Hamiltonicity in power graphs of a class of abelian groups

Abstract

The undirected power graph $\mathscr {G}(S)$ of a semigroup S is an undirected graph with vertex set S where two vertices $u,v\in S$ are adjacent if and only if there is a positive integer m such that $u^{m}=v$ or $v^{m}=u$. 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.

1 Introduction

A groupoid (magma) $(S,\cdot )$ is power-associative if the the subalgebra generated by any element is associative [2]. The undirected power graph ${\mathscr {G}}(S)$ of S has vertex set S, and uv is an edge in ${\mathscr {G}}(S)$ if there is an integer $m>1$ such that $u=v^m$ or $v=u^m$ [1, 4]. It is the underlying graph of the directed power graph $\mathbf {{\mathscr {G}}}(S)$ [5] with directed edges (u, v) whenever v is a power of u. At least in the case of Abelian groups, there is a very close connection between the group G and its power graph ${\mathscr {G}}(G)$: As shown in [3], two Abelian groups are isomorphic if and only of their power graphs are isomorphic.

A graph G is Hamiltonian if there is a cycle that contains all its vertices. The question which condition a power graph is Hamiltonian has attracted some attention in the past. We briefly review the main results relevant for this contribution. For finite p-groups, a complete characterization is available:

Proposition 1

[7] Let G be a finite p-group, then ${\mathscr {G}}(G)$ is Hamiltonian if and only if $\vert G \vert \ne 2$ and G is cyclic.

Furthermore, if p is an odd prime, then ${\mathscr {G}}(G)$ is 2-connected if and only if it is Hamiltonian [7]. For cyclic groups, ${\mathscr {G}}({\mathbb {Z}}_n)$ is Hamiltonian for all $n\ge 3$ [4]. On the other hand, ${\mathscr {G}}(G)$ is not Hamiltonian for Abelian groups of order $2^{m}$ with $m>1$ that are not cyclic [4]. Trivially, ${\mathscr {G}}(G)$ is Hamiltonian if it is a complete graph and $|G|\ge 3$. Therefore, the following characterization of power graphs that are complete is useful and interesting:

Proposition 2

[4] Let G be a finite group. Then, ${\mathscr {G}}(G)$ 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 $m\in {\mathbb {N}}$.

Characterizations of Hamiltonian power graphs for groups of the form $G=({\mathbb {Z}}_p)^{n}\oplus \mathbb ({Z}_q)^{m}$ with distinct primes p and q and $m, n \in {\mathbb {N}}$ can be found in [6].

2 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 $\langle x \rangle$ for the (abelian) subgroup generated by a single element $x\in G$. It is easy to see that the identity 0 is adjacent to all other group elements in ${\mathscr {G}}(G)$. The subgraph of ${\mathscr {G}}(G)$ obtained by removing 0 and its incident edges will be denoted by ${\mathscr {G}}^{*}(G)$, and we write ${\mathscr {G}}_{x}={\mathscr {G}}(\langle x \rangle )$. We denote the greatest common divisor of r natural numbers $q_{1}$, $q_{2}$,...,$q_{r}$ by $[q_{1},q_{2},\dots ,q_{r}]$. Furthermore, we write $q_{2}\vert q_{1}$ 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 $G={\mathbb {Z}}_{q_1}\oplus {\mathbb {Z}}_{q_2}\oplus \dots \oplus {\mathbb {Z}}_{q_{\ell }}$ of cyclic groups whose sizes $q_j$ are powers of prime numbers. Up to reordering, the number of factors $\ell$ 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 $j,k\in {\mathbb {N}}$, ${\mathbb {Z}}_{jk}\simeq {\mathbb {Z}}_{j}\oplus {\mathbb {Z}}_{k}$ 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:

Lemma 2

Let G be an abelian group with $\ell$ different cyclic groups ${\mathbb {Z}}_{q_i}$ in its product representation, where $q_i = p_{i}^{k_{i}}$ and $p_i\ge 3$ for all $1\le i \le \ell$. Then, ${\mathscr {G}}(G)$ is Hamiltonian whenever $[p_{i}, p_{j}]=1$ for all i, j.

Proof

Using Lemma 1, G can be written as $G={\mathbb {Z}}_{p_{1}^{k_{1}} p_{2}^{k_{2}} \ldots p_{r}^{k_{r}}}$ and by Proposition 2, ${\mathscr {G}}(G)$ is complete and thus Hamiltonian. $\square$

The Invariant factor decomposition allows us to write G in the form

$$\begin{aligned} G={\mathbb {Z}}_{n_{1}}\oplus \dots \oplus {\mathbb {Z}}_{n_{j_{1}}} \end{aligned}$$

(1)

where $j_{1}$ and $n_{i}$ are the unique integers satisfying $j_{1}\ge 1$, $n_{i}\ge 2$ and $n_{i}\vert n_{i+1}$ for $1\le i \le j_{1}-1$. The $n_{i}$’s in the last equation are called the invariant factor of G. For later reference, we note that for $j_{1}=2$, Eq. (1) reduces to

$$\begin{aligned} G= {\mathbb {Z}}_{p_{s}^{k_{i}}\dots p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}} p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}}\ \end{aligned}$$

(2)

where $p_{i}$ is prime and $k'_{i}\ge k_{i}$ for $i\in \{1,\dots ,s, s+1,\dots ,l\}$.

The following result, by [9], will be useful to identify paths in power graphs:

Theorem 2

[9] The system of congruences $a_{i}x \equiv b_{i} \mod m_{i}$, with $m_i \ne 0$ and $i\in \{1,\dots ,r\}$ admits a solution if and only if:

• (1) $[a_{i},m_{i}] \mid b_{i}$ for $i\in \{1,\dots ,r\}$

• (2) $[a_{i}m_{j},a_{j}m_{i}] \mid a_{i}b_{j}-a_{j}b_{i}$ for $i\in \{1,\dots ,r\}$.

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.

Lemma 3

Let $G= {\mathbb {Z}}_{p_{s}^{k_{s}}\dots p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}}p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}}$, $(x,y)\in G$ where

$x=p_{x}p_{s}^{k_{s}(x)}\dots p_{1}^{k_{1}(x)}$, $y=p_{y}p_{l}^{k_{l}(y)}\dots p_{1}^{k_{1}(y)}$, with $[p_{x},p_{1},\dots ,p_{s}]=1=[p_{y},p_{1},\dots ,p_{l}]$. Then, (x, y) and $(x,y+p_{l}^{q_{l}}\dots p_{1}^{q_{1}})$ are adjacent in ${\mathscr {G}}(G)$ where

$$\begin{aligned} q_{j} = \left\{ \begin{array}{lcl} k_{j}(y) &{} if &{} j>s \text { and } k_{j}(y)\le k'_{j}\\ k'_{j} &{} if &{} j>s \text { and } k_{j}(y)> k'_{j}\\ k_{j}-k_{j}(x)+ k_{j}(y) &{} if &{} 1\le j\le s \text { and } k_{j}-k_{j}(x)+ k_{j}(y)\le k'_{j}\\ k'_{j} &{} if &{} 1\le s\le i \text { and } k_{j}-k_{j}(x)+ k_{j}(y)> k'_{j} \end{array}\right. \end{aligned}$$

Proof

By construction, (x, y) and $(x,y+p_{l}^{q_{l}}\dots p_{1}^{q_{1}})$ are adjacent whenever the following system of congruences has a solution:

$$\begin{aligned} \begin{aligned} n x&\equiv x \mod (p_{s}^{k_{s}} \dots p_{1}^{k_{1}}) \\ n y&\equiv y+ p_{l}^{q_{l}}\dots p_{1}^{q_{1}} \mod (p_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}}p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}) \end{aligned} \end{aligned}$$

By Theorem 2, this system of congruences has solution whenever the following conditions are true.

• 1. $[x, p_{s}^{k_{s}}\dots p_{1}^{k_{1}}] \vert x$ and $[y,p_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}}p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}] \mid y+ p_{l}^{q_{l}}\dots p_{1}^{q_{1}}$

• 2. $[xp_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}}p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}, y p_{s}^{k_{s}}\dots p_{1}^{k_{1}}] \mid xp_{l}^{q_{l}}\dots p_{1}^{q_{1}}$

The first part of the first condition is clearly true. The second part is true because $[y,p_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}} p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}] = p_{l}^{k_{l}(y)}\dots p_{1}^{k_{1}(y)}$.

The second conditions are fulfilled, because we know that

$[xp_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}}p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}, y p_{s}^{k_{s}}\dots p_{1}^{k_{1}}] = p_{l}^{k_{l}(y)}\dots p_{s+1}^{k_{s+1}(y)} p_{s}^{q'_{s}}\dots p_{l}^{q'_{l}}$,

where $q'_{j}=$ min $\{k_{j}(x)+k'_{j},k_{j}(y)+k_{j}\}$ for $1\le j\le s$. $\square$

3 Hamiltonian graphs for $\varvec{G}={\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$

Let us consider groups of the form

$$\begin{aligned} G={\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}\ \end{aligned}$$

(3)

where the $p_{i}$ are prime numbers and pairwise coprime. For G, we define the sets

$$\begin{aligned} {\mathscr {H}}_{i} {:}{=}\{(x,y)\in G \vert k_{1}(x)=i, \wedge y\in {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}\} \end{aligned}$$

(4)

, and for a fixed $x\in {\mathbb {Z}}_{p_{1}^{k_{1}}}$ with $k_{1}(x)=i$, we set

$$\begin{aligned} N_{x}^{i}(y) {:}{=}\{(x,y+ mp_{1}^{q_{1}(y)})\vert 0\le m< p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}p_{1}^{k'_{1}-q_{1}(y)} \} \end{aligned}$$

(5)

First we note that for G, Lemma 3 specializes to

Corollary 1

Let $G= {\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$, $(x,y)\in G$, where $x=p_{x}p_{1}^{k_{1}(x)}$, $y=p_{y}p_{l}^{k_{l}(y)}\dots p_{1}^{k_{1}(y)}$, with $[p_{x},p_{1}]=1=[p_{y},p_{1},\dots ,p_{l}]$. Then, (x, y) and $(x,y+p_{l}^{q_{l}}\dots p_{1}^{q_{1}})$ are adjacent in ${\mathscr {G}}(G)$ where

$$\begin{aligned} q_{j}= \left\{ \begin{array}{lcl} k_{j}(y) &{} if &{} j>1 \text { and } k_{j}(y)\le k'_{j}\\ k'_{j} &{} if &{} j>1 \text { and } k_{j}(y)> k'_{j}\\ k_{1}-k_{1}(x)+ k_{1}(y) &{} if &{} j=1 \text { and } k_{1}-k_{1}(x)+ k_{1}(y)\le k'_{1}\\ k'_{1} &{} if &{} j=1 \text { and } k_{1}-k_{1}(x)+ k_{1}(y)> k'_{1} \\ \end{array} \right. \end{aligned}$$

(6)

Lemma 4

If $G= {\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$ and $(x,y)\in G$, then the subgraph induced by $N_{x}^{i}(y)$ is Hamiltonian.

Proof

For each $2\le j \le l$ and for every $(x,y')\in N_{x}^{i}(y)$, we define

$$\begin{aligned} r_{j}(y') {:}{=}\left\{ \ \begin{array}{lcl} k_{j}(y') &{} if &{} 0\le k_{j}(y')\le k'_{j}\\ k'_{j} &{} if &{} k_{j}(y')\ge k'_{j}\\ \end{array} \right. \end{aligned}$$

(7)

Furthermore, we define the following subsets

$$\begin{aligned} V_{p_{l}^{r_{l}}\dots p_{2}^{r_{2}}}{:}{=}\{(x,y')\in N_{x}^{i}(y)\vert r_{i}(y')=r_{i} \} \end{aligned}$$

in $N_{x}^{i}(y)$, where $0\le r_{j} \le k'_{j}$ for $2\le r \le l$.

First, we note that $k_{1}(y)=k_{1}(y')$ for all $(x,y')\in N_{x}^{i}(y)$. To prove this, let $p_{y}\in {\mathbb {N}}$ such that $[p_{y},p_{1},p_{2},\dots ,p_{l}]=1$. We compute $y'=y+mp_{1}^{q_{1}(y)}= p_{y}p_{l}^{k_{l}(y)}\dots p_{1}^{k_{1}(y)}+mp_{1}^{q_{1}(y)}= p_{1}^{k_{1}(y)}(p_{y}p_{l}^{k_{l}(y)}\dots p_{2}^{k_{2}(y)}+mp_{1}^{k_{1}-i})$. The assertion now follows from $[p_{y}p_{l}^{k_{l}(y)}\dots p_{2}^{k_{2}(y)},p_{1}]=1$.

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 $N_{x}^{i}(y)$.

Claim: If $s\vert r$, then each vertex in $V_{s}$ is adjacent to all the vertices in $V_{r}$.

To prove this claim, consider $(x,y_{1})\in V_{s}$ and $(x,y_{2})\in V_{r}$ with $y_{1}= p_{y_{1}}p_{l}^{k_{l}(y_{1})}\dots p_{1}^{k_{1}(y_{1})}$ and $y_{2}= p_{y_{2}}p_{l}^{k_{l}(y_{2})}\dots p_{1}^{k_{1}(y_{2})}$, where $0\le k_{i}(y_{1})\le k_{i}(y_{2})$. Using Lemma 3, we observer that $(x,y_{1})$ and $(x,y_{2})$ are adjacent if the following system of linear congruences has solution:

$$\begin{aligned} \begin{aligned} n x&\equiv x \mod (p_{s}^{k_{s}}\dots p_{1}^{k_{1}}) \\ n y_{1}&\equiv y_{2} \mod (p_{l}^{k'_{l}}\dots p_{s+1}^{k'_{s+1}}p_{s}^{k'_{s}}\dots p_{1}^{k'_{1}}) \end{aligned} \end{aligned}$$

The following two conditions are clearly sufficient for the existence of a solution:

• 1. $[x, p_{1}^{k_{1}}] \vert x$ and $[y_{1},p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid y_{2}$

• 2. $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y_{1} p_{1}^{k_{1}} ] \mid x(y_{2}-y_{1})$

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 $N_{x}^{i}(y)$.

Any Hamiltonian cycle in the subgraph within $V_{s}$ can be extended to a Hamiltonian cycle in $V_{r}\cup V_{s}$ 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 $e_{s}=(u_{s},v_{s})\in E(C_{s})$ and $e_{r}=(u_{r},v_{r})\in E(C_{r})$ 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 $E({\mathscr {G}}(G))$, so a Hamiltonian cycle in the subgraph induced by $V_{r}\cup V_{s}$ is:

$u_{s}\rightarrow T_{V_{s}}\rightarrow v_{s} \rightarrow v_{r} \rightarrow T_{V_{r}}\rightarrow u_{r}\rightarrow u_{s}.$

This construction is sketched in Fig. 1.

Fig. 1

Extension of the Hamiltonian cycle in $V_{j}$ to a Hamiltonian cycle in $V_{j}\cup V_{i}$

In order to construct a Hamiltonian cycle in the subgraph induced by $N_{x}^{i}(y)$, 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 $2\le i\le l$. 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 $V_{p_{i}^{k'_{i}}\dots p_{s}^{r_{s}}}$ has as predecessor the set $V_{p_{i}^{k'_{i}}\dots p_{s}^{r_{s}-1}}$ if $r_{s}\le 2$ and $V_{p_{i}^{k'_{i}}\dots p_{s-1}^{k'_{s-1}}}$ if $r_{s}=1$; the successor is $V_{p_{i}^{k'_{i}}\dots p_{s}^{r_{s}+1}}$ when $r_{s}< k'_{s}$ or $V_{p_{i}^{k'_{i}}\dots p_{s}^{k'_{s}}p_{s+1}}$ when $r_{s}=k'_{s}$. See Fig. 2.

Fig. 2

Tree arrangement of the sets $V_{q}$

The leaves of the tree are $V_{p_{l}^{k'_{l}}p_{l-1}^{k'_{l-1}}\dots p_{k'_{1}}}$, $V_{p_{l-1}^{k'_{l-1}}\dots p_{k'_{1}}}$,...,$V_{p_{2}^{k'_{2}}}$, $V_{p_{1}^{k'_{1}}}$. Note that the leaf $V_{p_{s}^{k'_{s}}\dots p_{k'_{1}}}$ 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 $N_{x}^{i}(y)$ 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 $v_{j}\in V_{j}$ therefore can be combined to a Hamiltonian cycles in the following manner: $W{:}{=}u_{1}\rightarrow u_{p_{l}}\rightarrow T_{V_{p_{l}}}\rightarrow v_{p_{l}}\rightarrow u_{p_{l}^{2}}\rightarrow T_{p_{l}^{2}} \rightarrow v_{p_{l}^{2}}\rightarrow \dots \rightarrow u_{p_{l}^{k'_{l}}}\rightarrow T_{V_{p_{l}^{k'_{l}}}} \rightarrow v_{p_{l}^{k'_{l}}}\rightarrow u_{p_{l}^{k'_{l}}p_{l-1}}\rightarrow T_{V_{p_{l}^{k'_{l}}p_{l-1}}}\rightarrow v_{p_{l}^{k'_{l}}p_{l-1}} \rightarrow \dots \rightarrow u_{p_{l}^{k'_{l}}p_{l-1}^{k'_{l-1}}.\dots p_{1}^{k'_{1}}} \rightarrow T_{V_{p_{l}^{k'_{l}}p_{l-1}^{k'_{l-1}}\dots p_{1}^{k'_{1}}}}\rightarrow v_{p_{l}^{k'_{l}}p_{l-1}^{k'_{l-1}}\dots p_{1}^{k'_{1}}}\rightarrow u_{p_{l-1}} \rightarrow T_{V_{p_{l-1}}}\rightarrow v_{p_{l-1}}\rightarrow u_{p_{l-1}^{2}}\rightarrow \dots \rightarrow u_{p_{l-1}^{k'_{l-1}}\dots p_{1}^{k'_{1}}}\rightarrow T_{V_{p_{l-1}^{k'_{l-1}}\dots p_{1}^{k'_{1}}}}\rightarrow v_{p_{l-1}^{k'_{l-1}}\dots p_{1}^{k'_{1}}} \rightarrow \dots v_{p_{3}^{k'_{3}}p_{2}^{k'_{2}}}\rightarrow u_{p_{2}}\rightarrow T_{V_{p_{2}}}\rightarrow v_{p_{2}}\rightarrow \dots \rightarrow v_{p_{2}^{k'_{2}}}\rightarrow v_{1}\rightarrow T_{V_{1}}\rightarrow u_{1}$.

The extension of the sets $V_{p_{2}^{k'_{2}}} \cup V_{1}$ is possible because each vertex in $V_{1}$ is adjacent with all the vertices in $N_{x}^{i}(y)$. 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 $N_{x}^{i}(y)$. $\square$

The Hamiltonian graph ${\mathscr {G}}_{N_{x}^{i}(y)}$ 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 ${\mathscr {H}}_{i}$. To this end, we will need the following technical result.

Lemma 5

Let $G= {\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$, $(x,y), (x',y')\in G$ with $k_{1}(x)=k_{1}(x')=i$, $y=p_{y}p_{1}^{k_{1}(y)}$ such that $[p_{y},p_{l},\dots ,p_{1}]=1$ and let $(x',y_{1})\in N_{x'}^{i}(y)$. If (x, y) and $(x',y')$ are adjacent in ${\mathscr {G}}(G)$, then (x, y) and $(x',y_{1})$ are adjacent in ${\mathscr {G}}(G)$.

Proof

Let $x=p_{x}p_{1}^{k_{1}(x)}$ with $[p_{x},p_{1}]=1$ and $p_{xy}=[p_{y},p_{x}]$. We know that $y_{1}=y'+mp_{1}^{q_{1}(y')}$, with $0\le m < p_{l}^{k'_{l}\dots p_{2}^{k'_{2}}p_{1}^{k'_{1}-q_{1}(y')}}$. 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:

$$\begin{aligned} \begin{aligned} n x&\equiv x' \mod p_{1}^{k_{1}} \\ n y&\equiv y'+ mp_{1}^{q_{1}(y')} \mod (p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}) \end{aligned} \end{aligned}$$

The following two conditions are clearly sufficient for the existence of a solution:

• 1. $[x, p_{1}^{k_{1}}] \vert x'$ and $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid y'+mp_{1}^{q_{1}(y')}$

• 2. $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ] \mid xy'+xmp_{1}^{q_{1}(y')}-yx'$

Since (x, y) and $(x',y')$ are adjacent, we know that $[x, p_{1}^{k_{1}}] \vert x'$, $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid y'$ and $p_{1}^{k_{1}(y)}\mid p_{1}^{q_{1}(y')}$. Taking this into account, the first part of the first condition is true, and the second part is fulfilled, because $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}]=p_{1}^{k_{1}(y)}$ and clearly $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid y'$ and $p_{1}^{k_{1}(y)}\vert p_{1}^{q_{1}(y')}$.

It is clear that $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ] \mid xy'-yx'$ because (x, y) and $(x',y')$ are adjacent, besides $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ]=p_{xy}p_{1}^{k_{1}(x)+k_{1}(y)}$ so $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ]\mid xp_{1}^{q_{1}(y')}$, and the second condition holds. $\square$

Let $S_{i}=\{x'\in {\mathbb {Z}}_{p_{1}^{k_{1}}} : k_{1}(x)=i\}=\{x'=mx : [m,p_{1}]=1 \}$. We can easily note that there is a path $P_{x}=(x,y) (x_{1},y_{1})\dots (x_{\vert S_{i}\vert }, y_{\vert S_{i}\vert })$ such that $(x_{r},y_{r})=(n_{1}^{r} x, n_{1}^{r} y)$ with $[n_{1},p_{1},p_{2},\dots ,p_{l}]=1$. With this notation, we define

$$\begin{aligned} N_{{\mathscr {H}}_{i}}(y){:}{=}\bigcup _{r=1}^{\vert S_{i}\vert } N_{x_{r}}^{i}(y_{r})\cup N_{x}^{i}(y) \end{aligned}$$

(8)

Lemma 6

Let $G= {\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$ and $(x,y)\in G$, with $y=p_{y}p_{1}^{k_{1}(y)}$ and $[p_{y},p_{l},\dots p_{2},p_{1}]=1$. Then, the subgraph induced by $N_{{\mathscr {H}}_{i}}(y)$ is Hamiltonian in ${\mathscr {G}}(G)$.

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 $N_{{\mathscr {H}}_{i}}(y)$. If $v=(x,y)$ and $v'=(x, y+mp_{1}^{q_{1}(y)})$, then $v'\in N_{x}^{i}(y)$ and v and $v'$ are adjacent in the Hamiltonian cycle in $N_{x}^{i}(y)$. Analogously, for $v_{r}=(x_{r}, y_{r}+m_{r}p_{1}^{q_{1}(y)})$ and $v'_{r}=(x_{r}, y_{r}+m'_{r}p_{1}^{q_{1}(y)})$ we choose $v_{r}$ and $v'_{r}$ to be adjacent in the Hamiltonian cycle in $N_{x_{r}}^{i}(y_{r})$. Moreover, if $v_{r+1}\in N_{x_{r+1}}^{i}(y_{r+1})$ with $v_{r+1}=n_{1}v'_{r}=(x_{r+1}, y_{r+1}+n_{1}m'_{r}p_{1}^{q_{1}(y)})$, then $v'_{r}$ and $v_{r+1}$ are adjacent.

Let $T_{v_{r}v'_{r}}$ be the Hamiltonian path in $N_{(x)}^{i}(y)$, with initial vertex $v_{r}$ and final vertex $v'_{r}$. Such a path exists because of Lemma 4. The Hamiltonian cycle in $N_{{\mathscr {H}}_{i}}(y)$ is:

$v=(x,y)\rightarrow T_{(v v')} \rightarrow v_{1}\rightarrow T_{(v_{1}v'_{1})}\rightarrow v_{2} \rightarrow T_{(v_{2}v'_{2})}\rightarrow \dots \rightarrow v_{\vert S_{i}\vert } \rightarrow T_{(v_{\vert S_{i}\vert }v'_{\vert S_{i}\vert } )} \rightarrow v$

$v'_{\vert S_{i}\vert }$ and v are adjacent by Lemma 5, taking into account that $v'_{\vert S_{i}\vert }\in N_{x_{\vert S_{i}\vert }}(y_{\vert S_{i}\vert })$ and v, $v_{\vert S_{i}\vert }$ are adjacent. $\square$

The next theorem counts the number of disjoint Hamiltonian cycles in the subgraphs generated by $N_{{\mathscr {H}}_{i}}(y)$, with $y=p_{y}p_{1}^{k_{1}(y)}$, such that their disjoint union is ${\mathscr {H}}_{i}$.

Theorem 3

Let $G=Z_{p_{1}^{k_{1}}}\oplus Z_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$ and $[p_{y},p_{1},p_{2},\dots ,p_{l}]=1$ then

• (1) For all $x\in {\mathbb {Z}}_{p_{1}^{k_{1}}}$, $N_{{\mathscr {H}}_{i}}(y) \cap N_{{\mathscr {H}}_{i}}(y')= \emptyset$ if and only if $(x,y)\notin N_{{\mathscr {H}}_{i}}(y')$.

• (2) For each $0\le k_{1}(y) \le k'_{1}-k_{1}+i-1$, there are $(p_{1}-1) p_{1}^{k_{1}-i-1}$ different $N_{{\mathscr {H}}_{i}}(y)$ sets with $y=p_{y}p_{1}^{k_{1}(y)}$.

• (3) For $k'_{1}-k_{1}+i\le k_{1}(y)\le k'_{1}-1$, there are $(p_{1}-1)p_{1}^{k'_{1}-k_{1}(y)-1}$ different $N_{{\mathscr {H}}_{i}}(y)$ sets with $y=p_{y}p_{1}^{k_{1}(y)}$.

• (4) For $k_{1}(y)=k'_{1}$, there is only a single $N_{{\mathscr {H}}_{i}}(y)$ set with $y=p_{y}p_{1}^{k'_{1}}$.

• (5) There are $(k'_{1}-k_{1}+i)(p_{1}-1)p_{1}^{k_{1}-i-1} + \sum _{k_{1}(y)=k'_{1}-k_{1}+i}^{k'_{1}-1} (p_{1}-1)p_{1}^{k'_{1}-k_{1}(y)-1} + 1$ disjoint Hamiltonian subgraphs whose union is $H_{i}$.

Proof

(1)   Suppose that $(x,y)\notin N_{{\mathscr {H}}_{i}}(y')$ and there is $(x_{1},y_{1})\in N_{{\mathscr {H}}_{i}}(y) \cap N_{{\mathscr {H}}_{l}}(x,y')$. Then, the following system of linear congruences has a solution:

$$\begin{aligned} \begin{aligned} x_{1}&= m_{2}p_{1}^{i}\\ y_{1}&= y+ mp_{1}^{q_{1}(y)}\\ y_{1}&= y'+ m'p_{1}^{q_{1}(y')}\\ \end{aligned} \end{aligned}$$

with $[m_{2},p_{1}]=1$, $0\le m\le p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}-q_{1}(y)}$ and $0\le m'\le p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}-q_{1}(y')}$. Without loss of generality, we assume that $\min \{q_{1}(y), q_{1}(y')\}=q_{1}(y)$, then

$y= y' +(m'p_{1}^{q_{1}(y')-q_{1}(y)}-m)p_{1}^{q_{1}(y)}$ with $(x,y)\in N_{{\mathscr {H}}_{i}}(y')$, which is a contradiction.

(2)   Let $0\le k_{1}(y)\le k'_{1}-k_{1}+i-1$. There are $(p_{1}-1)$ vertices of the form $(x,y)\in {\mathscr {H}}_{i}$, with a fixed x, such that $y=p_{y}p_{1}^{k_{1}(y)}$, in the discrete interval $[np_{1}^{k_{1}(y)+1}, (n+1)p_{1}^{k_{1}(y)+1}]$ with $[n,p_{1}]=1$ or $[n+1,p_{1}]=1$. Furthermore, there are $p_{1}^{q_{1}(y)}/p_{1}^{k_{1}(y)+1}= p_{1}^{k_{1}-i+k_{1}(y)}/p_{1}^{k_{1}(y)+1}= p_{1}^{k_{1}-i-1}$ disjoint intervals of the form $[np_{1}^{k_{1}(y)+1}, (n+1)p_{1}^{k_{1}(y)+1}]$ in the interval $[0,p_{1}^{k_{1}-i+k_{1}(y)}]$. We conclude that there are $(p_{1}-1) p_{1}^{k_{1}-i-1}$ different $N_{{\mathscr {H}}_{i}}(y)$ with $y=p_{y}p_{1}^{k_{1}(y)}$.

(3)   For $k'_{1}-k_{1}+i\le k_{1}(y)\le k'_{1}-1$, $q_{1}(y)=k'_{1}$, there are $(p_{1}-1)$ vertices of the form $(x,y)\in {\mathscr {H}}_{i}$ with $y=p_{y}p_{1}^{k_{1}(y)}$ and a fixed x, in the discrete interval $[np_{1}^{k_{1}(y)+1}, (n+1)p_{1}^{k_{1}(y)+1}]$ with $(n,p_{1})=1$ or $(n+1,p_{1})=1$. Moreover, there are $p_{1}^{k'_{1}}/p_{1}^{k_{1}(y)+1} = p_{1}^{k'_{1}-k_{1}(y)-1}$ intervals of the form $[np_{1}^{k_{1}(y)+1}, (n+1)p_{1}^{k_{1}(y)+1}]$ in the interval $[0,p_{1}^{k'_{1}}]$. Therefore, there are $(p_{1}-1) p_{1}^{k'_{1}-k_{1}(y)-1}$ different $N_{{\mathscr {H}}_{i}}(y)$ with $y=p_{y}p_{1}^{k_{1}(y)}$.

(4)   The vertices in ${\mathscr {H}}_{i}$ with $k_{1}(y)=k'_{1}$ are $(x,mp_{1}^{k'_{1}})$ with $0\le m\le p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}-1$ and $q_{1}(y)=p_{1}^{k'_{1}}$. Therefore, they are contained in $N_{{\mathscr {H}}_{i}}(x,0)$.

(5)   The first part of the result follows from statements (1)-(4). If we fix $x=p_{1}^{i}$ and the sum of the elements of the different sets $N_{{\mathscr {H}}_{i}}(y)$ with $y=p_{y}p_{1}^{k_{1}(y)}$ and $0\le k_{1}(y) \le k'_{1}$ is $p_{1}^{k'_{l}}\dots p_{2}^{k'_{2}}p_{1}^{k'_{1}}$, the result would be proved because for every $x=mp_{1}^{i}$ with $[m,p_{1}]=1$ the analysis is the same. Taking into account that each set $N_{(x)}^{i}(y)$ has $p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}p_{1}^{k'_{1}-q_{1}(y)}$ vertices, we obtain

$$\begin{aligned} \begin{aligned}&(p_{1}-1)p_{1}^{k_{1}-l-1} \sum _{r=0}^{k'_{1}-k_{1}+i-1} (p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}) p_{1}^{k'_{1}-q_{1}(p_{1}^{k_{1}(y)})}\\&\qquad + \sum _{k_{1}(y)=k'_{1}-k_{1}+i}^{k'_{1}-1} (p_{1}-1)p_{1}^{k'_{1}-k_{1}(y)-1}(p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}})p_{1}^{k'_{1}-q_{1}(p_{1}^{k_{1}(y)})}+ (p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}})\\&\quad = (p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}})(p_{1}-1) (\sum _{k_{1}(y)=k_{1}-i+1}^{k_{1}-1} p_{1}^{k_{1}(y)}+ \sum _{k_{1}(y)=0}^{k_{1}-i}p_{1}^{k_{1}(y)} + \frac{1}{p_{1}-1}) \\&\quad = (p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}})(p_{1}-1)(\sum _{k_{1}(y)=0}^{k'_{1}-1} p_{1}^{r}+\frac{1}{p_{1}-1})\\&\quad = (p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}})(p_{1}-1)(\frac{p_{1}^{k'_{1}}-1}{p_{1}-1} +\frac{1}{p_{1}-1}) \\&\quad = (p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}})p_{1}^{k'_{1}} \end{aligned} \end{aligned}$$

$\square$

Theorem 3 provides us with detailed information on the Hamiltonian cycles that cover ${\mathscr {H}}_{i}$. The next two results will identify vertices at which these Hamiltonian cycles can be joined together.

Definition 1

Let $(x,y)\in {\mathscr {H}}_{i}$ with $y=p_{y}p_{1}^{k_{1}(y)}$, $x=p_{x}p_{1}^{i}$, $[p_{y},p_{l},\dots ,p_{1}]=1$, $[p_{x},p_{1}]=1$ and $(x',y')\in {\mathscr {H}}_{r}$, with $r>i$, such that $(x',y')=(p_{x}p_{1}^{r}, p_{y}p_{1}^{k_{1}(y)+r-i})$. Clearly, (x, y) and $(x',y')$ are adjacent in ${\mathscr {G}}(G)$. We define the set

$$\begin{aligned} N_{i,r}(x,y) {:}{=}\{ (x',y'+mp_{1}^{q_{1}(y)}) \vert 0\le m\le p_{l}^{k'_{l}}\dots p_{2}p_{1}^{k'_{1}-q_{1}(y)}\} \end{aligned}$$

Theorem 4

Let $G={\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$, $y=p_{y}p_{1}^{k_{1}(y)}$, $x=p_{x}p_{1}^{i}$ with $\small {[p_{y},p_{2},\dots , p_{l}]=1}$, $[p_{x},p_{1}]=1$ and $p_{xy}=[p_{x},p_{y}]$. Then,

• (1) For each $(x,y)\in {\mathscr {H}}_{i}$, (x, y) is adjacent to all vertices in $N_{i,r}(x,y)$.

• (2) For $0\le k_{1}(y)< k'_{1}-r+i$

  $$\begin{aligned} N_{i,r}(x,y)= N_{i,r}(x,y+p_{1}^{q_{1}(y)-r+i})\,. \end{aligned}$$

• (3) For $k'_{1}-r+i\le k_{1}(y)\le k'_{1}$,

  $$\begin{aligned} N_{{\mathscr {H}}_{i,r}}(x,y)= \{(p_{x}p_{1}^{r},mp_{1}^{k'_{1}})\vert 0\le m\le p_{l}^{k'_{l}} \dots p_{2}^{k'_{2}}-1 \}\,. \end{aligned}$$

• (4) For $r=k_{1}$, $N_{i,k_{1}}(x,y)=\{(0,mp_{1}^{k'_{1}})\vert 0\le m\le p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}-1\}$.

Proof

(1)   Let $(x',y')\in {\mathscr {H}}_{r}$ such that (x, y) and $(x',y')$ adjacent, then (x, y) is adjacent to all vertices in $N_{i,r}(x,y)$ if the following system of linear congruences has solution:

$$\begin{aligned} \begin{aligned} nx&= x' \mod (p_{1}^{k_{1}}) \\ ny&= y'+mp_{1}^{q_{1}(y)} \mod (p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}})\\ \end{aligned} \end{aligned}$$

The following two conditions are clearly sufficient for the existence of a solution:

• 1. $[x, p_{1}^{k_{1}}] \vert x'$ and $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid y'+mp_{1}^{q_{1}(y)}$

• 2. $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ] \mid xy'+xmp_{1}^{q_{1}(y)}-yx'.$

Because (x, y) and $(x',y')$ are adjacent, we know that $[x, p_{1}^{k_{1}}] \vert x'$, $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid y'$ and $p_{1}^{k_{1}(y)}\mid p_{1}^{q_{1}(y)}$. Taking this into account, the first part of the first condition is true, and the second part is fulfilled, because $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}]=p_{1}^{k_{1}(y)}$ and clearly $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid y'$ and $p_{1}^{k_{1}(y)}\vert p_{1}^{q_{1}(y)}$.

It is clear that $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ] \mid xy'-yx'$ because (x, y) and $(x',y')$ are adjacent, besides $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ]=p_{xy}p_{1}^{k_{1}(x)+k_{1}(y)}$ so $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ]\mid xp_{1}^{q_{1}(y)}$, and the second condition holds.

(2)   Clearly, $(x,y+p_{1}^{q_{1}(y)-r+i})$ and $(x',y' + p_{1}^{q_{1}(y)})$ are adjacent due to $(x',y' + p_{1}^{q_{1}(y)})=p_{1}^{r-i}(x,y+p_{1}^{q_{1}(y)-r+i})$, and $(x',p_{1}^{r-i}y + p_{1}^{q_{1}(y)})\in N_{i,r}(x,y)$. Using statement (1), we obtain

$$\begin{aligned} N_{i,r}(x,y)= N_{i,r}(x,y+p_{1}^{q_{1}(y)-r+i}) \end{aligned}$$

(3)   If (x, y) and $(x',0)$ are adjacent, then the following system of linear congruences has a solution:

$$\begin{aligned} \begin{aligned} nx&= x' \mod (p_{^{k_{1}}}) \\ ny&= 0 \mod (p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}})\\ \end{aligned} \end{aligned}$$

Again, the following two conditions are clearly sufficient for the existence of a solution:

• 1. $[x, p_{1}^{k_{1}}] \vert x'$ and $[y,p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}] \mid 0$

• 2. $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ] \mid yx'$

Because (x, y) and $(x',y')$ are adjacent, we know that the second part $[x, p_{1}^{k_{1}}] \vert x'$ is clearly fulfilled.

For the second condition, we notice that $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ]=p_{xy}p_{1}^{k_{1}(x)+k_{1}(y)}$ so $[xp_{l}^{k'_{l}}\dots p_{1}^{k'_{1}} \varvec{,} y p_{1}^{k_{1}} ]\mid yx'$, due to $k_{1}(x')>k_{1}(x)$. The second condition holds.

On the other hand, we note that $k_{1}-i+k_{1}(y)\ge k'_{1}-r+i-i+k_{1}\ge k'_{1}-r+k_{1}\ge k'_{1}$. Therefore, $q_{1}(y)=k'_{1}$ and $N_{i,r}(x,y)=(x',mp_{1}^{k'_{1}})$.

(4) is now an immediate consequence of (3). $\square$

The subgraph induced by ${\mathscr {H}}_{k_{1}}$ is isomorphic to ${\mathscr {G}}({\mathbb {Z}}_{p_{1}^{k_{1}}})$, i.e., a clique. Moreover, we have ${\mathscr {H}}_{k_{1}}= \bigcup _{r=0}^{k'_{1}} N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{r})$ and $N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{r})= \big \{(0,p_{1}^{r}),(0,2p_{1}^{r}),\dots \dots , (0,(p_{1}-1)p_{1}^{r}),(0,(p_{1}+1)p_{1}^{r}),\dots , (0,(p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}p_{1}^{k'_{1}-r}-1)p_{1}^{r})\big \}$, and in particular $N_{{\mathscr {H}}_{k_{1}}} (p_{1}^{k'_{1}})=\{(0,0), (0,p_{1}^{k'_{1}}),\dots ,(0,(p_{l})^{k'_{l}}\dots p_{2}^{k'_{2}})p_{1}^{k'_{1}})\}$. Next we introduce some sets of vertices that will be useful in the subsequent discussion:

$$\begin{aligned} \begin{aligned} {\mathscr {T}}_{i}^{j}&{:}{=}\bigcup _{m=0}^{k_{1}-i} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y) \text { if } 0\le j \le k'_{1}-k_{1}+i \\ {\mathscr {T}^\prime }_{i}^{j}&{:}{=}\bigcup _{m=0}^{k'_{1}-j} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y) \text { if } k'_{1}-k_{1}+i+1\le j \le k'_{1}\\ \end{aligned} \end{aligned}$$

Theorem 5

Let $\displaystyle G={\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}}\dots p_{1}^{k'_{1}}}$. Then,

• (1) For $0\le j \le k'_{1}-k_{1}+i$, the subgraph generated by ${\mathscr {T}}_{i}^{j}$ is Hamiltonian.

• (2) For $k'_{1}-k_{1}+i+1\le j \le k'_{1}$, the subgraph generated by ${\mathscr {T}^\prime }_{i}^{j}$ is Hamiltonian.

Proof

Claim 1. $N_{r,r+1}(x,y)= N_{p_{1}x}^{r+1}(p_{1}y)\subseteq N_{{\mathscr {H}}_{r+1}}(p_{1}y)$.

This is true because of Definition 1 and that $q_{1}(y)=q_{1}(p_{1}y)$ with $(x,y)\in {\mathscr {H}}_{r}$ and $(p_{1}x,p_{1}y)\in {\mathscr {H}}_{r+1}$.

As an immediate consequence of claim 1 and Theorem 4(1), we obtain

Claim 2. The subgraph generated by the sets $N_{p_{1}^{r}x}^{r}(p_{1}^{r}y)\subset N_{{\mathscr {H}}_{r} (p_{1}^{r}y)}$ and $N_{r,r+1}(p_{1}^{r}x,p_{1}^{r}y)= N_{p_{1}^{r+1}x}^{r+1}(p_{1}^{r+1}y)\subset N_{{\mathscr {H}}_{r+1} (p_{1}^{r+1}y)}$ is complete and has cardinality $p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}p_{1}^{k'_{1}-k_{1}-i+r+j}$.

We now proceed to show the first statement.

Case 1: $j-i+k_{1}-1\le k'_{1}$.

Subcase 1: $0\le j \le k'_{1}-k_{1}+i-1$. Consider the vertex set

$$\begin{aligned} V^* {:}{=}\bigcup _{m=0}^{1} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y)\ \end{aligned}$$

By Theorem 3 (2), there are $(p_{1}-1)p_{1}^{k_{1}-i-1}$ disjoint Hamiltonian subgraphs generated by the set $\bigcup _{k_{1}(y)=j} N_{{\mathscr {H}}_{i}}(y)$, and by Theorem 4 (2), we can build a Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{i+1}}(p_{1}y)\cup (\bigcup _{r=0}^{p_{1}-1} N_{{\mathscr {H}}_{i}}(y+rp_{1}^{q_{1}(y)-1}))$.

To construct a Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{i+1}}(p_{1}y)\cup (\bigcup _{r=0}^{p_{1}-1} N_{{\mathscr {H}}_{i}}(y+rp_{1}^{q_{1}(y)-1}))$, let $v, v'\in N_{{\mathscr {H}}_{i}}(y)$ be two adjacent vertices in the Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{i}}(y)$ and let $v_{r}, v'_{r}\in N_{{\mathscr {H}}_{i}}(y+rp_{1}^{q_{1}(y)-1})$ be two adjacent vertices in the Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{i}}(y+rp_{1}^{q_{1}(y)-1})$. Let $u,u_{1},\dots ,u_{p_{1}}$ be consecutive vertices along the Hamiltonian path in $N_{p_{1}x}^{i+1}(p_{1}y)$ that are also contained in the Hamiltonian cycle in the subgraph generated by the set $N_{{\mathscr {H}}_{i}}(p_{1}y)$. Such a set of vertices exist by construction of the Hamiltonian cycle in $N_{{\mathscr {H}}_{i+1}}(p_{1}y)$ in Lemma 6 and because $\vert N_{p_{1}x}^{i+1}(p_{1}y) \vert > p_{1}$. Now, let $T_{y+rp_{1}^{q_{1}(y)-1}}$ be the Hamiltonian path in $N_{{\mathscr {H}}_{i}}(y+rp_{1}^{q_{1}(y)-1})$ 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 $N_{{\mathscr {H}}_{i+1}}(p_{1}y)$ with initial vertex $u_{p_{1}}$ and final vertex u. A Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{i+1}}(p_{1}y)\cup (\bigcup _{r=0}^{p_{1}-1} N_{{\mathscr {H}}_{i}}(y+rp_{1}^{q_{1}(y)-1}))$ can now be constructed as sketched in Figure 3). More formally, it is of the form $v\rightarrow T_{y} \rightarrow v' \rightarrow u_{1} \rightarrow v_{1}\rightarrow T_{y+p_{1}^{q_{1}(y)-1}} \rightarrow v'_{1} \rightarrow u_{2} \rightarrow v_{2} \rightarrow T_{y+2p_{1}^{q_{1}(y)-1}}\rightarrow v'_{2} \rightarrow u_{3}\rightarrow \dots \rightarrow u_{p_{1}} \rightarrow T_{p_{1}y} \rightarrow u\rightarrow v$. The adjacencies between u and v and between $u_{r}$ and $v_{r}$ are guaranteed by claim 2.

Fig. 3

Hamiltonian cycle in $N_{{\mathscr {H}}_{i+1}}(p_{1}y)\cup (\bigcup _{r=0}^{p_{1}-1} N_{{\mathscr {H}}_{i}}(y+rp_{1}^{q_{1}(y)-1}))$

An analogous construction for each set $N_{{\mathscr {H}}_{i+1}}(p_{1}y)$ with $k_{1}(y)=j+1$ yields $(p_{1}-1)p_{1}^{k_{1}-i-2}$ Hamiltonian cycles that cover all vertices of $V^*$. We then consider the $(p_{1}-1)p_{1}^{k_{1}-i-2}$ cycles that cover $V^*$ and perform the same construction for $N_{{\mathscr {H}}_{i+2}}(p_{1}^{2}y)$. This yields $(p_{1}-1)p_{1}^{k_{1}-i-3}$ disjoint Hamiltonian cycles that cover $\bigcup _{m=0}^{2} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y)$. Proceeding in this manner, we obtain at step $k_{1}-i-1$ a collection of $p-1$ disjoint cycles that cover $\bigcup _{m=0}^{k_{1}-i-1} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y)$. Similar to the previous paragraph, the resulting Hamiltonian cycle in ${\mathscr {T}}_{i}^{j}$ is: $u_{1} \rightarrow v_{1}\rightarrow T_{p_{1}^{k_{1}-i-1}y}\rightarrow v'_{1} \rightarrow u_{2}\rightarrow v_{2}\rightarrow T_{p_{1}^{k_{1}-i-1}y +p_{1}^{q_{1}(y)}}\rightarrow v'_{2}\rightarrow u_{3}\rightarrow \dots \rightarrow v_{p_{1}-1}\rightarrow T_{p_{1}^{k_{1}-i-1}y+(p_{1}-1)p_{1}^{q_{1}(y)}}\rightarrow v'_{p_{1}-1}\rightarrow u_{p_{1}}\rightarrow T_{p_{1}^{k_{1}-i}y} \rightarrow u_{1}$, where $v_{r}, v'_{r}\in N_{{\mathscr {H}}_{k_{1}-1}}(y+rp_{1}^{q_{1}(y)-1})$ are two adjacent vertices in the Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{k_{1}-1}}(y+rp_{1}^{q_{1}(y)-1})$. As before, let $u,u_{1},\dots ,u_{p_{1}}$ be consecutive vertices in the Hamiltonian path in $N_{p_{1}^{k_{1}-i}x}^{k_{1}-i}(p_{1}^{k_{1}-i}y)$ that are also part of the Hamiltonian cycle in the subgraph generated by the set $N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{k_{1}-i}y)$. Such a set of vertices exist because the subgraph generated by ${\mathscr {H}}_{k_{1}}$ is complete and because $\vert N_{{\mathscr {H}}_{k_{1}}}^{i+1}(p_{1}^{j+k_{1}-i}) \vert > p_{1}$. Let $T_{y+rp_{1}^{q_{1}(y)-1}}$ be the Hamiltonian path in $N_{{\mathscr {H}}_{k_{1}-1}}(y+rp_{1}^{q_{1}(y)-1})$ 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 ${\mathscr {H}}_{k_{1}}$, 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 $(p_{1}-1)p_{1}^{k'_{1}-j-1}=(p_{1}-1)p_{1}^{k_{1}-i-1}$.

Case 2: $j+k_{1}-i \ge k'_{1}$. First, we note that $k'_{1}-j\le k'_{1}-k'_{1}+k_{1}-i=k_{1}-i$. The initial steps are exactly the same as in case 1. In the step $k'_{1}-1-j$, therefore, we obtain $p_{1}-1$ disjoint cycles that cover the vertex set $\bigcup _{m=0}^{k'_{1}-1-j} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y)$. We next construct a Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{i+k'_{1}-j}}(p_{1}^{k'_{1}-j}y)\cup (\bigcup _{r=1}^{p_{1}-1} N_{r})$, where $N_{r}$ is one of the $p_{1}-1$ cycles that cover the set $\bigcup _{m=0}^{k'_{1}-2-j} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y)$. To this end, consider two vertices $v_{r}, v'_{r}\in N_{i+k'_{1}-1-j}(p_{1}^{k'_{1}-1-j}y+rp_{1}^{k'_{1}-1})\subset N_{r}$ that are adjacent along the cycle covering $N_{r}$. Furthermore, let $u_{1},\dots ,u_{p_{1}}$ be consecutive vertices in the Hamiltonian path in $N_{{\mathscr {H}}_{i+k'_{1}-j}}(p_{1}^{k'_{1}-j}y)$. Let $T_{r}$ be the Hamiltonian path in $N_{r}$, with initial vertex $v_{r}$ and final vertex $v'_{r}$ and T be a path, part of the Hamiltonian cycle in $N_{{\mathscr {H}}_{i+k'_{1}-j}}(p_{1}^{k'_{1}-j}y)$ with initial vertex $u_{p_{1}}$ and final vertex $u_{1}$. A Hamiltonian cycle in the subgraph generated by $N_{{\mathscr {H}}_{i+k'_{1}-j}}(p_{1}^{k'_{1}-j}y)\cup (\bigcup _{r=0}^{p_{1}-1} \bigcup _{m=0}^{k'_{1}-1-j} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y)$ can now be obtained as follows: $v_{1}\rightarrow T_{1} \rightarrow v'_{1} \rightarrow u_{1} \rightarrow v_{2}\rightarrow T_{2} \rightarrow v'_{2} \rightarrow u_{2} \rightarrow v_{3} \rightarrow T_{3}\rightarrow v'_{3} \rightarrow u_{3}\rightarrow \dots \rightarrow u_{p_{1}} \rightarrow T \rightarrow u_{1}\rightarrow v_{1}$. As a consequence of Theorem 3(4), we obtain a Hamiltonian cycle that covers the set of vertices $\bigcup _{m=0}^{k'_{1}-j} \bigcup _{k_{1}(p_{1}^{m}y)=j+m} N_{{\mathscr {H}}_{m+i}}(p_{1}^{m}y)={\mathscr {T}^\prime }_{i}^{j}$. $\square$

Theorem 6

If $G={\mathbb {Z}}_{p_{1}^{k_{1}}}\oplus {\mathbb {Z}}_{p_{l}^{k'_{l}} \dots p_{1}^{k'_{1}}}$, then ${\mathscr {G}}(G)$ is Hamiltonian.

Proof

It is easy to note that $G=(\bigcup _{r=0}^{k'_{1}-k_{1}} {\mathscr {T}}_{0}^{r})\cup (\bigcup _{r=k'_{1}-k_{1}+1}^{k'_{1}} {\mathscr {T}^\prime }_{0}^{r})\cup (\bigcup _{r=1}^{k_{1}}{\mathscr {T}}_{r}^{0})$.

Claim. The subgraphs induced by the vertex sets (i) ${\mathscr {T}}_{0}^{r}$ and ${\mathscr {T}}_{0}^{r+1}$ with $0\le r \le k'_{1}-k_{1}$, (ii) ${\mathscr {T}^\prime }_{0}^{r}$ and ${\mathscr {T}^\prime }_{0}^{r+1}$ with $k'_{1}-k_{1}+1 \le r \le k'_{1}$, and (iii) ${\mathscr {T}}_{r}^{0}$ and ${\mathscr {T}}_{r+1}^{0}$ with $0\le r\le k_{1}$ each contain a clique of cardinality $p_{l}^{k'_{l}}\dots p_{2}^{k'_{2}}$.

This is true because of the claim 2 in Theorem 5, and the fact that, for the first case, $N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{k_{1}+r})\subset {\mathscr {T}}_{0}^{r}$, $N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{k_{1}+r+1})\subset {\mathscr {T}}_{0}^{r}$, for the second case, $N_{{\mathscr {H}}_{k'_{1}-r}}(p_{1}^{k'_{1}})\subset {\mathscr {T}^\prime }_{r}^{0}$, $N_{{\mathscr {H}}_{k'_{1}-r-1}}(p_{1}^{k'_{1}})\subset {\mathscr {T}}_{r+1}^{0}$ and, for the third case, $N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{k_{1}-r})\subset {\mathscr {T}}_{0}^{r}$. $N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{k_{1}-r-1})\subset {\mathscr {T}}_{0}^{r}$.

Now, we can construct the Hamiltonian cycle in ${\mathscr {G}}(G)$ as follows. Let $v_{i}^{j}, {v'}_{i}^{j}\in {\mathscr {T}}_{i}^{j}$ or $v_{i}^{j}, {v'}_{i}^{j} \in {\mathscr {T}^\prime }_{i}^{j}$, be pairs of adjacent vertices in the Hamiltonian cycle in the subgraph generated by ${\mathscr {T}}_{i}^{j}$ or ${\mathscr {T}^\prime }_{i}^{j}$, respectively. Let $T_{i}^{j}$ be the Hamiltonian path in ${\mathscr {T}}_{i}^{j}$ with initial vertex $v_{i}^{j}$ and final vertex ${v'}_{i}^{j}$ and ${T'}_{i}^{j}$ be the Hamiltonian path in ${\mathscr {T}^\prime }_{i}^{j}$ with initial vertex $v_{i}^{j}$ and final vertex ${v'}_{i}^{j}$. For the first case, $v_{0}^{r},{v'}_{0}^{r}\in N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{k_{1}+r})\subset {\mathscr {T}}_{0}^{r}$, for the second case $v_{0}^{r},{v'}_{0}^{r}\in N_{{\mathscr {H}}_{k'_{1}-r}}(p_{1}^{k'_{1}})\subset {\mathscr {T}^\prime }_{0}^{r}$ and for the third case, $v_{0}^{r},{v'}_{0}^{r}\in N_{{\mathscr {H}}_{k_{1}}}(p_{1}^{k_{1}-r})\subset {\mathscr {T}}_{r}^{0}$, these three pair of vertices are adjacent in ${\mathscr {G}}(G)$ because of the claim. Finally, let T be the path in the subgraph generated by ${\mathscr {H}}_{k_{1}}$ with the vertices that were not used at the end of the process in the construction of the Hamiltonian cycle. A Hamiltonian cycle in ${\mathscr {G}}(G)$ can now be composed as follows:

$(0,0)\rightarrow v_{0}^{k'_{1}}\rightarrow {T'}_{0}^{k'_{1}}\rightarrow {v'}_{0}^{k'_{1}}\rightarrow v_{0}^{k'_{1}-1}\rightarrow {T'}_{0}^{k'_{1}-1}\rightarrow {v'}_{0}^{k'_{1}-1}\rightarrow \dots \rightarrow v_{0}^{k'_{1}-k_{1}+1 }\rightarrow {T'}_{0}^{k'_{1}-k_{1}+1}\rightarrow {v'}_{0}^{k'_{1}-k_{1}+1}\rightarrow v_{0}^{k'_{1}-k_{1}}\rightarrow {T'}_{0}^{k'_{1}-k_{1}}\rightarrow {v'}_{0}^{k'_{1}-k_{1}}\rightarrow \dots \rightarrow v_{0}^{1}\rightarrow T_{0}^{1}\rightarrow {v'}_{0}^{1}\rightarrow v_{0}^{0}\rightarrow T_{0}^{0}\rightarrow {v'}_{0}^{0}\rightarrow v_{1}^{0}\rightarrow T_{1}^{0}\rightarrow {v'}_{1}^{0}\rightarrow \dots \rightarrow v_{k_{1}}^{0} \rightarrow T_{k_{1}}^{0}\rightarrow {v'}_{k_{1}}^{0}\rightarrow T\rightarrow (0,0).$ $\square$