Open Access
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*Mathematics*
**2018**,
*6*(9),
160;
doi:10.3390/math6090160

Article

On Magnifying Elements in E-Preserving Partial Transformation Semigroups

^{1}

Department of Mathematics and Statistics, Prince of Songkla University, Hat Yai, Songkhla 90110, Thailand

^{2}

Algebra and Applications Research Unit, Department of Mathematics and Statistics, Prince of Songkla University, Hat Yai, Songkhla 90110, Thailand

^{3}

Centre of Excellence in Mathematics, CHE, Si Ayuthaya Road, Bangkok 10400, Thailand

^{*}

Author to whom correspondence should be addressed.

Received: 15 August 2018 / Accepted: 4 September 2018 / Published: 6 September 2018

## Abstract

**:**

Let S be a semigroup. An element a of S is called a right [left] magnifying element if there exists a proper subset M of S satisfying $S=Ma$ $[S=aM].$ Let E be an equivalence relation on a nonempty set X. In this paper, we consider the semigroup $P(X,E)$ consisting of all E-preserving partial transformations, which is a subsemigroup of the partial transformation semigroup $P\left(X\right)$. The main propose of this paper is to show the necessary and sufficient conditions for elements in $P(X,E)$ to be right or left magnifying.

Keywords:

magnifying elements; transformation semigroups; equivalence relations## 1. Introduction

An element a of a semigroup S is a right [left] magnifying element in S if there exists a proper subset M of S satisfying $S=Ma$ $[S=aM]$. The concepts of right and left magnifying elements of a semigroup were first introduced in 1963 by Ljapin [1]. Many initially significant results were later published by Migliorini in [2,3], where he also introduced the notion of minimal subset related to a magnifying element of S. In [4], Catino and Migliorini determined the existence of strong magnifying elements in a semigroup and the existence of magnifying elements in simple and bisimple semigroups as well as regular semigroups. Semigroups with strong and nonstrong magnifying elements were investigated by Gutan [5]. A year later, he showed in [6] that every semigroup which contains magnifying elements is factorizable; this solved a problem raised by Catino and Migliorini. Gutan also established in [7] the method for obtaining semigroups having good left magnifying elements such that none of those is very good.

Let X be a nonempty set. The full transformation semigroup on X is the set
of all transformations from X into itself, which is a semigroup under the composition of functions. In [8], Magill, Jr. characterized transformation semigroups with identities containing magnifying elements. Gutan and Kisielewicz solved in [9] a long-standing open problem by showing the existence of semigroups containing both good and bad magnifying elements.

$$T\left(X\right)=\{f:X\to X\mid f\phantom{\rule{4.pt}{0ex}}\mathrm{is}\phantom{\rule{4.pt}{0ex}}\mathrm{a}\phantom{\rule{4.pt}{0ex}}\mathrm{function}\}$$

Interesting properties, especially regularity and Green’s relations, on semigroups of transformations preserving relations have been widely conducted; see, e.g., [10,11,12,13,14]. In 2013, Huisheng and Weina [15] studied naturally orderd semigroups of partial transformations preserving an equivalence relation. Chinram and Baupradist have lately investigated right and left magnifying elements in some generalized transformation semigroups in [16,17].

Let E be an equivalence relation of a nonempty set X. We conventionally set $P\left(X\right)=\{\alpha :A\to X\mid A\subseteq X\}$. All functions will be written from the right, $\left(x\right)\alpha $ rather than $\alpha \left(x\right)$, and composed as $\left(x\right)\left(\alpha \beta \right)$ rather than $(\beta \circ \alpha )\left(x\right)$, for $\alpha ,\beta \in P\left(X\right)$. The semigroup of partial transformations preserving the equivalence relation E
is exactly a subsemigroup of $P\left(X\right)$. Furthermore, if $E=X\times X$, then $P(X,E)=P\left(X\right)$. In this paper, we study right and left magnifying elements in $P(X,E)$ and conclude necessary and sufficient conditions for elements of $P(X,E)$ to be left or right magnifying.

$$P(X,E)=\{\alpha \in P(X)\mid (x,y)\in E\phantom{\rule{4.pt}{0ex}}\mathrm{implies}\phantom{\rule{4.pt}{0ex}}(\left(x\right)\alpha ,\left(y\right)\alpha )\in E\}$$

## 2. Right Magnifying Elements

**Lemma**

**1.**

If α is a right magnifying element in $P(X,E)$, then α is onto.

**Proof.**

Suppose $\alpha $ is a right magnifying element in $P(X,E)$. According to the definition of right magnifying element, there exists a proper subset M of $P(X,E)$ with $M\alpha =P(X,E)$. Clearly, the identity map $i{d}_{X}$ on X belongs to $P(X,E)$. Thus, there exists $\beta \in M$ such that $\beta \alpha =i{d}_{X}$. This shows that $\alpha $ is onto. ☐

**Lemma**

**2.**

Let α be a right magnifying element in $P(X,E)$. For any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $.

**Proof.**

Suppose $\alpha $ is a right magnifying element in $P(X,E)$. Again, by definition, we obtain a proper subset M of $P(X,E)$ satisfying $M\alpha =P(X,E)$. Since $i{d}_{X}\in P(X,E)$, there exists $\beta \in M$ such that $\beta \alpha =i{d}_{X}$. Let $x,y\in X$ be such that $(x,y)\in E$. It follows that $\left(x\right)\beta \alpha =x$ and $\left(y\right)\beta \alpha =y$. Since $\beta \in P(X,E)$, we have $\left(\right(x)\beta ,(y\left)\beta \right)\in E$. We then choose $a=\left(x\right)\beta $ and $b=\left(y\right)\beta $. Therefore, the proof is complete. ☐

**Lemma**

**3.**

If $dom$ $\alpha =X$ and α is bijective in $P(X,E)$, then α is not right magnifying.

**Proof.**

Assume that $dom$ $\alpha =X$ and $\alpha $ is bijective in $P(X,E)$. By Lemma 2, ${\alpha}^{-1}\in P(X,E)$ such that $dom$ ${\alpha}^{-1}=X$. Suppose that $\alpha $ is right magnifying. By definition, there is a proper subset M of $P(X,E)$ with $M\alpha =P(X,E)$. Consequently, $M\alpha =P(X,E)\alpha $. Then $M=M\alpha {\alpha}^{-1}=P(X,E)\alpha {\alpha}^{-1}=P(X,E)$, which is a contradiction since M is a proper subset of $P(X,E)$. Hence $\alpha $ is not right magnifying. ☐

**Lemma**

**4.**

If $\alpha \in P(X,E)$ is onto but not one-to-one, $dom$ $\alpha =X$ and, for any $(x,y)\in E,$ there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $, then α is right magnifying.

**Proof.**

Let $\alpha \in P(X,E)$ be onto but not one-to-one and $dom$ $\alpha =X$. For any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $. Let $M=\{\beta \in P(X,E)\mid \beta $ is not onto}. Then, $M\ne P(X,E)$.

Let $\gamma $ be any function in $P(X,E)$. Since $\alpha $ is onto, we have for each $x\in dom$ $\gamma $, there exists ${a}_{x}\in dom\phantom{\rule{3.33333pt}{0ex}}\alpha $ such that $\left({a}_{x}\right)\alpha =\left(x\right)\gamma $ (if $\left({a}_{1}\right)\gamma =\left({a}_{2}\right)\gamma $, we must choose ${a}_{{x}_{1}}={a}_{{x}_{2}}$ and if $\left(\right(x)\gamma ,(y\left)\gamma \right)\in E$, we must choose $({a}_{x},{a}_{y})\in E$). Define $\beta \in P\left(X\right)$ by $\left(x\right)\beta ={a}_{x}$ for all $x\in dom$ $\gamma $. To show that $\beta \in P(X,E)$, let $x,y\in X$ be such that $(x,y)\in E$. Since $\gamma \in P(X,E)$, $\left(\right(x)\gamma ,(y\left)\gamma \right)\in E$. By assumption, we obtain $({a}_{x},{a}_{y})\in E$ such that $\left(x\right)\gamma =\left({a}_{x}\right)\alpha $ and $\left(y\right)\gamma =\left({a}_{y}\right)\alpha $. Hence, $\left(\right(x)\beta ,(y\left)\beta \right)\in E$. Since $\alpha $ is not one-to-one, $\beta $ is not onto either. Thus, $\beta \in M$ and we obtain, for all $x\in X$, that

$$\left(x\right)\beta \alpha =\left(\left(x\right)\beta \right)\alpha =\left({a}_{x}\right)\alpha =\left(x\right)\gamma .$$

Then, $\beta \alpha =\gamma $, hence $M\alpha =P(X,E)$. Therefore, $\alpha $ is right magnifying. ☐

**Example**

**1.**

Let $X=\mathbb{N}$. Define a relation E on X by

$$(x,y)\in E\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}\mathit{and}\phantom{\rule{4.pt}{0ex}}\mathit{only}\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}\lfloor {\displaystyle \frac{x}{3}}\rfloor =\lfloor {\displaystyle \frac{y}{3}}\rfloor .$$

Consider $X/E=\left\{\right\{1,2\left\}\right\}\cup \left\{\right\{x,x+1,x+2\}\mid x\in 3X\}=\left\{\right\{1,2\},\{3,4,5\},\{6,7,8\},\dots \}$. It is clear that E is an equivalence relation on X. Let $\alpha \in P(X,E)$ be defined by $\left(x\right)\alpha =x$ for all positive integers $x\le 5$ and $\left(x\right)\alpha =x-3$ for all positive integers $x>5$, that is,

$$\alpha =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ 1& 2& 3& 4& 5& 3& 4& 5& \cdots \end{array}\right).$$

Then, α is onto but not one-to-one and, for any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $. Let $M=\{\beta \in P(X,E)\mid \beta $ is not onto}. For any function $\gamma \in P(X,E)$, Lemma 4 ensures that there exists $\beta \in M$ such that $\beta \alpha =\gamma $.

We will illustrate these ideas by considering the element γ of $P(X,E)$, which is defined by $\left(1\right)\gamma =1$, $\left(2\right)\gamma =2$ and $\left(x\right)\gamma =x-3$ for positive integers $x>5$, that is,

$$\gamma =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ 1& 2& -& -& -& 3& 4& 5& \cdots \end{array}\right).$$

Define a function $\beta :dom\phantom{\rule{0.277778em}{0ex}}\gamma \to X$ by $\left(x\right)\beta =x$ for all $x\in dom\phantom{\rule{0.277778em}{0ex}}\gamma $, that is,

$$\beta =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ 1& 2& -& -& -& 6& 7& 8& \cdots \end{array}\right).$$

Thus $\beta \in M$ and we have

$$\begin{array}{cc}\hfill \beta \alpha & =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ 1& 2& -& -& -& 6& 7& 8& \cdots \end{array}\right)\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ 1& 2& 3& 4& 5& 3& 4& 5& \cdots \end{array}\right)\hfill \\ & =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ 1& 2& -& -& -& 3& 4& 5& \cdots \end{array}\right)=\gamma .\hfill \end{array}$$

**Lemma**

**5.**

If $\alpha \in P(X,E)$ is onto, $dom$ $\alpha \ne X$ and for any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $, then α is right magnifying.

**Proof.**

Let $\alpha \in P(X,E)$ be onto and $dom$ $\alpha \ne X$. For any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $. We follow the method of proof used in Lemma 4 and define $\beta :dom$ $\gamma \u27f6X$ by $\left(x\right)\beta ={a}_{x}$ for all $x\in dom\phantom{\rule{0.277778em}{0ex}}\gamma $. To show that $\beta \in P(X,E)$, let $x,y\in X$ be such that $(x,y)\in E$. Since $\gamma \in P(X,E)$, we have $\left(\right(x)\gamma ,(y\left)\gamma \right)\in E$. By assumption, there exists $({a}_{x},{a}_{y})\in E$ such that $\left(x\right)\gamma =\left({a}_{x}\right)\alpha $ and $\left(y\right)\gamma =\left({a}_{y}\right)\alpha $. Hence, $\left(\right(x)\beta ,(y\left)\beta \right)\in E$. Since $dom\phantom{\rule{3.33333pt}{0ex}}\alpha \ne X$, $\beta $ is not onto either.

Thus, $\beta \in M$ and we obtain, for all $x\in X$, that
$$\left(x\right)\beta \alpha =\left(\left(x\right)\beta \right)\alpha =\left({a}_{x}\right)\alpha =\left(x\right)\gamma .$$

Then, $\beta \alpha =\gamma $, hence $M\alpha =P(X,E)$. Therefore, $\alpha $ is right magnifying. ☐

**Example**

**2.**

Let $X=\mathbb{N}$. Define an equivalence relation E on X by
$$(x,y)\in E\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}\mathit{and}\phantom{\rule{4.pt}{0ex}}\mathit{only}\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}\lfloor {\displaystyle \frac{x}{3}}\rfloor =\lfloor {\displaystyle \frac{y}{3}}\rfloor .$$

We obtain $X/E=\left\{\right\{1,2\left\}\right\}\cup \left\{\right\{x,x+1,x+2\}\mid x\in 3X\}=\left\{\right\{1,2\},\{3,4,5\},\{6,7,8\},\dots \}$. Let $\alpha \in P(X,E)$ be defined by $\left(3\right)\alpha =1,\left(4\right)\alpha =2$ and $\left(x\right)\alpha =x-3$ for all positive integers $x>5$, that is,

$$\alpha =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ -& -& 1& 2& -& 3& 4& 5& \cdots \end{array}\right).$$

Then, α is onto, $dom$ $\alpha \ne X$ and for any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $. Let $M=\{\beta \in P(X,E)\mid \beta $ is not onto}. For any function $\gamma \in P(X,E)$, Lemma 5 ensures that there exists $\beta \in M$ such that $\beta \alpha =\gamma $.

We will illustrate these ideas by considering the element γ of $P(X,E)$, which is defined by $\left(x\right)\gamma =\lfloor {\displaystyle \frac{x+3}{3}}\rfloor $ for all $x>2$, that is,

$$\gamma =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ -& -& 2& 2& 2& 3& 3& 3& \cdots \end{array}\right).$$

To get the required result, define a function $\beta :dom$ $\gamma \to X$ by $\left(x\right)\beta =\lfloor {\displaystyle \frac{x+9}{3}}\rfloor $ if $x=3,4,5$ and $\left(x\right)\beta =\lfloor {\displaystyle \frac{x+12}{3}}\rfloor $ for all positive integers $x\ge 6$, that is,

$$\beta =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ -& -& 4& 4& 4& 6& 6& 6& \cdots \end{array}\right).$$

Thus, $\beta \in M$ and we have

$$\begin{array}{cc}\hfill \beta \alpha & =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ -& -& 4& 4& 4& 6& 6& 6& \cdots \end{array}\right)\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ -& -& 1& 2& -& 3& 4& 5& \cdots \end{array}\right)\hfill \\ & =\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& \cdots \\ -& -& 2& 2& 2& 3& 3& 3& \cdots \end{array}\right)=\gamma .\hfill \end{array}$$

We summarize those lemmas in a theorem as follows.

**Theorem**

**1.**

α is right magnifying in $P(X,E)$ if and only if α is onto, for any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $ and either

- 1.
- $dom$$\alpha \ne X$ or
- 2.
- $dom$$\alpha =X$ and α is not one-to-one.

**Proof.**

It follows by Lemmas 1–5. ☐

As a consequence, the following result holds for $E=X\times X.$

**Corollary**

**1.**

Let $\alpha \in P\left(X\right)$. Then, α is right magnifying in a semigroup $P\left(X\right)$ if and only if α is onto and either

- 1.
- $dom$$\alpha \ne X$ or
- 2.
- $dom$$\alpha =X$ and α is not one-to-one.

**Proof.**

It follows immediately from Theorem 1. ☐

## 3. Left Magnifying Elements

**Lemma**

**6.**

If α is a left magnifying element in $P(X,E)$, then α is one-to-one.

**Proof.**

Using the same argument as in Lemma 1, we obtain a proper subset M of $P(X,E)$ with $\alpha M=P(X,E)$ and there exists $\beta \in M$ such that $\alpha \beta =i{d}_{X}$. This shows that $\alpha $ is one-to-one. ☐

**Lemma**

**7.**

If α is a left magnifying element in $P(X,E)$, then $dom$ $\alpha =X$.

**Proof.**

Suppose $\alpha $ is a left magnifying in $P(X,E)$. Again, by definition, there is a proper subset M of $P(X,E)$ with $\alpha M=P(X,E)$. Let $\gamma \in P(X,E)$ be such that $dom$ $\gamma =X$. Then, $\gamma =\alpha \beta $ for some $\beta \in M$. Since $dom$ $\gamma =X$, we obtain $dom$ $\alpha =X$. ☐

**Lemma**

**8.**

Let α be a left magnifying element in $P(X,E)$. For any $x,y\in X$, if $\left(\right(x)\alpha ,(y\left)\alpha \right)\in E,$ then $(x,y)\in E$.

**Proof.**

Suppose $\alpha $ is a left magnifying element in $P(X,E)$. By Lemma 7, $dom$ $\alpha =X$. There exists a proper subset M of $P(X,E)$ satisfying $\alpha M=P(X,E)$. Since $i{d}_{X}\in P(X,E)$, there exists $\beta \in M$ such that $\alpha \beta =i{d}_{X}$. Let $x,y\in X$ be such that $\left(\right(x)\alpha ,(y\left)\alpha \right)\in E$. It follows that $\left(x\right)\alpha \beta =x$ and $\left(y\right)\alpha \beta =y$. Then, we obtain $(x,y)=\left(\right(x)\alpha \beta ,(y\left)\alpha \beta \right)\in E$ because $\beta \in P(X,E)$. ☐

**Lemma**

**9.**

If $\alpha \in P(X,E)$ is bijective and $dom$ $\alpha =X$, then α is not left magnifying.

**Proof.**

As in Lemma 3, the result holds by applying Lemma 8. ☐

**Lemma**

**10.**

If $\alpha \in P(X,E)$ is one-to-one but not onto, $dom$ $\alpha =X$ and for any $x,y\in X$, $\left(\right(x)\alpha ,(y\left)\alpha \right)\in E$ implies $(x,y)\in E$, then α is a left magnifying element in $P(X,E)$.

**Proof.**

Assume that $\alpha \in P(X,E)$ is one-to-one but not onto, $dom$ $\alpha =X$ and for any $x,y\in X$, $\left(\right(x)\alpha ,(y\left)\alpha \right)\in E$ implies $(x,y)\in E$. Let $M=\{\beta \in P(X,E)\mid dom$ $\beta \subseteq ran$ $\alpha \}$. Claim that $\alpha M=P(X,E).$ Let $\gamma \in P(X,E)$. For $x\in ran$ $\alpha $, then there exists ${a}_{x}\in dom$ $\gamma $ such that $\left({a}_{x}\right)\alpha =x$. Define $\beta \in P\left(X\right)$ by $\left(x\right)\beta =\left({a}_{x}\right)\gamma $ if $x\in ran$ $\alpha $ and ${a}_{x}\in dom$ $\gamma $. We see that $dom$ $\beta =\{x\in ran$ $\alpha \mid {a}_{x}\in dom$ $\gamma \}\subseteq ran$ $\alpha $. To show that $\beta \in P(X,E)$, assume $x,y\in ran$ $\alpha $, $\left({a}_{x}\right)\alpha =x,\left({a}_{y}\right)\alpha =y$, ${a}_{x},{a}_{y}\in dom$ $\gamma $ and $(x,y)\in E$. Then, $(\left({a}_{x}\right)\alpha ,\left({a}_{y}\right)\alpha )\in E$ and hence, by assumption, we obtain $({a}_{x},{a}_{y})\in E$. Thus, $(\left(x\right)\beta ,\left(y\right)\beta )=(\left({a}_{x}\right)\gamma ,\left({a}_{y}\right)\gamma )\in E$ because $\gamma \in P(X,E).$ Then, $\beta \in M$ and for $x\in X$, we have

$$\left(x\right)\alpha \beta =\left(\right(x\left)\alpha \right)\beta =\left(x\right)\gamma .$$

This gives us that $\alpha \beta =\gamma $ and $\alpha M=P(X,E)$. Hence, $\alpha $ is a left magnifying element in $P(X,E)$. ☐

**Example**

**3.**

Let $X=\mathbb{N}$. Define a relation E on X by

$$(x,y)\in E\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}\mathit{and}\phantom{\rule{4.pt}{0ex}}\mathit{only}\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}\lfloor {\displaystyle \frac{x}{2}}\rfloor \equiv \lfloor {\displaystyle \frac{y}{2}}\rfloor \phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}2).$$

It is obvious that E is an equivalence relation on X and, in addition, $X/E=\left\{\right\{1,4,5,8,9,\dots \},\{2,3,6,7,\dots \left\}\right\}$. Let $\alpha \in E$ be defined by $\left(x\right)\alpha =x+2$ for all positive integers $x\in X$. For convenience, we write α as

$$\alpha =\left(\begin{array}{ccccccccccccccc}1& 4& 5& 8& 9& 12& 13& \cdots & 2& 3& 6& 7& 10& 11& \cdots \\ 3& 6& 7& 10& 11& 14& 15& \cdots & 4& 5& 8& 9& 12& 13& \cdots \end{array}\right).$$

We now obtain that α is one-to-one but not onto and for any $x,y\in X$, $\left(\right(x)\alpha ,(y\left)\alpha \right)\in E$ implies $(x,y)\in E$. Let $M=\{\beta \in P(X,E)\mid dom$ $\beta \subseteq ran$ $\alpha \}$ and $\gamma \in P(X,E)$ be any function. By Lemma 10, there exists $\beta \in M$ such that $\alpha \beta =\gamma $.

We will illustrate these ideas by considering the element γ of $P(X,E)$, which is defined by $\left(x\right)\gamma =x-2$ for all positive integers $x>3$, that is,

$$\gamma =\left(\begin{array}{ccccccccccccccc}1& 4& 5& 8& 9& 12& 13& \cdots & 2& 3& 6& 7& 10& 11& \cdots \\ -& 2& 3& 6& 7& 10& 11& \cdots & -& -& 4& 5& 8& 9& \cdots \end{array}\right).$$

To get the required result, define a function $\beta \in P(X,E)$ by $\left(x\right)\beta =x-4$ for all $x>5$, that is,

$$\beta =\left(\begin{array}{ccccccccccccccc}1& 4& 5& 8& 9& 12& 13& \cdots & 2& 3& 6& 7& 10& 11& \cdots \\ -& -& -& 4& 5& 8& 9& \cdots & -& -& 2& 3& 6& 7& \cdots \end{array}\right).$$

Thus, $\beta \in M$ and we have

$$\begin{array}{c}\alpha \beta =\left(\begin{array}{ccccccccccccccc}1& 4& 5& 8& 9& 12& 13& \cdots & 2& 3& 6& 7& 10& 11& \cdots \\ 3& 6& 7& 10& 11& 14& 15& \cdots & 4& 5& 8& 9& 12& 13& \cdots \end{array}\right)\\ \text{\hspace{1em}}\left(\begin{array}{ccccccccccccccc}1& 4& 5& 8& 9& 12& 13& \cdots & 2& 3& 6& 7& 10& 11& \cdots \\ -& -& -& 4& 5& 8& 9& \cdots & -& -& 2& 3& 6& 7& \cdots \end{array}\right)\\ =\left(\begin{array}{ccccccccccccccc}1& 4& 5& 8& 9& 12& 13& \cdots & 2& 3& 6& 7& 10& 11& \cdots \\ -& 2& 3& 6& 7& 10& 11& \cdots & -& -& 4& 5& 8& 9& \cdots \end{array}\right)=\gamma .\end{array}$$

**Example**

**4.**

Let $X=\mathbb{Z}\times \mathbb{Z}$. Define a relation E on X by

$$\left(\right(a,b),(c,d\left)\right)\in E\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}\mathit{and}\phantom{\rule{4.pt}{0ex}}\mathit{only}\phantom{\rule{4.pt}{0ex}}\mathit{if}\phantom{\rule{4.pt}{0ex}}a=c.$$

Let $\alpha \in P(X,E)$ be defined by $(a,b)\alpha =(2a,2b)$ for all $a,b\in \mathbb{Z}$. Then, α is one-to-one but not onto and for any $(a,b),(c,d)\in X$, $\left(\right(a,b)\alpha ,(c,d\left)\alpha \right)\in E$ implies $\left(\right(a,b),(c,d\left)\right)\in E$. Let $M=\{\beta \in P(X,E)\mid dom$ $\beta \subseteq ran$ $\alpha \}$ and $\gamma \in P(X,E)$ be any function. By Lemma 10, there exists $\beta \in M$ such that $\alpha \beta =\gamma $. We will illustrate these ideas by considering the element γ of $P(X,E)$, which is defined by $(a,b)\gamma =(a+1,b+2)$ for all $a,b\in \mathbb{Z}$.

To get the required result, define a function $\beta \in P(X,E)$ by $(a,b)\beta =(k+1,l+2)$ if $a=2k$ and $b=2l$ for some $k,l\in \mathbb{Z}.$ Thus, $\beta \in M$ and we have $(a,b)\alpha \beta =\left(\right(a,b\left)\alpha \right)\beta =(2a,2b)\beta =(a+1,b+2)=(a,b)\gamma $ for all $a,b\in \mathbb{Z},$ which shows that $\alpha \beta =\gamma $.

We summarize those lemmas in a theorem as follows.

**Theorem**

**2.**

α is left magnifying in $P(X,E)$ if and only if α is one-to-one but not onto, $dom$ $\alpha =X$ and for any $x,y\in X$, $\left(\right(x)\alpha ,(y\left)\alpha \right)\in E$ implies $(x,y)\in E$.

**Proof.**

It follows from Lemmas 6–10. ☐

As a consequence, the following result holds for $E=X\times X.$

**Corollary**

**2.**

α is a left magnifying element in $P\left(X\right)$ if and only if α is one-to-one but not onto and $dom$ $\alpha =X$.

**Proof.**

It follows from Theorem 2. ☐

## 4. Conclusions

In this paper, necessary and sufficient conditions for elements in $P(X,E)$ to be right or left magnifying are established as follows:

- $\alpha $ is right magnifying in $P(X,E)$ if and only if $\alpha $ is onto, for any $(x,y)\in E$, there exists $(a,b)\in E$ such that $x=\left(a\right)\alpha ,y=\left(b\right)\alpha $ and either
- $dom$$\alpha \ne X$ or
- $dom$$\alpha =X$ and $\alpha $ is not one-to-one.

- Let $\alpha \in P\left(X\right)$. Then, $\alpha $ is right magnifying in a semigroup $P\left(X\right)$ if and only if $\alpha $ is onto and either
- $dom$$\alpha \ne X$ or
- $dom$$\alpha =X$ and $\alpha $ is not one-to-one.

- $\alpha $ is left magnifying in $P(X,E)$ if and only if $\alpha $ is one-to-one but not onto, $dom$ $\alpha =X$ and for any $x,y\in X$, $\left(\right(x)\alpha ,(y\left)\alpha \right)\in E$ implies $(x,y)\in E$.
- $\alpha $ is a left magnifying element in $P\left(X\right)$ if and only if $\alpha $ is one-to-one but not onto and $dom$ $\alpha =X$.

## Author Contributions

Conceptualization, T.K., M.P. and R.C.; Investigation, T.K. and M.P.; Writing—Original Draft Preparation, T.K. and M.P.; Writing—Review and Editing, M.P.; Supervision, R.C.

## Funding

This research received no external funding.

## Acknowledgments

This research was supported by the Algebra and Applications Research Unit, Department of Mathematics and Statistics, Faculty of Science, Prince of Songkla University.

## Conflicts of Interest

The authors declare no conflict of interest.

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