K of class [0,m,n] which involves the number of external hyperplanes through any point not belonging to the set. Let Q denote a point with Q∉K. For each integer i such that 0 ≤ i ≤ q^r−1+…+q + 1, let us denote by u_i(Q) the number of hyperplanes of PG(r,q) through Q meeting K in exactly i points.

Lemma 2.

A necessary condition for the existence of a k-set K of class [0,m,n] is that

$$\left\{\begin{array}{c}n-m|n{u}_0\left(Q\right)-q^{r-1}\\ n-m|m{u}_0\left(Q\right)-q^{r-1}\end{array}\right..$$

Proof. 

Double counting the number of hyperplanes through a point Q∉K, the number of pairs (Q,H), where Q∉K and H is a hyperplane through Q, we get the following equations on the integers u_i(Q):

$$\left\{\begin{array}{c}{u_0\left(Q\right)+u}_m\left(Q\right)+u_n\left(Q\right)=q^{r-1}+\cdots +q+1\\ {mu}_m\left(Q\right)+n{u}_n\left(Q\right)=k\left(q^{r-2}+\cdots +q+1\right)\end{array}\right.$$

Thus, we have that

$$\left\{\begin{array}{c}{u}_m\left(Q\right)={\displaystyle \frac{n-k}{n-m}}\left(q^{r-2}+\cdots +q+1\right)+{\displaystyle \frac{n}{n-m}}{q}^{r-1}-{\displaystyle \frac{n}{n-m}}{u}_0\left(Q\right)\\ u_n\left(Q\right)={\displaystyle \frac{k-m}{n-m}}\left(q^{r-2}+\cdots +q+1\right)-{\displaystyle \frac{m}{n-m}}{q}^{r-1}+{\displaystyle \frac{m}{n-m}}{u}_0\left(Q\right)\end{array}\right.$$

Therefore,

$$\left\{\begin{array}{c}{u}_m\left(Q\right)=v_m-{\displaystyle \frac{n{u}_0\left(Q\right)-q^{r-1}}{n-m}}\\ u_n\left(Q\right)=v_n+{\displaystyle \frac{m{u}_0\left(Q\right)-q^{r-1}}{n-m}}\end{array}\right..$$

Since u_i(Q) and v_i are integers, the claimed necessary condition is established. □

Unfortunately, neither any general formula nor general divisibility properties are known for the existence of sets of class [0,m,n]. Sets of class [0,m,n] have been studied in a variety of contexts with some special cases forming individual fields of study, cf. [7]. The most interesting cases appear to be two intersection sets, cf. [12,13]. As they have been extensively studied, in the next section, we show another large class of symmetric point sets.

3. Transitive Sets of PG(r,q) Are Symmetric

In this section, inspired by the work of [10], we continue to investigate symmetric sets by using the cyclic structure of a finite projective space. We derive a sufficient condition for the existence of such sets and determine enough information on intersection numbers. We report the cyclic construction here for the sake of completeness, cf. [14]. The projective space PG(r,q) admits a collineation $\sigma$ such that the group $〈\sigma 〉$ generated by $\sigma$ is transitive and hence regular on the set of all points. The cyclic group $〈\sigma 〉$ is, up to conjugacy, the only cyclic subgroup of PGL(r + 1,q) acting regularly on the set of points of PG(r,q), and it is called the Singer group of PG(r,q), cf. [14]. In order to show this, let us consider ${\mathbb{F}}_{q^{r+1}}$ as a r dimensional projective space PG(r,q) over 𝔽_q. The points of PG(r,q) are represented as elements of ${\mathbb{F}}_{q^{r+1}}^{\ast }/{\mathbb{F}}_q^{\ast }$, with two elements v₁ and v₂ of ${\mathbb{F}}_{q^{r+1}}^{\ast }$ representing the same point if and only if v₂ = λv₁ for some λ in ${\mathbb{F}}_q^{\ast }$. Let $\omega$ be a primitive element of ${\mathbb{F}}_{q^{r+1}}$. To construct the cyclic Singer group, we consider the following:

$${\mathbb{F}}_{q^{r+1}}=\left\{x_0+x_1\omega +\cdots +x_r{\omega }^r|x_i\in {\mathbb{F}}_q\right\},$$

$${\mathbb{F}}_q^{\ast }=\left\{{\left({\omega }^{q^r+q^{r-1}+\cdots +q+1}\right)}^i|1\le i\le q-1\right\},$$

By the function $f:{\mathbb{F}}_{q^{r+1}}^{\mathrm{*}}\to \mathrm{P}\mathrm{G}(r,q)$ defined as $f\left(x_0+x_1\omega +\cdots +x_r{\omega }^r\right)≔\left(x_0{,x}_1{,\dots ,x}_r\right)$, the points of PG(r,q) can be represented by the set $\Omega =\left\{{\omega }^i|0\le i\le q^r+q^{r-1}+\cdots +q\right\}$. Consider the action of the map $\sigma$ over $\Omega$, where $\sigma :\Omega \to \Omega$ $\sigma \left({\omega }^i\right)=\omega ·{\omega }^i={\omega }^{i+1\left(\mathrm{m}\mathrm{o}\mathrm{d}{q}^r+q^{r-1}+\cdots +q+1\right)}$. The map $\sigma$ is a collineation of PG(r,q), as $\sigma \left(x_0{,x}_1{,\dots ,x}_r\right)=f\left(\omega \left(x_0+x_1\omega +\cdots +x_r{\omega }^r\right)\right).$ If $p\left(x\right)=x^{r+1}-a_0-a_{1}x-\dots -a_r{x}^r$ is the minimum polynomial of $\omega$ over 𝔽_q, then $f\left(\omega \left(x_0+x_1\omega +\cdots +x_r{\omega }^r\right)\right)=f\left(x_0\omega +x_1{\omega }^2+\cdots +x_r{\omega }^{r+1}\right)=\left(a_0{x}_r,x_0+a_1{x}_2,\dots ,x_{r-1}+a_r{x}_r\right)$. Thus, $\sigma$ is defined by the companion matrix of $p\left(x\right),$ $C\left(p\right)=\left(\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & ⋮\\ ⋮& ⋮& & \ddots & \\ a_0& a_1& a_2& \cdots & a_r\end{array}\right)$, as

$$\sigma \left(x_0{,x}_1{,\dots ,x}_r\right)=\left(x_0{,x}_1{,\dots ,x}_r\right)\left(\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & ⋮\\ ⋮& ⋮& & \ddots & \\ a_0& a_1& a_2& \cdots & a_r\end{array}\right)=\left(a_0{x}_r,x_0+a_1{x}_r,\dots ,x_{r-1}+a_r{x}_r\right)$$

the group $〈\sigma 〉$ is the cyclic Singer group of PG(r,q). A transitive k-set of PG(r,q) is a k-set which consists of a point orbit under a subgroup of the cyclic Singer group of PG(r,q), cf. [13,14,15,16]. Then, k divides $q^r+q^{r-1}+\cdots +q+1$ and the group $〈{\sigma }^{{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}}〉$ has ${\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}$ orbits on points of PG(r,q), each of size k. Since the orbits are projectively equivalent, we will assume, without loss of generality, that $K=\left\{1,{\omega }^{{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}},\dots ,{\omega }^{{\displaystyle \frac{\left(k-1\right)\left(q^r+q^{r-1}+\cdots +q+1\right)}{k}}}\right\}.$ It is well-known that, under the action of the cyclic Singer group $〈\sigma 〉$ of PG(r,q), the point set and the hyperplane set have the same cyclic structure. Thus, the group $〈{\sigma }^{{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}}〉$ also has ${\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}$ orbits on hyperplanes of PG(r,q), each of size k. Therefore, the set $K=\left\{1,{\omega }^{{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}},\dots ,{\omega }^{{\displaystyle \frac{\left(k-1\right)\left(q^r+q^{r-1}+\cdots +q+1\right)}{k}}}\right\}$ is a set with at most ${\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}$ intersection numbers, m_i, $i\in \left\{1,\dots ,{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}\right\}$, with respect to the hyperplanes.

Since the points of PG(r,q) can be represented by the set $\Omega =\left\{{\omega }^i|0\le i\le q^r+q^{r-1}+\cdots +q\right\}$, any point ${\omega }^i$ can be identified with the exponent i, $0\le i\le q^r+q^{r-1}+\cdots +q$ and so with the elements of the additive cyclic group ${\mathbb{Z}}_{q^r+q^{r-1}+\cdots +q+1}$. By using this identification, the set $K=\left\{0,{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}},\dots ,{\displaystyle \frac{\left(k-1\right)\left(q^r+q^{r-1}+\cdots +q+1\right)}{k}}\right\}.$

The principal Lemma relevant to the present discussion is the following:

Lemma 3.

A transitive k-set is symmetric.

Proof. 

Let $\left\{i,j\right\}\subset \left\{0,{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}},\dots ,{\displaystyle \frac{\left(k-1\right)\left(q^r+q^{r-1}+\cdots +q+1\right)}{k}}\right\}$ be two points of a transitive k-set K. Therefore, there is an integer a, with $a\in \left\{1,2,\dots ,k-1\right\},$ such that $j=i+a{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}.$ Thus, if H is an m-hyperplane through i, then ${\sigma }^{a{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}}\left(H\right)$ is an m-hyperplane through j and the assertion follows. □

If ${\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{k}}$ is small, the set K is a k-set with few intersection numbers. Now, we give an explicit construction of a transitive k-set in the projective space PG(r,q) which has few intersection numbers. We prove

Lemma 4.

If q≡1 mod r + 1, then the projective space PG(r,q) contains a transitive (${\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}}$)-set with at most r + 1 intersection numbers.

Proof. 

If q ≡ 1 mod r + 1, then r + 1 is a divisor of $q^r+q^{r-1}+\dots +q+1$, the number of points of PG(r,q). The orbit of the point 0 of PG(r,q) under the unique subgroup ${\mathbb{Z}}_{r+1}$ of order r + 1 of the additive cyclic group ${\mathbb{Z}}_{q^r+q^{r-1}+\cdots +q+1}$ is the transitive ${\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}}$-set $\left\{0,r+1,2\left(r+1\right),\dots ,\left({\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}}-1\right)\left(r+1\right)\right\}$ with at most r + 1 intersection numbers. □

By using the above identification, the set K is the set $K_0=\left\{0,r+1,2\left(r+1\right),\dots ,\left({\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}}-1\right)\left(r+1\right)\right\}$ and any hyperplane H₀ is a (v,k,λ)-difference set, i.e., a k-set H₀ = {d₁,…,d_k} of residues modulo v such that for any residue α≠0 modulo v the congruence d_i − d_j≡α (mod v) has exactly λ solutions pairs (d_i,d_j) with d_i and d_j in H₀, where $v=q^r+q^{r-1}+\cdots +q+1$, $k=q^{r-1}+q^{r-2}+\cdots +q+1$ and $\lambda =q^{r-2}+q^{r-3}+\cdots +q+1$, cf. [17]. Let us arrange the hyperplane indexes in such a way that $H_i={\sigma }^i\left(H_0\right)$ and put $K_i={\sigma }^i\left(K_0\right)$. It follows that if m_i denotes the intersection number |K₀∩H_i|, we have that m_i is also |K_-i∩H₀|, i.e., the number of d_j in H₀ satisfying d_j≡-i mod r+1. Thus, we have that the family of intersection numbers {m_i}, $i\in \left\{0,{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}},\dots ,{\displaystyle \frac{r\left(q^r+q^{r-1}+\cdots +q+1\right)}{r+1}}\right\},$ is a multiset, cf. [18], in which the elements m_i are allowed to occur more than once, cf. [19]. The multiset is specified by its underlying set {m_i} and, for each element m∈{m_i}, its multiplicity μ(m), a nonnegative integer. Then, we have that $v_i=\mu \left(m_i\right)m_i$ for any $i\in \left\{0,{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}},\dots ,{\displaystyle \frac{r\left(q^r+q^{r-1}+\cdots +q+1\right)}{r+1}}\right\}.$

Therefore, we have proved that

Lemma 5.

The at most r + 1 intersection numbers, m_i, $i\in \left\{0,{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}},\dots ,{\displaystyle \frac{r\left(q^r+q^{r-1}+\cdots +q+1\right)}{r+1}}\right\},$ with respect to the hyperplanes of a transitive (${\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}}$)-set K are related to the at most r + 1 intersection numbers of any point P∈K by the equations  $v_i=\mu \left(m_i\right)m_i$.

Now, we prove that

Lemma 6.

The r + 1 intersection numbers, m_i, $i\in \left\{0,{\displaystyle \frac{q^r+q^{r-1}+\cdots +q+1}{r+1}},\dots ,{\displaystyle \frac{r\left(q^r+q^{r-1}+\cdots +q+1\right)}{r+1}}\right\},$ with respect to the hyperplanes, satisfy the system of equations

$$\left\{\begin{array}{c}{\displaystyle \sum _{i=0}^r}{m}_i=q^{r-1}+q^{r-2}+\cdots +q+1\\ {\displaystyle \sum _{i=0}^r}{m}_i^2={+q}^{r-1}+{\displaystyle \frac{\left(q^{r-2}+q^{r-3}+\cdots +q+1\right)\left(q^r+q^{r-1}+\cdots +q+1\right)}{r+1}}\\ {\displaystyle \sum _{i=0}^r}{m}_i{m}_{i-j\left(modr+1\right)}={\displaystyle \frac{\left(q^{r-2}+q^{r-3}+\cdots +q+1\right)\left(q^r+q^{r-1}+\cdots +q+1\right)}{r+1}}j=1,\dots ,r\end{array}\right.$$

Proof. 

Since any hyperplane H₀ is a (v,k,λ)-difference set H₀={d₁,…,d_k}, with $v=q^r+q^{r-1}+\dots +q+1$, $k=q^{r-1}+q^{r-2}+\dots +q+1$ and $\lambda =q^{r-2}+q^{r-3}+\dots +q+1$, then its Hall polynomial $\theta \left(x\right)=x^{d_1}+\dots +x^{d_k}$ satisfies the congruence $\theta \left(x\right)\theta \left(x^{-1}\right)\equiv k-\lambda +\lambda \left(1+x+\cdots +x^{v-1}\right)$ modulo $x^v-1$. Thus, for any divisor r + 1 of v, one gets

$$\theta \left(x\right)\equiv m_0+m_{1}x+\cdots +m_r{x}^r\mathrm{modulo}{x}^{r+1}-1$$

and

$$\theta \left(x\right)\theta \left(x^{-1}\right)\equiv k-\lambda +\lambda {\displaystyle \frac{v}{r+1}}\left(1+x+\cdots +x^r\right)\mathrm{modulo}{x}^{r+1}-1$$

where m_i is the number of d_j in H₀ satisfying d_j≡i mod r+1. By comparing the coefficients, the equations yield. □

4. Symmetric Sets with at Most Three Intersection Numbers

The simplest case we can consider is the plane PG(2,q) with q≡1 mod 3. Since 3 divides q²+q+1, the group $〈{\sigma }^3〉$ has three orbits $\left\{0,3,\dots ,q^2+q-2\right\}$, $\left\{1,4,\dots ,q^2+q-1\right\}$ and $\left\{2,5,\dots ,q^2+q\right\}$ on PG(2,q). Since the three orbits are projectively equivalent, we will assume without loss of generality that $K=\left\{0,3,\dots ,q^2+q-2\right\}$.

Since 4 is the smallest order with q≡1 mod 3, let us consider PG(2,4). The equations of Lemma 3.4 with r = 2 and q = 4 become

$$\left\{\begin{array}{c}x+y+z=5\\ x^2+y^2+z^2=11\\ xy+z\left(x+y\right)=7\end{array}\right.$$

Since z is an integer such that $z^2<11$, it follows that $z\in \left\{0,1,2,3\right\}$.

$$z=3\Rightarrow \left\{\begin{array}{c}x+y=2\\ x^2+y^2=2\\ xy+3\left(x+y\right)=7\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=2\\ x^2+y^2=2\\ xy=1\end{array}\right.\Rightarrow \left\{\begin{array}{c}x=1\\ y=1\\ z=3\end{array}\right..$$

$$z=2\Rightarrow \left\{\begin{array}{c}x+y=3\\ x^2+y^2=7\\ xy+2\left(x+y\right)=7\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=3\\ x^2+y^2=7\\ xy=1\end{array}\right.\Rightarrow nointegersolutions.$$

$$z=1\Rightarrow \left\{\begin{array}{c}x+y=4\\ x^2+y^2=10\\ xy+\left(x+y\right)=7\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=4\\ x^2+y^2=10\\ xy=3\end{array}\right.\Rightarrow \left\{\begin{array}{c}x=1\\ y=3\\ z=1\end{array}\right..$$

$$z=0\Rightarrow \left\{\begin{array}{c}x+y=5\\ x^2+y^2=11\\ xy=7\end{array}\right.\Rightarrow nointegersolutions.$$

Therefore, the multiset {1,1,3} is the unique solution. We explicitly construct the transitive two-intersection sets of type (1,3) of PG(2,4). The elements of the field 𝔽₄ are {0,1,α,α²} where α is a root of the irreducible polynomial t² + t + 1 in 𝔽₂[t]. In order to write the cyclic structure of ℙ²: = PG(2,4), let ω be a primitive element of 𝔽₄³ over 𝔽₄ and let $p\left(x\right)=x^3-a_0-a_{1}x-a_2{x}^2$ be its minimal polynomial over 𝔽₄. The companion matrix $C\left(p\right)$ of f is given by $C\left(p\right)=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ a_0& a_1& a_2\end{array}\right)$, and it induces a Singer cycle $\sigma$ of PG(2,4), cf. [14]. Let us consider a primitive polynomial of degree 3 over 𝔽₄, p(x) = x³ + α + αx + αx². The companion matrix C(p) is $\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 5& 4& 0\end{array}\right)$. Let us consider the point ${\omega }^0=\left(x_0,x_1,x_2\right)=\left(1,0,0\right)$. We get:

$${\omega }^1={\omega }^{0}C(p)=\left(1,0,0\right)\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ \alpha & \alpha & \alpha \end{array}\right)=\left(0,1,0\right),$$

$${\omega }^2={\omega }^{0}C(f{)}^2={\omega }^{1}C(f)=\left(0,1,0\right)\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ \alpha & \alpha & \alpha \end{array}\right)=\left(0,0,1\right),$$

$${\omega }^3={\omega }^{0}C(f{)}^3={\omega }^{2}C(f)=\left(0,0,1\right)\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ \alpha & \alpha & \alpha \end{array}\right)=\left(\alpha ,\alpha ,\alpha \right)=\alpha \left(1,1,1\right)=\left(1,1,1\right)$$

By continuing in this way and by denoting the points represented by ωⁱ simply by i, the points of PG(2,4) can be identified with the elements of the additive cyclic group ${\mathbb{Z}}_{21}$, the integers modulo 21 as shown in

Table 1. The cyclic structure of the points of PG(2,4).

ω0 = (1,0,0) = 0
ω1 = (0,1,0) = 1	ω2 = (0,0,1) = 2	ω3 = (1,1,1) = 3	ω4 = (1,α, α) = 4
ω5 = (1,α2,α) = 5	ω6 = (1,α2,0) = 6	ω7 = (0,1,α2) = 7	ω8 = (1,1,0) = 8
ω9 = (0,1,1) = 9	ω10 = (1,1,α) = 10	ω11 = (1,α2,α2) = 11	ω12 = (1,0,α) = 12
ω13 = (1,α2,1) = 13	ω14 = (1,α,α2) = 14	ω15 = (1,0,α2) = 15	ω16 = (1,0,1) = 16
ω17 = (1,α,1) = 17	ω18 = (1,α,0) = 18	ω19 = (0,1,α) = 19	ω20 = (1,1,α2) = 20

.

Since under the action of a cyclic collineation group of a finite projective plane, the point set and the line set have the same cyclic structure, select any line, for example, we choose the line 𝓁₀:= x₀ = 0, which contains the 5-set of points written in

Table 2. The starting line.

𝓁0

1

2

7

9

19

.

The remaining lines of the plane are found by adding 1 to each point of the preceding line beginning with 𝓁₀ and using addition modulo 21.

The lines of PG(2,4) consist of the points in the vertical columns of

Table 3. The lines of PG(2,4).

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	0
2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	0	1
7	8	9	10	11	12	13	14	15	16	17	18	19	20	0	1	2	3	4	5	6
9	10	11	12	13	14	15	16	17	18	19	20	0	1	2	3	4	5	6	7	8
19	20	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18

.

The cyclic structure of PG(2,4) is well represented in Figure 1, reprinted from [20].

Let us consider the orbit of the point 0 under the action of $〈{\sigma }^3〉$,

$$K=\left\{\mathbf{0},\mathbf{3},\mathbf{6},\mathbf{9},\mathbf{12},\mathbf{15},\mathbf{18}.\right\}$$

Since lines of an orbit have the same intersection number, in order to compute the intersection numbers of K, it is sufficient to consider only the three base lines written in

Table 4. The base lines.

𝓁0	1	2	7	9	19
𝓁1	2	3	8	10	20
𝓁2	0	3	4	9	11

:

Thus, the set K = $\left\{0,3,6,9,12,15,18\right\}$ is a 7-set with 14 1-lines and 7 3-lines, i.e., K is of type (1,3), a Baer subplane, as was noted by Brouwer in [10], see also [6]. By the projective equivalence of points and lines, K is a symmetric 7-set, i.e., any point P has v₁(P) = 2·1 = 2 and v₃(P) = 1·3 = 3.

The second order q with q≡1 mod 3 is 7; let us consider the field 𝔽₇. The equations of Lemma 3.2 with k − 1 = 2 and q = 7 become

$$\left\{\begin{array}{c}x+y+z=8\\ x^2+y^2+z^2=26\\ xy+xz+yz=19\end{array}\right.$$

Since z is an integer such that $z^2<26$, it follows that $z\in \left\{0,1,2,3,4,5\right\}$.

$$z=5\Rightarrow \left\{\begin{array}{c}x+y=3\\ x^2+y^2=1\\ xy+5\left(x+y\right)=19\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=3\\ x^2+y^2=1\\ xy=4\end{array}\right.\Rightarrow nointegersolutions.$$

$$z=4\Rightarrow \left\{\begin{array}{c}x+y=4\\ x^2+y^2=10\\ xy+4\left(x+y\right)=19\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=4\\ x^2+y^2=10\\ xy=3\end{array}\right.\Rightarrow \left\{\begin{array}{c}x=1\\ y=3\\ z=4\end{array}\right..$$

$$z=3\Rightarrow \left\{\begin{array}{c}x+y=5\\ x^2+y^2=17\\ xy+3\left(x+y\right)=19\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=5\\ x^2+y^2=17\\ xy=4\end{array}\right.\Rightarrow \left\{\begin{array}{c}x=1\\ y=4\\ z=3\end{array}\right..$$

$$z=2\Rightarrow \left\{\begin{array}{c}x+y=6\\ x^2+y^2=22\\ xy+2\left(x+y\right)=19\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=6\\ x^2+y^2=22\\ xy=7\end{array}\right.\Rightarrow nointegersolutions.$$

$$z=1\Rightarrow \left\{\begin{array}{c}x+y=7\\ x^2+y^2=25\\ xy+\left(x+y\right)=19\end{array}\right.\Rightarrow \left\{\begin{array}{c}x+y=7\\ x^2+y^2=25\\ xy=12\end{array}\right.\Rightarrow \left\{\begin{array}{c}x=3\\ y=4\\ z=1\end{array}\right..$$

$$z=0\Rightarrow \left\{\begin{array}{c}x+y=8\\ x^2+y^2=26\\ xy=19\end{array}\right.\Rightarrow nointegersolutions.$$

Therefore, (1,3,4) is the unique solution. We explicitly construct the transitive three-intersection sets of type (1,3,4) of PG(2,7). In 𝔽₇, 3 is a root of x²+x+2. In order to write the cyclic structure of ℙ² :=PG(2,7), let ω be a primitive element of 𝔽₇³ over 𝔽₇ and let $p\left(x\right)=x^3-a_0-a_{1}x-a_2{x}^2$ be its minimal polynomial over 𝔽₇. The companion matrix $C\left(p\right)$ of f is given by $C\left(p\right)=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ a_0& a_1& a_2\end{array}\right)$, and it induces a Singer cycle $\sigma$ of PG(2,7), cf. [14]. Let us consider a primitive polynomial with minimal weight, i.e., the minimal number of nonzero coefficients, among all primitives of that degree over 𝔽₇, p(x) = x³ + 2 + 3x, cf. [21]. The companion matrix C(p) is $\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 5& 4& 0\end{array}\right)$. Let us consider the point ${\omega }^0=\left(x_0,x_1,x_2\right)=\left(1,0,0\right)$.

We get:

$${\omega }^1={\omega }^{0}C(f)=\left(1,0,0\right)\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 5& 4& 0\end{array}\right)=\left(0,1,0\right),$$

$${\omega }^2={\omega }^{0}C(f{)}^2={\omega }^{1}C(f)=\left(0,1,0\right)\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 5& 4& 0\end{array}\right)=\left(0,0,1\right),$$

$${\omega }^3={\omega }^{0}C(f{)}^3={\omega }^{2}C(f)=\left(0,0,1\right)\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 5& 4& 0\end{array}\right)=\left(5,4,0\right)=5\left(1,5,0\right)=\left(1,5,0\right)$$

By continuing in this way and by denoting the points represented by ωⁱ simply by i, we obtain the cyclic structure of PG(2,7), as shown in

Table 5. The cyclic structure of the points of PG(2,7).

ω0 = (1,0,0) = 0
ω1 = (0,1,0) = 1	ω2 = (0,0,1) = 2	ω3 = (1,5,0) = 3	ω4 = (0,1,5) = 4
ω5 = (1,5,2) = 5	ω6 = (1,3,4) = 6	ω7 = (1,4,4) = 7	ω8 = (1,4,3) = 8
ω9 = (1,6,4) = 9	ω10 = (1,4,1) = 10	ω11 = (1,1,5) = 11	ω12 = (1,0,2) = 12
ω13 = (1,3,0) = 13	ω14 = (0,1,3) = 14	ω15 = (1,5,1) = 15	ω16 = (1,1,1) = 16
ω17 = (1,1,3) = 17	ω18 = (1,6,1) = 18	ω19 = (1,1,4) = 19	ω20 = (1,4,6) = 20
ω21 = (1,2,2) = 21	ω22 = (1,3,3) = 22	ω23 = (1,6,3) = 23	ω24 = (1,6,6) = 24
ω25 = (1,2,3) = 25	ω26 = (1,6,2) = 26	ω27 = (1,3,2) = 27	ω28 = (1,3,1) = 28
ω29 = (1,1,2) = 29	ω30 = (1,3,5) = 30	ω31 = (1,0,6) = 31	ω32 = (1,2,0) = 32
ω33 = (0,1,2) = 33	ω34 = (1,5,5) = 34	ω35 = (1,0,3) = 35	ω36 = (1,6,0) = 36
ω37 = (0,1,6) = 37	ω38 = (1,5,4) = 38	ω39 = (1,4,2) = 39	ω40 = (1,3,6) = 40
ω41 = (1,2,5) = 41	ω42 = (1,0,4) = 42	ω43 = (1,4,0) = 43	ω44 = (0,1,4) = 44
ω45 = (1,5,6) = 45	ω46 = (1,2,6) = 46	ω47 = (1,2,1) = 47	ω48 = (1,1,6) = 48
ω49 = (1,2,4) = 49	ω50 = (1,4,5) = 50	ω51 = (1,0,1) = 51	ω52 = (1,1,0) = 52
ω53 = (0,1,1) = 53	ω54 = (1,5,3) = 54	ω55 = (1,6,5) = 55	ω56 = (1,0,5) = 56

. Thus, the Singer group is isomorphic to the additive group Z₅₇, the integers modulo 57.

Since under the action of a cyclic collineation group of a finite projective plane the point set and the line set have the same cyclic structure, select any line. For example, we choose the line 𝓁₀: = x₀ = 0, which contains the 8-set of points written in

Table 6. The starting line.

𝓁0

1

2

4

14

33

37

44

53

.

The remaining lines of the plane are found by adding 1 to each point of the preceding line beginning with 𝓁₀ and using addition modulo 57. Let us consider the orbit of the point 0 under the action of $〈{\sigma }^3〉$,

$$K=\left\{\mathbf{0},\mathbf{3},\mathbf{6},\mathbf{9},\mathbf{12},\mathbf{15},\mathbf{18},\mathbf{21},\mathbf{24},\mathbf{27},\mathbf{30},\mathbf{33},\mathbf{36},\mathbf{39},\mathbf{42},\mathbf{45},\mathbf{48},\mathbf{51},\mathbf{54}\right\}.$$

Since lines of an orbit have the same intersection number, in order to compute the intersection numbers of K, it is sufficient to consider only the three base lines written in

Table 7. The base lines, with in bold the elements of K.

𝓁0	1	2	4	14	33	37	44	53
𝓁1	2	3	5	15	34	38	45	54
𝓁2	3	4	6	16	35	39	46	55

:

Thus, the set

$$K=\left\{\mathbf{0},\mathbf{3},\mathbf{6},\mathbf{9},\mathbf{12},\mathbf{15},\mathbf{18},\mathbf{21},\mathbf{24},\mathbf{27},\mathbf{30},\mathbf{33},\mathbf{36},\mathbf{39},\mathbf{42},\mathbf{45},\mathbf{48},\mathbf{51},\mathbf{54}\right\}$$

is a 19-set with 19 1-lines, 19 3-lines and 19 4-lines, i.e., K is of type (1,3,4), the largest minimal blocking set of PG(2,7). By the projective equivalence of points and lines, K is a symmetric 19-set, i.e., any point P has v₁(P) = 1, v₃(P) = 3 and v₄(P) = 4.

5. Symmetric Sets with at Most Four Intersection Numbers

The second case we can consider is the space PG(3,q) with q ≡ 1 mod 4. Since 4 divides q³ + q² + q + 1, the group $〈{\sigma }^4〉$ has four orbits $\left\{0,4,\dots ,q^3+q^2+q-3\right\}$, $\left\{1,5,\dots ,q^3+q^2+q-2\right\}$, $\left\{2,6,\dots ,q^3+q^2+q-1\right\}$ and $\left\{3,7,\dots ,q^3+q^2+q\right\}$ on PG(3,q), each of size ${\displaystyle \frac{{\mathit{q}}^3+{\mathit{q}}^2+\mathit{q}+1}{4}}$. Since the four orbits are projectively equivalent, we will assume without loss of generality that $K=\left\{0,4,\dots ,q^3+q^2+q-3\right\}$. Since 5 is the smallest order with q≡1 mod 4, let us consider PG(3,5). The equations of Lemma 3.3 with r = 3 and q = 5 become

$$\left\{\begin{array}{c}\left(x+z\right)+\left(y+t\right)=31\\ x^2+y^2+z^2+t^2=259\\ \begin{array}{c}\left(x+z\right)\left(y+t\right)=234\\ 2\left(xz+yt\right)=234\end{array}\end{array}\right.\Rightarrow \left\{\begin{array}{c}\left(x+z\right)=13\\ {\left(z-x\right)}^2+{\left(t-y\right)}^2=25\\ \begin{array}{c}\left(y+t\right)=18\\ xz+yt=117\end{array}\end{array}\right.\Rightarrow \left\{\begin{array}{c}\left(x+z\right)=13\\ z-x=3\\ \begin{array}{c}\left(y+t\right)=18\\ t-y=4\end{array}\end{array}\right.\Rightarrow \left\{\begin{array}{c}x=5\\ y=7\\ \begin{array}{c}z=8\\ t=11\end{array}\end{array}\right.$$

Therefore, (5,7,8,11) is the unique solution. We explicitly construct the transitive four-intersection sets of type (5,7,8,11) of PG(3,5). In order to write the cyclic structure of ℙ³: = PG(3,5), let ω be a primitive element of 𝔽₅⁴ over 𝔽₅ and let $p\left(x\right)=x^4-a_0-a_{1}x-a_2{x}^2-a_3{x}^3$ be its minimal polynomial over 𝔽₅. The companion matrix $C\left(p\right)$ of f is given by $C\left(p\right)=\left(\begin{array}{ccc}0& 1& \begin{array}{cc}0& 0\end{array}\\ 0& 0& \begin{array}{cc}1& 0\end{array}\\ \begin{array}{c}0\\ a_0\end{array}& \begin{array}{c}0\\ a_1\end{array}& \begin{array}{cc}\begin{array}{c}0\\ a_2\end{array}& \begin{array}{c}1\\ a_3\end{array}\end{array}\end{array}\right)$, and it induces a Singer cycle $\sigma$ of PG(3,5), cf. [14]. Let us consider a primitive polynomial with minimal weight, i.e., the minimal number of nonzero coefficients, among all primitives of that degree over 𝔽₅, p(x) = x⁴ + 2 + 2x + x², cf. [21]. The companion matrix C(p) is $\left(\begin{array}{ccc}0& 1& \begin{array}{cc}0& 0\end{array}\\ 0& 0& \begin{array}{cc}1& 0\end{array}\\ \begin{array}{c}0\\ 3\end{array}& \begin{array}{c}0\\ 3\end{array}& \begin{array}{cc}\begin{array}{c}0\\ 4\end{array}& \begin{array}{c}1\\ 0\end{array}\end{array}\end{array}\right)$.

Let us consider the point ${\omega }^0=\left(x_0,x_1,x_2,x_3\right)=\left(1,0,0,0\right).$ We get:

$${\omega }^1={\omega }^{0}C(f)=\left(1,0,0,0\right)\left(\begin{array}{ccc}0& 1& \begin{array}{cc}0& 0\end{array}\\ 0& 0& \begin{array}{cc}1& 0\end{array}\\ \begin{array}{c}0\\ 3\end{array}& \begin{array}{c}0\\ 3\end{array}& \begin{array}{cc}\begin{array}{c}0\\ 4\end{array}& \begin{array}{c}1\\ 0\end{array}\end{array}\end{array}\right)=\left(0,1,0,0\right),$$

$${\omega }^2={\omega }^{0}C(f{)}^2={\omega }^{1}C(f)=\left(0,1,0,0\right)\left(\begin{array}{ccc}0& 1& \begin{array}{cc}0& 0\end{array}\\ 0& 0& \begin{array}{cc}1& 0\end{array}\\ \begin{array}{c}0\\ 3\end{array}& \begin{array}{c}0\\ 3\end{array}& \begin{array}{cc}\begin{array}{c}0\\ 4\end{array}& \begin{array}{c}1\\ 0\end{array}\end{array}\end{array}\right)=\left(0,0,1,0\right),$$

$${\omega }^3={\omega }^{0}C(f{)}^3={\omega }^{2}C(f)=\left(0,0,1,0\right)\left(\begin{array}{ccc}0& 1& \begin{array}{cc}0& 0\end{array}\\ 0& 0& \begin{array}{cc}1& 0\end{array}\\ \begin{array}{c}0\\ 3\end{array}& \begin{array}{c}0\\ 3\end{array}& \begin{array}{cc}\begin{array}{c}0\\ 4\end{array}& \begin{array}{c}1\\ 0\end{array}\end{array}\end{array}\right)=\left(0,0,0,1\right)$$

$${\omega }^4={\omega }^{0}C(f{)}^4={\omega }^{3}C(f)=\left(0,0,0,1\right)\left(\begin{array}{ccc}0& 1& \begin{array}{cc}0& 0\end{array}\\ 0& 0& \begin{array}{cc}1& 0\end{array}\\ \begin{array}{c}0\\ 3\end{array}& \begin{array}{c}0\\ 3\end{array}& \begin{array}{cc}\begin{array}{c}0\\ 4\end{array}& \begin{array}{c}1\\ 0\end{array}\end{array}\end{array}\right)=\left(3,3,4,0\right)=2\left(3,3,4,0\right)=\left(1,1,3,0\right)$$

By continuing in this way and by denoting the points represented by ωⁱ simply by i, we obtain the cyclic structure of PG(3,5), as shown in

Table 8. The cyclic structure of the points of PG(3,5).

ω0 = (1,0,0,0) = 0
ω1 = (0,1,0,0) = 1	ω2 = (0,0,1,0) = 2	ω3 = (0,0,0,1) = 3	ω4 = (1,1,3,0) = 4	ω5 = (0,1,1,3) = 5
ω6 = (1,1,2,4) = 6	ω7 = (1,4,1,1) = 7	ω8 = (1,3,1,2) = 8	ω9 = (1,2,1,1) = 9	ω10 = (1,3,2,2) = 10
ω11 = (1,2,1,2) = 11	ω12 = (1,2,0,1) = 12	ω13 = (1,3,2,0) = 13	ω14 = (0,1,3,2) = 14	ω15 = (1,1,4,3) = 15
ω16 = (1,0,2,1) = 16	ω17 = (1,3,3,4) = 17	ω18 = (1,4,2,4) = 18	ω19 = (1,4,0,1) = 19	ω20 = (1,3,1,0) = 20
ω21 = (0,1,3,1) = 21	ω22 = (1,1,0,1) = 22	ω23 = (1,3,0,0) = 23	ω24 = (0,1,3,0) = 24	ω25 = (0,0,1,3) = 25
ω26 = (1,1,3,4) = 26	ω27 = (1,4,1,4) = 27	ω28 = (1,4,0,3) = 28	ω29 = (1,0,4,0) = 29	ω30 = (0,1,0,4) = 30
ω31 = (1,1,1,0) = 31	ω32 = (0,1,1,1) = 32	ω33 = (1,1,0,2) = 33	ω34 = (1,2,4,0) = 34	ω35 = (0,1,2,4) = 35
ω36 = (1,1,1,1) = 36	ω37 = (1,3,0,2) = 37	ω38 = (1,2,1,0) = 38	ω39 = (0,1,2,1) = 39	ω40 = (1,1,0,4) = 40
ω41 = (1,4,1,0) = 41	ω42 = (0,1,4,1) = 42	ω43 = (1,1,0,3) = 43	ω44 = (1,0,2,0) = 44	ω45 = (0,1,0,2) = 45
ω46 = (1,1,4,0) = 46	ω47 = (0,1,1,4) = 47	ω48 = (1,1,1,3) = 48	ω49 = (1,0,2,4) = 49	ω50 = (1,4,3,1) = 50
ω51 = (1,3,1,1) = 51	ω52 = (1,3,4,2) = 52	ω53 = (1,2,1,4) = 53	ω54 = (1,4,4,3) = 54	ω55 = (1,0,4,1) = 55
ω56 = (1,3,3,3) = 56	ω57 = (1,0,0,2) = 57	ω58 = (1,2,3,0) = 58	ω59 = (0,1,2,3) = 59	ω60 = (1,1,2,3) = 60
ω61 = (1,0,2,3) = 61	ω62 = (1,0,3,3) = 62	ω63 = (1,0,3,2) = 63	ω64 = (1,2,3,3) = 64	ω65 = (1,0,1,2) = 65
ω66 = (1,2,3,1) = 66	ω67 = (1,3,2,1) = 67	ω68 = (1,3,4,4) = 68	ω69 = (1,4,2,2) = 69	ω70 = (1,2,2,2) = 70
ω71 = (1,2,0,2) = 71	ω72 = (1,2,0,0) = 72	ω73 = (0,1,2,0) = 73	ω74 = (0,0,1,2) = 74	ω75 = (1,1,3,1) = 75
ω76 = (1,3,0,1) = 76	ω77 = (1,3,4,0) = 77	ω78 = (0,1,3,4) = 78	ω79 = (1,1,1,4) = 79	ω80 = (1,4,1,3) = 80
ω81 = (1,0,4,4) = 81	ω82 = (1,4,3,2) = 82	ω83 = (1,2,2,3) = 83	ω84 = (1,0,1,3) = 84	ω85 = (1,0,3,4) = 85
ω86 = (1,4,3,4) = 86	ω87 = (1,4,0,4) = 87	ω88 = (1,4,0,0) = 88	ω89 = (0,1,4,0) = 89	ω90 = (0,0,1,4) = 90
ω91 = (1,1,3,3) = 91	ω92 = (1,0,2,2) = 92	ω93 = (1,2,3,2) = 93	ω94 = (1,2,0,3) = 94	ω95 = (1,0,1,0) = 95
ω96 = (0,1,0,1) = 96	ω97 = (1,1,0,0) = 97	ω98 = (0,1,1,0) = 98	ω99 = (0,0,1,1) = 99	ω100 = (1,1,3,2) = 100
ω101 = (1,2,4,3) = 101	ω102 = (1,0,1,1) = 102	ω103 = (1,3,3,2) = 103	ω104 = (1,2,1,3) = 104	ω105 = (1,0,1,4) = 105
ω106 = (1,4,3,3) = 106	ω107 = (1,0,4,2) = 107	ω108 = (1,2,3,4) = 108	ω109 = (1,4,4,4) = 109	ω110 = (1,4,0,2) = 110
ω111 = (1,2,2,0) = 111	ω112 = (0,1,2,2) = 112	ω113 = (1,1,4,2) = 113	ω114 = (1,2,4,4) = 114	ω115 = (1,4,4,2) = 115
ω116 = (1,2,2,4) = 116	ω117 = (1,4,4,1) = 117	ω118 = (1,3,1,3) = 118	ω119 = (1,0,0,4) = 119	ω120 = (1,4,3,0) = 120
ω121 = (0,1,4,3) = 121	ω122 = (1,1,2,1) = 122	ω123 = (1,3,0,4) = 123	ω124 = (1,4,2,0) = 124	ω125 = (0,1,4,2) = 125
ω126 = (1,1,4,4) = 126	ω127 = (1,4,1,2) = 127	ω128 = (1,2,2,1) = 128	ω129 = (1,3,2,4) = 129	ω130 = (1,4,2,1) = 130
ω131 = (1,3,1,4) = 131	ω132 = (1,4,2,3) = 132	ω133 = (1,0,4,3) = 133	ω134 = (1,0,3,1) = 134	ω135 = (1,3,3,1) = 135
ω136 = (1,3,4,1) = 136	ω137 = (1,3,4,3) = 137	ω138 = (1,0,0,1) = 138	ω139 = (1,3,3,0) = 139	ω140 = (0,1,3,3) = 140
ω141 = (1,1,2,2) = 141	ω142 = (1,2,4,2) = 142	ω143 = (1,2,0,4) = 143	ω144 = (1,4,4,0) = 144	ω145 = (0,1,4,4) = 145
ω146 = (1,1,1,2) = 146	ω147 = (1,2,4,1) = 147	ω148 = (1,3,2,3) = 148	ω149 = (1,0,0,3) = 149	ω150 = (1,0,3,0) = 150
ω151 = (0,1,0,3) = 151	ω152 = (1,1,2,0) = 152	ω153 = (0,1,1,2) = 153	ω154 = (1,1,4,1) = 154	ω155 = (1,3,0,3) = 155

. Thus, the Singer group is isomorphic to the additive group Z₁₅₆, the integers modulo 156.

Since under the action of a cyclic collineation group of a finite projective space the point set and the plane set have the same cyclic structure, select any plane. For example, we choose the plane π₀: = x₀=0, which contains the 31-set of points written in

Table 9. The starting plane.

π0	1	2	3	5	14	21	24	25	30	32	35	39	42	45	47	59
	73	74	78	89	90	96	98	99	112	121	125	140	145	151	153

.

The remaining planes of the space are found by adding 1 to each point of the preceding plane beginning with π₀ and using addition modulo 156. Let us consider the orbit of the point 0 under the action of $〈{\sigma }^4〉$,

$$K=\left\{\mathbf{0},\mathbf{4},\mathbf{8},\dots ,\mathbf{152}\right\}.$$

Since planes of an orbit have the same intersection number, in order to compute the intersection numbers of K, it is sufficient to consider only the four base planes written in

Table 10. The base planes, with in bold the elements of K.

π0	1	2	3	5	14	21	24	25	30	32	35	39	42	45	47	59
	73	74	78	89	90	96	98	99	112	121	125	140	145	151	153
π1	2	3	4	6	15	22	25	26	31	33	36	40	43	46	48	60
	74	75	79	90	91	97	99	100	113	122	126	141	146	152	154
π2	3	4	5	7	16	23	26	27	32	34	37	41	44	47	49	61
	75	76	80	91	92	98	100	101	114	123	127	142	147	153	155
π3	0	4	5	6	8	17	24	27	28	33	35	38	42	45	48	50
	62	76	77	81	92	93	99	101	102	115	124	128	143	148	154

:

Thus, the set

$$K=\left\{\mathbf{0},\mathbf{4},\mathbf{8},\dots ,\mathbf{152}\right\}$$

is a 39-set with 39 5-planes, 39 7-planes, 39 8-planes and 39 11-planes, i.e., K is of type (5,7,8,11). By the projective equivalence of points and planes, K is a symmetric 39-set, i.e., any point P has v₅(P) = 5, v₇(P) = 7, v₈(P) = 8 and v₁₁(P) = 11.

6. Conclusions

In this work, we investigated symmetric sets in finite projective spaces. Symmetric sets are interesting not only in themselves but also through becoming associated with objects in which an interest already exists, such as sets with few intersection numbers, cf. [7]. Symmetry of sets of class [0,m,n] means that the family of symmetric sets is very large, including caps and two-character sets widely studied in the literature to construct projective linear codes with two weights, cf. [4,5]. We provided necessary conditions for the existence of such sets which, to our knowledge, seem to be novel and involve the number of external hyperplanes through the points nonbelonging to the set. Since caps and two-character sets have been extensively studied, we directed our attention to transitive sets which represent another large class of symmetric sets. We generalized the construction of Brouwer [10], derived a sufficient condition for the existence of such sets and determined enough information on their intersection numbers. After reporting the cyclic construction for the sake of completeness, we reproduced and verified examples that can be calculated by hand. All these examples together seem to reveal a close and interesting relationship between several geometric and combinatorial structures. We expect that the computer-literate reader can extend the techniques used to larger cases.