Magnifying Elements in Semigroups of Fixed Point Set Transformations Restricted by an Equivalence Relation

Abstract

Let be a nonempty set and be an equivalence relation on . For a nonempty subset of , we denote the semigroup of transformations restricted by an equivalence relation fixing pointwise by . In this paper, magnifying elements in will be investigated. Moreover, we will give the necessary and sufficient conditions for elements in to be right or left magnifying elements.

1. Introduction

Magnifying elements of a semigroup were first mentioned in 1963 by Ljapin in [1]. An element of a semigroup is called a right (left) magnifying element if there exists a proper subset of such that (). Many studies on magnifying elements in semigroups were conducted by many authors in various aspects between 1971 and 2003. In 1971, Migliorini [2] introduced the new notion of a minimal subset relative to a magnifying element of . By means of this, he constructed an infinite chain of minimal subsets properly contained in the preceding ones, which gives rise to an infinite number of magnifying elements of the form , where is a magnifying element and is any positive integer. Migliorini also investigated the structure of a semigroup with minimal subsets in [3]. According to Tolo [4], if the proper subset of relative to a magnifying element is a subsemigroup of , then is called a strong magnifying element. It was shown that if a semigroup contains a strong magnifying element, then is factorizable, i.e., for some proper subsemigroups of . In other words, the existence of strong magnifying elements plays an important role in factorizing a semigroup. In 1992, Catino and Migliorini [5] provided the example of semigroups with nonstrong magnifying elements which are factorizable and determined the existence of magnifying elements in simple, bisimple, and regular semigroups by improving Tolo’s results. Moreover, they proved that the magnifying elements of the natural partial-order semigroups are maximal. Two years later, Magill [6] provided the necessary and sufficient conditions for elements in a semigroup with identity to be left or right magnifying elements and applied the result to the semigroup of linear transformations of a vector space and the semigroup of all continuous self-maps of a topological space. In 1996, Gutan [7] constructed the semigroup containing both strong and nonstrong magnifying elements, which turned out to be a positive answer to the question, posed in [5, 6], of whether or not there is a semigroup containing both strong and nonstrong magnifying elements. A year later, he showed in [8] the necessary and sufficient conditions for a semigroup containing a magnifying element to be factorizable. The results point out that every semigroup containing magnifying elements, which is not necessarily strong, is factorizable; and this improved the results by Tolo, and Catino and Migliorini. In 1999, Gutan characterized semigroups containing left strong magnifying elements with a minimal subsemigroup and proposed the method for obtaining such a semigroup in [9]. In 2000, Gutan [10] introduced the definition of very good magnifying elements in a semigroup. If such a set relative to a magnifying element is a ring ideal, then is a very good magnifying element. Furthermore, he established a characterization of semigroups in which all the left magnifying elements are very good. In 2003, Gutan and Kisielewicz [11] presented the notion of primitive semigroups and further constructed semigroups having both good and bad magnifying elements. In addition, some general properties of semigroups containing magnifying elements with its minimal subsemigroup were established. At that time, many researchers focused on the minimal subsets relative to magnifying elements. Recently, some researchers have paid attention to magnifying elements in various transformation semigroups. For instance, Luangchaisri et al. [12] generalized Magill’s results in partial transformation semigroups, and Prakitsri [13] investigated magnifying elements in linear transformation semigroups with infinite nullity and in those with infinite co-rank. He showed that linear transformation semigroups with infinite nullity have no right magnifying elements. However, all left magnifying elements in these semigroups are strong magnifying elements. Contrarily, linear transformation semigroups with infinite co-rank have no left magnifying elements but all right magnifying elements are strong. In [14], the conditions for elements in the semigroup of transformations with a fixed point set to be magnifying elements have been established by Petapirak, Kaewnoi, and Chinram. In this paper, efforts have been made to extend the results obtained in [14] by showing the necessary and sufficient conditions for elements in the semigroup of transformations restricted by an equivalence relation with a fixed point set to be right or left magnifying elements.

2. Preliminaries

In this section, we first provide the reader with some basic but essential definitions.

A semigroup is a system consisting of a nonempty set together with the binary associative operation , i.e., belongs to and for all elements . For convenience, we write instead of and let stand for for any . A subset of a semigroup is called a subsemigroup of if is a semigroup under the operation of . A nonempty set of a semigroup is a subsemigroup of if for all . The intersection of any set of subsemigroups of is either an empty set or a subsemigroup of .

The next finding is an initial factor in this study.

Theorem 1 (see [1], pp. 118–119). The following statements are true:(1)No element of a semigroup is simultaneously a left and a right magnifying element.(2)Finite semigroups have no magnifying elements.(3)Commutative semigroups have no magnifying elements.(4)Semigroups with two-sided cancellation have no magnifying elements, and hence groups do not contain magnifying elements.

Therefore, we will focus our attention on infinite noncommutative semigroups without two-sided cancellation.

Let be the set of all functions from a nonempty set into itself. As is well known, the composition of functions is closed, and the associative law holds. Therefore, is a semigroup under the composition of functions. We then call the full transformation semigroup. Throughout this paper, the identity function on is denoted by . The range of a function is denoted by for all elements . Moreover, we write functions from the right, rather than , and compose from the left to the right, rather than for all elements .

Let be an identity relation on , and let be an equivalence relation on . For each , we denote the equivalence class of containing by and . For any , as in [14], we set . The conditions for elements in to be magnifying elements are demonstrated as follows.

Theorem 2 (see [14], Theorem 2.5). A function is a left magnifying element if and only if is one-to-one but not onto.

Theorem 3 (see [14], Theorem 2.14). A function is a right magnifying element if and only if is onto but not one-to-one.

Denote the transformation semigroup restricted by an equivalence relation by . It is widely known that is a subsemigroup of . Let be the intersection of and . Evidently, if is a nonempty set, then is a subsemigroup of and . We then call the semigroup of transformations restricted by an equivalence relation fixing pointwise.

3. Propositions for

In this section, we illustrate the following propositions which are characterizations of .

Proposition 1. if and only if .

Proof. Assume that . Let such that . Clearly, the identity function . By assumption, . Hence, . This shows that . Conversely, assume that . Clearly, . Let . Then, for all , . So since . This shows that . Therefore, .

Note that if and , then for all , . Hence, if . By the proof of Proposition 1, if and only if .

Proposition 2. if and only if .

Proof. Assume that . Let be such that . Clearly, the identity function . By assumption, . Hence, . This shows that . Conversely, assume that . It is clear that . By Proposition 1, we have . So we have .

Proposition 3. The identity function belongs to if and only if .

Proof. Assume that the identity function belongs to . Let such that . Then, . This shows that for all , if , then , which implies that is an identity relation on . Conversely, assume that . By Proposition 2, . It is clear that . So .

Clearly, is a proper subset of . It is easy to verify that if , then is the set of all constant functions, and if and , then .

Proposition 4. If and , then is empty.

Proof. Assume that and . Then there are distinct elements such that . Suppose that there exists an element . Then , which is a contradiction. Therefore, is empty.

Proposition 5. If and , then .

Proof. Assume that and . Then there is only one element . Hence, there is a function defined by for all . Suppose that , and let . By assumption, for all , and hence, for all . This shows that . Therefore, .

By Propositions 4 and 5, left and right magnifying elements do not exist in if .

4. Main Results

In this section, we will give the necessary and sufficient conditions for elements in to be right or left magnifying elements.

4.1. Right Magnifying Elements

By Proposition 2, we obtain the next theorem.

Theorem 4 (see [14]). Suppose that and . A function is right magnifying in if and only if is onto but not one-to-one.

Our next two results are related to the existence of magnifying elements in .

Lemma 1. If for some , then is empty.

Proof. Let . Assume that . Suppose that there is an element belonging to . By assumption, there are two distinct elements . Then . So , which is a contradiction.

Lemma 2. If for all , then .

Proof. If and for all , then , and hence, . So . Next, assume that and for all . Then , and hence for each , there is a unique element such that . Define a function byfor all . Clearly, . Next, we will show that is the only element in . Let be a function in . For all , . By assumption, for each , there is a unique such that , and hence, . Therefore, and .

By Lemmas 1 and 2, if and either for all or for some , then there exists no magnifying element in .

Example 1. Consider , , and if and only if .
Clearly, is an equivalence relation on and . Let be a function in defined by, for all ,By Lemma 2, the function is the only function in . So has no nonempty proper subset, and hence, there exists no magnifying element in .
For the rest of this section, we focus on nontrivial cases, i.e., and , and establish the existence of right magnifying elements in . We thus assume now that .

Lemma 3. Suppose that is countably infinite. If there are infinite such that , then there is a surjective function in . Consequently, if is a right magnifying element, then is onto.

Proof. Suppose that there are infinite such that . Hence, is infinite. Let . By assumption, is infinite. Then there is a bijective function from to . For each such that , there exists a unique such that , by Lemma 1. Define a function byClearly, . Moreover, is onto. Let be a right magnifying element in . There is a proper subset of such that . Then for some . This implies that is onto.

Suppose that is uncountably infinite and . It is easy to prove that if and have the same cardinality, then and have the same cardinality as well. The proof of our next result is similar to the proof of Lemma 3 and so will be omitted.

Lemma 4. Suppose that is uncountably infinite and . If and have the same cardinality, then there is a surjective function in . Consequently, if is a right magnifying elements, then is onto.

Theorem 5. Suppose that and . If for all and there are infinite such that , then a function is a right magnifying element if and only if is onto.

Proof. By Lemmas 3 and 4, the necessity is clear. Conversely, assume that is onto. Let . Clearly, is a proper subset of since . Let be a function in . Since is onto, for each , there exists an element such that . For each , if , then there is a unique element such that ; and if , we choose only one to define the function asfor all . It is readily seen that fixes every element in . Next, we let such that . Then for some . If , then . If , then . Therefore, . Since and , the function is not one-to-one. Then there are distinct elements such that . So at least one of and does not belong to , and hence, is not onto. So . For all , . This shows that . Hence, , which implies that is a right magnifying element.

Example 2. Consider and . Let be an equivalence relation on such that . Assume that is a function in defined by, for all ,That is,Clearly, the function is onto. By Theorem 5, the function is a right magnifying element. Let . Then there is a proper subset of such that . Let be a function in defined by, for all ,That is,Then there is a function such that . For more details, define a function by, for all ,That is,We obtain that and , as required.

Corollary 1. The following statements hold:(1)Suppose that and . Then is a right magnifying element if and only if is one-to-one but not onto.(2)Suppose that and . Let be infinite and . If and have the same cardinality, and for all , , then is a right magnifying element if and only if is onto.

4.2. Left Magnifying Elements

By Proposition 2, we obtain the next theorem.

Theorem 6 (see [14]). Suppose that , ,and there are infinite such that . The function is a left magnifying element in if and only if is one-to-one but not onto.

Next, we will consider the case in which the equivalence relation and .

Lemma 5. Suppose that and . Let be a function in . Then is a left magnifying element if the following conditions hold:(1)For each such that , there exists a unique element ,(2)For all , there is a unique such that for all .

Proof. Let be a function in satisfying(1)For each with , there exists a unique element ,(2)For all , there is a unique such that for all .Since , there is a unique such that . Fix an element and let . Clearly, is a proper subset of . Let be a function in . We then define a function by, for all ,Clearly, , and for all , . Hence, , and so . Therefore, is a left magnifying element.

Lemma 6. Suppose that and . If is a left magnifying element, then the following conditions hold:(1)For each such that , there exists a unique element ,(2)For all , there is a unique such that for all .

Proof. Since and , . Let be a left magnifying element. Then there exists a proper subset of such that . Suppose that there exists such that there are two distinct elements . Since and , there are two distinct equivalence classes such that for all and for all .
Case 1 ( and ): Then . This implies , which is impossible, by Lemma 1.
Case 2 ( and ): Note that, for any , by Lemma 1, if , then there exists a unique such that . Let be distinct elements and let be a function in defined by, for all ,Then there is a function such that . Hence, , which is a contradiction.
Case 3 (WLOG, and ): By Lemma 1, if , then there exists a unique such that , and hence, . Let such that , and let be a function in defined by, for all ,Then there is a function such that . Since and , we obtain , which is a contradiction. Therefore, for each such that , there exists a unique element .
Let . Suppose that there are two distinct equivalence classes such that for all .
Case 1 ( and ): Then we have and . Hence, and , which is a contradiction.
Case 2 ( and ): For any , by Lemma 1, if , then there exists a unique such that . Let and define a function in byfor all . Then there is a function such that . Hence, , which is a contradiction.
Case 3 (WLOG, and ): By Lemma 1, if , then there exists a unique such that . Since and for all , there exists a unique element such that . Define a function in byfor all . Then there is a function such that . Hence, . Then , and hence , which is a contradiction. All the cases show that there is a unique such that for all . Therefore, the proof is complete.

By Lemmas 5 and 6, we obtain the following theorem.

Theorem 7. Suppose that and . A function is a left magnifying element in if and only if the following conditions hold:(1)For each such that , there exists a unique element ,(2)For all , there is a unique such that for all .

Example 3. Consider and . Let be an equivalence relation on such that . Assume that is a function in defined by, for all ,That is,Clearly, and . We can see that , , , , and . So for each such that , there exists a unique element . Moreover, for all , there is a unique such that for all . By Theorem 7, is a left magnifying element. Let . Then there is a proper subset of such that . Let be a function in defined by, for all ,That is,Then there is a function such that . Define a function by, for all ,That is,So , and hence,

5. Conclusion

In this paper, some particular properties of are investigated in Section 3. The main task of this paper is to establish the necessary and sufficient conditions for an elements to be right or left magnifying elements, which are summarized as follows:

If and , the following statements are true:(1)The function is a right magnifying element if and only if is onto but not one-to-one.(2)Suppose that there are infinite equivalence classes such that . Then is a left magnifying element if and only if is one-to-one but not onto.

If and , the following statements are true:(1)If the set and the set of all equivalence classes such that have the same cardinality, and for all , , then is a right magnifying element if and only if is onto.(2)The function is a left magnifying element if and only if(i)If is nonempty, then it is singleton,(ii)For all , there is a unique equivalence class whose members are all mapped to .

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 work was supported by the Faculty of Science Research Fund, Prince of Songkla University (Contract No. 264006).