Hamiltonicity in power graphs of a class of abelian groups

The undirected power graph G(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathscr {G}(S)$$\end{document} of a semigroup S is an undirected graph with vertex set S where two vertices u,v∈S\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$u,v\in S$$\end{document} are adjacent if and only if there is a positive integer m such that um=v\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$u^{m}=v$$\end{document} or vm=u\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$v^{m}=u$$\end{document}. Here, we investigate the power graphs of a class of abelian groups and answer, in this case, the question whether or not the power graph is Hamiltonian.

other hand, G (G) is not Hamiltonian for Abelian groups of order 2 m with m > 1 that are not cyclic [4]. Trivially, G (G) is Hamiltonian if it is a complete graph and |G| ≥ 3. Therefore, the following characterization of power graphs that are complete is useful and interesting: Proposition 2 [4] Let G be a finite group. Then, G (G) is complete if and only if G is a cyclic group of order 1 or p m , for some prime number p and for some m ∈ N.
Characterizations of Hamiltonian power graphs for groups of the form G = (Z p ) n ⊕ (Z q ) m with distinct primes p and q and m, n ∈ N can be found in [6].

Power graphs of Abelian groups
We denote the identity element of G by 0 since we are interested here in abelian groups. Furthermore, we write x for the (abelian) subgroup generated by a single element x ∈ G. It is easy to see that the identity 0 is adjacent to all other group elements in G (G). The subgraph of G (G) obtained by removing 0 and its incident edges will be denoted by G * (G), and we write G x = G ( x ). We denote the greatest common divisor of r natural numbers q 1 , q 2 ,…,q r by [q 1 , q 2 , . . . , q r ]. Furthermore, we write q 2 |q 1 if q 2 divides q 1 .
Naturally, the investigation of the power graphs of Abelian groups will make use of the Fundamental Theorem for Abelian Groups, see, e.g. [8]: Theorem 1 Every finite Abelian group G is the direct product G = Z q 1 ⊕ Z q 2 ⊕ · · · ⊕ Z q of cyclic groups whose sizes q j are powers of prime numbers. Up to reordering, the number of factors and the powers q j are uniquely determined by G.
We will also make use of the following well-known consequence of the Chinese remainder theorem: Lemma 1 For j , k ∈ N, Z jk Z j ⊕ Z k if and only if j and k are coprime.
The fundamental theorem for abelian groups implies some simple sufficient conditions for the existence of Hamiltonian cycles: The Invariant factor decomposition allows us to write G in the form G = Z n 1 ⊕ · · · ⊕ Z n j 1 (1) where j 1 and n i are the unique integers satisfying j 1 ≥ 1, n i ≥ 2 and n i |n i+1 for 1 ≤ i ≤ j 1 − 1. The n i 's in the last equation are called the invariant factor of G. For later reference, we note that for j 1 where p i is prime and k i ≥ k i for i ∈ {1, . . . , s, s + 1, . . . , l}.
The following result, by [9], will be useful to identify paths in power graphs: Theorem 2 [9] The system of congruences a i x ≡ b i mod m i , with m i = 0 and i ∈ {1, . . . , r } admits a solution if and only if: The next lemma provides key adjacencies that will be part on the main structure of the Hamiltonian cycle in the power graphs of some abelian groups. Let us consider groups of the form where the p i are prime numbers and pairwise coprime. For G, we define the sets , and for a fixed x ∈ Z p k 1 First we note that for G, Lemma 3 specializes to and (x, y) ∈ G, then the subgraph induced by Proof For each 2 ≤ j ≤ l and for every (x, y ) ∈ N i x (y), we define Furthermore, we define the following subsets where 0 ≤ r j ≤ k j for 2 ≤ r ≤ l. First, we note that k 1 (y) = k 1 (y ) for all (x, y ) ∈ N i x (y). To prove this, let p y ∈ N such that [ p y , p 1 , p 2 , . . . , p l ] = 1. We compute y = y + mp , p 1 ] = 1. Therefore, using the last note and Corollary 1, we can conclude that the subgraph induced by each of the sets V q is complete and, of course, Hamiltonian. Furthermore, it is easy to see that the sets V q are pairwise disjoint and their disjoint union equals N i x (y). Claim: If s|r , then each vertex in V s is adjacent to all the vertices in V r .
To prove this claim, consider (x, y 1 ) ∈ V s and (x, y 2 ) ∈ V r with y 1 = p y 1 p Using Lemma 3, we observer that (x, y 1 ) and (x, y 2 ) are adjacent if the following system of linear congruences has solution: The following two conditions are clearly sufficient for the existence of a solution: Since the two conditions are clearly satisfied, the claim is true.
It is important to note that each element in V 1 is adjacent to any other element in N i x (y). Any Hamiltonian cycle in the subgraph within V s can be extended to a Hamiltonian cycle in V r ∪ V s in the following way: Let C s and C r be the Hamiltonian cycles in the subgraphs induced by V r and V s , respectively, and let e s = (u s , v s ) ∈ E(C s ) and e r = (u r , v r ) ∈ E(C r ) edges in each of Hamiltonian cycles. If we remove the two edges in each cycle, we get two Hamiltonian trajectories in the subgraphs induced by V s and V r , respectively. We denote these by T V s and T V r . By the claim, there are edges (v s , v r ) and (u s , u r ) in E(G (G)), so a Hamiltonian cycle in the subgraph induced by V r ∪ V s is: This construction is sketched in Fig. 1.
In order to construct a Hamiltonian cycle in the subgraph induced by N i x (y), we can arrange the different sets V q as a tree. Each of the vertices in the tree is the Hamiltonian sets V q , its root is V 1 , and the edges in the tree are the extension of the Hamiltonian cycles that we show in Fig. 1. The tree has l branches, and the first element in the branches adjacent to V 1 is the set V p i , where 2 ≤ i ≤ l. If we take into account the natural partial order carried by a tree, when we are in the branch with initial vertex V p i , then an element in this branch V is adjacent by the claim to the vertex V p s−1 ; therefore, it can serve as extension that connect these two vertices. Constructing the Hamiltonian cycle in N i x (y) therefore follows a depth-first traversal of the tree. The Hamiltonian trajectories T V j in V j with starting point u j and end point v j ∈ V j therefore can be combined to a Hamiltonian cycles in the following manner: Since the T V p l−1 traverse different vertex sets, no edge appears twice. Furthermore, W covers all vertices of the V j , and no inner vertex of W appears more than once. Thus, W is a Hamiltonian cycle in N i x (y). The Hamiltonian graph G N i x (y) contains all vertices (x, y), with a fixed x and a fixed k 1 (y). The next step is to extend the construction of these Hamiltonian cycles to a single Hamiltonian cycle on H i . To this end, we will need the following technical result.
We know that In order to prove that (x , y 1 ) and (x, y) are adjacent, we check whether the following system of linear congruences has a solution: The following two conditions are clearly sufficient for the existence of a solution: , and the second condition holds.
We can easily note that there is a path P x = (x, y)(x 1 , y 1 ) . . . (x |S i | , y |S i | ) such that (x r , y r ) = (n r 1 x, n r 1 y) with [n 1 , p 1 , p 2 , . . . , p l ] = 1. With this notation, we define Proof We introduce the following notation for a class of vertices that will be useful to construct a Hamiltonian cycle in the subgraph generated by ) we choose v r and v r to be adjacent in the Hamiltonian cycle in N i ), then v r and v r +1 are adjacent.
Let T v r v r be the Hamiltonian path in N i (x) (y), with initial vertex v r and final vertex v r . Such a path exists because of Lemma 4. The Hamiltonian cycle in The next theorem counts the number of disjoint Hamiltonian cycles in the subgraphs generated by N H i (y), with y = p y p k 1 (y) 1 , such that their disjoint union is H i .
disjoint Hamiltonian subgraphs whose union is H i . N H l (x, y ). Then, the following system of linear congruences has a solution: Without loss of generality, we assume that min{q 1 (y), q 1 (y )} = q 1 (y), then y = y + (m p with (x, y) ∈ N H i (y ), which is a contradiction.

(4) The vertices in H
Theorem 3 provides us with detailed information on the Hamiltonian cycles that cover H i . The next two results will identify vertices at which these Hamiltonian cycles can be joined together.
). Clearly, (x, y) and (x , y ) are adjacent in G (G). We define the set The following two conditions are clearly sufficient for the existence of a solution:  , and the second condition holds.
(2) Clearly, (x, y + p q 1 (y)−r +i 1 ) and (x , y + p q 1 (y) 1 ) are adjacent due to (x , y + p q 1 (y) 1 r (x, y). Using statement (1), we obtain (x, y) and (x , 0) are adjacent, then the following system of linear congruences has a solution: Again, the following two conditions are clearly sufficient for the existence of a solution: Because (x, y) and (x , y ) are adjacent, we know that the second part [x, p k 1 1 ]|x is clearly fulfilled.
For the second condition, we notice that [x p The second condition holds. On the other hand, we note that k 1 −i +k 1 (y) ≥ k 1 −r +i −i +k 1 ≥ k 1 −r +k 1 ≥ k 1 . 2 ) p k 1 1 )}. Next we introduce some sets of vertices that will be useful in the subsequent discussion: . This is true because of Definition 1 and that q 1 (y) = q 1 ( p 1 y) with (x, y) ∈ H r and ( p 1 x, p 1 y) ∈ H r +1 .
As an immediate consequence of claim 1 and Theorem 4(1), we obtain Claim 2. The subgraph generated by the sets N r We now proceed to show the first statement.
By Theorem 3 (2), there are ( p 1 −1) p k 1 −i−1 1 disjoint Hamiltonian subgraphs generated by the set k 1 (y)= j N H i (y), and by Theorem 4 (2), we can build a Hamiltonian cycle in the subgraph generated by N H i+1 ( p 1 y) ∪ ( To construct a Hamiltonian cycle in the subgraph generated by N H i+1 ( p 1 y) ∪ ( ) with initial vertex v r and final vertex v r and T p 1 y be a path that is part of the Hamiltonian cycle in N H i+1 ( p 1 y) with initial vertex u p 1 and final vertex u. A Hamiltonian cycle in the subgraph generated by N H i+1 ( p 1 y) ∪ ( can now be constructed as sketched in Figure 3). More formally, it is of the form The adjacencies between u and v and between u r and v r are guaranteed by claim 2. An analogous construction for each set N H i+1 ( p 1 y) with k 1 (y) = j + 1 yields Hamiltonian cycles that cover all vertices of V * . We then consider the ( p 1 − 1) p k 1 −i−2 1 cycles that cover V * and perform the same construction for N H i+2 ( p 2 1 y). This yields ( p 1 − 1) p k 1 −i−3 1 disjoint Hamiltonian cycles that cover  ) with initial vertex v r and final vertex v r and T p 1 y be a path, part of the Hamiltonian cycle in the subgraph generated by H k 1 , with initial vertex u p 1 and final vertex u 1 . Subcase 2: j = k 1 − k 1 + i. We can use exactly the same procedure because ( p 1 − 1) p The subgraphs induced by the vertex sets (i) T r 0 and T r +1 0 with 0 ≤ r ≤ k 1 − k 1 , (ii) T r 0 and T r +1 0 with k 1 − k 1 + 1 ≤ r ≤ k 1 , and (iii) T 0 r and T 0 r +1 with 0 ≤ r ≤ k 1 each contain a clique of cardinality p k l l . . . p k 2 2 . This is true because of the claim 2 in Theorem 5, and the fact that, for the first case, N H k 1 ( p k 1 +r ) ⊂ T 0 r , these three pair of vertices are adjacent in G (G) because of the claim. Finally, let T be the path in the subgraph generated by H k 1 with the vertices that were not used at the end of the process in the construction of the Hamiltonian cycle. A Hamiltonian cycle in G (G) can now be composed as follows:

Data availability
The data used to support the findings of this study are included within the paper.
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