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`Journal of MathematicsVolume 2018, Article ID 9304645, 6 pageshttps://doi.org/10.1155/2018/9304645`
Research Article

## The Structure of Isomorphic Digraph from Powers Modulo

1School of Mathematical Sciences, Inner Mongolia University, Hohhot 010021, China
2College of Mathematics and Statistics, Guangxi Teachers Education University, Nanning 530023, China
3Institute of Water Resources for Pastoral Ministry of Water Resources, Hohhot 010020, China

Correspondence should be addressed to Tiejun Liu; moc.621@jtlskm

Received 30 August 2018; Accepted 12 November 2018; Published 2 December 2018

Copyright © 2018 Jinxing Zhao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

For positive integers and , let denote the digraph whose set of vertices is and there is a directed edge from to if . It is great value to consider conditions of . In this paper, we obtain necessary and sufficient conditions for when

#### 1. Introduction

Let be a nonempty set and be a map from to itself. Let denote the digraph whose set of vertices is and there is a directed edge from to if . In the case that is a commutative ring or a finite abelian group (written multiplicative) and maps into its -th power in , we denote by . In particular, let denote the digraph whose set of vertices is and there is a directed edge from to if , or equivalently . The investigation of graphs of this sort is important, in particular, because they have valuable applications, as explained, for example, in the book [1] and survey [2].

There are two particular subdigraphs of . Let be the induced subdigraph of on the set of vertices which are coprime to and be the induced subdigraph of on the set of vertices which are not coprime to . We observe that and are disjoint and that . It is clear that , where is the unit group of .

Many interesting properties of are obtained by people. The distribution of indegree and structure of cycles of are obtained in [39]. Somer and Křížek [6, 10] determined when is semiregular. Carlip and Mincheva [11] called a digraph symmetric of order if the component of can be partitioned into subsets of size , each containing isomorphic components. They also proved that is symmetric of order 2 for and . Somer and Křížek obtained all which are symmetric of order 2. Deng and Yuan [12] determined all which are symmetric of order , where has an odd prime divisor. Meemark and Wiroonsri [13, 14] investigated the structure of , where is the quotient ring of polynomials over finite fields.

In this paper, we are mainly interested in determining when . Lucheta, Miller, and Reiter [4] showed that if and only if . They also found that and thus two digraphs and can be isomorphic without the condition. Wilson [9] also obtained that if and only if , where is the Carmichael lambda-function. Deng and Yuan [15] gave a necessary and sufficient condition for . They also determined that for an abelian group by using graph theory and group theory. In this paper, we obtain a necessary and sufficient condition for .

This paper is organized as follows. In Section 2, we review some basic properties of including indegrees, cycles, and structure of components. We point out that is an induced subdigraph of . In Section 3, we review the properties of product of digraphs. In Section 4, we investigate when is a power of a prime and obtain a necessary and sufficient condition for that .

#### 2. Some Preliminary Results on and

In this section we review some basic properties of and , where is a finite abelian group. The structure of and hence is well understood in [16]. It has highly symmetric structure.

Before proceeding further, we need to review some properties of the Carmichael lambda-function and the exponent of an abelian group.

Definition 1. Let be a positive integer. Then the Carmichael-lambda-function is defined as follows: , where are distinct primes and for all .

The following theorem generalizes the well-known Euler’s theorem which says that if and only if .

Theorem 2 (Carmichael). Let . Then if and only if . Moreover, there exists an integer such that , where denotes the multiplicative order of modulo .

For the proof see [17, p. 21].

Let be a finite group. The exponent of , denoted by , is the minimal positive integer such that for any It is well known that if is the prime factorization of , then and hence Thus, is exactly the exponent of the unit group of .

We have the following result about and .

Theorem 3 ([16] Theorem 3.2-3.4). Let be an abelian group and let be the exponent of . Then if and only if the following conditions hold:
(i) ,
(ii) we have a factorization of , where is the maximal divisor of relatively prime to and is also the maximal divisor of relatively prime to ; for any divisor of ,
Moreover, let be the component of containing the identity element 1 of , . Condition is equivalent to .

Definition 4. We define a height function on the vertices and components of . Let be a vertex of . Then is defined to be the minimal nonnegative integer such that is congruent modulo to a cycle vertex in . If is a component of , then we define .

The indegree and height function play an important role in the structure of . We need the following results concerning the indegrees and heights.

Lemma 5 ([10] Theorem 3.1). Let be the prime factorization of . Let be a vertex of positive indegree in . Then where if and , and otherwise.

Lemma 6 ([10] Theorem 3.2). Let be a prime. Let be a vertex with positive indegree in . Suppose that and . Then for some positive integer and where if , and , and otherwise.

Lemma 7 ([18] Lemma 3.2). Let be a prime and , be two positive integers. Then

Lemma 8. Let be a prime and , be two positive integers. Let be the unique positive integer such that . Then .

Proof. It is clear that and . Moreover, if and only if . The proof is complete.

Lemma 9. Let be a prime and , be two positive integers. Let , where is the maximal divisor of relatively prime to . If is the component of containing 1, then

Proof. Let . Then there exists a divisor of such that is not a divisor of . By Theorem 2, there exists a vertex such that . Let . Then , , and . Hence, by the definition of the height function.
Conversely, if , then there exists such that , so . Since , we have , so . That is, .

#### 3. Properties of Digraphs Products

Given two digraphs and . Let denote the digraph whose vertices are the ordered pairs , where and there is a directed edge from to if there is a directed edge from to and a directed edge from to . Somer and Křížek [18] noted the following fact: if and , then . The canonical isomorphism is given by where , . In general,if is the prime factorization of . We use the following notation:where each is a digraph. We need this fact and the following lemmas.

Lemma 10 ([19] Lemma 3.1). Let where . Let be a component of with cycle length . Then is a subdigraph of consisting of components, each having cycles of length .

Corollary 11. Let , , and be digraphs. Suppose that . Then .

Proof. Let be the number of -cycles contained in , be the number of -cycles contained in , and be the number of -cycles contained in . Using Lemma 10 to compare the number of -cycles of , we have Let ; we see that . Now assume that for any . We have Thus and we finish the proof by induction.

Lemma 12 ([19] Lemma 3.12). Let be the prime factorization of . Let and be two vertices in . If and are in the same cycle, then and are in the same cycle for each .

Let denote the number of vertices of a digraph . Let , , and . The following lemma is immediately from the properties of product of digraphs.

Lemma 13. Let and be two digraphs, and and . Then . Moreover, , , and . If and , then .

Definition 14. For any positive integers and , we define to be the digraph which satisfies the following: (i) it has vertices and a -cycle, (ii) if is a cycle vertex and otherwise. In particular, let be the -cycle.

Remark 15. Let be a component of with . By Lemma 12, each cycle vertex of has the same indegree. Thus, , where is the cycle length of and is the indegree of a cycle vertex of . We use to denote the indegree of a cycle vertex of .

#### 4. The Main Result When

In this section we study , where is prime. It is clear that consists of a single component. We begin with the following Lemma.

Lemma 16. Let be an odd prime and , two positive integers. Suppose that . Then

Proof. Let denote the cyclic group of order . Recall that for an abelian group , is the digraph whose vertex set is . There exists a directed edge from to if and only if . Note that the unit group is a cyclic group of order . Thus, for any
By hypothesis we have , and hence . Since , we have

Recall that a digraph is called if there exists a positive integer such that each vertex of has either indegree or 0. In this case, is called -semiregular.

Lemma 17 ([10] Theorem 4.4). Let be a prime and , two positive integers.

(1)If is odd, then we can obtain that is semiregular if and only if or and , where . In the first case is -semiregular, and in the second case is -semiregular. Thus, is semiregular if and only if .(2)If , then is semiregular if and only if one of the following conditions hold:(a) and ;(b) and ;(c) and , where .Moreover, is semiregular if and only if one of the following conditions hold:(a) and ;(b) and ;(c) and , where .

Lemma 18. Let be a prime. Suppose that . Then for any positive integers , , we have , except the case .

Proof. In fact, for , we have and , where if and , and otherwise. It is clear that , since .
Now, let . Then . Assume that is odd. Suppose that . Let . Then . If , then . Then . If , we still have . Otherwise we have . Since , then , which is a contradiction.
Suppose that . By Lemmas 5 and 7, we know that . In the following, we assume that . Suppose that . Let . Then . If , then . Then . If , by easy computations we still have since .

We use the following notation. Let be a fixed prime. Let the mapping be defined by such that for , where is the set of nonnegative integers and is the set of positive integers. Furthermore, we define a function from the set of digraphs to the set of subsets of as follows: for any digraph . Let and be two nonempty subsets of . We denote . Since , we have for any digraphs and .

Lemma 19. Let be a prime and , be positive integers. Suppose that and . Then or .

Proof. If , then since for any . Suppose that . Then . But . It contradicts the fact that . In the following we always assume that .
Case 1. First we assume that is an odd prime. We show that . Let . We need to show that . By Lemmas 5, 6, and 7, we see thatWe have , and then for any . Hence for any . It implies that But . So . We get .
Now if , by (13) we have . Then , . If , then . We have .
Subcase 1. If , then and .
Subcase 2. If , then and . But we have and , . Let , . We haveWe get from this equation and thus .
Subcase 3. If , we have . Using the same notation we still have (16); thus .
Case 2. Now we treat the case . The case is trivial. Assume that ; in this case we still have . Let . It suffices to show that . By Lemmas 5, 6, and 7, we see that where if and , and otherwise. If , then . We have . Suppose that . Then . If , . Sowhere if , and otherwise. We have Thus , .
Now if , we have . Then , . If , then , where if , and otherwise. Assume that .
Subcase 1. . Then, we have .
(i) If , then , .
(ii) If , then . We have , let , . ThenWe get , . If , then . That is, one of , is odd and the other one is even. It is contrary to . If , we have or and . It implies that , . But it is contrary to .
(iii) If , then . Using the same notations we still have (20). It is impossible.
Subcase 2. . Then we have .
(i) If and , then , .
(ii) If and , then . We have . Let , . Then We must have , .
(iii) If and , then . It is a contradiction.
Subcase 3. . We have .
(i) If , then , . One of and is odd and the other one is even. It is contrary to .
(ii) If , then , and we still have . The lemma is proved.

Corollary 20. Let be a prime and be a positive integer. Let , be two positive integers. Then if and only if and .

Proof. Suppose that . It suffices to show that . If , then . We assume that and . If , by Theorem 3 we have and where is the component of containing the vertex 1. By Lemma 17 we have or , or . It implies that and therefore since . We have , which is a contradiction.
If , , one of , is 2. Then by easy computations we know that if and only if . Now let ; we have that , , and are semiregular. Then we have By Lemma 18, Combining these formulas, we obtain and is -semiregular, .
If , then . If , by Lemma 17 we know that . We still have . The other direction is trivial.

Theorem 21. Let be an odd prime and , , positive integers. Then if and only if one of the following conditions holds:
(i) If , then .
(ii) If , then , and we have a factorization of , where is the maximal divisor of relatively prime to and is also the maximal divisor of relatively prime to . Moreover, for any divisor of ,

Proof. By Corollary 20, we know that if and only if , . But if is odd, and , where we use to denote the cyclic group of order .
So we have and for some . The proof is complete by Lemma 18 and Theorem 3.

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

#### Acknowledgments

This research was supported by the Guangxi Natural Science Foundation (2015 GXNSFBA139012) and Inner Mongolia Natural Science Foundation (2018MS01017).

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