Abstract
We study the dynamics of endomorphisms on a finite abelian group. We obtain the automorphism group for these dynamical systems. We also give criteria and algorithms to determine whether it is a fixed point system.
1. Introduction
A discrete dynamical system is consisting of a pair , where is a nonempty set and is a function. The functional graph of is a directed graph with vertex set and an arrow from to if and only if . We denote the functional graph of by .
Finite discrete dynamical systems appeared in engineering, control systems computer science, and genetic networks; see [1, 2]. One of the most important examples of discrete dynamical systems is given by , where is the vector space over a finite field and is a linear transformation. The structure these dynamical systems can be described by the elementary divisors of the corresponding matrix; see [3, 4]. By using the same method, Brown and Vaughan [5] studied the cycle structure of linear dynamical system over . Bach and Bridy [6] gave an upper and a lower bound for the distinct number of functional graph defined by affine mapping over finite dimensional linear space over finite fields. But the study of linear dynamical systems over a general finite commutative ring has difficulties since the factorization of polynomials in is not unique. Brown and Vaughan [5] showed that can be decompose into by Fitting’s Lemma and is the product of the dynamical systems induced on and . They also provided an algorithm to compute the minimal positive integer in the above decomposition, but the algorithm is not efficient. Xu and Zou [7] obtained an efficient algorithm to compute under a reasonable assumption on the size of the ring . Deng [8] gave a criterion to determine whether a linear dynamical system is over the ring of integer modulo .
Let be a finite abelian group written additively and be the endomorphism ring of . In this paper we are interested in the dynamical systems defined by . These dynamical systems have nice structure and were studied in many special cases; see [9]. In this paper we are interested in the case in which is a finite abelian group written additive, and is an endomorphism of .
The paper is organized as follows. In Section 2, we give some notations and basic results on the structure of functional graphs . In Section 3, we obtain necessary and sufficient conditions for and . We also give an explicit form of the automorphism group of . In Section 4, we obtain a criterion to determine whether is a fixed point system.
2. Elementary Properties of
A component of a digraph is a maximal connected subgraph of the associated nondirected graph. The indegree of a vertex of , denoted by , is the number of directed edges coming into . The outdegree of is the number of directed edges leaving the vertex . It is clear that the outdegree of any vertex in is equal to 1, and the indegree of a vertex is the number of vertices coming to . A vertex in is called a cycle vertex if for some positive integers . In particular, a cycle vertex is called a fixed point of if . Let denote the number of -cycles contained in . Let denote the set of positive integers which is the length of the cycle which appeared in . We need the following results about and . In particular, let be the number of fixed point in .
Definition 1. We define a height function on the vertices and components of . Let be a vertex of . Then we define to be the minimal nonnegative integer such that is a cycle vertex in . We set if is a component of .
Definition 2. For any integer and any vertex , defineLet be the subdigraph of with the vertices set , and there is a directed arrow from to in if and . Hence, is a directed tree with root .
Let be a finite abelian group and The basic properties of in this section are proved in some special cases; see [10–14]. We give some basic properties of these digraphs. The proof is similar, but we still present it.
Lemma 3. Let . Then(1)the indegree of any vertex in is or .(2)the number of the vertices with indegree is .
Proof. Let be a vertex with positive indegree. Then is a coset of and they have the same number of elements. The statement (2) follows immediately from (1).
We present two results on the tree structure attached to a vertex.
Lemma 4. Let be a cycle vertex of . Then .
Proof. For any there exists an unique cycle vertex such that since is a cycle vertex. Now if , we define a map as follows: Then . We observe that is not a cycle vertex, but is a cycle vertex; then is not a cycle vertex since the set of all cycle vertices of is a subgroup of . The map is well-defined and clearly injective. If , then and . is also surjective.
Now if and with , thenFinally we set : if . By (3) is an isomorphism.
Corollary 5. Let and be two components of . Then if and only if they have the same cycle length.
Proof. If follows immediately from Lemma 4.
Lemma 6. Let , be two finite abelian groups and . Let with positive height. Then if and only if for any .
Proof. Let . First we assume that and let be an isomorphism. Since the length of the longest directed path in is and this path ends at , it follows that . Thus, and .
Conversely, suppose that for any . Then . We induct on . The case is trivial. Note that for any vertex with positive height, by Definition 2. Since is also a group homomorphism, by Lemma 3 we have or for any . Let and . Let and . Since is a disjoint union for any and if and only if , one has . It follows thatWe derive that for and . We obtain by replacing by in the above arguments. After reordering we may assume that for any . Then for any fixed and . By induction, there exists digraph isomorphism . We set , where and if . Then is also a digraph isomorphism. This finishes the proof.
3. Automorphism Group of
We first show when is isomorphic to for .
Lemma 7. Let , be two finite abelian groups and . Let be the component of containing the vertex 0. Then if and only if(1);(2) for any .
Proof. It follows immediately from Lemma 4 and Corollary 5.
Theorem 8. Let , be two finite abelian groups and . Then if and only if for any positive integer (1);(2).
Proof. Let be the component of containing the vertex . Since if and only if for some , so is a disjoint union. It follows that for any . Thus, for any if and only if for any . By Lemma 6, condition (1) is equivalent to (1) of Lemma 7. Note that . Thus, condition (2) is equivalent to (2) of Lemma 7. The proof is finished by Lemma 7.
Let be a finite abelian group. By the structure of finite abelian group, we have where runs over the set of prime divisor of and is the unique Sylow -subgroup of . Moreover, if there is another factorization of in this way, it must have for each .
We also recall the product of graphs. Let and be two digraphs. The digraph product is the digraph whose vertices are the set of order pairs , where is a vertex of . There is a directed edge from to in if and only if there is a directed edge from to in and there is a directed edge from to in .
Lemma 9. Let be a finite abelian group and . Suppose that is prime factorization. Then
Proof. It follows immediately from the fact that and .
Theorem 10. Let , be two abelian groups and . Then if and only if for any prime .
Proof. First we assume that . By Lemma 9, we haveSo we obtain .
Conversely, suppose that . Let . By Theorem 8, and for any . It follows that and . By Theorem 8 again, . The proof is finished.
Next we would determine the automorphism group of the digraph . We need the wreath product of groups; see [4, 15]. We rewrite in a simple way. Let be a group and be a permutation group on ; then the wreath product is generated by the direct product of copies of , together with the elements of action on these copies of . The wreath product is associative, so we write for .
Theorem 11 (see [16]). Let be a graph. Suppose that the connected components of consist of copies and copies of , where are pairwise nonisomorphic. Then where is the symmetric group on a finite set of order .
By the above Theorem and Corollary 5, we have where is a component of with cycle length Let be the component containing in . Sha [9] showed that , where is the cyclic group of order . Thus, But he did not obtain the structure of in general. We would determine the automorphism group of .
Lemma 12. Let be a vertex with . Suppose that and for . Then where .
Proof. By induction on , let for . Then . Thus,We derive that and for Let for and . By Lemma 6, we have if and . By Theorem 11 and by induction we have
Theorem 13. Suppose that and . Then
Proof. The proof is similar to Lemma 12. We induct on . Let for . Note that since is a fixed point. Thus, It follows that and for Let for and . By induction,
4. Fixed Point Systems
A dynamical system is called a fixed point system if its cycle vertices are all fixed points. In this section we would describe when is a fixed point system.
Definition 14. Let and be two dynamical systems. A morphism is a map such that .
Lemma 15. Let be a morphism of dynamical systems. Suppose that is a fixed point system and is surjective. Then is also a fixed point system.
Proof. It follows immediately from Lemma  3.1 in [8].
Lemma 16. Let be a finite abelian group and . Let be the affine mapping for any . Then(1) if and only if has a fixed point;(2) is a fixed point system if and only if it has a fixed point and is a fixed point system.
Proof. It follows immediately from Lemma  3.3 in [8].
The following lemma shows why we need to consider the affine system.
Lemma 17. Let , be two finite abelian groups and . Suppose that is a group homomorphism and is also a surjective morphism of dynamical systems. Let be a fixed point in and . Then , where for some .
Proof. It is clear that is a coset of . We define , . Then is bijective and . Thus, is an isomorphism of dynamical systems.
Recall that denotes the number of fixed points of .
Lemma 18. Let , be two finite abelian groups and . Suppose that is a surjective morphism of dynamical systems. Let . Then is a fixed point system if and only if the following two conditions are satisfied:(1) is a fixed point system;(2).
Proof. We write and . Since maps cycle vertices into cycle vertices, so the cycle vertices of are contained in , and . Thus, is a fixed point system if and only if is a fixed point system for any . By Lemma 17, is a fixed point system if and only if and is a fixed point system. This finishes the proof.
By Theorem 10 we only need to deal with the case that is a -group. Let such that . Recall that any endomorphism of of has a matrix representation , where for . If then . We also write
We recall the Smith normal form and invariant factor of matrices over a principal ideal ring.
Definition 19. Let be a principal ideal ring and . There exist invertible matrices such that where for . is called the Smith normal form of . The diagonal entries are called the invariant factors of and are unique up to units.
Theorem 20 ([8], Theorem  5.2). Let be a prime and and . Then is a fixed point system if and only if the following conditions are satisfied.(1);(2)Each invariant factors of over is a unit or .
Corollary 21. Let and , where and for . Suppose that and each invariant factor of over is a unit or . Then is a fixed point system.
Proof. Since the canonical mapping is a surjective morphism of dynamical systems, it follows from Lemma 18 and Theorem 20 that is a fixed point system.
However, the converse of Corollary 21 is not true in general. Consider the following example.
Example 22. Let and , , and . By a direct computation, is a fixed point system with three fixed points , , and , and has three fixed points , , and , and two -cycle and , and is a fixed point system with fixed points , , and .
Let be an integer matrix and . We write if for any .
Theorem 23. Let with . Let be a matrix with partition such that for . Let and for . Let and . Then is a fixed point system if and only if(1) is a fixed point system for each ;(2).
Proof. We induct on . The case is trivial. Let and Define where . Since , one hasIt follows that is a surjective morphism of dynamical systems. By Lemma 18, is a fixed point system if and only if is a fixed point system and . By induction, is a fixed point system if and only if is a fixed point system for each and . It remains to show that , where .
We set . where . It is clear that is injective and . We have Thus, . The proof is finished by induction.
Remark 24. By Theorem 23, we can determine whether is a fixed point system and also can compute be using the Smith normal form of . So the problem for determining whether is a fixed point system is now reduced to solve the congruence equation
Example 25. Let and . By Theorem 23, let , and let It is straightforward to check that is a fixed point system with two fixed points , is a fixed point system with one fixed point , and is a fixed point system with two fixed points . Now we compute the number of fixed points of . Let be a fixed point in . Then and We derive that and by the first and third equation. Hence, the second equation is equivalent to . It follows that if , and if . Thus, has two fixed points and . By Theorem 23 again, is a fixed point system.
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 National Natural Science Foundation of China (Grant no. 11371343).