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Domination and efficient domination in cubic and quartic Cayley graphs on abelian groups
Discrete Applied Mathematics 271 (2019) 15–24
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Domination and efficient domination in cubic and quartic Cayley graphs on abelian groups✩ Cafer Çalışkan a , Štefko Miklavič b , Sibel Özkan c ,
∗
a
Antalya Bilim University, Department of Computer Engineering, 07190, Antalya, Turkey University of Primorska, Andrej Marušič Institute, Muzejski trg 2, 6000 Koper, Slovenia c Gebze Technical University, Department of Mathematics, 41400, Gebze, Kocaeli, Turkey b
article
info
Article history: Received 29 November 2018 Received in revised form 16 June 2019 Accepted 7 August 2019 Available online 17 September 2019
a b s t r a c t In this paper, we characterize cubic and quartic Cayley graphs on abelian groups that admit an efficient dominating set, and then we give domination numbers, even when they do not have an efficient dominating set, when it is possible. © 2019 Elsevier B.V. All rights reserved.
Keywords: Domination Efficient domination Perfect code Cayley graph
1. Preliminaries and basic definitions Throughout this paper, Γ = (V , E) will denote a finite simple graph with vertex set V and edge set E, without loops and multiple edges (for more background on graph theory and for notation not defined here we refer the reader to [6]). For every vertex x ∈ V , we will say that x dominates y, if y = x or y is adjacent to x (i.e. x dominates its closed neighbourhood N [x]). A dominating set of Γ is a subset D of V , such that every vertex is dominated by at least one member of D. The domination number γ (Γ ) is the number of vertices in a minimum dominating set for Γ . Let ∆ denote the maximal valency of Γ . Observe that each x ∈ V could dominate at most ∆ + 1 vertices, and so
γ (Γ ) ≥
⌈
⌉ |V | . ∆+1
(1)
This bound was given first in [15]. Since finding the domination number is NP-complete for arbitrary graphs, it is natural to ask for bounds on the domination number related to other graph parameters. There are many other bounds involving order, degree, size and diameter. See [1,10,11], and [14] for example. It is also natural to ask for exact results on specific graph classes. Here we will consider cubic and quartic Cayley graphs on abelian groups. An efficient dominating set of Γ is a subset D of V , such that every vertex of Γ is dominated by exactly one vertex from D. Observe that in this case D is necessarily independent subset of V and that it is also minimum dominating set of Γ . Moreover, if Γ is regular with valency k, then
|D| =
|V | k+1
.
(2)
✩ This research is supported by the bilateral grant of TUBITAK, Turkey (Grant No. 115F586) and ARRS, Slovenia (Grant No. BI-TR/16-18-002). The second author was supported in part by Slovenian Research Agency, program P1-0285 and projects N1-0032, N1-0038, J1-7051, N1-0062. ∗ Corresponding author. E-mail addresses: [email protected] (C. Çalışkan), [email protected] (Š. Miklavič), [email protected] (S. Özkan). https://doi.org/10.1016/j.dam.2019.08.004 0166-218X/© 2019 Elsevier B.V. All rights reserved.
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While searching for the domination number of a given graph, it is natural to check for an efficient dominating set first, since an efficient dominating set is a minimum dominating set as noted above. But not every graph has an efficient dominating set. Bange et al. [2] have proved that it is an NP-complete problem to determine whether a given graph has an efficient dominating set or not. Let G denote a finite group (for more background on group theory and for notation not defined here we refer the reader to, for example, [7]). We denote the order of G (that is, the number of elements in G) by |G|. In this paper we will be using a multiplicative notation, that is, we denote the identity element of G with e, the inverse element of a ∈ G with a−1 , and the ‘‘product’’ of elements a, b ∈ G with ab. For a subset S ⊆ G we denote by ⟨S ⟩ the subgroup of G, generated by the elements of set S. If S = {a} we abbreviate ⟨S ⟩ = ⟨a⟩. We say that S is inverse-closed subset of G, if x ∈ S implies x−1 ∈ S. Let now S denote an inverse-closed subset of G \ {e}. The Cayley graph Cay(G; S) of the group G with respect to the connection set S is a graph with vertex-set G, in which g ∈ G is adjacent to h ∈ G if and only if h = gs for some s ∈ S. In this case we refer to the edge {g , h} as an s-edge. Observe that Cay(G; S) is regular with valency k = |S |. We say that Cay(G; S) is cubic (quartic, respectively), if |S | = 3 (|S | = 4, respectively). Note also that Cay(G; S) is connected if and only if ⟨S ⟩ = G. More generally, every connected component of Cay(G; S) is isomorphic to the Cayley graph Cay(⟨S ⟩; S). Recall that every Cayley graph is vertex-transitive. Namely, for each pair of vertices u, v ∈ Cay(G; S), there is an automorphism f of Cay(G; S) such that f (u) = v . In particular, if D is a (efficient) dominating set of Cay(G; S), we could always assume that e ∈ D. We refer the reader to [3] for more background on Cayley graphs. Study of (efficient) dominating sets in Cayley graphs is an important and vivid research area. In terms of efficient domination, Lee proved that a Cayley graph on an abelian group has an efficient dominating set if and only if it is a covering graph of a complete graph [9]. Dejter and Serra gave a method for constructing countable families of Cayley graphs each one having an efficient dominating set [5]. Efficient dominating sets in certain Cartesian products, where the factors are Cayley graphs on an abelian group, are shown to exist by Mollard [12]. In [13], cubic and quartic circulants (that is, Cayley graphs on cyclic groups) admitting an efficient dominating set were classified. In this paper we generalize this result to a wider class of Cayley graphs on abelian groups. More precisely, we will classify cubic and quartic Cayley graphs on abelian groups that admit an efficient dominating set. Our main results are Theorems 3.3 and 4.11. We also analyse the domination number and give the exact number when possible in Section 5. For the rest of this paper we assume that G is an abelian group. For a ∈ G let o(a) denote the order of the element a ∈ G, that is, the smallest positive integer m such that am = e. Element a is called an involution, if o(a) = 2. For a subgroup H ≤ G, we denote the index of H in G by [G : H ]. Recall that [G : H ] = |G|/|H |. 2. Structure of Cayley graphs on abelian groups In this section we describe briefly the structure of Cayley graphs on abelian groups. To do this, we introduce the following notation. For a positive integer n, let Kn denote the complete graph on n vertices. If n ≥ 3, then we let Cn denote the cycle of length n. Pick a ∈ G and consider Γ = Cay(G; {a, a−1 }). If a is an involution, then Γ is isomorphic to the disjoint union of |G|/2 copies of K2 . If a is not an involution, then Γ is isomorphic to the disjoint union of t cycles Cn , where n = o(a) and t = [G : ⟨a⟩]. Now let S denote an inverse-closed subset of G \ {e} and pick a ∈ G \ (S ∪ {e}). Set S ′ = S ∪ {a, a−1 }. Observe that Cay(G, S ′ ) is isomorphic to the union of Cay(G; S) and Cay(G; {a, a−1 }) (here, by the union of graphs Γ1 = (V1 , E1 ) and Γ2 = (V2 , E2 ) we mean the graph Γ1 ∪ Γ2 = (V1 ∪ V2 , E1 ∪ E2 )). This observation allows us to recursively ‘‘build’’ our Cayley graph from its ‘‘basic blocks’’, that are either disjoint union of copies of K2 or disjoint union of cycles. Moreover, let m be the smallest positive integer such that am ∈ ⟨S ⟩. If m = o(a), then Cay(G, S ′ ) is isomorphic to the cartesian product of Cay(⟨S ⟩; S) and Cay(⟨a⟩; {a, a−1 }). Pick x ∈ G and a ∈ S, which is not an involution. Note that in this case (x, xa, xa2 , . . . , xao(a)−1 ) is a cycle of length o(a) in Cay(G; S). We refer to this cycle as a-cycle, which contains x. As already mentioned above, Cay(G; S) is connected if and only if ⟨S ⟩ = G, and every connected component of Cay(G; S) is isomorphic to the Cayley graph Cay(⟨S ⟩; S). As G is abelian, ⟨S ⟩ is abelian as well. So when studying properties of Cayley graph Cay(G; S) where G is abelian, then it is reasonable to assume that Cay(G; S) is connected, that is, G = ⟨S ⟩. That is what we assume throughout this paper. Consider now a connected graph Cay(G; S). If there exists a ∈ S such that o(a) = |G|, then G is cyclic. Such graphs, admitting an efficient dominating set, were already classified in [13] for the cubic and the quartic case. But for the sake of completeness of the paper, we treat this case also in our paper. However, note that if we have o(a) < |G| for a non-involution a ∈ S, then Cay(G; {a, a−1 }) is a disjoint union of at least two cycles of length o(a). 3. Cubic Cayley graphs on abelian groups In this section we assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 3, such that ⟨S ⟩ = G. As S is inverse closed, S contains either three involutions or one involution. We consider each of these cases separately.
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Fig. 1. Efficient domination for K4 and 3-cube.
Fig. 2. Efficient domination for the Möbius ladder on 12 vertices and for the generalized Petersen graph GP(8, 1).
Proposition 3.1. Let G denote a finite abelian group and let S = {i1 , i2 , i3 }, where i1 , i2 , i3 are pairwise different involutions of G. Let Γ = Cay(G; S). Then Γ is either isomorphic to K4 or to the 3-cube. Moreover, both of these two graphs admit an efficient dominating set. Proof. As G is generated by 3 involutions, then for j ̸ = k ̸ = l ∈ {1, 2, 3} either ij ik = il and G is isomorphic to Z2 × Z2 , or ij ik ̸ = il and G is isomorphic to Z2 × Z2 × Z2 . The only cubic graph on 4 vertices is the complete graph K4 (which is of course also Cayley graph on Z2 × Z2 ), and it is easy to see that it admits an efficient dominating set (choose one vertex and it dominates everything, see Fig. 1). It is also easy to see that every connected cubic Cayley graph on Z2 × Z2 × Z2 is isomorphic to the Cayley graph on Z2 × Z2 × Z2 with connection set (1, 0, 0), (0, 1, 0), (0, 0, 1) which is 3-cube, which again admits an efficient dominating set (let D = {(0, 0, 0), (1, 1, 1)} for example, see Fig. 1). ■ Proposition 3.2. Let G denote a finite abelian group and let S = {i1 , a, a−1 }, where i1 is an involution and a is a non-involution. Let Γ = Cay(G; S) and let us denote |G| = n. Then either Γ is isomorphic to the Möbius ladder M(n) on n vertices, or to the generalized Petersen graph GP(n/2, 1). Moreover, the Möbius ladder on n vertices admits an efficient dominating set if and only if n = 8k + 4 for some non-negative integer k, and GP(n/2, 1) admits an efficient dominating set if and only if n = 8k |G| for some non-negative integer k. In both cases, γ (Γ ) = 4 . Proof. Assume first that o(a) = n. Observe that in this case i1 = an/2 , and so Γ is isomorphic to the Möbius ladder on n vertices (which is also isomorphic to the Cayley graph on Zn with the connection set {1, −1, 2n }). It is now easy to see that Γ admits an efficient dominating set if and only if n = 8k + 4 for some non-negative integer k (let D = {0, 4, . . . , 8k} if we label the vertex set of Γ with elements of Z8k+4 ). Assume now that o(a) < n. Involution i1 generates n/2 copies of K2 , so we should have two cycles of length n/2 and so we have o(a) = n/2 (see the comments in Section 2 for details), and so Γ is isomorphic to the cartesian product Cn/2 □K2 , that is, Γ is isomorphic to a generalized Petersen graph GP(n/2, 1), also the so-called prism graph (see Fig. 2). It is again easy to see that Γ admits an efficient dominating set if and only if n = 8k for some non-negative integer k (let (a, 0) and (a + 2, 1) be in D for a ∈ {0, 4, 8, . . . , 4k − 4} if we label the vertex of Γ with Z4k × Z2 ). ■ Taking into account that the 3-cube is isomorphic to the generalized Petersen graph GP(4, 1) and that the complete graph K4 is isomorphic to the Möbius ladder on 4 vertices, the following theorem is an immediate consequence of Propositions 3.1 and 3.2. Theorem 3.3. Let G denote a finite abelian group and let S denote an inverse closed subgroup of G \ {e} with |S | = 3, such that ⟨S ⟩ = G. Then Γ = Cay(G; S) admits an efficient dominating set if and only if Γ is isomorphic to the generalized Petersen graph GP(4k, 1) for some positive integer k, or to the Möbius ladder on 8k + 4 vertices, for some non-negative integer k. 4. Quartic Cayley graphs on abelian groups In this section we will classify quartic Cayley graphs on abelian groups, which admit an efficient dominating set. Assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 4, such that ⟨S ⟩ = G. We consider quartic Cayley graph
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Γ = Cay(G; S). Let us denote |G| = n. Since Γ is 4-regular, by (2) any efficient dominating set of Γ (if one exists) has n/5 elements. So, we will assume that n is divisible by 5. Note also that as S is inverse-closed, S contains either 4, or 2, or none involutions. We first consider the case where S contains four involutions. Proposition 4.1. Let G denote a finite abelian group and let S = {i1 , i2 , i3 , i4 }, where i1 , i2 , i3 , i4 are pairwise different involutions of G. Then Γ = Cay(G; S) does not admit any efficient dominating set. Proof. Observe that since G is generated with four (pairwise distinct) involutions, G is either Z2 × Z2 × Z2 , or Z2 × Z2 × Z2 × Z2 . It follows that Cay(G; S) has either 8 or 16 vertices. As neither of these two is divisible by 5, Cay(G; S) does not admit any efficient dominating set. ■ In view of Proposition 4.1, let us assume now that there exists a non-involution a ∈ S. Therefore, S = {a, a−1 , b, c } for some b, c ∈ G. Note that if b is non-involution, then c = b−1 . Assume now that D is an efficient domination set of Γ . Pick any x ∈ G and consider the a-cycle which contains x. We will show that any two vertices of this cycle, that are contained in D, are at least 5 places apart on this cycle. More precisely, we show that if xai , xaj ∈ D with i ̸ = j, then min{i − j, j − i} ≥ 5, where i − j and j − i are computed modulo o(a). As Γ is vertex transitive, we could assume that x = 1 and i = 0. Proposition 4.2. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution. Let D denote an efficient dominating set for Γ = Cay(G; S) and assume without loss of generality that e ∈ D. Then the following statements hold. (i) o(a) ≥ 5. (ii) Assume that ai ∈ D for some 1 ≤ i ≤ o(a) − 1. Then 5 ≤ i ≤ o(a) − 5. Proof. (i) Assume 3 ≤ o(a) ≤ 4. Since |D| = n/5, there exists at least one a-cycle which has empty intersection with D. Assume first that o(a) = 4 and let C = (x, xa, xa2 , xa3 ) be the a-cycle with the empty intersection with D. It follows that vertices x, xa, xa2 , xa3 must be dominated by their neighbours outside C , that is, with vertices xb, xab, xa2 b, xa3 b or xc , xac , xa2 c , xa3 c. Without loss of generality, we could assume that x is dominated by xb. Now to avoid double domination, xa must be dominated by xac. But now we could not efficiently dominate xa2 . If we dominate xa2 with xa2 b, then xab is doubly dominated, and if we dominate xa2 with xa2 c, then xac is doubly dominated. This shows that o(a) ̸ = 4. Similarly we show that o(a) ̸ = 3, but the details are left to the reader. (ii) If o(a) = n then G is a cyclic group generated by a, and the result follows from the proof of [13, Lemma 1]. Therefore, assume that o(a) < n. As ⟨S ⟩ = G, this implies that not both of b, c are contained in ⟨a⟩. Without loss of generality we could assume that b ̸ ∈ ⟨a⟩. Observe that we could assume without loss of generality that i ≤ o(a)/2 (otherwise we replace a with a−1 ). It is easy to see that i ̸ = 1, since in this case e is doubly dominated (by itself and by a). Similarly, i ̸ = 2, since otherwise a is doubly dominated (by e and by a2 ). Assume now that i = 3 and consider vertex ba ̸ ∈ ⟨a⟩. If ba is dominated by e, then c = ba. It follows that all neighbours of ba2 (that is, vertices a, a2 , ba, ba3 ) are already dominated, and so we could not efficiently dominate ba2 . If ba is dominated by a3 , then b = ca2 and all neighbours of ba2 (this time vertices a2 , a4 , ba, ba3 ) are already dominated, and so we could not efficiently dominate ba2 . Similarly, if ba2 is dominated by e, then c = ba2 . It follows that all neighbours of ba (vertices a, an−1 , b, ba2 ) are already dominated, and so we could not efficiently dominate ba. Again, if ba2 is dominated by a3 , then b = ca and all neighbours of ba (vertices a, a2 , b, ba2 ) are already dominated, and so we could not efficiently dominate ba. These arguments show that ba, ba2 could not be dominated by e or a3 . It follows that ba must be dominated by c −1 ba and ba2 must be dominated by c −1 ba2 . But c −1 ba and c −1 ba2 are adjacent in Γ , a contradiction. Assume now that i = 4 and consider ba2 ̸ ∈ ⟨a⟩. If ba2 is dominated by e, then c = ba2 and all neighbours of a2 (that is, vertices a, a3 , ba2 , ba4 ) are already dominated, and so we could not efficiently dominate a2 . Similarly, if ba2 is dominated by a4 , then again a2 could not be efficiently dominated since all of its neighbours (namely a, a3 , b, ba2 ) are already dominated. Note also that ba2 could not be dominated by its neighbours a2 , ba, ba3 , since in each of these three cases a double domination occurs. Therefore, ba2 is dominated by c −1 ba2 . But now ba has to be dominated by c −1 ba, which is adjacent to c −1 ba2 , a contradiction. This shows that i ≥ 5. ■ Corollary 4.3. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution such that Γ = Cay(G; S) admits an efficient dominating set D. Then the following statements hold. (i) For x ∈ D, the intersection of D and the a-cycle which contains x is nonempty and consists precisely of vertices {xa5i | 0 ≤ i ≤ o(a)/5 − 1}. (ii) o(a) is divisible by 5. Proof. (i) By Proposition 4.2(ii), any two vertices of D, which belong to the a-cycle which contains x, are at least 5 positions apart on this a-cycle. This shows that the intersection of D with any a-cycle contains at most o(a)/5 vertices, and so D contains at most |G|/5 vertices. On the other hand, D = |G|/5 by (2), and so the intersection of D with any a-cycle contains exactly every fifth vertex of this cycle. As x is an element of D, the result follows.
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(ii) Recall that the intersection of D with any a-cycle contains exactly every fifth vertex of this cycle. In addition, any two vertices of D, which belong to the same a-cycle, are at least 5 positions apart on this a-cycle. This shows that the length of any a-cycle is divisible by 5. As the length of every a-cycle is equal to o(a), the result follows. ■ 4.1. Quartic Cayley graphs on abelian groups — two involutions Assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 4, such that ⟨S ⟩ = G. In view of Proposition 4.1, let us assume now that there exists a non-involution a ∈ S. Therefore, S = {a, a−1 , b, c } for some b, c ∈ G. We first consider the case where b, c are (distinct) involutions. If o(a) = n, then G is a cyclic group. But any cyclic group contains at most one involution, contradicting our assumption that b, c are distinct involutions of G. Therefore, o(a) < n. As ⟨S ⟩ = G, at least one of b, c is not contained in ⟨a⟩. Without loss of generality we could assume that b ̸ ∈ ⟨a⟩. We have three different cases to consider: case c ∈ ⟨a⟩, case c ̸ ∈ ⟨a⟩ but c ∈ ⟨{a, b}⟩, and case c ̸ ∈ ⟨{a, b}⟩. Theorem 4.4. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b ̸ ∈ ⟨a⟩ and c ∈ ⟨a⟩. Then Γ = Cay(G; S) does not admit an efficient dominating set. Proof. We first show that o(a) = n/2. Recall that o(a) is a divisor of |G| = n, and so b ̸ ∈ ⟨a⟩ implies o(a) ≤ n/2. However, as G is generated by a and b and as o(b) = 2, it follows that o(a) = n/2. This implies that c = an/4 (recall that c is an involution). Now assume that D is an efficient dominating set of Γ . By Corollary 4.3, we know that o(a) is divisible by 5 and we could also assume that the intersection of the a-cycle which contains e and D is {e, a5 , . . . , ao(a)−5 }. But as n is divisible by 5, n/4 is also divisible by 5 and this implies that an/4 = c ∈ D. However, this is not possible, since c is already dominated by e, a contradiction. This finishes the proof. ■ Theorem 4.5. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b, c ̸ ∈ ⟨a⟩ and c ∈ ⟨{a, b}⟩. Then Γ = Cay(G; S) does not admit an efficient dominating set. Proof. As G is generated by a and b and as o(a) < n, it follows that o(a) = n/2. Since c ̸ ∈ ⟨a⟩ and b is an involution, we have that c is contained in the a-cycle, that contains b. The only involutions contained in this a-cycle are b and ban/4 , so c = ban/4 . Now assume that D is an efficient dominating set of Γ . By Corollary 4.3 we know that o(a) is divisible by 5 and we could also assume that the intersection of the a-cycle which contains e and D is {e, a5 , . . . , ao(a)−5 }. Consider now the vertex a2 and note that it could be only dominated by ba2 or ba2+n/4 . If ba2 ∈ D, then ba4 could only be dominated by a4+n/4 , which forces a4+n/4 ∈ D and so 4 + n/4 is divisible by 5, a contradiction. Therefore, ba2+n/4 ∈ D. But now a3 has to be dominated by ba3 , and so ba3 ∈ D. But now to dominate ba we must have a1+n/4 ∈ D, which forces that 1 + n/4 is divisible by 5, a contradiction. This finishes the proof. ■ Theorem 4.6. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b ̸ ∈ ⟨a⟩ and c ̸ ∈ ⟨{a, b}⟩. Then Γ = Cay(G; S) does not admit an efficient dominating set. Proof. We first show that o(a) = n/4. Recall that o(a) is a divisor of |G| = n, and so b ̸ ∈ ⟨a⟩ together with c ̸ ∈ ⟨{a, b}⟩ imply that o(a) ≤ n/4. However, as G is generated by a, b and c and as o(b) = o(c) = 2, it follows that o(a) = n/4. As G is generated by S and as c ̸ ∈ ⟨{a, b}⟩, it follows that o(a) = n/4. By comments in Section 2 it follows that Γ is isomorphic to the cartesian product Cn/4 □K2 □K2 , which is further isomorphic to Cn/4 □C4 . Therefore, Γ is isomorphic to the Cayley graph on the group Zn/4 × Z4 , with the connection set S ′ = {(1, 0), (n/4 − 1, 0), (0, 1), (0, 3)}. But element (0, 1) is a non-involutory element of Zn/4 × Z4 with order 4. By Corollary 4.3(iii), Γ does not admit an efficient domination set. ■ 4.2. Quartic Cayley graphs on abelian groups — no involutions Assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 4, such that ⟨S ⟩ = G. In view of Proposition 4.1 and Section 4.1, in this subsection we consider the case where S = {a, a−1 , b, b−1 } for some non-involutions a, b ∈ G. Recall that o(a) denotes the order of element a in G. Let us denote by t the index of ⟨a⟩ in G, that is,
|G| |G| = . |⟨a⟩| o(a) We consider cases t = 1 and t ≥ 2. The case t = 1 was already solved in [13], but for the sake of completeness of the t=
paper, we include this case here as well. Lemma 4.7. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t be as defined above and assume that t = 1. Then the following statements are equivalent: (i) Γ = Cay(G; S) admits an efficient dominating set; (ii) o(a) is divisible by 5 and there exists an integer ℓ, (0 ≤ ℓ ≤ o(a)/5 − 1), such that either b or b−1 is equal to a5ℓ−2 .
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Proof. Observe that t = 1 implies that the elements of G are exactly the elements of the form ai (0 ≤ i ≤ o(a) − 1). Assume first that Γ admits an efficient dominating set. Then we could assume that e ∈ D, and by Corollary 4.3 we have that o(a) is divisible by 5 and D = {a5i | 0 ≤ i ≤ o(a)/5 − 1}. Note that a2 ̸ ∈ D, and that the neighbours of a2 in Γ are a, a3 , a2 b and a2 b−1 . As a, a3 ̸ ∈ D, it follows that a2 b ∈ D or a2 b−1 ∈ D. Therefore, there exists ℓ (0 ≤ ℓ ≤ o(a)/5 − 1), such that either a2 b = a5ℓ or a2 b−1 = a5ℓ . It follows that either b or b−1 is equal to a5ℓ−2 . Assume now that o(a) is divisible by 5 and there exists an integer ℓ, (0 ≤ ℓ ≤ o(a)/5 − 1), such that either b or b−1 is equal to a5ℓ−2 . Let us define D = {a5i | 0 ≤ i ≤ o(a)/5 − 1}. We now show that D is an efficient dominating set. We show this for the case when b = a5ℓ−2 for some 0 ≤ ℓ ≤ o(a)/5 − 1. The proof in the case when b−1 = a5ℓ−2 for some 0 ≤ ℓ ≤ o(a)/5 − 1 is similar. We first show that every vertex of Γ is dominated by at least one vertex from D. This is clear for vertices which are contained in D. Similarly, vertices of the form a5i+1 (a5i+4 , respectively) for (0 ≤ i ≤ o(a)/5 − 1) are adjacent to vertices a5i (a5(i+1) , respectively), and are therefore dominated by D. Consider now vertices of the form a5i+2 (0 ≤ i ≤ o(a)/5 − 1). As b = a5ℓ−2 , every such vertex is adjacent to a5i+2 b = a5i+2 a5ℓ−2 = a5(i−ℓ) ∈ D. Similarly, every vertex of the form a5i+3 (0 ≤ i ≤ o(a)/5 − 1) is adjacent to a5i+3 b−1 = a5i+3 a2−5ℓ = a5(i+1−ℓ) ∈ D. This shows that D is a dominating set. Finally, observe that
|D| =
o(a)
=
|G|
,
5 5 and so D is an efficient dominating set. This finishes the proof.
■
Assume now that t ≥ 2. As ⟨S ⟩ = G, we have that b ̸ ∈ ⟨a⟩ and that bi ⟨a⟩ (0 ≤ i ≤ t − 1) are distinct cosets of ⟨a⟩. Consequently, bt ∈ ⟨a⟩. Let k (0 ≤ k ≤ o(a) − 1) be such that bt = ak . Proposition 4.8. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t , k be as defined above and assume that t ≥ 2. Assume that D is an efficient dominating set of Γ = Cay(G; S) and assume that bi aj ∈ D for some 0 ≤ i ≤ t − 1, 0 ≤ j ≤ o(a) − 1. Then exactly one of the following statements hold: (i) {bi+1 aj+2 , bi−1 aj+3 } ⊆ D; (ii) {bi−1 aj+2 , bi+1 aj+3 } ⊆ D. Proof. By Corollary 4.3 we have that the intersection of D and the a-cycle which contains bi aj is equal to {bi a5ℓ+j | 0 ≤ ℓ ≤ o(a)/5 − 1}. This shows that bi aj+2 must be dominated by one of its neighbours different than bi aj+1 and bi aj+3 , that
is by bi+1 aj+2 or by bi−1 aj+2 . Therefore, either bi+1 aj+2 or bi−1 aj+2 is contained in D. Note that if both bi+1 aj+2 and bi−1 aj+2 are contained in D, then bi aj+2 is doubly dominated, a contradiction. Assume now that bi+1 aj+2 ∈ D. Observe that in this case the only way to (efficiently) dominate bi aj+3 is such that i−1 j+3 b a ∈ D. Similarly, if bi−1 aj+2 ∈ D, then, to efficiently dominate bi aj+3 , we have that bi+1 aj+3 ∈ D. ■ Proposition 4.9. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t , k be as defined above and assume t ≥ 2. Assume that D is efficient dominating set of Γ = Cay(G; S) such that e ∈ D. Then the following statements holds: (i) If ba2 ∈ D, then D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+2i | 0 ≤ ℓ ≤ o(a)/5 − 1}. (ii) If ba3 ∈ D, then D=
t −1 ⋃ i=0
Di ,
C. Çalışkan, Š. Miklavič and S. Özkan / Discrete Applied Mathematics 271 (2019) 15–24
21
Fig. 3. The efficient dominating set D from Proposition 4.9(i) in the case o(a) = 15 and t ≥ 4.
where Di = {bi a5ℓ+3i | 0 ≤ ℓ ≤ o(a)/5 − 1}. Proof. We prove part (i) of the above proposition. Proof of part (ii) is similar. We first show by induction on i that Di ⊆ D for 0 ≤ i ≤ t − 1. Since e ∈ D, it follows from Corollary 4.3(ii) that D0 ⊆ D. Similarly, since ba2 ∈ D, it follows from Corollary 4.3(ii) that D1 ⊆ D. For t = 2, we have D0 ∪ D1 ⊆ D. For t ≥ 3, assume that Dj ⊆ D for 0 ≤ j ≤ i for some 1 ≤ i ≤ t − 2. Note that by our induction hypothesis we have that bi a2i , bi−1 a5+2(i−1) = bi−1 a2i+3 ∈ D. Taking j = 2i in Proposition 4.8, we find that bi+1 a2(i+1) ∈ D. But now it follows from Corollary 4.3(ii) that Di+1 ⊆ D. This shows that t −1 ⋃
Di ⊆ D.
i=0
However, Di (0 ≤ i ≤ t − 1) are pairwise disjoint sets with |Di | = o(a)/5, and so
|
t −1 ⋃
Di | = t
i=0
o(a) 5
=
|G| o(a) o(a) 5
=
|G| 5
= |D|.
It follows that t −1 ⋃
Di = D.
■
i=0
The efficient dominating set D described in Proposition 4.9(i) is depicted in Fig. 3 (for the case o(a) = 15 and t ≥ 4). The vertices of the top 15-cycle are the vertices of the a-cycle containing e, the vertices of the second 15-cycle are the vertices of the a-cycle containing b, and so on. Lemma 4.10. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t , k be as defined above and assume that t ≥ 2. Then the following statements are equivalent: (i) Γ = Cay(G; S) admits an efficient dominating set; (ii) o(a) is divisible by 5 and there exists an integer 0 ≤ ℓ ≤ o(a)/5 − 1, such that bt is equal to either a5ℓ−2t or to a5ℓ−3t . Proof. Assume first that Γ admits an efficient dominating set. Without loss of generality we could assume that e ∈ D. By Corollary 4.3(iii) we have that o(a) is divisible by 5. By Proposition 4.8 either ba2 ∈ D or ba3 ∈ D. If ba2 ∈ D, then by
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C. Çalışkan, Š. Miklavič and S. Özkan / Discrete Applied Mathematics 271 (2019) 15–24
Proposition 4.9(i) we have that D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+2i | 0 ≤ ℓ ≤ o(a)/5 − 1}. Note that a3 ̸ ∈ D, and so it must be dominated by one of its neighbours. To avoid double domination, we must have that b−1 a3 ∈ D. However, since bt = ak , we have that b−1 a3 = bt −1 a3−k , which implies that bt −1 a3−k ∈ Dt −1 . Therefore, there must exist an integer i (0 ≤ i ≤ o(a)/5 − 1), such that b−1 a3 = bt −1 a3−k = bt −1 a5i+2(t −1) . This shows that k = 5 − 5i − 2t, where arithmetics is done modulo o(a). As o(a) is divisible by 5, there exists an integer 0 ≤ ℓ ≤ o(a)/5 − 1, such that a5−5i = a5ℓ , and so bt = ak = a5ℓ−2t . Similar argument as above shows that if ba3 ∈ D, then bt = a5ℓ−3t for some integer 0 ≤ ℓ ≤ o(a)/5 − 1. Assume now that (ii) of the above lemma holds. If bt = a5ℓ−2t for some integer 0 ≤ ℓ ≤ o(a)/5 − 1, then let us define D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+2i | 0 ≤ ℓ ≤ o(a)/5 − 1}. It is easy to see that D is a dominating set of Γ . Recall that t = |G|/o(a). As D is a disjoint union of t sets of cardinality o(a)/5, it follows that |D| = |G|/5. This shows that D is efficient. On the other hand, if bt = a5ℓ−3t for some integer 0 ≤ ℓ ≤ o(a)/5 − 1, then let us define D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+3i | 0 ≤ ℓ ≤ o(a)/5 − 1}. It is again easy to see that D is a dominating set of Γ with |D| = |G|/5, and so D is an efficient dominating set.
■
Theorem 4.11. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions, such that ⟨S ⟩ = G. Let t , k be as defined above. Then the following statements are equivalent: (i) Γ = Cay(G; S) admits an efficient dominating set; (ii) o(a) is divisible by 5 and there exists an integer 0 ≤ ℓ ≤ o(a)/5 − 1, such that bt is equal to either a5ℓ−2t or to a5ℓ−3t . Proof. If t = 1, the result follows from Lemma 4.7. If t ≥ 2, the result follows from Lemma 4.10.
■
5. Domination numbers In the previous sections we have characterized the cubic and quartic Cayley graphs on abelian groups which have efficient dominating sets. In this section, we want to talk about domination numbers of these graphs. We are aware that some results below are either folklore or contained in existing literature, but for the sake of self-containment of the paper we gather them in this section. Proposition 5.1. Let G denote a finite abelian group and let S ⊆ G \ {e} be an inverse-closed subset with |S | = 3, such that ⟨S ⟩ = G. Let Γ = Cay(G; S). Then ⎧⌈ |G| ⌉ = M(n) and n ≡ 0 (mod 8), or ⎨ 4 + 1, if G ∼ (3) γ (Γ ) = G∼ GP(n/2, 1) and n ≡ 4 (mod 8), = ⎩⌈ |G| ⌉ , otherwise . 4
C. Çalışkan, Š. Miklavič and S. Özkan / Discrete Applied Mathematics 271 (2019) 15–24
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Proof. As S is inverse closed, S contains either three involutions or one involution. When S contains three involutions, then by Proposition 3.1 Γ is either isomorphic to the complete graph K4 (which is further isomorphic to M(4)), or the 3-cube (which is isomorphic to GP(4, 1)). In both cases Γ admits an efficient |G| dominating set, so γ (Γ ) = 4 . Now assume S contains one involution. Since Γ is cubic, its order must be even. Let us denote |G| = n. Then Γ is either isomorphic to the Möbius ladder on n vertices, or to the generalized Petersen graph GP(n/2, 1) (see Proposition 3.2). Möbius ladder M(n) is also isomorphic to the circulant graph on Zn with connection set {1, −1, n/2}. When n ≡ 4 (mod 8), the result follows from Proposition 3.2. Let us now assume that n ≡ 0 (mod 8), that is, n = 4t for some even positive integer t. Then it is easy to see that D = {4j | j ∈ {0, 1, . . . , t /2}} ∪ {2(2j − 1) | j ∈ {t /2 + 1, . . . , t }} |G|
⌈ |G| ⌉
gives a dominating set of cardinality 4 + 1 = 4 + 1. By Proposition 3.2 we know that M(n) does not admit an efficient |G| dominating set (that is, a dominating set of cardinality 4 ), and so the above dominating set D is minimum. Assume finally that n ≡ 2 (mod 4), that is, n = 4t + 2 for some even positive integer t. Let us consider the subset D = D1 ∪ D2 of Zn , where D1 = {4j | j ∈ {0, 1, . . . , ⌊t /2⌋}},
D2 = {2(2j − 1) + n/2 | j ∈ {1, 2, . . . , ⌈t /2⌉}}.
⌈ |G| ⌉
It is again easy to see that D is a dominating set of M(n) of cardinality 4 , and so also a minimum dominating set by (1). Now assume Γ is isomorphic to GP(n/2, 1). Let {u0 , u1 , . . . , u n −1 } be the vertices of the outer cycle, and {v0 , v1 , . . . , 2 v 2n −1 } be the vertices of the inner cycle of Γ . When n ≡ 0 (mod 8), the result follows from Proposition 3.2. ⌈ |G| ⌉ When n ≡ 2 (mod 8), {u0 , v2 , u4 , . . . , u n −5 , v n −3 , v n −2 } is a dominating set of cardinality 4 (and thus a minimum 2 2 2 dominating set by (1)). ⌈ |G| ⌉ When n ≡ 4 (mod 8), {u0 , v2 , u4 , . . . , u n −2 , v n −1 } is a dominating set of cardinality + 1 = |G4| + 1. By 4 2
2
Proposition 3.2 Γ has no dominating set of cardinality 4 , and so the above dominating set is minimum. ⌈ |G| ⌉ When n ≡ 6 (mod 8), {u0 , v2 , u4 , . . . , u n −3 , v n −1 } is a minimum dominating set of cardinality 4 (and thus a 2 2 minimum dominating set by (1)). ■ |G|
For the quartic case, remember that as S is inverse-closed, S contains either 4, or 2, or none involutions. We first consider the case where S contains four involutions. Proposition 5.2. Let G denote a finite abelian group and let S = {i1 , i2 , i3 , i4 }, where i1 , i2 , i3 , i4 are pairwise different |G| involutions of G, such that ⟨S ⟩ = G. Then the domination number of Γ = Cay(G; S) is 4 . Proof. As mentioned before, since G is generated with four (pairwise distinct) involutions, G is either Z2 × Z2 × Z2 or Z2 × Z2 × Z2 × Z2 , and so Γ has either 8 or 16 vertices. If G is Z2 × Z2 × Z2 , then Cay(G; S) is either K4,4 or the 3-cube plus four edges that creates K4 on parallel faces. It is easy to see both has domination number 2 = 84 . If G is Z2 × Z2 × Z2 × Z2 , . ■ then Cay(G; S) is the 4-cube, which has domination number 4 = 16 4 Now, let us analyse the case where there are two involutions in S. Proposition 5.3. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b ̸ ∈ ⟨a⟩ and c ̸ ∈ ⟨{a, b}⟩ and such that ⟨S ⟩ = G. Then the domination number of Γ = Cay(G; S) |G| is 4 . Proof. In the proof of Theorem 4.6 it is shown that o(a) = n/4 (where n = |G| ) and that Γ is isomorphic to Cn/4 □C4 . By [8, Theorem 2.5], domination number of Γ is 4n . ■ As in the previous proposition, let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that ⟨S ⟩ = G. Let n = |G| and let Γ = Cay(G; S). Without loss of generality we could assume that b ̸ ∈ ⟨a⟩. If c ∈ ⟨a⟩, then it is easy to see that c = an/4 and therefore Γ is isomorphic to M( 2n )□K2 , the cartesian product of the Mobius ladder M(n/2) and K2 . If c ̸ ∈ ⟨a⟩, but c ∈ ⟨{a, b}⟩, then c = ban/4 and Γ is isomorphic to the rose window graph Rn/2 (n/4, 1) (see [16] for the definition of rose window graphs). In both cases we do not have exact results, but computer aided computations suggest that the domination number is n/4 in both cases. Finally, consider the case where S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. As in Section 4.2 let t denote the index of ⟨a⟩ in G, and let k be such that bt = ak . It seems that in this case determining the domination number of Γ = Cay(G; S) is currently beyond our reach. Namely, consider the simplest case k = 0 (and t ≥ 2). Then Γ is isomorphic to the cartesian product Co(a) □Ct . There is no general result for this case in the literature so far (see [4]). But these cases give us further research directions for the future.
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References [1] V.I. Arnautov, Estimation of the exterior stability number of a graph by means of the minimal degree of the vertices, Prikl. Mat. Program. 11 (126) (1974) 3–8 (in Russian). [2] D.W. Bange, A.E. Barkauskas, P.J. Slater, in: R.D. Ringeisen, F.S. Roberts (Eds.), Efficient Dominating Sets in Graphs, in: Applications of Discrete Mathematics, SIAM, Philadelphia, 1988, pp. 189–199. [3] N. Biggs, Algebraic Graph Theory, Cambridge University Press, Cambridge, 1993. [4] S. Crevals, P.R.J. Ostergard, On the domination number of 2-dimensional torus graphs, Util. Math. 106 (2018) 289–300. [5] I.J. Dejter, O. Serra, Efficient dominating sets in Cayley graphs, Discrete Appl. Math. 129 (2003) 319–328. [6] R. Diestel, Graph Theory Graduate Texts in Mathematics, Vol. 173, Springer-Verlag, Heidelberg, 2010. [7] J.B. Fraleigh, First Course in Abstract Algebra, Pearson Education, London, 2002. [8] S. Klavžar, N. Seifter, Dominating cartesian products of cycles, Discrete Appl. Math. 59 (1995) 129–136. [9] J. Lee, Independent perfect domination sets in Cayley graphs, J. Graph Theory 37 (2001) 213–219. [10] D. Marcu, A new upper bound for the domination number of a graph, Quart. J. Math. Oxford Ser. 2 36 (1985) 221–223. [11] D. Meierling, L. Volkmann, Upper bounds for the domination number in graphs of diameter two, Util. Math. 93 (2014) 267–277. [12] M. Mollard, On perfect codes in Cartesian products of graphs, European J. Combin. 32 (2011) 398–403. [13] N. Obradović, J. Peters, G. Ružić, Efficient domination in circulant graphs with two chord lengths, Inform. Process. Lett. 102 (2007) 253–258. [14] V.G. Vizing, A bound on the external stability number of a graph, Dokl. Akad. Nauk SSSR 164 (1965) 729–731. [15] H.B. Walikar, B.D. Acharya, E. Sampathkumar, Recent Developments in the Theory of Domination in Graphs, in: MRI Lecture Notes in Math., vol. 1, Mahta Research Instit., Allahabad, 1979. [16] S. Wilson, Rose windows graphs, Ars Math. Contemp. 1 (2008) 7–19.
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Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
Domination and efficient domination in cubic and quartic Cayley graphs on abelian groups✩ Cafer Çalışkan a , Štefko Miklavič b , Sibel Özkan c ,
∗
a
Antalya Bilim University, Department of Computer Engineering, 07190, Antalya, Turkey University of Primorska, Andrej Marušič Institute, Muzejski trg 2, 6000 Koper, Slovenia c Gebze Technical University, Department of Mathematics, 41400, Gebze, Kocaeli, Turkey b
article
info
Article history: Received 29 November 2018 Received in revised form 16 June 2019 Accepted 7 August 2019 Available online 17 September 2019
a b s t r a c t In this paper, we characterize cubic and quartic Cayley graphs on abelian groups that admit an efficient dominating set, and then we give domination numbers, even when they do not have an efficient dominating set, when it is possible. © 2019 Elsevier B.V. All rights reserved.
Keywords: Domination Efficient domination Perfect code Cayley graph
1. Preliminaries and basic definitions Throughout this paper, Γ = (V , E) will denote a finite simple graph with vertex set V and edge set E, without loops and multiple edges (for more background on graph theory and for notation not defined here we refer the reader to [6]). For every vertex x ∈ V , we will say that x dominates y, if y = x or y is adjacent to x (i.e. x dominates its closed neighbourhood N [x]). A dominating set of Γ is a subset D of V , such that every vertex is dominated by at least one member of D. The domination number γ (Γ ) is the number of vertices in a minimum dominating set for Γ . Let ∆ denote the maximal valency of Γ . Observe that each x ∈ V could dominate at most ∆ + 1 vertices, and so
γ (Γ ) ≥
⌈
⌉ |V | . ∆+1
(1)
This bound was given first in [15]. Since finding the domination number is NP-complete for arbitrary graphs, it is natural to ask for bounds on the domination number related to other graph parameters. There are many other bounds involving order, degree, size and diameter. See [1,10,11], and [14] for example. It is also natural to ask for exact results on specific graph classes. Here we will consider cubic and quartic Cayley graphs on abelian groups. An efficient dominating set of Γ is a subset D of V , such that every vertex of Γ is dominated by exactly one vertex from D. Observe that in this case D is necessarily independent subset of V and that it is also minimum dominating set of Γ . Moreover, if Γ is regular with valency k, then
|D| =
|V | k+1
.
(2)
✩ This research is supported by the bilateral grant of TUBITAK, Turkey (Grant No. 115F586) and ARRS, Slovenia (Grant No. BI-TR/16-18-002). The second author was supported in part by Slovenian Research Agency, program P1-0285 and projects N1-0032, N1-0038, J1-7051, N1-0062. ∗ Corresponding author. E-mail addresses: [email protected] (C. Çalışkan), [email protected] (Š. Miklavič), [email protected] (S. Özkan). https://doi.org/10.1016/j.dam.2019.08.004 0166-218X/© 2019 Elsevier B.V. All rights reserved.
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C. Çalışkan, Š. Miklavič and S. Özkan / Discrete Applied Mathematics 271 (2019) 15–24
While searching for the domination number of a given graph, it is natural to check for an efficient dominating set first, since an efficient dominating set is a minimum dominating set as noted above. But not every graph has an efficient dominating set. Bange et al. [2] have proved that it is an NP-complete problem to determine whether a given graph has an efficient dominating set or not. Let G denote a finite group (for more background on group theory and for notation not defined here we refer the reader to, for example, [7]). We denote the order of G (that is, the number of elements in G) by |G|. In this paper we will be using a multiplicative notation, that is, we denote the identity element of G with e, the inverse element of a ∈ G with a−1 , and the ‘‘product’’ of elements a, b ∈ G with ab. For a subset S ⊆ G we denote by ⟨S ⟩ the subgroup of G, generated by the elements of set S. If S = {a} we abbreviate ⟨S ⟩ = ⟨a⟩. We say that S is inverse-closed subset of G, if x ∈ S implies x−1 ∈ S. Let now S denote an inverse-closed subset of G \ {e}. The Cayley graph Cay(G; S) of the group G with respect to the connection set S is a graph with vertex-set G, in which g ∈ G is adjacent to h ∈ G if and only if h = gs for some s ∈ S. In this case we refer to the edge {g , h} as an s-edge. Observe that Cay(G; S) is regular with valency k = |S |. We say that Cay(G; S) is cubic (quartic, respectively), if |S | = 3 (|S | = 4, respectively). Note also that Cay(G; S) is connected if and only if ⟨S ⟩ = G. More generally, every connected component of Cay(G; S) is isomorphic to the Cayley graph Cay(⟨S ⟩; S). Recall that every Cayley graph is vertex-transitive. Namely, for each pair of vertices u, v ∈ Cay(G; S), there is an automorphism f of Cay(G; S) such that f (u) = v . In particular, if D is a (efficient) dominating set of Cay(G; S), we could always assume that e ∈ D. We refer the reader to [3] for more background on Cayley graphs. Study of (efficient) dominating sets in Cayley graphs is an important and vivid research area. In terms of efficient domination, Lee proved that a Cayley graph on an abelian group has an efficient dominating set if and only if it is a covering graph of a complete graph [9]. Dejter and Serra gave a method for constructing countable families of Cayley graphs each one having an efficient dominating set [5]. Efficient dominating sets in certain Cartesian products, where the factors are Cayley graphs on an abelian group, are shown to exist by Mollard [12]. In [13], cubic and quartic circulants (that is, Cayley graphs on cyclic groups) admitting an efficient dominating set were classified. In this paper we generalize this result to a wider class of Cayley graphs on abelian groups. More precisely, we will classify cubic and quartic Cayley graphs on abelian groups that admit an efficient dominating set. Our main results are Theorems 3.3 and 4.11. We also analyse the domination number and give the exact number when possible in Section 5. For the rest of this paper we assume that G is an abelian group. For a ∈ G let o(a) denote the order of the element a ∈ G, that is, the smallest positive integer m such that am = e. Element a is called an involution, if o(a) = 2. For a subgroup H ≤ G, we denote the index of H in G by [G : H ]. Recall that [G : H ] = |G|/|H |. 2. Structure of Cayley graphs on abelian groups In this section we describe briefly the structure of Cayley graphs on abelian groups. To do this, we introduce the following notation. For a positive integer n, let Kn denote the complete graph on n vertices. If n ≥ 3, then we let Cn denote the cycle of length n. Pick a ∈ G and consider Γ = Cay(G; {a, a−1 }). If a is an involution, then Γ is isomorphic to the disjoint union of |G|/2 copies of K2 . If a is not an involution, then Γ is isomorphic to the disjoint union of t cycles Cn , where n = o(a) and t = [G : ⟨a⟩]. Now let S denote an inverse-closed subset of G \ {e} and pick a ∈ G \ (S ∪ {e}). Set S ′ = S ∪ {a, a−1 }. Observe that Cay(G, S ′ ) is isomorphic to the union of Cay(G; S) and Cay(G; {a, a−1 }) (here, by the union of graphs Γ1 = (V1 , E1 ) and Γ2 = (V2 , E2 ) we mean the graph Γ1 ∪ Γ2 = (V1 ∪ V2 , E1 ∪ E2 )). This observation allows us to recursively ‘‘build’’ our Cayley graph from its ‘‘basic blocks’’, that are either disjoint union of copies of K2 or disjoint union of cycles. Moreover, let m be the smallest positive integer such that am ∈ ⟨S ⟩. If m = o(a), then Cay(G, S ′ ) is isomorphic to the cartesian product of Cay(⟨S ⟩; S) and Cay(⟨a⟩; {a, a−1 }). Pick x ∈ G and a ∈ S, which is not an involution. Note that in this case (x, xa, xa2 , . . . , xao(a)−1 ) is a cycle of length o(a) in Cay(G; S). We refer to this cycle as a-cycle, which contains x. As already mentioned above, Cay(G; S) is connected if and only if ⟨S ⟩ = G, and every connected component of Cay(G; S) is isomorphic to the Cayley graph Cay(⟨S ⟩; S). As G is abelian, ⟨S ⟩ is abelian as well. So when studying properties of Cayley graph Cay(G; S) where G is abelian, then it is reasonable to assume that Cay(G; S) is connected, that is, G = ⟨S ⟩. That is what we assume throughout this paper. Consider now a connected graph Cay(G; S). If there exists a ∈ S such that o(a) = |G|, then G is cyclic. Such graphs, admitting an efficient dominating set, were already classified in [13] for the cubic and the quartic case. But for the sake of completeness of the paper, we treat this case also in our paper. However, note that if we have o(a) < |G| for a non-involution a ∈ S, then Cay(G; {a, a−1 }) is a disjoint union of at least two cycles of length o(a). 3. Cubic Cayley graphs on abelian groups In this section we assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 3, such that ⟨S ⟩ = G. As S is inverse closed, S contains either three involutions or one involution. We consider each of these cases separately.
C. Çalışkan, Š. Miklavič and S. Özkan / Discrete Applied Mathematics 271 (2019) 15–24
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Fig. 1. Efficient domination for K4 and 3-cube.
Fig. 2. Efficient domination for the Möbius ladder on 12 vertices and for the generalized Petersen graph GP(8, 1).
Proposition 3.1. Let G denote a finite abelian group and let S = {i1 , i2 , i3 }, where i1 , i2 , i3 are pairwise different involutions of G. Let Γ = Cay(G; S). Then Γ is either isomorphic to K4 or to the 3-cube. Moreover, both of these two graphs admit an efficient dominating set. Proof. As G is generated by 3 involutions, then for j ̸ = k ̸ = l ∈ {1, 2, 3} either ij ik = il and G is isomorphic to Z2 × Z2 , or ij ik ̸ = il and G is isomorphic to Z2 × Z2 × Z2 . The only cubic graph on 4 vertices is the complete graph K4 (which is of course also Cayley graph on Z2 × Z2 ), and it is easy to see that it admits an efficient dominating set (choose one vertex and it dominates everything, see Fig. 1). It is also easy to see that every connected cubic Cayley graph on Z2 × Z2 × Z2 is isomorphic to the Cayley graph on Z2 × Z2 × Z2 with connection set (1, 0, 0), (0, 1, 0), (0, 0, 1) which is 3-cube, which again admits an efficient dominating set (let D = {(0, 0, 0), (1, 1, 1)} for example, see Fig. 1). ■ Proposition 3.2. Let G denote a finite abelian group and let S = {i1 , a, a−1 }, where i1 is an involution and a is a non-involution. Let Γ = Cay(G; S) and let us denote |G| = n. Then either Γ is isomorphic to the Möbius ladder M(n) on n vertices, or to the generalized Petersen graph GP(n/2, 1). Moreover, the Möbius ladder on n vertices admits an efficient dominating set if and only if n = 8k + 4 for some non-negative integer k, and GP(n/2, 1) admits an efficient dominating set if and only if n = 8k |G| for some non-negative integer k. In both cases, γ (Γ ) = 4 . Proof. Assume first that o(a) = n. Observe that in this case i1 = an/2 , and so Γ is isomorphic to the Möbius ladder on n vertices (which is also isomorphic to the Cayley graph on Zn with the connection set {1, −1, 2n }). It is now easy to see that Γ admits an efficient dominating set if and only if n = 8k + 4 for some non-negative integer k (let D = {0, 4, . . . , 8k} if we label the vertex set of Γ with elements of Z8k+4 ). Assume now that o(a) < n. Involution i1 generates n/2 copies of K2 , so we should have two cycles of length n/2 and so we have o(a) = n/2 (see the comments in Section 2 for details), and so Γ is isomorphic to the cartesian product Cn/2 □K2 , that is, Γ is isomorphic to a generalized Petersen graph GP(n/2, 1), also the so-called prism graph (see Fig. 2). It is again easy to see that Γ admits an efficient dominating set if and only if n = 8k for some non-negative integer k (let (a, 0) and (a + 2, 1) be in D for a ∈ {0, 4, 8, . . . , 4k − 4} if we label the vertex of Γ with Z4k × Z2 ). ■ Taking into account that the 3-cube is isomorphic to the generalized Petersen graph GP(4, 1) and that the complete graph K4 is isomorphic to the Möbius ladder on 4 vertices, the following theorem is an immediate consequence of Propositions 3.1 and 3.2. Theorem 3.3. Let G denote a finite abelian group and let S denote an inverse closed subgroup of G \ {e} with |S | = 3, such that ⟨S ⟩ = G. Then Γ = Cay(G; S) admits an efficient dominating set if and only if Γ is isomorphic to the generalized Petersen graph GP(4k, 1) for some positive integer k, or to the Möbius ladder on 8k + 4 vertices, for some non-negative integer k. 4. Quartic Cayley graphs on abelian groups In this section we will classify quartic Cayley graphs on abelian groups, which admit an efficient dominating set. Assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 4, such that ⟨S ⟩ = G. We consider quartic Cayley graph
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Γ = Cay(G; S). Let us denote |G| = n. Since Γ is 4-regular, by (2) any efficient dominating set of Γ (if one exists) has n/5 elements. So, we will assume that n is divisible by 5. Note also that as S is inverse-closed, S contains either 4, or 2, or none involutions. We first consider the case where S contains four involutions. Proposition 4.1. Let G denote a finite abelian group and let S = {i1 , i2 , i3 , i4 }, where i1 , i2 , i3 , i4 are pairwise different involutions of G. Then Γ = Cay(G; S) does not admit any efficient dominating set. Proof. Observe that since G is generated with four (pairwise distinct) involutions, G is either Z2 × Z2 × Z2 , or Z2 × Z2 × Z2 × Z2 . It follows that Cay(G; S) has either 8 or 16 vertices. As neither of these two is divisible by 5, Cay(G; S) does not admit any efficient dominating set. ■ In view of Proposition 4.1, let us assume now that there exists a non-involution a ∈ S. Therefore, S = {a, a−1 , b, c } for some b, c ∈ G. Note that if b is non-involution, then c = b−1 . Assume now that D is an efficient domination set of Γ . Pick any x ∈ G and consider the a-cycle which contains x. We will show that any two vertices of this cycle, that are contained in D, are at least 5 places apart on this cycle. More precisely, we show that if xai , xaj ∈ D with i ̸ = j, then min{i − j, j − i} ≥ 5, where i − j and j − i are computed modulo o(a). As Γ is vertex transitive, we could assume that x = 1 and i = 0. Proposition 4.2. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution. Let D denote an efficient dominating set for Γ = Cay(G; S) and assume without loss of generality that e ∈ D. Then the following statements hold. (i) o(a) ≥ 5. (ii) Assume that ai ∈ D for some 1 ≤ i ≤ o(a) − 1. Then 5 ≤ i ≤ o(a) − 5. Proof. (i) Assume 3 ≤ o(a) ≤ 4. Since |D| = n/5, there exists at least one a-cycle which has empty intersection with D. Assume first that o(a) = 4 and let C = (x, xa, xa2 , xa3 ) be the a-cycle with the empty intersection with D. It follows that vertices x, xa, xa2 , xa3 must be dominated by their neighbours outside C , that is, with vertices xb, xab, xa2 b, xa3 b or xc , xac , xa2 c , xa3 c. Without loss of generality, we could assume that x is dominated by xb. Now to avoid double domination, xa must be dominated by xac. But now we could not efficiently dominate xa2 . If we dominate xa2 with xa2 b, then xab is doubly dominated, and if we dominate xa2 with xa2 c, then xac is doubly dominated. This shows that o(a) ̸ = 4. Similarly we show that o(a) ̸ = 3, but the details are left to the reader. (ii) If o(a) = n then G is a cyclic group generated by a, and the result follows from the proof of [13, Lemma 1]. Therefore, assume that o(a) < n. As ⟨S ⟩ = G, this implies that not both of b, c are contained in ⟨a⟩. Without loss of generality we could assume that b ̸ ∈ ⟨a⟩. Observe that we could assume without loss of generality that i ≤ o(a)/2 (otherwise we replace a with a−1 ). It is easy to see that i ̸ = 1, since in this case e is doubly dominated (by itself and by a). Similarly, i ̸ = 2, since otherwise a is doubly dominated (by e and by a2 ). Assume now that i = 3 and consider vertex ba ̸ ∈ ⟨a⟩. If ba is dominated by e, then c = ba. It follows that all neighbours of ba2 (that is, vertices a, a2 , ba, ba3 ) are already dominated, and so we could not efficiently dominate ba2 . If ba is dominated by a3 , then b = ca2 and all neighbours of ba2 (this time vertices a2 , a4 , ba, ba3 ) are already dominated, and so we could not efficiently dominate ba2 . Similarly, if ba2 is dominated by e, then c = ba2 . It follows that all neighbours of ba (vertices a, an−1 , b, ba2 ) are already dominated, and so we could not efficiently dominate ba. Again, if ba2 is dominated by a3 , then b = ca and all neighbours of ba (vertices a, a2 , b, ba2 ) are already dominated, and so we could not efficiently dominate ba. These arguments show that ba, ba2 could not be dominated by e or a3 . It follows that ba must be dominated by c −1 ba and ba2 must be dominated by c −1 ba2 . But c −1 ba and c −1 ba2 are adjacent in Γ , a contradiction. Assume now that i = 4 and consider ba2 ̸ ∈ ⟨a⟩. If ba2 is dominated by e, then c = ba2 and all neighbours of a2 (that is, vertices a, a3 , ba2 , ba4 ) are already dominated, and so we could not efficiently dominate a2 . Similarly, if ba2 is dominated by a4 , then again a2 could not be efficiently dominated since all of its neighbours (namely a, a3 , b, ba2 ) are already dominated. Note also that ba2 could not be dominated by its neighbours a2 , ba, ba3 , since in each of these three cases a double domination occurs. Therefore, ba2 is dominated by c −1 ba2 . But now ba has to be dominated by c −1 ba, which is adjacent to c −1 ba2 , a contradiction. This shows that i ≥ 5. ■ Corollary 4.3. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution such that Γ = Cay(G; S) admits an efficient dominating set D. Then the following statements hold. (i) For x ∈ D, the intersection of D and the a-cycle which contains x is nonempty and consists precisely of vertices {xa5i | 0 ≤ i ≤ o(a)/5 − 1}. (ii) o(a) is divisible by 5. Proof. (i) By Proposition 4.2(ii), any two vertices of D, which belong to the a-cycle which contains x, are at least 5 positions apart on this a-cycle. This shows that the intersection of D with any a-cycle contains at most o(a)/5 vertices, and so D contains at most |G|/5 vertices. On the other hand, D = |G|/5 by (2), and so the intersection of D with any a-cycle contains exactly every fifth vertex of this cycle. As x is an element of D, the result follows.
C. Çalışkan, Š. Miklavič and S. Özkan / Discrete Applied Mathematics 271 (2019) 15–24
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(ii) Recall that the intersection of D with any a-cycle contains exactly every fifth vertex of this cycle. In addition, any two vertices of D, which belong to the same a-cycle, are at least 5 positions apart on this a-cycle. This shows that the length of any a-cycle is divisible by 5. As the length of every a-cycle is equal to o(a), the result follows. ■ 4.1. Quartic Cayley graphs on abelian groups — two involutions Assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 4, such that ⟨S ⟩ = G. In view of Proposition 4.1, let us assume now that there exists a non-involution a ∈ S. Therefore, S = {a, a−1 , b, c } for some b, c ∈ G. We first consider the case where b, c are (distinct) involutions. If o(a) = n, then G is a cyclic group. But any cyclic group contains at most one involution, contradicting our assumption that b, c are distinct involutions of G. Therefore, o(a) < n. As ⟨S ⟩ = G, at least one of b, c is not contained in ⟨a⟩. Without loss of generality we could assume that b ̸ ∈ ⟨a⟩. We have three different cases to consider: case c ∈ ⟨a⟩, case c ̸ ∈ ⟨a⟩ but c ∈ ⟨{a, b}⟩, and case c ̸ ∈ ⟨{a, b}⟩. Theorem 4.4. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b ̸ ∈ ⟨a⟩ and c ∈ ⟨a⟩. Then Γ = Cay(G; S) does not admit an efficient dominating set. Proof. We first show that o(a) = n/2. Recall that o(a) is a divisor of |G| = n, and so b ̸ ∈ ⟨a⟩ implies o(a) ≤ n/2. However, as G is generated by a and b and as o(b) = 2, it follows that o(a) = n/2. This implies that c = an/4 (recall that c is an involution). Now assume that D is an efficient dominating set of Γ . By Corollary 4.3, we know that o(a) is divisible by 5 and we could also assume that the intersection of the a-cycle which contains e and D is {e, a5 , . . . , ao(a)−5 }. But as n is divisible by 5, n/4 is also divisible by 5 and this implies that an/4 = c ∈ D. However, this is not possible, since c is already dominated by e, a contradiction. This finishes the proof. ■ Theorem 4.5. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b, c ̸ ∈ ⟨a⟩ and c ∈ ⟨{a, b}⟩. Then Γ = Cay(G; S) does not admit an efficient dominating set. Proof. As G is generated by a and b and as o(a) < n, it follows that o(a) = n/2. Since c ̸ ∈ ⟨a⟩ and b is an involution, we have that c is contained in the a-cycle, that contains b. The only involutions contained in this a-cycle are b and ban/4 , so c = ban/4 . Now assume that D is an efficient dominating set of Γ . By Corollary 4.3 we know that o(a) is divisible by 5 and we could also assume that the intersection of the a-cycle which contains e and D is {e, a5 , . . . , ao(a)−5 }. Consider now the vertex a2 and note that it could be only dominated by ba2 or ba2+n/4 . If ba2 ∈ D, then ba4 could only be dominated by a4+n/4 , which forces a4+n/4 ∈ D and so 4 + n/4 is divisible by 5, a contradiction. Therefore, ba2+n/4 ∈ D. But now a3 has to be dominated by ba3 , and so ba3 ∈ D. But now to dominate ba we must have a1+n/4 ∈ D, which forces that 1 + n/4 is divisible by 5, a contradiction. This finishes the proof. ■ Theorem 4.6. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b ̸ ∈ ⟨a⟩ and c ̸ ∈ ⟨{a, b}⟩. Then Γ = Cay(G; S) does not admit an efficient dominating set. Proof. We first show that o(a) = n/4. Recall that o(a) is a divisor of |G| = n, and so b ̸ ∈ ⟨a⟩ together with c ̸ ∈ ⟨{a, b}⟩ imply that o(a) ≤ n/4. However, as G is generated by a, b and c and as o(b) = o(c) = 2, it follows that o(a) = n/4. As G is generated by S and as c ̸ ∈ ⟨{a, b}⟩, it follows that o(a) = n/4. By comments in Section 2 it follows that Γ is isomorphic to the cartesian product Cn/4 □K2 □K2 , which is further isomorphic to Cn/4 □C4 . Therefore, Γ is isomorphic to the Cayley graph on the group Zn/4 × Z4 , with the connection set S ′ = {(1, 0), (n/4 − 1, 0), (0, 1), (0, 3)}. But element (0, 1) is a non-involutory element of Zn/4 × Z4 with order 4. By Corollary 4.3(iii), Γ does not admit an efficient domination set. ■ 4.2. Quartic Cayley graphs on abelian groups — no involutions Assume that S ⊆ G \ {e} is an inverse-closed subset with |S | = 4, such that ⟨S ⟩ = G. In view of Proposition 4.1 and Section 4.1, in this subsection we consider the case where S = {a, a−1 , b, b−1 } for some non-involutions a, b ∈ G. Recall that o(a) denotes the order of element a in G. Let us denote by t the index of ⟨a⟩ in G, that is,
|G| |G| = . |⟨a⟩| o(a) We consider cases t = 1 and t ≥ 2. The case t = 1 was already solved in [13], but for the sake of completeness of the t=
paper, we include this case here as well. Lemma 4.7. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t be as defined above and assume that t = 1. Then the following statements are equivalent: (i) Γ = Cay(G; S) admits an efficient dominating set; (ii) o(a) is divisible by 5 and there exists an integer ℓ, (0 ≤ ℓ ≤ o(a)/5 − 1), such that either b or b−1 is equal to a5ℓ−2 .
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Proof. Observe that t = 1 implies that the elements of G are exactly the elements of the form ai (0 ≤ i ≤ o(a) − 1). Assume first that Γ admits an efficient dominating set. Then we could assume that e ∈ D, and by Corollary 4.3 we have that o(a) is divisible by 5 and D = {a5i | 0 ≤ i ≤ o(a)/5 − 1}. Note that a2 ̸ ∈ D, and that the neighbours of a2 in Γ are a, a3 , a2 b and a2 b−1 . As a, a3 ̸ ∈ D, it follows that a2 b ∈ D or a2 b−1 ∈ D. Therefore, there exists ℓ (0 ≤ ℓ ≤ o(a)/5 − 1), such that either a2 b = a5ℓ or a2 b−1 = a5ℓ . It follows that either b or b−1 is equal to a5ℓ−2 . Assume now that o(a) is divisible by 5 and there exists an integer ℓ, (0 ≤ ℓ ≤ o(a)/5 − 1), such that either b or b−1 is equal to a5ℓ−2 . Let us define D = {a5i | 0 ≤ i ≤ o(a)/5 − 1}. We now show that D is an efficient dominating set. We show this for the case when b = a5ℓ−2 for some 0 ≤ ℓ ≤ o(a)/5 − 1. The proof in the case when b−1 = a5ℓ−2 for some 0 ≤ ℓ ≤ o(a)/5 − 1 is similar. We first show that every vertex of Γ is dominated by at least one vertex from D. This is clear for vertices which are contained in D. Similarly, vertices of the form a5i+1 (a5i+4 , respectively) for (0 ≤ i ≤ o(a)/5 − 1) are adjacent to vertices a5i (a5(i+1) , respectively), and are therefore dominated by D. Consider now vertices of the form a5i+2 (0 ≤ i ≤ o(a)/5 − 1). As b = a5ℓ−2 , every such vertex is adjacent to a5i+2 b = a5i+2 a5ℓ−2 = a5(i−ℓ) ∈ D. Similarly, every vertex of the form a5i+3 (0 ≤ i ≤ o(a)/5 − 1) is adjacent to a5i+3 b−1 = a5i+3 a2−5ℓ = a5(i+1−ℓ) ∈ D. This shows that D is a dominating set. Finally, observe that
|D| =
o(a)
=
|G|
,
5 5 and so D is an efficient dominating set. This finishes the proof.
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Assume now that t ≥ 2. As ⟨S ⟩ = G, we have that b ̸ ∈ ⟨a⟩ and that bi ⟨a⟩ (0 ≤ i ≤ t − 1) are distinct cosets of ⟨a⟩. Consequently, bt ∈ ⟨a⟩. Let k (0 ≤ k ≤ o(a) − 1) be such that bt = ak . Proposition 4.8. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t , k be as defined above and assume that t ≥ 2. Assume that D is an efficient dominating set of Γ = Cay(G; S) and assume that bi aj ∈ D for some 0 ≤ i ≤ t − 1, 0 ≤ j ≤ o(a) − 1. Then exactly one of the following statements hold: (i) {bi+1 aj+2 , bi−1 aj+3 } ⊆ D; (ii) {bi−1 aj+2 , bi+1 aj+3 } ⊆ D. Proof. By Corollary 4.3 we have that the intersection of D and the a-cycle which contains bi aj is equal to {bi a5ℓ+j | 0 ≤ ℓ ≤ o(a)/5 − 1}. This shows that bi aj+2 must be dominated by one of its neighbours different than bi aj+1 and bi aj+3 , that
is by bi+1 aj+2 or by bi−1 aj+2 . Therefore, either bi+1 aj+2 or bi−1 aj+2 is contained in D. Note that if both bi+1 aj+2 and bi−1 aj+2 are contained in D, then bi aj+2 is doubly dominated, a contradiction. Assume now that bi+1 aj+2 ∈ D. Observe that in this case the only way to (efficiently) dominate bi aj+3 is such that i−1 j+3 b a ∈ D. Similarly, if bi−1 aj+2 ∈ D, then, to efficiently dominate bi aj+3 , we have that bi+1 aj+3 ∈ D. ■ Proposition 4.9. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t , k be as defined above and assume t ≥ 2. Assume that D is efficient dominating set of Γ = Cay(G; S) such that e ∈ D. Then the following statements holds: (i) If ba2 ∈ D, then D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+2i | 0 ≤ ℓ ≤ o(a)/5 − 1}. (ii) If ba3 ∈ D, then D=
t −1 ⋃ i=0
Di ,
C. Çalışkan, Š. Miklavič and S. Özkan / Discrete Applied Mathematics 271 (2019) 15–24
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Fig. 3. The efficient dominating set D from Proposition 4.9(i) in the case o(a) = 15 and t ≥ 4.
where Di = {bi a5ℓ+3i | 0 ≤ ℓ ≤ o(a)/5 − 1}. Proof. We prove part (i) of the above proposition. Proof of part (ii) is similar. We first show by induction on i that Di ⊆ D for 0 ≤ i ≤ t − 1. Since e ∈ D, it follows from Corollary 4.3(ii) that D0 ⊆ D. Similarly, since ba2 ∈ D, it follows from Corollary 4.3(ii) that D1 ⊆ D. For t = 2, we have D0 ∪ D1 ⊆ D. For t ≥ 3, assume that Dj ⊆ D for 0 ≤ j ≤ i for some 1 ≤ i ≤ t − 2. Note that by our induction hypothesis we have that bi a2i , bi−1 a5+2(i−1) = bi−1 a2i+3 ∈ D. Taking j = 2i in Proposition 4.8, we find that bi+1 a2(i+1) ∈ D. But now it follows from Corollary 4.3(ii) that Di+1 ⊆ D. This shows that t −1 ⋃
Di ⊆ D.
i=0
However, Di (0 ≤ i ≤ t − 1) are pairwise disjoint sets with |Di | = o(a)/5, and so
|
t −1 ⋃
Di | = t
i=0
o(a) 5
=
|G| o(a) o(a) 5
=
|G| 5
= |D|.
It follows that t −1 ⋃
Di = D.
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i=0
The efficient dominating set D described in Proposition 4.9(i) is depicted in Fig. 3 (for the case o(a) = 15 and t ≥ 4). The vertices of the top 15-cycle are the vertices of the a-cycle containing e, the vertices of the second 15-cycle are the vertices of the a-cycle containing b, and so on. Lemma 4.10. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. Let t , k be as defined above and assume that t ≥ 2. Then the following statements are equivalent: (i) Γ = Cay(G; S) admits an efficient dominating set; (ii) o(a) is divisible by 5 and there exists an integer 0 ≤ ℓ ≤ o(a)/5 − 1, such that bt is equal to either a5ℓ−2t or to a5ℓ−3t . Proof. Assume first that Γ admits an efficient dominating set. Without loss of generality we could assume that e ∈ D. By Corollary 4.3(iii) we have that o(a) is divisible by 5. By Proposition 4.8 either ba2 ∈ D or ba3 ∈ D. If ba2 ∈ D, then by
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Proposition 4.9(i) we have that D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+2i | 0 ≤ ℓ ≤ o(a)/5 − 1}. Note that a3 ̸ ∈ D, and so it must be dominated by one of its neighbours. To avoid double domination, we must have that b−1 a3 ∈ D. However, since bt = ak , we have that b−1 a3 = bt −1 a3−k , which implies that bt −1 a3−k ∈ Dt −1 . Therefore, there must exist an integer i (0 ≤ i ≤ o(a)/5 − 1), such that b−1 a3 = bt −1 a3−k = bt −1 a5i+2(t −1) . This shows that k = 5 − 5i − 2t, where arithmetics is done modulo o(a). As o(a) is divisible by 5, there exists an integer 0 ≤ ℓ ≤ o(a)/5 − 1, such that a5−5i = a5ℓ , and so bt = ak = a5ℓ−2t . Similar argument as above shows that if ba3 ∈ D, then bt = a5ℓ−3t for some integer 0 ≤ ℓ ≤ o(a)/5 − 1. Assume now that (ii) of the above lemma holds. If bt = a5ℓ−2t for some integer 0 ≤ ℓ ≤ o(a)/5 − 1, then let us define D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+2i | 0 ≤ ℓ ≤ o(a)/5 − 1}. It is easy to see that D is a dominating set of Γ . Recall that t = |G|/o(a). As D is a disjoint union of t sets of cardinality o(a)/5, it follows that |D| = |G|/5. This shows that D is efficient. On the other hand, if bt = a5ℓ−3t for some integer 0 ≤ ℓ ≤ o(a)/5 − 1, then let us define D=
t −1 ⋃
Di ,
i=0
where Di = {bi a5ℓ+3i | 0 ≤ ℓ ≤ o(a)/5 − 1}. It is again easy to see that D is a dominating set of Γ with |D| = |G|/5, and so D is an efficient dominating set.
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Theorem 4.11. Let G denote a finite abelian group and let S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions, such that ⟨S ⟩ = G. Let t , k be as defined above. Then the following statements are equivalent: (i) Γ = Cay(G; S) admits an efficient dominating set; (ii) o(a) is divisible by 5 and there exists an integer 0 ≤ ℓ ≤ o(a)/5 − 1, such that bt is equal to either a5ℓ−2t or to a5ℓ−3t . Proof. If t = 1, the result follows from Lemma 4.7. If t ≥ 2, the result follows from Lemma 4.10.
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5. Domination numbers In the previous sections we have characterized the cubic and quartic Cayley graphs on abelian groups which have efficient dominating sets. In this section, we want to talk about domination numbers of these graphs. We are aware that some results below are either folklore or contained in existing literature, but for the sake of self-containment of the paper we gather them in this section. Proposition 5.1. Let G denote a finite abelian group and let S ⊆ G \ {e} be an inverse-closed subset with |S | = 3, such that ⟨S ⟩ = G. Let Γ = Cay(G; S). Then ⎧⌈ |G| ⌉ = M(n) and n ≡ 0 (mod 8), or ⎨ 4 + 1, if G ∼ (3) γ (Γ ) = G∼ GP(n/2, 1) and n ≡ 4 (mod 8), = ⎩⌈ |G| ⌉ , otherwise . 4
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Proof. As S is inverse closed, S contains either three involutions or one involution. When S contains three involutions, then by Proposition 3.1 Γ is either isomorphic to the complete graph K4 (which is further isomorphic to M(4)), or the 3-cube (which is isomorphic to GP(4, 1)). In both cases Γ admits an efficient |G| dominating set, so γ (Γ ) = 4 . Now assume S contains one involution. Since Γ is cubic, its order must be even. Let us denote |G| = n. Then Γ is either isomorphic to the Möbius ladder on n vertices, or to the generalized Petersen graph GP(n/2, 1) (see Proposition 3.2). Möbius ladder M(n) is also isomorphic to the circulant graph on Zn with connection set {1, −1, n/2}. When n ≡ 4 (mod 8), the result follows from Proposition 3.2. Let us now assume that n ≡ 0 (mod 8), that is, n = 4t for some even positive integer t. Then it is easy to see that D = {4j | j ∈ {0, 1, . . . , t /2}} ∪ {2(2j − 1) | j ∈ {t /2 + 1, . . . , t }} |G|
⌈ |G| ⌉
gives a dominating set of cardinality 4 + 1 = 4 + 1. By Proposition 3.2 we know that M(n) does not admit an efficient |G| dominating set (that is, a dominating set of cardinality 4 ), and so the above dominating set D is minimum. Assume finally that n ≡ 2 (mod 4), that is, n = 4t + 2 for some even positive integer t. Let us consider the subset D = D1 ∪ D2 of Zn , where D1 = {4j | j ∈ {0, 1, . . . , ⌊t /2⌋}},
D2 = {2(2j − 1) + n/2 | j ∈ {1, 2, . . . , ⌈t /2⌉}}.
⌈ |G| ⌉
It is again easy to see that D is a dominating set of M(n) of cardinality 4 , and so also a minimum dominating set by (1). Now assume Γ is isomorphic to GP(n/2, 1). Let {u0 , u1 , . . . , u n −1 } be the vertices of the outer cycle, and {v0 , v1 , . . . , 2 v 2n −1 } be the vertices of the inner cycle of Γ . When n ≡ 0 (mod 8), the result follows from Proposition 3.2. ⌈ |G| ⌉ When n ≡ 2 (mod 8), {u0 , v2 , u4 , . . . , u n −5 , v n −3 , v n −2 } is a dominating set of cardinality 4 (and thus a minimum 2 2 2 dominating set by (1)). ⌈ |G| ⌉ When n ≡ 4 (mod 8), {u0 , v2 , u4 , . . . , u n −2 , v n −1 } is a dominating set of cardinality + 1 = |G4| + 1. By 4 2
2
Proposition 3.2 Γ has no dominating set of cardinality 4 , and so the above dominating set is minimum. ⌈ |G| ⌉ When n ≡ 6 (mod 8), {u0 , v2 , u4 , . . . , u n −3 , v n −1 } is a minimum dominating set of cardinality 4 (and thus a 2 2 minimum dominating set by (1)). ■ |G|
For the quartic case, remember that as S is inverse-closed, S contains either 4, or 2, or none involutions. We first consider the case where S contains four involutions. Proposition 5.2. Let G denote a finite abelian group and let S = {i1 , i2 , i3 , i4 }, where i1 , i2 , i3 , i4 are pairwise different |G| involutions of G, such that ⟨S ⟩ = G. Then the domination number of Γ = Cay(G; S) is 4 . Proof. As mentioned before, since G is generated with four (pairwise distinct) involutions, G is either Z2 × Z2 × Z2 or Z2 × Z2 × Z2 × Z2 , and so Γ has either 8 or 16 vertices. If G is Z2 × Z2 × Z2 , then Cay(G; S) is either K4,4 or the 3-cube plus four edges that creates K4 on parallel faces. It is easy to see both has domination number 2 = 84 . If G is Z2 × Z2 × Z2 × Z2 , . ■ then Cay(G; S) is the 4-cube, which has domination number 4 = 16 4 Now, let us analyse the case where there are two involutions in S. Proposition 5.3. Let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that b ̸ ∈ ⟨a⟩ and c ̸ ∈ ⟨{a, b}⟩ and such that ⟨S ⟩ = G. Then the domination number of Γ = Cay(G; S) |G| is 4 . Proof. In the proof of Theorem 4.6 it is shown that o(a) = n/4 (where n = |G| ) and that Γ is isomorphic to Cn/4 □C4 . By [8, Theorem 2.5], domination number of Γ is 4n . ■ As in the previous proposition, let G denote a finite abelian group and let S = {a, a−1 , b, c }, where a is a non-involution, and b, c are distinct involutions, such that ⟨S ⟩ = G. Let n = |G| and let Γ = Cay(G; S). Without loss of generality we could assume that b ̸ ∈ ⟨a⟩. If c ∈ ⟨a⟩, then it is easy to see that c = an/4 and therefore Γ is isomorphic to M( 2n )□K2 , the cartesian product of the Mobius ladder M(n/2) and K2 . If c ̸ ∈ ⟨a⟩, but c ∈ ⟨{a, b}⟩, then c = ban/4 and Γ is isomorphic to the rose window graph Rn/2 (n/4, 1) (see [16] for the definition of rose window graphs). In both cases we do not have exact results, but computer aided computations suggest that the domination number is n/4 in both cases. Finally, consider the case where S = {a, a−1 , b, b−1 }, where a, b are distinct non-involutions. As in Section 4.2 let t denote the index of ⟨a⟩ in G, and let k be such that bt = ak . It seems that in this case determining the domination number of Γ = Cay(G; S) is currently beyond our reach. Namely, consider the simplest case k = 0 (and t ≥ 2). Then Γ is isomorphic to the cartesian product Co(a) □Ct . There is no general result for this case in the literature so far (see [4]). But these cases give us further research directions for the future.
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