A characterization of 2-dimensional Cayley graphs on dihedral groups

TL;DR

This paper characterizes all 2-dimensional Cayley graphs on dihedral groups, providing a complete description of their structure and properties in terms of metric dimension.

Contribution

It offers a comprehensive characterization of Cayley graphs on dihedral groups with metric dimension two, filling a gap in algebraic graph theory.

Findings

All such Cayley graphs are characterized explicitly.

The metric dimension for these graphs is determined.

Structural properties of these graphs are elucidated.

Abstract

In this paper we characterize all of Cayley graphs on dihedral groups with metric dimension two.

Equations

$$di{m}_M(P_m\times C_n)=\{\begin{array}{ll}2& nodd,\\ 3& neven.\end{array}$$

$$⟨a^{i}b,a^{j}b⟩=\{a^{(i-j)t},a^{(i-j)t+i}b,a^{(i-j)t+j}b|t\in \mathbb{Z}\}.$$

$$d(u_1,v)=d-1,d(u_2,v)=d,d(u_3,v)=d+1.$$

$$a^{ki}b,a^{(k+1)i}b,...,a^{(k+n-2)i}b,a^{(k+n-1)i}b$$

$$e\sim y\sim xy\sim yxy\sim (xy{)}^2\sim y(xy{)}^2\sim \dots \sim y(xy{)}^{n-1}\sim (xy{)}^n=e.$$

$$(xy{)}^{\ell +n/2}(xy{)}^{-\ell }=(xy{)}^{n/2}=(a^{i-j}{)}^{n/2}=a^{n/2}\in S$$

$$e\sim y\sim xy\sim yxy\sim (xy{)}^2\sim y(xy{)}^2\sim \dots \sim y(xy{)}^{n/2-1}\sim (xy{)}^{n/2}=e,$$

$$a^{n/2}\sim y{a}^{n/2}\sim (xy)a^{n/2}\sim y(xy)a^{n/2}\sim (xy{)}^2{a}^{n/2}\sim \dots \sim (xy{)}^{n/2}{a}^{n/2}=a^{n/2}.$$

$$e$$

Full text

Keywords: Metric dimension, Cayley graph, Dihedral group.

MSC(2010): Primary: 05C25, 05C75, 20B10.

1 Introduction

Let $\Gamma=(V,E)$ be a simple and connected graph with vertex set $V$ and edge set $E$. The distance between two vertices $x,y\in V$ is the length of a shortest path between them and is denoted by $d(x,y)$. If $d(x,y)=1$, then for convenient we write $x\sim y$. The neighborhood of $x$ is $N(x)=\{y:~{}x\sim y\}$. For an ordered subset $W=\{w_{1},w_{2},\ldots,w_{k}\}$ of vertices and a vertex $v\in V$, the $k$-vector $r(v|W):=(d(v,w_{1}),d(v,w_{2}),\ldots,d(v,w_{k}))$ is called the metric representation of $v$ with respect to $W$. The set $W$ is called a resolving set for $\Gamma$ if distinct vertices of $\Gamma$ have distinct representations with respect to $W$. Each minimum resolving set is a basis and the metric dimension of $\Gamma$, $\dim_{M}(\Gamma)$, is the cardinality of a basis for $\Gamma$. This concept has applications in many areas including network discovery and verification [2], robot navigation [11], problems of pattern recognition and image processing [12], coin weighing problems [15], strategies for the Mastermind game [5], combinatorial search and optimization [15]. These concepts were introduced by Slater in 1975 when he was working with U.S. Sonar and Coast Guard Loran stations and he described the usefulness of these concepts, [16]. Independently, Harary and Melter [9] discovered these concepts. Finding families of graphs with constant metric dimension or characterizing $n$-vertex graphs with a specified metric dimension are fascinating problems and atracts the attention of many researchers. The problem of finding metric dimension is NP-Complete for general graphs but the metric dimension of trees can be obtained using a polynomial time algorithm. In [13] the metric dimension for two different models of random forests is investigated. It is not hard to see that for each $n$-vertex graph $\Gamma$ we have $1\leq\dim_{M}(\Gamma)\leq n-1$. Chartrand et al. [6] proved that for $n\geq 2$, $\dim_{M}(\Gamma)=n-1$ if and only if $\Gamma$ is the complete graph $K_{n}$. The metric dimension of each complete $t$-partite graph with $n$ vertices is $n-t$. They also provided a characterization of graphs of order $n$ with metric dimension $n-2$, see [6]. Graphs of order $n$ with metric dimension $n-3$ are characterized in [10]. Khuller et al. [11] and Chartrand et al. [6] proved that $\dim_{M}(\Gamma)=1$ if and only if $\Gamma$ is a path $P_{n}$. The metric dimension of the power graph of a cyclic group is determined in [8]. Salman $et~{}al.$ studied this parameter for the Cayley graphs on cyclic groups [14]. Each cycle graph $C_{n}$ is a $2$-dimensional graph ($\dim_{M}(C_{n})=2$). All of 2-trees with metric dimension two are characterized in [3] and $2$-dimensional Cayley graphs on Abelian groups are characterized in [18]. Moreover, in [17] some properties of $2$-dimensional graphs are obtained as below.

Theorem 1.1**.**

[17]* Let $\Gamma$ be a $2$-dimensional graph. If $\{u,v\}$ is a basis for $\Gamma$, then*

there is a unique shortest path $P$ between $u$ and $v$, 2. 2.

the degrees of $u$ and $v$ are at most three, 3. 3.

the degree of each internal vertex on $P$ is at most five.

The $M\ddot{o}bius~{}Ladder$ graph $M_{n}$ is a cubic circulant graph with an even number $n$ of vertices formed from an $n$-cycle by connecting opposite pairs of vertices in the cycle. For the metric dimension of $M\ddot{o}bius~{}Ladders$ we have the following result.

Theorem 1.1**.**

[1]** Let $n\geq 8$ be an even number. The metric dimension of each Mobius Ladder $M_{n}$ is 3 or 4. Specially, $dim_{M}(M_{n})=3$ when $n\equiv 2~{}(\!\!\!\!\!\mod 8)$.

C$\acute{a}$ceres $et~{}al.$ studied the metric dimension of the Cartesian product of graphs.

Theorem 1.2**.**

[4]** Let $P_{m}$ be a path on $m\geq 2$ vertices and $C_{n}$ be a cycle on $n\geq 3$ vertices. Then the metric dimension of each prism $P_{m}\times C_{n}$ is given by

[TABLE]

Let $G$ be a group and let $S$ be a subset of $G$ that is closed under taking inverse and does not contain the identity element, say $e$. Recall that the Cayley graph $Cay(G,S)$ is a graph whose vertex set is $G$ and two vertices $u$ and $v$ are adjacent in it when $uv^{-1}\in S$. Since $S$ is inverse-closed ($S=S^{-1}$) and does not contain the identity, $Cay(G,S)$ is a simple graph. It is well known that $Cay(G,S)$ is a connected graph if and only if $S$ is a generating set for $G$. In this paper, we study the metric dimension of Cayley graphs on dihedral groups and we characterize all of Cayley graphs on dihedral groups whose metric dimension is two.

2 Main results

At first, we provide two useful lemmas on dihedral groups and a sharp lower bound for the metric dimension of $3$-regular bipartite graphs which will be used in the sequel.

Lemma 2.1**.**

The subset $\{a^{i}b,a^{j}b\}$ is a generating set for dihedral group $D_{2n}=\langle a,b|~{}a^{n}=b^{2}=(ab)^{2}=e\rangle$ if and only if $\gcd(n,i-j)=1$.

Proof.

It is strightforward using the following fact about the subgroup generated by these elements.

[TABLE]

∎

Lemma 2.2**.**

If $4\mid n$ and $\gcd(i-j,n)=2$, then $\{a^{n\over 2},a^{i}b,a^{j}b\}$ is not a generating set for $D_{2n}$.

Proof.

Since $\langle a^{i-j}\rangle$ and $\langle a^{2}\rangle$ are two cyclic subgroups of order ${n\over 2}$ in the cyclic group $\langle a\rangle$, we have $\langle a^{2}\rangle=\langle a^{i-j}\rangle\subseteq\langle\{a^{n\over 2},a^{i}b,a^{j}b\}\rangle$. Since $4\mid n$, $a^{n\over 2}\in\langle a^{2}\rangle$ and hence, $\langle\{a^{n\over 2},a^{i}b,a^{j}b\}\rangle=\langle\{a^{i}b,a^{j}b\}\rangle$. Now the result follows from Lemma 2.1. ∎

Lemma 2.3**.**

Let $\Gamma$ be a $3$-regular bipartite graph on $n$ vertices. Then $dim_{M}(\Gamma)\geq 3$.

Proof.

Since $\Gamma$ is not a path, $dim_{M}(\Gamma)$ is at least two. Suppose that $dim_{M}(\Gamma)=2$ and let $W=\{u,v\}$ be a resolving set for $\Gamma$. Assume that $d(u,v)=d$ and $N(u)=\{u_{1},u_{2},u_{3}\}$. It is easy to see that $d(u_{i},v)\in\{d-1,d,d+1\}$, for each $1\leq i\leq 3$. If there exist $1\leq i<j\leq 3$ such that $d(u_{i},v)=d(u_{j},v)$, then $r(u_{i}|W)=r(u_{j}|W)$, which is a contradiction. Hence, without loss of generality, we can assume that

[TABLE]

Let $\sigma_{1}$ be a (shortest) path between two vertices $u$ and $v$ of lengh $d$, and $\sigma_{2}$ be a (shortest) path between two vertices $u_{2}$ and $v$. Two paths $\sigma_{1}$ and $\sigma_{2}$ using the edge $uu_{2}$ produce an old closed walk of lengh $2d+1$ in $\Gamma$ which contradicts the fact that $\Gamma$ is a bipartite graph. For the sharpness of this bound, consider the hypercube $Q_{3}=K_{2}\times K_{2}\times K_{2}$. ∎

In Theorem 2.4 we characterize all of Cayley graphs on dihedral groups whose metric dimension is two. Recall that the center of $D_{2n}$ is $\langle a^{n\over 2}\rangle$ when $n$ is even, otherwise it is the trivial subgroup $\{e\}$.

Theorem 2.4**.**

Let $S$ be a generating subset of $D_{2n}=\langle a,b|~{}a^{n}=b^{2}=(ab)^{2}=e\rangle$ such that $e\notin S=S^{-1}$. Then we have $dim_{M}(Cay(D_{2n},S))=2$ if and only if one of the following situations occures.

• a)

$n=|S|=2$,

• b)

$S=\{a^{i}b,a^{j}b\}$* and $\gcd(i-j,n)=1$,*

• c)

$n$* is odd and $S=\{a^{i},a^{-i},a^{j}b\}$ with $gcd(i,n)=1$ and $j\in\{1,2,...,n\}$.*

Proof.

Since $D_{2n}$ is not a cyclic group, we have $|S|\geq 2$. Assume that $|S|=2$ and $S=\{x,y\}$. Since $S=S^{-1}$ and $D_{2n}$ is not cyclic, we have $y\neq x^{-1}$ and $x^{2}=y^{2}=e$. If $S=\{a^{n\over 2},a^{j}b\}$ for some $1\leq j\leq n$, then the condition $D_{2n}=\langle S\rangle$ implies that $n=2$ and $D_{2n}=D_{4}$. Otherwise, $S=\{a^{i}b,a^{j}b\}$ and using Lemma 2.1 we have $\gcd(i-j,n)=1$. Hence, $Cay(D_{2n},S)$ is a connected $2$-regular graph (a cycle) and $dim_{M}(Cay(D_{2n},S))=2$. If $|S|\geq 4$, then the degree of each vertex in $Cay(D_{2n},S)$ is at least $4$ and part (2) of Theorem 1.1 implies that $dim_{M}(Cay(D_{2n},S))\geq 3$. Now assume that $|S|=3$. Since $S$ is a generating set and $e\notin S=S^{-1}$, we consider the following cases.

Case 1. $S=\{a^{i},a^{-i},a^{j}b\}$.

Since $(a^{j}b)(a^{i})^{t}(a^{j}b)=a^{-it}$, the order of $a^{i}$ is ${n\over\gcd(i,n)}$ and $S$ is a generating set, we have $gcd(i,n)=1$. Thus $O(a^{i})=n$ and vertices $a^{ni},a^{(n-1)i},...,a^{2i},a^{i}$ induce an $n$-cycle in $Cay(D_{2n},S)$. Since $a^{j}\in\langle a^{i}\rangle$, there exists $k\in\{1,2,...,n\}$ such that $a^{j}=a^{ki}$. Therefore $n$ vertices

[TABLE]

induce another cycle in $Cay(D_{2n},S)$. Now for each $1\leq\ell\leq n$ let $M_{\ell}=\{a^{\ell i},a^{(k+n-\ell)i}b\}$. Note that $a^{ni}=e$ and $M_{s}\cap M_{k}=\emptyset$ for each $s\neq k$. Since $a^{\ell i}(a^{(k+n-\ell)i}b)^{-1}=a^{ki}b=a^{j}b\in S$, two vertices $a^{\ell i}$ and $a^{(k+n-\ell)i}b$ are adjacent in $Cay(D_{2n},S)$. Thus, the edges $M_{1},M_{2},...,M_{n}$ provide a perfect matching in $Cay(D_{2n},S)$. Consequently, $Cay(D_{2n},S)$ is isomorphic to $P_{2}\times C_{n}$. Now Theorem 1.2 implies that $dim_{M}(Cay(D_{2n},S))=2$ if and only if $n$ is odd.

Case 2. $S=\{a^{n/2},a^{i}b,a^{j}b\}$ where $n$ is an even number.

Let $x=a^{i}b$ and $y=a^{j}b$. Since $a^{n/2}$ is in the center of $D_{2n}$ and $O(a^{n\over 2})=2$, $\langle S\rangle=\langle a^{i}b,a^{j}b\rangle\cup a^{n/2}\langle a^{i}b,a^{j}b\rangle$. Thus, $a\in\langle a^{i}b,a^{j}b\rangle$ or $a\in a^{n/2}\langle a^{i}b,a^{j}b\rangle$. Note that $|\langle a^{n/2},a^{i}b\rangle|=|\langle a^{n/2},a^{j}b\rangle|=4$. Thus, $a\notin\langle a^{n/2},a^{i}b\rangle$ and $a\notin\langle a^{n/2},a^{j}b\rangle$.

Subcase 2.1. $a\in\langle a^{i}b,a^{j}b\rangle$.

In this case, using Lemma 2.1 we have $gcd(i-j,n)=1$. Thus, $O(xy)=O(a^{i-j})=n$ and $Cay(D_{2n},S)$ contains a Hamiltonian cycle (on $2n$ vertices) as below.

[TABLE]

For each divisor $d$ of $n$ the cyclic group $\mathbb{Z}_{n}$ has unique cyclic subgroup of order $d$. Since $\langle a^{i-j}\rangle=\langle a\rangle$ and $|\langle a^{(i-j){n\over 2}}\rangle|=|\langle a^{n\over 2}\rangle|=2$ , we have $a^{n/2}=(a^{i-j})^{n/2}$. For each $1\leq\ell\leq{n\over 2}$ let $M_{\ell}=\left\{(xy)^{\ell},(xy)^{\ell+n/2}\right\}$ and $T_{\ell}=\left\{y(xy)^{\ell},y(xy)^{\ell+n/2}\right\}$. Note that $M_{s}\neq M_{k}$ and $T_{s}\neq T_{k}$ for each $s\neq k$. Also, each $M_{\ell}$ is an edge in $Cay(D_{2n},S)$ because

[TABLE]

Thus, $\{M_{1},M_{2},...,M_{n\over 2}\}$ is a matching in $X(D_{n},S)$. Similarly, $\{T_{1},T_{2},...,T_{n\over 2}\}$ is a matching and hence, $\{M_{1},M_{2},...,M_{n\over 2},T_{1},T_{2},...,T_{n\over 2}\}$ provides a perfect matching for $Cay(D_{2n},S)$. Therefore, we have a cycle on $2n$ vertices in which its opposite pairs of vertices are adjacent (see Figure 1 (i)). This implies that $Cay(D_{2n},S)$ is a $M\ddot{o}bius~{}Ladder$ and by Theorem 1.1, $dim_{M}(Cay(D_{2n},S))\neq 2$.

Subcase 2.2. $a\in a^{n/2}\langle a^{i}b,a^{j}b\rangle$.

In this case, there exists $k\in\mathbb{Z}$ such that $a=a^{n\over 2}(a^{i-j})^{k}$. Hence, $a^{{n\over 2}+1}\in\langle a^{i-j}\rangle$ and $a^{2}=(a^{{n\over 2}+1})^{2}\in\langle a^{i-j}\rangle$. Thus, $|\langle a^{i-j}\rangle|\geq|\langle a^{2}\rangle|={n\over 2}$. Hence, $O(a^{i-j})={n\over 2}$ or $O(a^{i-j})=n$. The situation $O(a^{i-j})=n$ is considered in Subcase 2.1 and we can assume that $O(xy)=O(a^{i-j})=n/2$. Therefore, $Cay(D_{2n},S)$ contains two $n$-cycles as below.

[TABLE]

The fact $O(xy)=n/2$ implies that $n$ vertices apeared in each cycle are distinct. Also, using the appearence of $b$, it is easy to see that $(xy)^{t}\neq y(xy)^{s}a^{n\over 2}$ and $y(xy)^{t}\neq(xy)^{s}a^{n\over 2}$ for each $t,s\in\mathbb{Z}$. If there exist $t,s\in\mathbb{Z}$ such that $(xy)^{t}=(xy)^{s}a^{n\over 2}$ or $y(xy)^{t}=y(xy)^{s}a^{n\over 2}$, then $a^{n\over 2}=(a^{i-j})^{(t-s)}\in\langle a^{i-j}\rangle=\langle a^{2}\rangle$. Thus, $4\mid n$ which is a contradiction (see Lemma 2.2). Therefore, all of $2n$ vertices appeared in these cycles are distinct. Since $a^{n\over 2}\in S$, corresponding vertices of two cycles are adjacent (see Figure 1 (ii)). Hence, $Cay(D_{2n},S)$ is isomorphic to $P_{2}\times C_{n}$ and Theorem 1.2 implies that $dim_{M}(Cay(D_{2n},S))=3$.

Case 3. $S=\{a^{i}b,a^{j}b,a^{t}b\}$.

Let $H=\langle a\rangle$ and hence, $V(Cay(D_{2n},S))=H\cup Hb$. If $a^{s},a^{t}\in H$, then $a^{s}a^{-t}=a^{s-t}\notin S$. Thus, the subset $H$ of vertices induces an independent set in $Cay(D_{2n},S)$. Similarly, $Hb$ is an independent set. Consequently, $Cay(D_{2n},S)$ is a $3$-regular bipartite graph on $2n$ vertices. Now Lemma 2.3 implies that $dim(Cay(D_{2n},S))$ is at least three. This completes the proof. ∎

Corollary 2.5**.**

If $S\subseteq D_{2n}$ such that $e\notin S=S^{-1}$ and $|S|\geq 4$, then $\dim_{M}(Cay(D_{2n},S))\geq 3$.

We declare that we have no competing interests