** Open Journal of Discrete Mathematics** Vol.2 No.1(2012), Article ID:17157,11 pages DOI:10.4236/ojdm.2012.21006

On the Line Graph of the Complement Graph for the Ring of Gaussian Integers Modulo n

^{1}Department of Mathematics, Irbid National University, Irbid, Jordan

^{2}Department of Mathematics, Palestine Technical University (Kadoorie), Tulkarm, Palestine

Email: dr_mghanem@yahoo.com, k.nazzal@ptuk.edu.ps

Received October 13, 2011; revised November 2, 2011; accepted November 22, 2011

**Keywords:** Complement of a graph; Chromatic index; Diameter; Domination number; Eulerian graph; Gaussian integers modulo n; Hamiltonian graph; Line graph; radius; Zero divisor graph

ABSTRACT

The line graph for the complement of the zero divisor graph for the ring of Gaussian integers modulo n is studied. The diameter, the radius and degree of each vertex are determined. Complete characterization of Hamiltonian, Eulerian, planer, regular, locally and locally connected is given. The chromatic number when is a power of a prime is computed. Further properties for and are also discussed.

1. Introduction

The line graph of a graph is defined to be the graph whose vertex set constitutes of the edges of, Where two vertices are adjacent if the corresponding edges have a common vertex in. The importance of line graphs stems from the fact that the line graph transforms the adjacency relations on edges to adjacency relations on vertices. For example, the chromatic index of a graph leads to the chromatic number of its line graph. The zero divisor graph of a commutative ring, denoted by, is defined as the graph whose vertex set is the set of all non-zero zero divisors of and edge set. This type of graphs provides an example showing that algebraic methods could be applied to problems about graphs. The set of Gaussian integers, denoted by, is defined as the set of complex numbers, where . If is a prime Gaussian integer, then is either 1) or, or 2) q where q is a prime integer and, or 3), where, is a prime integer and.

Throughout this paper, and denote prime integers which are congruent to 1 modulo 4, while and and denote prime integers which are congruent to 3 modulo 4. All rings in this paper are assumed to be commutative with unity. The zero divisor graph for the ring of Gaussian integers modulo is studied in [1] and [2], the complement of this graph is discussed in [3]. While the line graph of the zero divisor graph for the ring of Gaussian integers modulo n is investigated in [4]. In this paper it should be kept in mind that

, and hence, its line graph is,

is an integral domain, so. Further,

is a complete graph whose complement is totally disconnected and thus its line graph is. While

, so its complement is disconnected with two components each of which is isomorphic to. Finally, note that the graph is bipartite, [1] and.

In this paper, we investigate properties of the graph

. We find the diameter, the radius of

. We determine which is Eulerian, Hamiltonian, regular, locally, locally connected or planer. Furthermore, the chromatic index and the edge domination number of where is a power of a prime are computed. While the domination number of is given. On the other hand, a formula which gives the degree of each vertex in is derived, thus the degree of its complement as well as its line graph could easily be found.

2. When Is Eulerian or Planner

If is a connected graph. Then is Eulerian if and only if every vertex of has even degree. For a finite ring, the line graph of a connected graph is Eulerian if and only if all vertices of have the same parity ( see the proof of Lemma 3.10, [5]). On the other hand, if has both even and odd vertices, then so is its complement. So, for a connected graph, the graph is Eulerian if and only if all vertices in are either even or all vertices in are all odd. But is connected if [3] and is Eulerian if or is a product of distinct odd primes [1]. It is easy to show that all vertices of are odd if and only if. This proves the following theorem.

**Theorem 2.1 ** is Eulerian if and only if is a product of distinct odd primes.

A planar graph is a graph that can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their endpoints.

Next we determine when the graph is planar.

In a graph the maximum vertex degree and the minimum vertex degree will be denoted by and, respectively.

The following theorem characterizes graphs whose line graph is planer.

**Theorem 2.2** [6]

A nonempty graph has a planer line graph if and only if 1) is planer.

2), and 3) if, then is a cut vertex.

The graph is planer if and only if

or [3]. For, ,. While for, , this graph is regular of degree 3.

Thus we obtain the following.

**Theorem 2.3** The graph is planer if and only is.

3. The Diameter of

For a connected graph, the distance, , between two vertices and is the minimum of the lengths of all paths of. The eccentricity of a vertex in is the maximum distance from to any vertex in. The diameter of, , is the maximum eccentricity among the vertices of. Since

is connected if and each of and is the union of two complete graphs, while and are the union of a nullgraph and a connected graph [3], we have the following.

**Theorem 3.1** is connected if and only if.

**Theorem 3.2** If or, then

.

**Proof**. 1) Assume that and

are two nonadjacent vertices in. Since for every, and are both even or odd [1], we have three cases:

**Case I**: for, and are odd. Then we have the path.

**Case II**: for, or is odd(even) and or is even (odd). Assume that are even and are odd. Then we have the path .

**Case III**: for, and are even.

Then and

where

are odd and for. If or

, say, then or, say.

So, we have the path. Now suppose that is odd. Then a) If, for or 2, say for

, then or, say. Hence, we have the path.

b) If or and, for or 2, say for, then we have a path or .

c) If, for or 2, say for

, then implies that. Otherwise or. Then we have a path

or .

2) Assume that and

Then or, say. Hence or

, say. Then we have the path

.

**Theorem 3.3** Let be a ring that is a product of two rings and with at least one of them is not ID with more than one regular element and the other has more than two regular elements. Then .

**Proof**. Suppose that and is not ID, and. Let and. Clearly,

in. So,

. Now, let

then or and or. So, we have three cases:

**Case I**: and. Then

implies that

.

And or, say implies that

where.

**Case II**: and. Then there exists and hence

.

**Case III**: and or and. Let and. Then

implies that

and

or.

And if, then or

and or.

For, [7] and for

with,.

Moreover and for

. An immediate consequence of Theorem 3.3 is the following.

**Theorem 3.4** Let or n is a composite such that. Then

.

4. The Radius and the Girth of the Graph

For a connected graph, the radius of, , is the minimum eccentricity among the vertices of. So,. Since for any

, and are non adjacent,. Using **Theorem 3.2** gives for or,

.

**Theorem 4.1** If or where

, is prime integer, andthen.

**Proof**. Since to show that

it is enough to find a vertex

with eccentricity 2. If

, then

for every

. So.

Now, assume that and

.

Then we have four cases:

**Case I**:. Then

.

**Case II**:. Then

.

**Case III**: and. Then and hence there exists. So,

.

**Case IV**:. Then

.

**Theorem 4.2** if and only if

or.

Vising [8], proved that for a connected simple graph with n-vertices and radius 2, the upper bound of the number of edges of is. Then Golberg [9]

proved that the lower bound of numbers of edges of a simple connected graph with radius 2 is.

So we can conclude the following.

**Theorem 4.3** For or,

implies that

.

The girth of a graph, is the length of a shortest cycle contained in the graph. If the graph does not contain any cycles (i.e.. it’s an acyclic graph), its girth is defined to be infinity. If is a cycle of length three in. Then is a cycle of length 3 in. So, whenever. In [3] it is proved that the girth of

equals 3 for. So, we have the following.

**Theorem 4.4** For,.

5. The Locally Connected Property of the Graphs and

We say that a vertex is locally connected if the neighborhood of, , is connected; and is locally connected if every vertex of is locally connected.

**Theorem 5.1** If for and either or is not ID, then is locally connected.

**Proof**. Suppose that is not ID and

. Then we have two cases:

**Case I**: or. If, then there exists. So for all

. And if, then there exists

such that. Therefore, for every

. So is connected.

**Case II**: and. Then there exist

, and

such that and

. Moreover,

. And for every, or. So is connected.

**Theorem 5.2** If for

and either or is not ID, then is locally connected.

**Proof**. Suppose that is not ID,

and, then we have three cases:

**Case I**:. Then

.

**Case II**:. If, then

.

Otherwise there exists. So,

.

**Case III**: or. Assume that, then implies that there exists satisfies

.

While implies that that there exists

satisfies

.

From Theorem 5.1 and Theorem 5.2 we conclude the following.

**Theorem 5.3** If or is a composite integer such that, then both and

are locally connected.

6. When Is Hamiltonian?

A Hamiltonian cycle is a cycle that visits each vertex exactly once (except the vertex which is both the start and end, and so is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph. The line graph of a graph with more than 4 vertices and diameter 2 is Hamiltonian [10]. But is disconnected with one isolated vertex and the other component, call this component, with diameter 2 [3]. So,. Similarly,

has a connected subgraph with diameter 2 and. Hence, the following result is obtained.

**Theorem 6.1** If or, then

is Hamiltonian.

Oberly and Sumner [11] proved that every connected, locally connected claw free graph (i.e. it does not contain a complete bipartite graph) is hamiltonian. Since the line graph is claw free, using Theorem 5.3, we get the following.

**Theorem 6.2** If or is a composite integer such that, then is hamiltonian.

7. The Chromatic Number of the Graph

The edge coloring of a graph is an assignment of colors to the edges of the graph so that no two adjacent edges have the same color. The minimum required number of colors for the edges of a given graph is called the chromatic index of the graph denoted by.

**Lemma 7.1** [12]

If has order and, then .

**Theorem 7.2** If, then

.

**Proof**. Note that in, the induced subgraph,

, with is connected, , [1] and

. Since the vertex is adjacent to all other vertices in, we have

. Using Lemma 6.1,

.

Since is empty graph and

is edgeless with vertices, we consider the case.

**Theorem 7.3** If, then

**Proof.** Let

Then is the set of all isolated vertices in.

So the induced subgraph, , with the vertices

is a connected graph,

. Clearly the vertex is adjacent to all other vertices in and hence,

. Using Lemma 7.1,

Finally we find the chromatic index of

.

A subset of the vertex set is said to be independent if no two vertices in this set are adjacent. A clique of a graph is a maximal complete subgraph. A graph is said to be split if it’s vertex set can be partitioned into two subsets and such that induces a clique and is independent in.

**Lemma 7.4** [13] Let be a split graph. If is odd, then.

**Theorem 7.5** If, then

.

**Proof**. Since, it is enough to find

. First, we’ll show that is a split graph. Let

.

Clearly, , induces a clique and is independent. Therefore, is a split graph. Moreover,

is odd. From Lemma 7.4,

.

A graph is said to be critical if is connected and and for every edge of, we have. The well-known Vizing’s theorem states that for a simple graph, or.

**Lemma 7.6** [14]

If is a critical graph, then has at least of vertices of maximum degree.

Therefore, if is a simple graph such that for every vertex of maximum degree there exists an edge such that is more than the number of vertices with maximum degree in, we have [13].

**Theorem 7.7** If, then

.

**Proof**. Let and.

Then the vertices of with maximum degree have the form or where and and

and

So,. And the vertices of with minimum degree have the form or where

and

. So

.

Therefore,

.

But the graph has only

vertices of maximum degree. So,

.

Since, the result holds.

Since the edge coloring of any graph leads to a vertex coloring of its line graph, we obtain the following.

Corollary 7.8 1) If, then

.

2) If, then

.

3) If, then

.

8. The Domination Number of

A subset of the vertex set of a graph is a dominating set in if each vertex of, not in, is adjacent to at least one vertex of. The minimum cardinality of all dominating sets in, , is called the domination number of.

In, the vertex is an isolated vertex while the vertex dominates all vertices in the second component. Therefore,

. The graphthus. In the vertices are isolated while the vertex is adjacent to all other vertices in

so. Since

and,

.

The set is a minimum dominating set for. And if, where

, then. This graph is connected and the set is a minimum dominating set for.

**Theorem 8.1** 1) If, then

.

2) and

9. The Domination Number of

The independence number of, , is the maximum cardinality of all independent sets in. A subset of the edge set of a graph is an edge dominating set in if each edge of, not in, is adjacent to at least one edge of. The minimum cardinality of all edge dominating sets in, , is called the edge domination number of. The minimum cardinality of all independent edge dominating sets, , is called the independence edge domination number of. The study of the domination number of the line graph of leads to the study of edge or line domination number of, i.e.. On the other hand, for any graph, [15].

If is an independent set in, then induces a complete graph in. While if induces a complete graph in, then it is independent in. Recall that [2]. Then the sets,

, form a partition for the set. Clearly, the set is the maximum independent set in

, while the set induces a maximum complete subgraph in. There are some edges joining to, no other adjacency exists in

. Any edge dominating set for must contain at least element in order to dominate

. On the other hand, this dominating set for dominates all other edges in. Since

, then and, could easily be computed to get the following theorem.

**Theorem 9.1** For.

1).

2).

3)

To study the graph, consider the partition of given by

and not both. The set

is the maximum independent set, while induces a maximum complete subgraph in. There are some edges joining to, and has no other adjacency. Easy calculations give

when,

and when

. While and

.

Thus we obtain the following theorem.

Theorem 9.2 If, then 1).

2) if is even and if is odd.

3)

Now, we move to the case. Let

.

Clearly, the sets where and not both or 0, partition the vertices of

and. Let

Note that induces a complete graph in

. Vertices in are adjacent to all vertices except some vertices in . Similarly, vertices in are adjacent to all vertices except some vertices in, and vertices in are adjacent to all vertices except vertices in. On the other hand induces a complete subgraph and vertices in this set are adjacent to all other vertices except those of. Clearly induces a complete subgraph. Vertices in form an independent set, and are adjacent to some vertices in. Each of and induces a complete subgraph and are adjacent to some vertices in. Besides, there are some edges between and. On the other hand,

The above argument shows that

10. The Degree of the Vertices in and

Now, we determine the cardinality of the annihilator of the element, in. This helps find the degree of each vertex in, its complement, as well as the degree of each vertex in their corresponding line graphs.

**Theorem 10.1** If, then

where.

Proof. Let and. Then

.

.

But. So,

and hence there exists such that.

Since where and the norm of is less than the norm of,

. By Theorem 2 of [7], , so the result holds.

**Theorem 10.2** Let and . Then

.

The order of can be easily computed using formulas given in [1]. Thus we can find the degree of each vertex in the complement of, here we give the degree of each vertex in the line graph of, an analogous formula for the degree of vertices in could be obtained.

**Corollary 10.3** Let,

and. Then

.

**Proof**. Note that, for any graph and,

.

In the following we determine the degree of every vertex in the graphs when and.

**Theorem 10.4** Let and are odd. Then in1).

2)

.

3).

**Proof**. 1) Note that, if

or and

if and only if and. Moreover

if and only if.

2) Obvious.

3) Note that if are odd, then .

Theorem 10.5 Let, are relatively prime with. Then in,

.

**Theorem 10.6** Let, and. Then in,

11. When is, Regular?

A graph in which all vertices have the same degree is called regular graph.

Regularity of was studied in [1]. However, we provide our own proof, since it comes as an immediate consequence of Theorem 10.2. Clearly, if, then is regular. If or, then the graph has a vertex which is adjacent to all other vertices and it is not complete graph, thus is not regular.

Now, we show that is regular if and only if.

**Theorem 11.1** If where are distinct Gaussian primes and and , then is not regular.

**Proof**. Choose two vertices and such that, then. So, the result follows.

Next, we discuss regularity of the graph

and. Clearly, if is regular, then is also regular, so if, then the graph is regular. On the other hand, if is the complete bipartite graph, then

for all vertices in. Thus

is regular. While is a bipartite graph with partite sets

and

Moreover, , and. Thus,

and hence, is not regular.

**Theorem 11.2** If, is a prime and, then the graph is not regular.

**Proof.** If, then

If

, then.

And if, , , then

**Theorem 11.3** Let where and are commutative rings with unity with at least one of them is not ID. Then is not regular.

**Proof**. Suppose that is not ID and, for. Let. If, then

and if

, hence

. And if,

and if, hence

. But

. So is not regular.

So as a consequence of Theorem 11.2 and Theorem 11.3, we conclude the following.

**Theorem 11.4** The graph is regular if and only if.

Observe that, for, is the empty graph., so the line graph

is regular. While

which is regular, so is.

And in,

. So, the graph

is not regular for, is a prime and.

**Theorem 11.5** Let where and are commutative rings with unity such that

, for. If

and, then is not regular.

**Proof**. Since, for, there exist and. Therefore

. Since

,

.

So, is not regular.

**Theorem 11.6** The graph is regular if and only if or.

12. When is, Locally H?

A simple graph is said to be locally if the neighborhood of each vertex in induces the same graph. The cartesian product of two graphs and is the graph with vertex set and two vertices in

are adjacent if and only if they are equal in one coordinate and adjacent in the other. Before we proceed, we give the following lemma.

**Lemma 12.1** 1) If, then is locally .

2) If, then is locally .

Proof. 1) Let, then

each of the sets and

induces a copy of and since we deal with an undirected graphs, then for a fixed, and are adjacent. Thus the result holds.

3) Let, with partite sets and and with,. Then

.

Each set induces a complete graph, respectively. And has no other edges. Thus induces.

In order for a graph to be locally, it should be regular graph. Thus for the graph, it suffices to check the cases, and for

, we consider only the cases. Since and,

is locally and

is locally. In the same manner we can show that is locally , is locally and

is locally.

**Theorem 12.2** The following statements are equivalent.

1) The graph is regular2) The graph is locally.

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