A Characterization of 2-Dimensional Cayley Graphs on Dihedral Groups

Home , Cycle graph (algebra)

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Theorem 1.1. [17] Let Γ be a 2-dimensional graph. If {u,v} is a basis for Γ, then 1. there is a unique shortest path P between u and v, 2. the degrees of u and v are at most three, 3. the degree of each internal vertex on P is at most ﬁve.

The Mobius¨ Ladder graph Mn 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 Mobius¨ Ladders we have the following result.

Theorem 1.1. [1] Let n ≥ 8 be an even number. The metric dimension of each Mobius Ladder Mn is 3 or 4. Specially, dimM(Mn)= 3 when n ≡ 2 (mod 8). C´aceres et al. studied the metric dimension of the Cartesian product of graphs.

Theorem 1.2. [4] Let Pm bea pathon m ≥ 2 vertices and Cn be a cycle on n ≥ 3 vertices. Then the metric dimension of each prism Pm ×Cn is given by

2 n odd, dimM(Pm ×Cn)= (3 n even. 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 ∈ 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 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.

i j n 2 Lemma 2.1. The subset {a b,a b} is a generating set for dihedral group D2n = ha,b| a = b = (ab)2 = ei if and only if gcd(n,i − j)= 1. Proof. It is strightforward using the following fact about the subgroup generated by these elements.

haib,a jbi = {a(i− j)t , a(i− j)t+ib, a(i− j)t+ jb | t ∈ Z}.

n i j Lemma 2.2. If 4 | n and gcd(i − j,n)= 2, then {a 2 ,a b,a b} is not a generating set for D2n. i− j 2 n Proof. Since ha i and ha i are two cyclic subgroups of order 2 in the cyclic group hai, we have n n n ha2i = hai− ji⊆h{a 2 ,aib,a jb}i. Since 4 | n, a 2 ∈ ha2i and hence, h{a 2 ,aib,a jb}i = h{aib,a jb}i. Now the result follows from Lemma 2.1.

Lemma 2.3. Let Γ be a 3-regular bipartite graph on n vertices. Then dimM(Γ) ≥ 3.

Proof. Since Γ is not a path, dimM(Γ) is at least two. Suppose that dimM(Γ)= 2 and let W = {u,v} be a resolving set for Γ. Assume that d(u,v)= d and N(u)= {u1,u2,u3}. It is easy to see that d(ui,v) ∈{d −1,d,d +1},foreach1 ≤ i ≤ 3. Ifthereexist1 ≤ i < j ≤ 3 such that d(ui,v)= d(u j,v), then r(ui|W)= r(u j|W), which is a contradiction. Hence, without loss of generality, we can assume that d(u1,v)= d − 1, d(u2,v)= d, d(u3,v)= d + 1.

Let σ1 be a (shortest) path between two vertices u and v of lengh d, and σ2 be a (shortest) path between two vertices u2 and v. Two paths σ1 and σ2 using the edge uu2 produce an old closed walk of lengh 2d + 1 in Γ which contradicts the fact that Γ is a bipartite graph. For the sharpness of this bound, consider the hypercube Q3 = K2 × K2 × K2. In Theorem 2.4 we characterize all of Cayley graphs on dihedral groups whose metric dimension n is two. Recall that the center of D2n is ha 2 i when n is even, otherwise it is the trivial subgroup {e}. n 2 2 Theorem 2.4. Let S be a generating subset of D2n = ha,b| a = b = (ab) = ei such that e ∈/ S = −1 S . Then we have dimM(Cay(D2n,S)) = 2 if and only if one of the following situations occures. a) n = |S| = 2, b) S = {aib,a jb} and gcd(i − j,n)= 1, c) nisoddandS = {ai,a−i,a jb} with gcd(i,n)= 1 and j ∈{1,2,...,n}.

Proof. Since D2n is not a cyclic group, we have |S|≥ 2. Assume that |S| = 2 and S = {x,y}. Since −1 −1 2 2 n j S = S and D2n is not cyclic, we have y 6= x and x = y = e. If S = {a 2 ,a b} for some 1 ≤ j ≤ n, i j then the condition D2n = hSi implies that n = 2 and D2n = D4. Otherwise, S = {a b,a b} and using Lemma 2.1 we have gcd(i − j,n)= 1. Hence, Cay(D2n,S) is a connected 2-regular graph (a cycle) and dimM(Cay(D2n,S)) = 2. If |S|≥ 4, then the degree of each vertex in Cay(D2n,S) is at least 4 and part (2) of Theorem 1.1 implies that dimM(Cay(D2n,S)) ≥ 3. Now assume that |S| = 3. Since S is a generating set and e ∈/ S = S−1, we consider the following cases.

3 Case 1. S = {ai,a−i,a jb}. j i t j −it i n Since (a b)(a ) (a b)= a , theorderof a is gcd(i,n) and S is a generating set, we have gcd(i,n)= 1. i ni (n−1)i 2i i j i Thus O(a )= n and vertices a ,a ,...,a ,a induce an n-cycle in Cay(D2n,S). Since a ∈ ha i, there exists k ∈{1,2,...,n} such that a j = aki. Therefore n vertices

akib,a(k+1)ib,...,a(k+n−2)ib,a(k+n−1)ib

ℓi (k+n−ℓ)i induce another cycle in Cay(D2n,S). Now for each 1 ≤ ℓ ≤ n let Mℓ = {a ,a b}. Note that ni ℓi (k+n−ℓ)i −1 ki j ℓi a = e and Ms ∩Mk = /0for each s 6= k. Since a (a b) = a b = a b ∈ S, two vertices a and (k+n−ℓ)i a b are adjacent in Cay(D2n,S). Thus, the edges M1,M2,...,Mn provide a perfect matching in Cay(D2n,S). Consequently, Cay(D2n,S) is isomorphic to P2 ×Cn. Now Theorem 1.2 implies that dimM(Cay(D2n,S)) = 2 if and only if n is odd.

Case 2. S = {an/2,aib,a jb} where n is an even number. i j n/2 n i j Let x = a b and y = a b. Since a is in the center of D2n and O(a 2 )= 2, hSi = ha b,a bi ∪ an/2haib,a jbi. Thus, a ∈ haib,a jbi or a ∈ an/2haib,a jbi. Note that |han/2,aibi| = |han/2,a jbi| = 4. Thus, a ∈/ han/2,aibi and a ∈/ han/2,a jbi.

Subcase 2.1. a ∈ haib,a jbi. i− j In this case, using Lemma 2.1 we have gcd(i− j,n)= 1. Thus, O(xy)= O(a )= n and Cay(D2n,S) contains a Hamiltonian cycle (on 2n vertices) as below.

e ∼ y ∼ xy ∼ yxy ∼ (xy)2 ∼ y(xy)2 ∼ ... ∼ y(xy)n−1 ∼ (xy)n = e.

i− j For each divisor d of n the cyclic group Zn has unique cyclic subgroup of order d. Since ha i = hai n n (i− j) 2 2 n/2 i− j n/2 n ℓ ℓ+n/2 and |ha i| = |ha i| = 2,wehave a = (a ) . Foreach1 ≤ ℓ ≤ 2 let Mℓ = (xy) ,(xy) and T = y(xy)ℓ,y(xy)ℓ+n/2 . Note that M 6= M and T 6= T for each s 6= k. Also, each M is an ℓ s k s k ℓ edge in Cay(D2n,S) because (xy)ℓ+n/2(xy)−ℓ = (xy)n/2 = (ai− j)n/2 = an/2 ∈ S

Thus, {M ,M ,...,M n } is a matchingin X(D ,S). Similarly, {T ,T ,...,Tn } is a matching and hence, 1 2 2 n 1 2 2 {M ,M ,...,M n ,T ,T ,...,Tn } provides a perfect matching for Cay(D ,S). Therefore,we havea cy- 1 2 2 1 2 2 2n cle on 2n vertices in which its opposite pairs of vertices are adjacent (see Figure 1 (i)). This implies that Cay(D2n,S) is a Mobius¨ Ladder and by Theorem 1.1, dimM(Cay(D2n,S)) 6= 2.

Subcase 2.2. a ∈ an/2haib,a jbi. n n 1 n 1 In this case, there exists k ∈ Z such that a = a 2 (ai− j)k. Hence, a 2 + ∈ hai− ji and a2 = (a 2 + )2 ∈ i− j i− j 2 n i− j n i− j i− j ha i. Thus, |ha i| ≥ |ha i| = 2 . Hence, O(a )= 2 or O(a )= n. The situation O(a )= n is i− j considered in Subcase 2.1 and we can assume that O(xy)= O(a )= n/2. Therefore, Cay(D2n,S) contains two n-cycles as below.

e ∼ y ∼ xy ∼ yxy ∼ (xy)2 ∼ y(xy)2 ∼ ... ∼ y(xy)n/2−1 ∼ (xy)n/2 = e, an/2 ∼ yan/2 ∼ (xy)an/2 ∼ y(xy)an/2 ∼ (xy)2an/2 ∼ ... ∼ (xy)n/2an/2 = an/2.

The fact O(xy)= n/2 implies that n vertices apeared in each cycle are distinct. Also, using the n n appearence of b, it is easy to see that (xy)t 6= y(xy)sa 2 and y(xy)t 6= (xy)sa 2 for each t,s ∈ Z. If there t s n t s n n i j t s i j 2 exist t,s ∈ Z such that (xy) = (xy) a 2 or y(xy) = y(xy) a 2 , then a 2 = (a − )( − ) ∈ ha − i = ha i.

4 Thus, 4 | n which is a contradiction (see Lemma 2.2). Therefore, all of 2n vertices appeared in these n cycles are distinct. Since a 2 ∈ S, correspondingvertices of two cycles are adjacent (see Figure 1 (ii)). Hence, Cay(D2n,S) is isomorphic to P2 ×Cn and Theorem 1.2 implies that dimM(Cay(D2n,S)) = 3. Case 3. S = {aib,a jb,at b}. s t s −t s−t Let H = hai and hence, V(Cay(D2n,S)) = H ∪ Hb. If a ,a ∈ H, then a a = a ∈/ S. Thus, the subset H of vertices induces an independent set in Cay(D2n,S). Similarly, Hb is an independent set. Consequently, Cay(D2n,S) is a 3-regular bipartite graph on 2n vertices. Now Lemma 2.3 implies that dim(Cay(D2n,S)) is at least three. This completes the proof.

un/2 yxy xy y(xy)n/2−1un/2 yun/2 (xy)2 y e y(xy)n/2−1 y (xy)n/2 e (xy)n/2−1un/2 (xy)n/2−1 xy xyun/2 yxy y(xy)n/2 (xy)n/2+2 (xy)n/2+1 y(xy)n/2+1 yxyun/2

(i) (ii)

Figure 1: (i) A Mobius¨ Ladder graph, and (ii) the Cartesian product P2 ×Cn.

−1 Corollary 2.5. If S ⊆ D2n such that e ∈/ S = S and |S|≥ 4, then dimM(Cay(D2n,S)) ≥ 3. We declare that we have no competing interests

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