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`ISRN CombinatoricsVolume 2013, Article ID 398641, 5 pageshttp://dx.doi.org/10.1155/2013/398641`
Research Article

## Isomorphic Digraphs from Affine Maps of Finite Cyclic Groups

School of Mathematical Science, Guangxi Teachers Education University, Nanning 530001, China

Received 27 June 2013; Accepted 28 July 2013

Copyright © 2013 Guixin Deng. 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

Let be a positive integer. For any pair of integers and , let be the digraph whose set of vertices is , and there exists a directed edge from vertex to vertex if . In this paper, we obtain a necessary and sufficient condition for which .

#### 1. Introduction

Let be a nonempty set and let be a mapping from into itself. Then, we call the iteration digraph from of , where the set of vertices of is and there is a directed edge from a vertex to a vertex if . In particular, there are many interesting structures on the digraphs when there is an algebraic structure on .

Many people study the iteration graphs from endomorphisms of finite abelian groups. The authors of [1] studied the digraph from the endomorphism of , where is a prime. These results were generalized to the digraph from any endomorphism of in [2]. Sha and Hu [3] studied some elementary properties of the digraph from any endomorphism of . They also obtained many elementary properties of the digraph from any endomorphism of finite cyclic groups in [3].

Let be a positive integer and lrt be two integers. In this paper, we denote by the digraph obtained from the affine mapping of , the ring of integers modulo . In particular, we set if , which is from an endomorphism .

We are mainly interested in the problem when . The authors in [2] gave an example . Let be a prime. Sha [4] showed that if and only if . In [5], Deng and Yuan gave a necessary and sufficient condition for , where is a prime. In this paper, we obtain a necessary and sufficient condition for when .

The paper is organized as follows. In Section 2, we study the , where is a power of a prime. The main results are proved in Section 3. We also determined the digraph from projective map on one-dimensional projective space over a finite field in the last section.

#### 2. Digraph Product and the Local Case

Let be a digraph and let be a vertex in . A component of a digraph is a subdigraph which is a maximal connected subgraph. Each component of a finite iteration graph contains a unique cycle. Let denote this -cycle. Let denote the number of cycle of length 1 in .

Let be iteration digraphs. The product is the digraph whose vertices are the ordered pairs , where and there is a directed edge from to if and only if there is a directed edge in from to for each . We use the following notation:

Lemma 1 (see [6]). is isomorphic to , which is a union of cycles, each having length .

Theorem 2. Let be the prime factorization of and let be two integers. Then

Proof. Let and . Assume that is the canonical isomorphism of rings. Then induces an isomorphism of digraphs.

##### 2.1. The Case of Linear Maps

The structure of has been studied in [2, 4, 5]. We present some results here.

Lemma 3 (see [4]). Let be positive integers. Then one has the following.(i)If for any prime , then is a tree attached to the fixed point 0.(ii)If , then , where and is the Möbius function.

Let be the -adic valuation of for any prime and integer . It is the largest power of dividing for any prime and integer . We set as usual. If is a set of primes and is a positive integer, we denote that

Lemma 4. Let be a prime. Let be positive integers such that . The following statements are equivalent:(i);(ii);(iii);(iv) has a fixed point.

Proof. By hypothesis the statements , , and are equivalent to that the congruence equation that has a solution. Suppose that . Then has a fixed point since has a fixed point 0. So . On the other hand, if , then is an isomorphism of and .

Remark 5. Many people showed that the tree attached to any two cycle vertices in is isomorphic. Moreover, Deng and Yuan [5] showed that the tree attached to cycle vertices in is totally determined by , where is a prime. In fact, if we replace by in the arguments in [5], then the result is also valid.

##### 2.2. The Case of Affine Maps

Lemma 6. Let be a positive integer, where is a prime and , .(i)If or , then for any integers .(ii)If and , assuming that , then

Proof. If , then it is clear that . Firstly, we assume that . By computation, for any and . But thus . It is clear that . So ; considering that . So if . Suppose that . If or , then Therefore, for any . Hence, for and . Thus, in this case.
Suppose that , . It remains to show that for . Let , where . By the discussion above we get and . Therefore, for . This finishes the proof.

Lemma 7. Let be a prime. Let be two positive integers such that . Then , where

Proof. Let . Then . We consider the following congruence equation Equation (8) has a solution if and only if . In this case, and , since . Since is the minimal positive integer such that , we get . Moreover, each integer is a solution of (8). Thus, is a union of cycles, each having length . The rest of the proof follows from Lemma 6.

The structure of digraph is clear now. We classify it as follows.Type : is a tree attached to a fixed point, which is determined by .Type : is a union of cycles, and the minimal length of these cycles is 1.Type : , where .

Let . By the properties of digraph product we see that where is a tree attached to a fixed point, is a union of cycles, and the minimal length of these cycles is 1, and .

#### 3. The Main Results

We begin with some lemmas before we prove our main results.

Lemma 8. Suppose that is a permutation of a finite set and . Then for each .

Proof. Assume that is a cycle of length in such that , for and . Suppose that is contained in a cycle of length in . Then . Hence, splits into cycles with the same length . This completes the proof.

Lemma 9. Let be a permutation of a finite set , . Then(i) if and only if the numbers of fixed point of and are equal for each ;(ii) if and only if for each .

Proof. Suppose that and .
Assume that is an isomorphism of digraphs. Then for any . By induction, . Hence, is also an isomorphism of and . Thus, . Conversely, if , then . Assume that for any . By Lemma 8, and . Hence, and . The proof of statement (i) is complete by induction.
It remains to prove (ii). Assume that . Then , where maps to for any . Let . By Lemma 8,
If , then for each . Let in (10). We get . Conversely, if , then by (10) if . If , then Hence, if . Therefore, for any . By statement (i), .

In [3], Sha and Hu showed that if and only if , where is a prime and . We generalized this result to the following.

Lemma 10. Let , and be positive integers such that , . Then if and only if for any divisor of .

Proof. Let . By Lemma 3 and Möbius inverse transform for any . Suppose that for any divisor of . If , let ; then . Thus . By symmetry, we have if . Thus, . By Lemma 9, .
Conversely, if , then and for any . If , let ; then . Hence, . By symmetry, . Thus, . The proof is complete.

Theorem 11. Let be the prime factorization of and let , be integers, . Suppose that and that the minimal length of cycles in is . Then if and only if the following two conditions are satisfied.(i).(ii), , and for any divisor of .

Proof. Suppose that and that . We have the following decomposition: where is a tree attached to a fixed point, is a union of cycles, and the minimal length of these cycles is 1.
Therefore, if and only if the following two conditions hold.(iii).(iv) and .
However, by Remark 5, is equivalent to . It remains to show that is equivalent to . If is satisfied, then and by Lemma 10  . By Lemma 9, . Conversely, if holds, we have since the minimal length of cycles in is . So , , , and . The rest of proof follows from Lemmas 9 and 10.

Remark 12. Suppose that . By Theorem 11, we see that both and are the same types for any prime . Moreover, they are isomorphic if they are Type (1) or (3).

Example 13. Consider , , and . One has where . Hence, , but . This shows that even if , it may happen that . But we also see that , , and . In fact, we have the following result if is fixed.

Theorem 14. Let be the prime factorization of . The following three statements are equivalent for any integers , , and :(i);(ii);(iii) for each .

Proof. Let , , and .
. By Theorem 11  , , and . If , then , . If , Suppose for any , where . By Lemma 7, . Thus, for any . The result follows from that , .
. Assume that for each with . If , then by Lemma 4  . If , then . By Lemma 7 again , where .
follows immediately from (2).

#### 4. The Digraph from Projective Map

Let be the one-dimensional projective space over , where if and only if there exists such that . With any invertible matrix over , we associate the Möbius transformation of as . Let be a polynomial over with . Recall that , the order of , is the minimal positive integer such that . We classify the digraph from Möbius transformation of as follows.

Theorem 15. Let . Suppose that and are the characteristic polynomial and minimal polynomial of , respectively. Then(i)if and , then ;(ii)if , then ;(iii)if and , then , where is the multiplicative order of ;(iv)if is irreducible over , then , where .

Proof. (i)–(iii). We see that is similar to , and in cases (i), (ii), and , respectively. But the digraphs are isomorphic from similar matrices. The results follows from a simple calculation.
It remains to prove . Let . We can consider as a subdigraph of under the natural inclusion . Then with some in . By (iii), , where is the multiplicative order of in . By a direct computation, we see that for if and only if . Hence, is a union of cycles of . However, has no fixed point since is irreducible. So the length of each cycle in is , and . Note that , and that . Therefore, . This finishes the proof.

Corollary 16. if and only if they have the same number of fixed points.

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