**1. Introduction and Preliminaries**

Let *S* be a semigroup. An element *x* of *S* is called left regular if *x* = *yx*<sup>2</sup> for some *y* ∈ *S*. A right regular element is defined dually. Denote by *LReg*(*S*) and *RReg*(*S*) the sets of all left regular elements and right regular elements of *S*, respectively. Left and right regular elements are important in semigroup theory because they play a key role in the study of regular semigroups, which are semigroups in which every element is both left and right regular. Regular semigroups have many interesting properties and are used in various areas of mathematics, including algebra, topology and theoretical computer science. An element *x* of *S* is called left (right) magnifying if there is a proper subset *M* of *S* satisfying *xM* = *S* (*Mx* = *S*). In 1963, Ljapin [1] studied the notion of right and left magnifying elements of a semigroup. Several years later, Migliorini introduced the concepts of the minimal subset related to a magnifying element of *S* in [2,3]. Gutan [4] researched semigroups with strong and nonstrong magnifying elements in 1996. Later, he proved that every semigroup with magnifying elements is factorizable in [5].

Let *T*(*X*) be the total transformation semigroup on a nonempty set *X*. It is well known that *T*(*X*) is a regular semigroup. Moreover, every semigroup is isomorphic to a subsemigroup of some total transformation semigroups. The most basic mathematical structures are transformation semigroups. In 1952, Malcev [6] characterized ideals of *T*(*X*). Later, Miller and Doss [7] studied its group H-classes and its Green's relations. The generalization of these studies is the focus of this paper.

In 1975, Symons [8] considered a subsemigroup of *T*(*X*) defined by

$$T(X,Y) = \{ \mathfrak{a} \in T(X) : X\mathfrak{a} \subseteq Y \}$$

where *Y* is a nonempty subset of *X*. He determined all the automorphisms of *T*(*X*,*Y*). In 2005, Nenthein et al. [9] described regular elements in *T*(*X*,*Y*) and counted the number

**Citation:** Sripon, K.; Laysirikul, E.; Sommanee, W. Left (Right) Regular Elements of Some Transformation Semigroups. *Mathematics* **2023**, *11*, 2230. https://doi.org/10.3390/ math11102230

Academic Editors: Irina Cristea and Hashem Bordbar

Received: 1 March 2023 Revised: 28 April 2023 Accepted: 8 May 2023 Published: 10 May 2023

**Copyright:** © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

of all regular elements of *T*(*X*,*Y*) when *X* is a finite set. They described such a number in terms of |*X*|, |*Y*| and their related Stirling numbers. A few years later, Sanwong and Sommanee [10] studied regularity and Green's relations for the semigroup *T*(*X*,*Y*). They determined when *T*(*X*,*Y*) becomes a regular semigroup. Moreover, they gave a class of maximal inverse subsemigroups of *T*(*X*,*Y*) in 2008. After that, they proved that the set *F*(*X*,*Y*) = {*α* ∈ *T*(*X*,*Y*) : *Xα* = *Yα*} is the largest regular subsemigroup of *T*(*X*,*Y*) and determined its Green's relations. In [11], Sanwong described Green's relations and found all maximal regular subsemigroups of *F*(*X*,*Y*). In 2009, maximal and minimal congruences on *T*(*X*,*Y*) were considered. Sanwong et al. [12] found that *T*(*X*,*Y*) has only one maximal congruence if *X* is a finite set. They generalized [13] Theorem 3.4 for *Y* being infinite. Furthermore, characterizations of all minimal congruences on *T*(*X*,*Y*) were given. In the same year, Sun [14] proved that while the semigroup *T*(*X*,*Y*) is not left abundant, it is right abundant. Later in 2016, Lei Sun and Junling Sun [15] investigated the natural partial order on *T*(*X*,*Y*). Moreover, they determined the maximal elements and the minimal elements of *T*(*X*,*Y*).

Consider the semigroup

$$S(X,Y) = \{ \mathfrak{a} \in T(X) : \mathrm{Ya} \subseteq Y \}.$$

of transformations that leave *Y* invariant. In 1966, Magill [16] constructed and discussed the semigroup *S*(*X*,*Y*). In fact, if *Y* = *X*, then *S*(*X*,*Y*) = *T*(*X*). Later in [8], automorphism groups of a semigroup *S*(*X*,*Y*) were given by Symons. In 2005, Nenthein et al. [9] characterized regularity for *S*(*X*,*Y*). In addition, they found the number of regular elements in *S*(*X*,*Y*) for a finite set *X*. Honyam and Sanwong [17] studied its ideals, group H-classes and Green's relations on *S*(*X*,*Y*). Furthermore, they described when *S*(*X*,*Y*) is isomorphic to *T*(*A*) for some set *A*. A few years later, the left, right regular and intra-regular elements of a semigroup *S*(*X*,*Y*) were discussed by Choomanee, Honyam and Sanwong [18]. Moreover, when *X* is finite, they calculated the number of left regular elements in *S*(*X*,*Y*). In [19], natural partial orders on the semigroup *S*(*X*,*Y*) were considered by Sun and Wang. Moreover, they investigated left and right compatible elements with respect to this partial oder. Finally, they described the abundance of *S*(*X*,*Y*). In [20], all elements in the semigroup *S*(*X*,*Y*) that are left compatible with the natural partial order were studied. Left and right magnifying elements of *S*(*X*,*Y*) were given by Chiram and Baupradist in [21]. In a recent study, Punkumkerd and Honyam [22] provided a characterization of left and right magnifying elements on the semigroup *PT*(*X*,*Y*). *PT*(*X*,*Y*) denotes the set of all partial transformations *α* from a subset of *X* to *X* and (*domα* ∩ *Y*)*α* ⊆ *Y*, where *domα* is the domain of *α*. Their results have shown to be more general than the previous findings from [21].

In Section 2, we consider left and right magnifying elements of *T*(*X*,*Y*) and *S*(*X*,*Y*). We prove that each left magnifying element in *T*(*X*,*Y*) is not regular. Furthermore, we show that every left and right magnifying element in *S*(*X*,*Y*) is regular. In Sections 3 and 4, we focus on left and right regularity on *T*(*X*,*Y*) and *S*(*X*,*Y*). We show that every left magnifying element in *T*(*X*,*Y*) is a right regular element. Every right magnifying element is a left regular element. As [10] determined when *T*(*X*,*Y*) becomes a regular semigroup, we also characterize whenever *T*(*X*,*Y*) and *S*(*X*,*Y*) is a left (right) regular semigroup.

Note that throughout this paper, we will write mappings from the right, *xα* rather than *α*(*x*) and compose that the left to the right, *x*(*αβ*)=(*xα*)*β* rather than (*αβ*)(*x*) = *α*(*β*(*x*)) where *α*, *β* ∈ *T*(*X*) and *x* ∈ *X*. For each *α* ∈ *T*(*X*), we denote the set {*zα*−<sup>1</sup> : *z* ∈ *Xα*} by *π*(*α*) and *πY*(*α*) is the set {*P* ∈ *π*(*α*) : *P* ∩ *Y* = ∅} for a subset *Y* ⊆ *X*. Then, it is obvious that *π*(*α*) is a partition of *X*.

#### **2. Magnifying Elements**

In this section, we focus on characterizations of left magnifying elements and right magnifying elements in *T*(*X*,*Y*) and *S*(*X*,*Y*). The relationships between magnifying elements and regular elements are given.

**Theorem 1** ([23])**.** *Let α* ∈ *T*(*X*,*Y*)*. Then, α is right magnifying if and only if α is surjective but not injective and is such that yα*−<sup>1</sup> ∩ *Y* = ∅ *for all y* ∈ *Y and* |*yα*−<sup>1</sup> ∩ *Y*| > 1 *for some y* ∈ *Y.*

**Theorem 2** ([23])**.** *A mapping α* ∈ *T*(*X*) *is left magnifying if and only if α is injective but not surjective.*

**Theorem 3** ([23])**.** *Let α* ∈ *T*(*X*,*Y*)*. If* |*Y*| = |*X*| *and X* = *Y, then α is left magnifying if and only if α is injective.*

**Remark 1.** *For each α* ∈ *T*(*X*,*Y*)*, we note from Theorem 1 that α is right magnifying if and only if Yα* = *Y and α*|*<sup>Y</sup> is not injective.*

**Theorem 4.** *Let α* ∈ *T*(*X*,*Y*) *and X* = *Y. Then, α is left magnifying in a semigroup T*(*X*,*Y*) *if and only if α is injective.*

**Proof.** Assume that *α* is left magnifying. Suppose that *M* is a proper subset of *T*(*X*,*Y*) satisfying *αM* = *T*(*X*,*Y*). Let *a*, *b* ∈ *X* be such that *aα* = *bα*. If |*Y*| = 1, then *T*(*X*,*Y*) contains exactly one element. Thus, *T*(*X*,*Y*) has no proper subset *M* such that *αM* = *T*(*X*,*Y*). This is a contradiction. Therefore, |*Y*| > 1. Let *y* ∈ *Y* \ {*aα*}. Define *γ* : *X* → *X* by

$$
\mathfrak{x}\gamma = \begin{cases} a\mathfrak{a} & \text{if } \mathfrak{x} = a, \\ y & \text{otherwise.} \end{cases}
$$

It is verifiable that *γ* ∈ *T*(*X*,*Y*). From *αM* = *T*(*X*,*Y*), there exists *β* ∈ *M* such that *αβ* = *γ*. Suppose that *a* = *b*. Then, *aα* = *aγ* = *aαβ* = *bαβ* = *bγ* = *y*, which is a contradiction. Hence, *a* = *b* and so *α* is injective.

Suppose that *α* is injective. Then, we choose *y* ∈ *Y* and let *M* be the set {*γ* ∈ *T*(*X*,*Y*) : (*X* \ *Y*)*γ* = {*y*}}. From *X* = *Y*, we have *M* = ∅. It follows from our assumption that every *x* ∈ *Xα*, there is a unique *x* ∈ *X* satisfying *x α* = *x*. Let *β* ∈ *T*(*X*,*Y*). We define *γ* : *X* → *X* by

$$\mathfrak{x}\gamma = \begin{cases} \mathfrak{x}'\beta & \text{if } \mathfrak{x} \in \mathsf{Xa}\_{\prime} \\ \mathfrak{y} & \text{otherwise.} \end{cases}$$

It is verifiable that *γ* ∈ *M*. Now, let *x* ∈ *X*. Thus, *xαγ* = (*xα*) *β*. From *α* being injective and (*xα*) *α* = *xα*, we obtain (*xα*) = *x*. Therefore, *xαγ* = (*xα*) *β* = *xβ*. Hence, *β* = *αγ* and so *αM* = *T*(*X*,*Y*). This implies that *α* is left magnifying.

The set of natural numbers is represented by the letter N. Additionally, we denote the set of even natural numbers and the set of all odd natural numbers greater than 3 by 2N and 2N + 1, respectively.

**Example 1.** *Let X* = N *and Y* = 2N*. Define α* : *X* → *Y by xα* = 2*x for all x* ∈ *X. Then, Xα* = *Y. Clearly, α is injective. From Theorem 4, we obtain that α is left magnifying. We will show that α is not a regular element. Suppose that α is a regular element in T*(*X*,*Y*)*. Thus, there exists β* ∈ *T*(*X*,*Y*) *such that αβα* = *α. Consider* 3*αβα* = 3*α* = 6*. Thus,* 3*αβ* ∈ 6*α*−<sup>1</sup> = {3}*. Hence,* 3*αβ* = 3*, which is a contradiction. So α is not regular element.*

From the above example, we will verify that in *T*(*X*,*Y*), each left magnifying element is not a regular element.

**Theorem 5.** *If X* = *Y, then every left magnifying element of T*(*X*,*Y*) *is not regular.*

**Proof.** Assume that *X* = *Y*. Let *α* be a left magnifying element. From Theorem 4, we obtain that *α* is injective. Suppose that *α* is a regular element in *T*(*X*,*Y*). Then, there exists

*β* ∈ *T*(*X*,*Y*) such that *α* = *αβα*. Since *X* = *Y*, we choose *x* ∈ *X* \ *Y*. Therefore, *xαβα* = *xα* and then *xαβ* ∈ (*xα*)*α*<sup>−</sup>1. Since *α* is injective, we have *xαβ* = *x* ∈/ *Y*. This is a contradiction with *Xβ* ⊆ *Y*. So *α* is not regular.

**Theorem 6.** *Every right magnifying element of a semigroup T*(*X*,*Y*) *is regular.*

**Proof.** Let *α* be a right magnifying element of *T*(*X*,*Y*). By Remark 1, we obtain that *Yα* = *Y*. It follows that *Xα* ⊆ *Y* = *Yα*. From *Yα* ⊆ *Xα*, we have *Xα* = *Yα*. This means that *α* ∈ *F*(*X*,*Y*) and so *α* is a regular element of *T*(*X*,*Y*).

The following example shows that there exists an element in some *T*(*X*,*Y*) which is regular but it is not right magnifying.

**Example 2.** *Let X* = N *and Y* = 2N*. Define α* : *X* → *Y by*

*xα* = *x* + 2 *if x* ∈ 2N, *x* + 3 *otherwise.*

*Note that α* ∈ *T*(*X*,*Y*)*. It is easy to see that Yα* = 2N \ {2} = *Y and Xα* = 2N \ {2} = *Yα. Thus, α is regular but it is not right magnifying.*

In the rest of this section, we consider magnifying elements in *S*(*X*,*Y*).

**Lemma 1** ([21])**.** *Let α* ∈ *S*(*X*,*Y*)*. Then, α is a right magnifying element if and only if α is surjective but not injective such that yα*−<sup>1</sup> ∩ *Y* = ∅ *for all y* ∈ *Y.*

**Lemma 2** ([21])**.** *Let α* ∈ *S*(*X*,*Y*)*. Then, α is a left magnifying element if and only if α is injective but not surjective such that yα*−<sup>1</sup> ⊆ *Y for all y* ∈ *Y* ∩ *Xα.*

**Lemma 3** ([9])**.** *Let α* ∈ *S*(*X*,*Y*)*. Then, α is a regular element if and only if Yα* = *Xα* ∩ *Y.*

**Theorem 7.** *Every left magnifying element of a semigroup S*(*X*,*Y*) *is regular.*

**Proof.** Suppose that *α* is left magnifying. We will show that *Xα* ∩ *Y* = *Yα*. Clearly, *Yα* ⊆ *Xα* ∩ *Y*. Let *y* ∈ *Xα* ∩ *Y*. Then, there exists *y* ∈ *X* such that *y* = *y α*. Thus, *y* ∈ *yα*−<sup>1</sup> ⊆ *Y* by Lemma 2. This implies that *Xα* ∩ *Y* = *Yα*. From Lemma 3, we obtain that *α* is regular.

**Theorem 8.** *Every right magnifying element of a semigroup S*(*X*,*Y*) *is regular.*

**Proof.** Suppose that *α* is right magnifying. We will show that *Xα* ∩ *Y* = *Yα*. Clearly, *Yα* ⊆ *Xα* ∩ *Y*. Let *y* ∈ *Xα* ∩ *Y*. By Lemma 1, we have *yα*−<sup>1</sup> ∩ *Y* = ∅. Thus, there exists *y* ∈ *yα*−<sup>1</sup> ∩ *Y*. Hence, *y* = *y α* ∈ *Yα*. Therefore, *Xα* ∩ *Y* = *Yα*. From Lemma 3, we obtain *α* is regular.

**Example 3.** *Let α be defined in Example 2. It is clear that α* ∈ *S*(*X*,*Y*) *and α is neither injective nor surjective. Since Xα* = *Yα, this means that Xα* ∩ *Y* = *Yα. Hence, α is regular, while it is neither a left nor right magnifying element in S*(*X*,*Y*)*.*
