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Research Article | Open Access

Volume 2021 |Article ID 5595620 | https://doi.org/10.1155/2021/5595620

Shikun Ou, Yanqi Fan, Qunfang Li, "Automorphism Group and Other Properties of Zero Component Graph over a Vector Space", Journal of Mathematics, vol. 2021, Article ID 5595620, 8 pages, 2021. https://doi.org/10.1155/2021/5595620

# Automorphism Group and Other Properties of Zero Component Graph over a Vector Space

Revised20 Mar 2021
Accepted24 Mar 2021
Published10 Apr 2021

#### Abstract

In this paper, we introduce an undirected simple graph, called the zero component graph on finite-dimensional vector spaces. It is shown that two finite-dimensional vector spaces are isomorphic if and only if their zero component graphs are isomorphic, and any automorphism of a zero component graph can be uniquely decomposed into the product of a permutation automorphism and a regular automorphism. Moreover, we find the dominating number, as well as the independent number, and characterize the minimum independent dominating sets, maximum independent sets, and planarity of the graph. In the case that base fields are finite, we calculate the fixing number and metric dimension of the zero component graphs and determine vector spaces whose zero component graphs are Hamiltonian.

#### 1. Introduction

A binary relation on a vector space, group, or ring can be studied with the help of considering the associating graph defined by this relation. Symplectic graphs, orthogonal graphs, subspace inclusion graphs, and nonzero component graphs are such examples which have been considered recently (see ).

In this paper, we always assume , is a field, , and is an -dimensional vector space over . Let be a basis of . Then, any can be uniquely written as with , and (resp., ) is the set of all basis vectors whose coefficients are nonzero (resp., zero) in the linear expression of with respect to (i.e., with respect to) , namely,

Clearly, and .

All graphs considered in this paper are undirected simple graphs (without loops and multiedges). The nonzero component graph of , introduced by Das , is a graph with vertex set , in which for distinct , there exists an edge joining and if and only if . In , the author studied the diameter, dominating number, and dependent number of . When the base field is finite, Das  considered the clique number, edge connectivity, and chromatic number of and showed that is Hamiltonian, but not Eulerian. Murtaza et al.  considered two parameters, called the locating-dominating number and identifying number, of . Also, the nonzero component union graph of has as its vertex set, but two vertices are adjacent if and only if, where stands for the subspace of generated by a subset . In , Das studied some properties of , such as the connectedness, dominating number as well as maximal cliques and showed that is weakly perfect when the base field is finite.

It is well known that the automorphism group of a graph reflects the symmetry of the graph. Generally, it is an important but difficult work to describe the full automorphism group both in graph theory and in algebra. In , the author characterizes the automorphisms of . For the automorphisms of symplectic graphs, orthogonal graphs, and subspace inclusion graphs, the reader is referred to [3, 4, 8].

The fixing number and metric dimension of a graph (see Section 2 for the definitions) are two parameters which “destroy” the automorphisms and symmetry. Moreover, the fixing number has been used to consider the problem of programming a robot to manipulate objects (see ), while the metric dimension has been applied to various areas, such as pharmaceutical chemistry, robot navigation, diverse as combinatorial optimization, and sonar (see ). Recently, several authors considered the fixing number and metric dimension of some associating graphs defined on a vector space. Fazil  computed the fixing number of , and Benish  studied the relation between fixing number and another parameter, the fixed number, of . Ali et al.  calculated the metric dimension of . For a positive integer , the Kneser graph is a graph with all -subsets of as vertex set, in which two distinct -subsets are connected by an edge if they are disjoint. Also, the vertex set of the Johnson graph consists of all -subsets of , but two distinct -subsets are adjacent if their intersection is a -subset. In , Boutin obtained sharp bounds for the fixing number of Kneser graphs and determined all Kneser graphs whose fixing number is 2, 3, or 4. In , using various algebraic, combinatorial, and geometric approaches, Bailey et al. studied the constructions of resolving sets of Kneser and Johnson graphs and provided bounds on their metric dimension. For more works on the fixing number and metric dimension, the reader is referred to [9, 1621] and the references therein.

If is a basis of , then it is clear that can be partitioned into classes:

On the zero component graph of with respect to , we mean it is a graph defined as follows: has as its vertex set, in which distinct are adjacent if and only if . Motivated by all the above, we intend to investigate the properties of .

The rest of this paper is structured as follows: in Section 2, we introduce some definitions and notation in graph theory. In Section 3, we show that the zero component graph does not depend on the choice of the bases and thus restate the definition of the zero component graph (see Definition 1). In Section 4, we prove that two finite-dimensional vector spaces are isomorphic if and only if their zero component graphs are isomorphic. The automorphisms of the zero component graph are determined in Section 5, and the fixing number and metric dimension are computed in Section 6. Finally, we consider other properties of the zero component graph, such as the girth, minimum independent dominating sets, maximum independent sets, planarity, and hamiltonicity in Section 7.

#### 2. Notation and Preliminaries

Let be a graph with a vertex set . Two vertices are said to be adjacent and written as if there exists an edge joining and , otherwise, nonadjacent and . A graph is said to be complete if any pair of its vertices are adjacent; as usual, stands for the complete graph of order . Let (resp., ) be the neighbors (resp., degree) of , that is, (resp., ). If or ; then, and are said to be twins. We use to denote the twin set of , which consists of all twins of in . In particular, a vertex is called a singleton whenever . Clearly, the relation “” is equivalent if it is defined as ; and then is a union of some disjoint twin sets of . The number of twin sets of is written as .

For , it is called a dominating set of if any vertex in is adjacent to at least one vertex in , and it is called an independent set of if any pair vertices in are nonadjacent in . Also, a dominating set is called an independent dominating set if it is an independent set. The minimum dominating set (resp., maximum independent set, minimum independent dominating set) of is a dominating set (resp., independent set, independent dominating set) with the minimum (resp., maximum, minimum) cardinality, and the cardinality of a minimum dominating set (resp., maximum independent set) is called the dominating number (resp., independent number) of .

A trail in is an alternating sequence of vertices and edges, for some positive integer , where ’s are not necessarily distinct; it is called a path if ’s are distinct except for possibly the first and last vertices; and this path (resp., trail) is said to be joining and . A graph is called connected if, for any pair of vertices, there exists a path joining them; it is called -connected if each induced subgraph obtained by the graph from deleting vertices is connected. The connectivity of is the maximum value of for which is -connected. A cycle of length is a path with ; also, a cycle of length 3 is called a triangle. The girth of is the minimum of the lengths of all cycles in . Moreover, a graph is called Eulerian (resp., Hamiltonian) if there exists a trail (resp., cycle) which contains all edges (resp., all vertices) in it exactly once.

For graphs and , they are called isomorphic if there exists an isomorphism from to , that is, is a bijection satisfying: in if and only if in . An automorphism of is an isomorphism from to itself. The set of all automorphisms of forms a group under the composition of the transformations. is said to be vertex transitive if, for any , there exists such that . A subset is called a fixing set (or determining set) of if the only automorphism of that fixes every vertex in is the identity, and the fixing number (or determining number) of , denoted by , is the smallest size of such a set. For , the distance between them, written as , is the length of the shortest path joining them, and the diameter of is the largest distance between pairs of vertices of . If is an ordered subset of , then the -dimensional vector is called the representation of with respect to , and is said to be resolved by if for any other vertex . Moreover, is called a resolving set (or locating set) of if distinct vertices of have distinct representations with respect to , and the metric dimension (or locating number) of , denoted by , is the minimum cardinality of such a set. A graph is called an -graph if . All other unexplained notations and definitions on graph theory are standard (see  or  for details).

Here, we list some results on graphs, which will be used in this paper.

Proposition 1 (Handshaking lemma, , Theorem 1.1). The number of edges in a graph is .

Proposition 2 (see , p. 18). A graph as well as its complement has the same automorphisms.

Proposition 3 (see , Theorem 1). A finite graph and its complement are of the same fixing number.

Proposition 4 (see , Lemma 2.8). Let be a connected graph. If has no singleton, then it is an -graph, and .

Proposition 5 (see , Theorem 3.5). A connected graph is Eulerian if and only if each vertex is of even degree.

Proposition 6 (Kuratowski’s theorem, , Theorem 10.30). A graph is planar if and only if it contains no subdivisions of or , where stands for the balanced complete bipartite graph of order .

Proposition 7 (see , Corollary 18.12). If is a connected graph in which all vertices are of odd degree, then it is Hamiltonian.

Proposition 8 (Chvátal–Erdős theorem, , Theorem 18.10). Let be a connected graph on at least 3 vertices with independent number and connectivity . If , then is Hamiltonian.

Proposition 9 (see ). Let be a 2-connected graph with minimum degree and independent number . If , then is Hamiltonian.

#### 3. Zero Component Graph of a Vector Space

We first show that the zero component graph of does not depend on the choice of the bases.

Theorem 1. Let and be bases of , then and are isomorphic.

Proof. Clearly, there exists an invertible linear transformation on such that for any . Considering the restriction of on , we see that is also a bijection from to . For any , , we obtain that in if and only if there exists some such that , if and only if , or equivalently in . Therefore, is an isomorphism from to .

Remark 1. In case the zero component graph of does not depend on the choice of the bases, two vectors may be adjacent with respect to one basis but not adjacent with respect to another basis. For example, suppose that is a 3-dimensional vector space over , and is a basis of . Then, is also a basis of . It is easily seen that in , but in .
As usual, denote by , , the vector with the -th component 1 and the others 0. It is well known that constructs a basis of . Theorem 1 shows that we need only to consider the zero component graph of . For the convenience of writing, we set and use , , , for , as well as for , instead of , , , , as well as , respectively. Then, by Theorem 1, we can restate the definition of the zero component graph as follows.

Definition 1. The zero component graph of is a graph defined as follows: has as its vertex set, and for distinct , if and only if .
Next, we describe the connectedness and diameter of .

Lemma 1. (i)If , then is a union of two complete graphs(ii)If , then is a connected graph with diameter 2

Proof. The result is clear for . Let . Then, by , we see that is not complete. On the other hand, for any distinct , if , then there exist such that and , and further . Consequently, is a connected graph with diameter 2.

#### 4. Zero Component Graph and Graph Isomorphism Problem

In this section, we investigate the interrelationship between the isomorphism of two vector spaces and that of their zero component graphs. Before that, we study the twins and the number of twin sets of the zero component graph.

Lemma 2. Two vectors are twins in if and only if (or equivalently ).

Proof. The sufficiency is clear; thus, we need only to prove the necessity. Let be twins in . If , then or . Without loss of the generalization, assume ; say for some . It follows and , a contradiction. Thus, . Similarly, . Therefore, .

Lemma 3. Let . Then,(i)The number of twin sets of is (ii)Any vertex in is a singleton when , but not a singleton when

Proof. (i) follows from and the fact that is a union of twin sets in (see Lemma 2), where .
For (ii), we see that the case of is clear according to Lemma 2. Assume . For any and any , it is obvious that and are twins. This shows that is not a singleton in , which confirms (ii).
The following result is a direct consequence of Lemma 1 (i).

Theorem 2. Let ; let and be, respectively, an -dimensional vector space and an -dimensional vector space over . Then, is isomorphic to as vector spaces if and only if is isomorphic to .

In the following, we will prove that two vector spaces over distinct finite fields are isomorphic if and only if their zero component graphs are isomorphic. To this end, we give the vertex degree of .

Lemma 4. Let , , and . Then, the degree of in is .

Proof. Assume with and denoteClearly, any pair of do not intersect with each other, and . Then, by , we getFor any , we write with and . Take a permutation on such that for , and define the map by for any . Then, the restriction of on is an automorphism of , and . Therefore, .
By Lemma 4, we get the following results immediately.

Corollary 1. Let , and . Then, the order and size of are and , respectively.

Proof. It is easily seen that the order of is . For the size of , by Handshaking lemma (see Proposition 1), Lemma 4, and for , we havefrom which it follows the result.

Corollary 2. Let , and . Then, the maximum and minimum degree of are and , respectively.

Corollary 3. Let , and let be a finite field. Then, is Eulerian if and only if is even.

Proof. When is even (resp., odd), by Lemma 4, we obtain that each vertex in is of even degree (resp., odd degree), which shows is (resp., not) Eulerian.
Finally, we show that two vector spaces over distinct finite fields are isomorphic if and only if their zero component graphs are isomorphic.

Theorem 3. Let ; an -dimensional vector space over a field with elements; and an -dimensional vector space over a field with elements. Then, is isomorphic to as vector spaces if and only if is isomorphic to .

Proof. The necessity is obvious; thus, it suffices to prove the sufficiency. Assume that is isomorphic to . If , then by Lemma 1 (i), we know that is a union of two complete graphs each of which is of order , and so is . Again, by Lemma 1 (i), we have and . Similarly, one may get and when . Hence, and are isomorphic as vector spaces when or . Now, let . Lemma 3 (i) shows that the numbers of twin sets of and are and , respectively, which follows . On the other hand, the minimum degrees of and are, respectively, and according to Corollary 2. Then, by , we obtain . Thus, and are isomorphic as vector spaces.

#### 5. Automorphism of

We first construct two types of standard automorphisms of as follows.

Definition 2. Let be a graph and a bijection on . If permutates the twins in , then it is called a regular automorphism of .

Definition 3. Denote by the symmetric group consisting of all permutations on . If , then the map is an invertible linear transformation on . By the proof of Theorem 1, we know that the restriction of on is an automorphism of , called a permutation automorphism of .

Lemma 5. (i)Let and be defined as above. Then, is a regular automorphism of .(ii)If an automorphism of is a regular automorphism but also a permutation automorphism, then is the identity.

Proof. For any with , it is clear that and are twins. By Lemma 2, we assume with . Then,Again, by Lemma 2, we know that and are twins. By the arbitrariness of , we prove (i).
Assume with . For , by applying on , we have , which shows that and are twins. It follows for any . Hence, , and further or . This confirms (ii).
Next, we need to present a result on the zero-divisor graph. Let be a commutative ring, then the zero-divisor graph of is a graph defined as follows: has vertex set consisting of all nonzero zero-divisors of , and two nonzero zero-divisors are adjacent in if and only if their product in is 0. If , we denote by the ring which has as an additive group and the multiplication of two elements with are defined naturally:It is clear that is also the vertex set of . In , the automorphisms of were described as follows.

Proposition 10 (see , Theorem 1). Any automorphism of is a permutation automorphism, and .

The following result shows that the relation between , for , and .

Lemma 6. If , then the complement graph of is isomorphic to .

Proof. Define by . It is clear that is a bijection on . Moreover, for any , we see that in (or equivalently ) if and only if , if and only if . Thus, in if and only if in , and therefore, is an isomorphism from to
By Proposition 2, Proposition 10, and Lemma 6, we can immediately determine the automorphisms of for .

Lemma 7. If , then any automorphism of is a permutation automorphism, and .

In the following, we consider the case for an arbitrary finite-dimensional vector space. Before that, we introduce the quotient graph of , which is defined as follows: has all twin sets of as its vertex set, and for any , in if and only if as well as . It is easily seen that the definition is well defined since for any and (see Lemma 2), and is isomorphic to , where is the vector space over the field . Let be an automorphism of . We define on by

Lemma 8. Let be an automorphism of ; then, defined as in (8) is an automorphism of .

Proof. If for , then are twins in . Furthermore, and are also twins in , following . Hence, is well defined. It is clear that is a bijection. Moreover, from the fact that in if and only if in , it follows that in if and only if in . Thus, is an automorphism of .
Now, we can consider the case for an arbitrary finite-dimensional vector space.

Theorem 4. Let . Then, a bijection on is an automorphism of if and only if it can be uniquely decomposed into the product of a permutation automorphism and a regular automorphism of .

Proof. The uniqueness of decomposition is clear according to Lemma 5 (ii); thus, it suffices to prove the existence. When , we get the result by Lemma 7. Now, assume .
For any automorphism of , define as in (8). Then, by Lemma 8, we know that is an automorphism of , which is isomorphic to (where is the vector space over the field ). Applying Lemma 7, we see that there exists a permutation such thatBy the definition of , we have , or, equivalently, and are twins. Consequently, there exists a regular automorphism of such that for any , and therefore .
The following is a direct consequence of Lemma 1 (i) and Theorem 4.

Corollary 4. is vertex transitive if and only if .

Denote by and the sets consisting of all regular automorphisms and all permutation automorphisms of , respectively. Then, and are subgroups of the automorphism group of . Finally, by Theorem 4, we can describe and its order as follows.

Corollary 5. (i). In particular, when .(ii)If , then with .

Proof. Theorem 4 shows . Then, by (see Lemma 5 (ii)) and the fact that is a normal subgroup of (see Lemma 5 (i)), we obtain the result of (i).
(ii) follows from (i), , and .

#### 6. Fixing Number and Metric Dimension of

In this section, we always assume is a finite field with elements. If , nothing needs to do for the metric dimension of according to Lemma 2; the fixing number is 1 when , and otherwise. In the following, we consider the case for .

Firstly, by Proposition 4, Lemma 3, and Corollary 1, we have the following result immediately.

Lemma 9. Let and . Then, is an -graph, and .

Next, we give the value of the fixing number of for and .

Lemma 10. Let and . Then, .

Proof. By Proposition 3 and Lemma 6, we have . Furthermore, from (see , Theorem 3.5), it follows that .
The vector with is said to be -type, wherefor . The following result is trivial.

Lemma 11. If , then all vectors in are of different types.

Now, we can calculate the metric dimension of for and .

Lemma 12. Let and . Then, when , and when .

Proof. When , it is clear that as shown in Figure 1.
Moreover, is a resolving set of . So, . Now, let . We first assert that is a resolving set of . For any , it is of -type, and its presentation with respect to is , where is defined as in (10). Since the types of all vectors in are different (see Lemma 11), the same is for their representation with respect to . Hence, is a resolving set of , and therefore . Next, we need to show . Suppose and that is a resolving set of . Define the map asLemma 1 (ii) shows that is well defined. Moreover, for any , since their representation with respect to is different, we see that . Consequently, is injective, and further . It follows that . Thus, .
Finally, combining Lemmas 9, 10, and 12, we get the following theorem.

Theorem 5. Let and . Then, is an -graph if and only if , or as well as . Moreover, when ; when , ; and as well as when , .

#### 7. Other Properties of

We first consider the girth, minimum independent dominating sets, maximum independent sets, and planarity of when is an arbitrary field.

Theorem 6. Each vertex in belongs to some triangle in . Furthermore, the girth of is 3.

Proof. The case for is clear according to Lemma 1 (i). Now, let . For any , if , assume for some . Choose distinct ; then ; if , say for some , then . Therefore, in any case, belongs to a triangle in .

Lemma 13. Let . Then, the dominating number of is 2.

Proof. This is a direct consequence of the fact that any vertex is adjacent to at least one of .

Theorem 7. Let . Then, is a minimum independent dominating set of if and only if as well as .

Proof. Let be an independent dominating set of . It is clear that . If , say for some , then and . This implies that is not a dominating set of , a contradiction. Hence, .
Conversely, assume and for . By , we see that is an independent set of . On the other hand, for any , since implies , we obtain that leads to , which shows that is a dominating set. Therefore, is an independent dominating set, and further a minimum independent dominating set of .

Lemma 14. The independence number of is (i.e., the dimension of ).

Proof. Let be the independence number of . Clearly, is an independent set of . This implies . Now, it suffices to show . Suppose to the contrary, , and assume that is an independent set of . Then, are disjoint nonempty sets. From , it follows that , a contradiction.

Theorem 8. The set is a maximum independent set of if and only if , and are disjoint.

Proof. The sufficiency is obvious; thus, we need only to prove the necessity. Let be an independent set of . Clearly, are disjoint. It suffices to show ; that is, has exactly one zero component, . The case for is clear. Let , and suppose, to the contrary, there exists some such that . Without loss of the generalization, assume . For any , by , we get , and consequently . Hence, , a contradiction.

Theorem 9. is planar if and only if and , or and .

Proof. When (by Lemma 1 (i)), we know that is a union of two complete graphs each of which is of order . This follows that is planar if and only if (see Proposition 6). When and , we obtain that is planar as shown in Figure 1. When (resp., and ), the subgraph of induced by (resp., ) is and further nonplanar.
Next, we provide a lower bound for the connectivity of when is a finite field.

Lemma 15. Let and . Then, the connectivity of is greater than or equal to .

Proof. Let and be two nonadjacent vertices in and suppose that such that and are in different components of , the subgraph of induced by . It is easily seen that ; indeed, if , then in for any , a contradiction. Assume , and setFrom , it follows that and , which implies . Furthermore, , and thus