Skew Pairs of Idempotents in Partial Transformation Semigroups

Abstract

Let S be a regular semigroup. A pair $(e,f)$ of idempotents of S is said to be a skew pair of idempotents if $fe$ is idempotent, but $ef$ is not. T. S. Blyth and M. H. Almeida (T. S. Blyth and M. H. Almeida, skew pair of idempotents in transformation semigroups, Acta Math. Sin. (English Series), 22 (2006), 1705–1714) gave a characterization of four types of skew pairs—those that are strong, left regular, right regular, and discrete—existing in a full transformation semigroup $T\left(X\right)$. In this paper, we do in this line for partial transformation semigroups.

1. Introduction

In [1], the authors defined a skew pair of idempotents in a regular semigroup as follows:

Definition 1.

Let S be a regular semigroup and $e,f$ be idempotents of S. A pair $(e,f)$ is said to be a skew pair of idempotents of S if $fe$ is idempotent, but $ef$ is not.

It was then proved that there are four distinct types of skew pairs of idempotents. Indeed:

Theorem 1.

Let S be a regular semigroup. Then, there are four distinct types of skew pairs $(e,f)$ of idempotents, namely those that are:

(1)

strong if $fe={\left(ef\right)}^2[=efe=fef]$;

(2)

left regular if $fe=fef\ne efe[={\left(ef\right)}^2]$;

(3)

right regular if $fe=efe\ne fef[={\left(ef\right)}^2]$;

(4)

discrete if$fe,fef,efe,{\left(ef\right)}^2$are distinct.

If X is a nonempty set, then the full transformation semigroup (under the composition) will be denoted by $T\left(X\right)$. A lot of information on transformation semigroups is contained in Clifford and Preston’s text [2]. Suschkewitsch showed in [3] that every semigroup can be embedded in $T\left(X\right)$ for some nonempty set X. This means that transformation semigroups play the same role for semigroup theory as permutation groups do for group theory. In [4], the same authors as in [1] determined whether the regular semigroup $T\left(X\right)$ contains skew pairs of idempotents. They proved the following theorem:

Theorem 2.

A transformation semigroup $T\left(X\right)$ contains skew pairs of idempotents that are:

(1)

strong if and only if $\left|X\right|\ge 3$;

(2)

left regular if and only if $\left|X\right|\ge 4$;

(3)

right regular if and only if $\left|X\right|\ge 5$;

(4)

discrete if and only if $\left|X\right|\ge 6$.

In this line, for a vector space V, T. Changphas and R. P. Sullivan [5] determined whether the regular subsemigroup $T\left(V\right)$ of linear transformations on V contains a skew pair of idempotents of each of the types. The purpose of this paper is to perform a similar task for a regular semigroup consisting of partial transformations on a nonempty set.

2. Results

Let X be a nonempty set. A partial transformation on X is a function $\alpha :A\to X$, mapping a subset A of X into X. Unlike the standard practice, we write all mappings on the left (for example, $\alpha \beta \left(x\right)=\alpha \left(\beta \right(x\left)\right)$). For a partial transformation $\alpha$, let $\mathrm{dom}\alpha$ and $\mathrm{ran}\alpha$ denote the domain of $\alpha$ and the image of $\alpha$, respectively. Let $P\left(X\right)$ be the semigroup (under the composition) of all partial transformations on X. The product of idempotents on $P\left(X\right)$ was considered widely (see e.g., [6,7]). For each $\alpha ,\beta \in P\left(X\right)$, it was observed that:

$$\begin{array}{c}\hfill \mathrm{dom}\alpha \beta ={\beta }^{-1}(\mathrm{ran}\beta \cap \mathrm{dom}\alpha )\subseteq \mathrm{dom}\beta \\ \hfill \mathrm{ran}\alpha \beta =\alpha (\mathrm{ran}\beta \cap \mathrm{dom}\alpha )\subseteq \mathrm{ran}\alpha .\end{array}$$

Moreover, it is known that for each $\alpha \in P\left(X\right)$, $\alpha$ is idempotent if and only if $\mathrm{ran}\alpha \subseteq \mathrm{dom}\alpha$ and $\alpha \left(x\right)=x$ for all $x\in \mathrm{ran}\alpha$, and that $P\left(X\right)$ is regular.

Hereafter, we shall prove the following main results:

Theorem 3.

Let X and Y be nonempty sets with $\left|X\right|\le \left|Y\right|$. If $P\left(X\right)$ contains a strong (respectively, left regular, right regular, discrete) skew pair of idempotents, then $P\left(Y\right)$ contains a strong (respectively, left regular, right regular, discrete) skew pair of idempotents.

Corollary 1.

Let X and Y be nonempty sets with $\left|X\right|\le \left|Y\right|$. If $P\left(Y\right)$ does not contain a strong (respectively, left regular, right regular, discrete) skew pair of idempotents, then $P\left(X\right)$ does not contain a strong (respectively, left regular, right regular, discrete) skew pair of idempotents.

Theorem 4.

Let X be a nonempty set. Then $P\left(X\right)$ contains a skew pair that is:

(1)

strong if and only if $\left|X\right|\ge 2$;

(2)

left regular if and only if $\left|X\right|\ge 4$;

(3)

right regular if and only if $\left|X\right|\ge 4$;

(4)

discrete if and only if $\left|X\right|\ge 5$.

Firstly, we state the following useful lemma, and the proof is clear:

Lemma 1.

$P\left(X\right)$ contains a skew pair of idempotents if and only if $\left|X\right|\ge 2$.

For a regular semigroup S, the set of all idempotents of S will be denoted by $E\left(S\right)$, and the set of all inverses of $a\in S$ will be denoted by $V\left(a\right)$. That is $V\left(a\right)=\{x\in S:a=axa,x=xax\}$.

Let $(e,f)$ be a skew pair of idempotents in $P\left(X\right)$. We first set $\mathrm{fix}ef=\{x\in X\mid ef(x)=x\}$. Then, $\mathrm{ran}ef\backslash \mathrm{fix}ef\ne \varnothing$. For every $x\in \mathrm{ran}ef\backslash \mathrm{fix}ef$, we choose and fix an element ${\alpha }_x$ such that $ef\left({\alpha }_x\right)=x$; if $x\in \mathrm{fix}ef$, we choose ${\alpha }_x=x.$ Therefore, we have $ef\left({\alpha }_x\right)=x$ for every $x\in \mathrm{ran}ef$.

Theorem 5.

Let $(e,f)$ be a skew pair of idempotents in $P\left(X\right)$. For each $x\in \mathrm{ran}ef$, ${\alpha }_x\in \mathrm{ran}ef$ if and only if $x\in \mathrm{fix}ef$.

Proof. 

Let $x\in \mathrm{ran}ef$. If ${\alpha }_x\in \mathrm{ran}ef$, then there exists $y\in \mathrm{dom}ef$ such that $ef\left(y\right)={\alpha }_x$, and so:

$$x=ef\left({\alpha }_x\right)={\left(ef\right)}^2\left(y\right)\in \mathrm{fix}ef.$$

The opposite direction is clear.    □

Corollary 2.

Let $(e,f)$ be a skew pair of idempotents in $P\left(X\right)$. If $x,{\alpha }_x\in \mathrm{ran}ef$, then ${\alpha }_{{\alpha }_x}={\alpha }_x$.

We now prove the main theorem by separating them into the following theorems:

Theorem 6.

If $(e,f)$ is a skew pair of idempotents in $P\left(X\right)$, then there exists $\gamma \in V\left(ef\right)\cap E\left(P\right(X\left)\right)$ such that the skew pair $(ef\gamma ,\gamma ef)$ is strong.

Proof. 

Let $(e,f)$ be a skew pair of idempotents in $P\left(X\right)$. Define the following:

$$\gamma :(\mathrm{dom}ef\cup \mathrm{ran}ef)\to X$$

by

$$\gamma \left(x\right)=\left\{\begin{array}{cc}{\alpha }_x\hfill & \mathrm{if}x\in \mathrm{ran}ef,\hfill \\ {\alpha }_{ef\left(x\right)}\hfill & \mathrm{if}x\notin \mathrm{ran}ef.\hfill \end{array}\right.$$

We then have that $\gamma \in E\left(P\right(X\left)\right)$. Indeed, we can see that $\mathrm{ran}\gamma \subseteq \mathrm{dom}ef\subseteq \mathrm{dom}\gamma$. Next, we let $x\in \mathrm{ran}\gamma$, then there exists $y\in \mathrm{dom}\gamma$ such that $\gamma \left(y\right)=x$.

Case 1: $y\in \mathrm{ran}ef$. Then $x=\gamma \left(y\right)={\alpha }_y$, and $\gamma \left(x\right)=\gamma \left({\alpha }_y\right).$ If ${\alpha }_y\in \mathrm{ran}ef$, then $\gamma \left({\alpha }_y\right)={\alpha }_{{\alpha }_y}={\alpha }_y=x$. If ${\alpha }_y\notin \mathrm{ran}ef$, then $\gamma \left({\alpha }_y\right)={\alpha }_{ef\left({\alpha }_y\right)}={\alpha }_y=x$.

Case 2: $y\notin \mathrm{ran}ef$. Then $x=\gamma \left(y\right)={\alpha }_{ef\left(y\right)}$, and $\gamma \left(x\right)=\gamma \left({\alpha }_{ef\left(y\right)}\right).$ If ${\alpha }_{ef\left(y\right)}\in \mathrm{ran}ef$, then $\gamma \left({\alpha }_{ef\left(y\right)}\right)={\alpha }_{{\alpha }_{ef\left(y\right)}}={\alpha }_{ef\left(y\right)}=x$. If ${\alpha }_{ef\left(y\right)}\notin \mathrm{ran}ef$, then $\gamma \left({\alpha }_{ef\left(y\right)}\right)={\alpha }_{ef\left({\alpha }_{ef\left(y\right)}\right)}={\alpha }_{ef\left(y\right)}=x$.

Next, $\gamma$ is an inverse of $ef$. Hence, we have the following:

$$\begin{array}{cc}\hfill \mathrm{dom}ef& ={\left(ef\right)}^{-1}\left(\mathrm{ran}ef\right)\hfill \\ \hfill & ={\left(ef\right)}^{-1}(\mathrm{ran}ef\cap \mathrm{dom}\gamma )\hfill \\ \hfill & ={\left(ef\right)}^{-1}(\mathrm{ran}ef\cap {\gamma }^{-1}\left(\mathrm{ran}\gamma \right))\hfill \\ \hfill & ={\left(ef\right)}^{-1}(\mathrm{ran}ef\cap {\gamma }^{-1}(\mathrm{ran}\gamma \cap \mathrm{dom}ef))\hfill \\ \hfill & ={\left(ef\right)}^{-1}(\mathrm{ran}ef\cap \mathrm{dom}ef\gamma )\hfill \\ \hfill & =\mathrm{dom}ef\gamma ef.\hfill \end{array}$$

Since $ef\gamma ef\left(x\right)=ef\left({\alpha }_{ef\left(x\right)}\right)=ef\left(x\right)$ for all $x\in \mathrm{dom}ef$, it follows that $ef=ef\gamma ef$. We consider $\mathrm{dom}\gamma$ and $\mathrm{dom}\gamma ef\gamma$:

$$\begin{array}{cc}\hfill \mathrm{dom}\gamma & ={\gamma }^{-1}\left(\mathrm{ran}\gamma \right)\hfill \\ \hfill & ={\gamma }^{-1}(\mathrm{ran}\gamma \cap \mathrm{dom}ef)\hfill \\ \hfill & ={\gamma }^{-1}(\mathrm{ran}\gamma \cap {\left(ef\right)}^{-1}\left(\mathrm{ran}ef\right))\hfill \\ \hfill & ={\gamma }^{-1}(\mathrm{ran}\gamma \cap {\left(ef\right)}^{-1}(\mathrm{ran}ef\cap \mathrm{dom}\gamma ))\hfill \\ \hfill & =\mathrm{dom}\gamma ef\gamma .\hfill \end{array}$$

Let $y\in \mathrm{dom}\gamma$. If $y\in \mathrm{ran}ef$, then $\gamma ef\gamma \left(y\right)=\gamma ef\left({\alpha }_y\right)=\gamma \left(y\right)$. If $y\notin \mathrm{ran}ef$, then $\gamma ef\gamma \left(y\right)=\gamma ef\left({\alpha }_{ef\left(y\right)}\right)=\gamma \left(ef\left(y\right)\right)={\alpha }_{ef\left(y\right)}=\gamma \left(y\right)$. Thus, $\gamma =\gamma ef\gamma$.

Finally, we show that $(ef\gamma ,\gamma ef)$ is strong by showing that $\gamma {\left(ef\right)}^2={\left(ef\right)}^2$ and ${\left(ef\right)}^2\gamma ={\left(ef\right)}^2$. Let $x\in \mathrm{dom}{\left(ef\right)}^2\gamma$, then we have two cases to consider.

Case 1: $x\in \mathrm{ran}ef$. Then we have:

$${\alpha }_x=\gamma \left(x\right)\in \mathrm{dom}{\left(ef\right)}^2=\mathrm{dom}{\left(ef\right)}^3.$$

Thus, $x=ef\left({\alpha }_x\right)\in \mathrm{dom}{\left(ef\right)}^2$.

Case 2: $x\notin \mathrm{ran}ef$. Then $x\in \mathrm{dom}ef$, and we have the following:

$${\alpha }_{ef\left(x\right)}=\gamma \left(x\right)\in \mathrm{dom}{\left(ef\right)}^2=\mathrm{dom}{\left(ef\right)}^3.$$

Thus,

$$\begin{array}{cc}\hfill x\in {\left(ef\right)}^{-1}\left(ef\left(x\right)\right)& \subseteq {\left(ef\right)}^{-1}(\mathrm{ran}ef\cap \mathrm{dom}{\left(ef\right)}^2)\hfill \\ \hfill & =\mathrm{dom}{\left(ef\right)}^3\hfill \\ \hfill & =\mathrm{dom}{\left(ef\right)}^2.\hfill \end{array}$$

Hence, $\mathrm{dom}{\left(ef\right)}^2\gamma \subseteq \mathrm{dom}{\left(ef\right)}^2$. Let $x\in \mathrm{dom}{\left(ef\right)}^2$, then $x\in \mathrm{dom}ef\subseteq \mathrm{dom}\gamma$.

Case 1: $x\in \mathrm{ran}ef$. Then $\gamma \left(x\right)={\alpha }_x\in \mathrm{dom}{\left(ef\right)}^2$, and it follows that:

$$x\in {\gamma }^{-1}\left({\alpha }_x\right)\subseteq {\gamma }^{-1}(\mathrm{ran}\gamma \cap \mathrm{dom}{\left(ef\right)}^2)=\mathrm{dom}{\left(ef\right)}^2\gamma .$$

Case 2: $x\notin \mathrm{ran}ef$. Since $x\in \mathrm{dom}{\left(ef\right)}^2$, it follows that:

$$\begin{array}{cc}\hfill {\alpha }_{ef\left(x\right)}\in {\left(ef\right)}^{-1}\left(ef\left(x\right)\right)& \subseteq {\left(ef\right)}^{-1}(\mathrm{ran}ef\cap \mathrm{dom}ef)=\mathrm{dom}{\left(ef\right)}^2.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill x\in {\gamma }^{-1}\left({\alpha }_{ef\left(x\right)}\right)& \subseteq {\gamma }^{-1}(\mathrm{ran}\gamma \cap \mathrm{dom}{\left(ef\right)}^2)=\mathrm{dom}{\left(ef\right)}^2\gamma .\hfill \end{array}$$

Hence, $\mathrm{dom}{\left(ef\right)}^2\subseteq \mathrm{dom}{\left(ef\right)}^2\gamma$. These imply that $\mathrm{dom}{\left(ef\right)}^2=\mathrm{dom}{\left(ef\right)}^2\gamma$. For each $x\in \mathrm{dom}{\left(ef\right)}^2\gamma$, we have the following:

$${\left(ef\right)}^2\gamma \left(x\right)=\left\{\begin{array}{cc}{\left(ef\right)}^2\left({\alpha }_x\right)={\left(ef\right)}^3\left({\alpha }_x\right)={\left(ef\right)}^2\left(x\right)\hfill & \mathrm{if}x\in \mathrm{ran}ef,\hfill \\ {\left(ef\right)}^2\left({\alpha }_{ef\left(x\right)}\right){\left(ef\right)}^2\left(x\right)\hfill & \mathrm{if}x\notin \mathrm{ran}ef.\hfill \end{array}\right.$$

We infer that ${\left(ef\right)}^2\gamma ={\left(ef\right)}^2$. Next, we obtain $\mathrm{dom}\gamma {\left(ef\right)}^2=\mathrm{dom}{\left(ef\right)}^2$ by:

$$\mathrm{dom}\gamma {\left(ef\right)}^2\subseteq \mathrm{dom}{\left(ef\right)}^2=\mathrm{dom}ef\gamma {\left(ef\right)}^2\subseteq \mathrm{dom}\gamma {\left(ef\right)}^2.$$

Let $x\in \mathrm{dom}{\left(ef\right)}^2$. Then $y={\left(ef\right)}^2\left(x\right)$ for some $y\in \mathrm{ran}{\left(ef\right)}^2\subseteq \mathrm{ran}ef$. Since we have:

$$ef\left(y\right)={\left(ef\right)}^3\left(x\right)={\left(ef\right)}^2\left(x\right)=y,$$

we obtain ${\alpha }_y=y$. Thus, $\gamma {\left(ef\right)}^2\left(x\right)=\gamma \left(y\right)={\alpha }_y=y={\left(ef\right)}^2\left(x\right)$. Therefore, $\gamma {\left(ef\right)}^2={\left(ef\right)}^2$. Hence, $(ef\gamma ,\gamma ef)$ is strong.    □

Corollary 3.

$P\left(X\right)$ contains a strong skew pair of idempotents if and only if $\left|X\right|\ge 2$.

Proof. 

Assume that $\left|X\right|⩾2$. Let $x,y$ be distinct elements of X. Define functions e and f by the following:

$$\begin{array}{c}\hfill e=\left(\begin{array}{cc}x& y\\ y& y\end{array}\right),f=\left(\begin{array}{c}x\\ x\end{array}\right).\end{array}$$

We can see that

$ef=\left(\begin{array}{c}x\\ y\end{array}\right)$ and $fe=\varnothing$.

Thus, $(e,f)$ is a skew pair of idempotents. According to Theorem 6, $P\left(X\right)$ contains a strong skew pair of idempotents. The opposite direction follows Lemma 1.    □

Corollary 4.

$P\left(X\right)$ contains a left regular skew pair of idempotents if and only if $\left|X\right|⩾4$.

Proof. 

Assume that $P\left(X\right)$ contains a left regular skew pair of idempotents $(e,f)$. Since we have:

$$\mathrm{dom}fe=\mathrm{dom}fefe\subseteq \mathrm{dom}efe\subseteq \mathrm{dom}fe,$$

we let $x\in \mathrm{dom}fe=\mathrm{dom}efe$ such that

$$a=fe\left(x\right)\ne efe\left(x\right)=b.$$

(1)

If $X=\{a,b,c\}$, then we have the following equations:

$$\begin{array}{cc}\hfill fe\left(a\right)& =a\hfill \\ \hfill fe\left(b\right)& =a\hfill \\ \hfill efe\left(b\right)& =b\hfill \\ \hfill e\left(a\right)& =b\hfill \\ \hfill e\left(b\right)& =b\hfill \\ \hfill efe\left(a\right)& =b\hfill \\ \hfill f\left(b\right)& =a\hfill \\ \hfill f\left(a\right)& =a.\hfill \end{array}$$

From these equations, we obtain $a\notin \mathrm{ran}e$ since e is an idempotent. Hence, we have:

$$\begin{array}{cc}\hfill ef\left(a\right)& =b\hfill \\ \hfill ef\left(b\right)& =b.\hfill \end{array}$$

We require two other elements, as $ef$ is not an idempotent of $P\left(X\right)$. Therefore, $\left|X\right|⩾4.$ Conversely, assume that $\left|X\right|⩾4$. Let $x,y,z,w$ be distinct elements of X. Define functions e and f by the following equation:

$$\begin{array}{c}\hfill e=\left(\begin{array}{cccc}x& y& z& w\\ y& y& z& z\end{array}\right),f=\left(\begin{array}{cccc}x& y& z& w\\ x& y& y& y\end{array}\right).\end{array}$$

Then, $(e,f)$ is a left regular skew pair of idempotents of $P\left(X\right)$.    □

Lemma 2.

If $P\left(X\right)$ contains a right regular skew pair of idempotents $(e,f)$ such that $\mathrm{dom}{\left(ef\right)}^2\subset \mathrm{dom}ef$, then $\left|X\right|⩾4$.

Proof. 

Assume that x is an element of $\mathrm{dom}{\left(ef\right)}^2\backslash \mathrm{dom}ef$. Then, there exists $y\in X$ such that $y=ef\left(x\right)\notin \mathrm{dom}ef$, and $x\notin \mathrm{ran}e$. Otherwise, there exists $z\in X$ such that:

$$ef\left(x\right)=efe\left(z\right)=efefe\left(z\right)=efef\left(x\right).$$

Moreover, $y\notin \mathrm{dom}f$ because

$$x\notin \mathrm{dom}efef\supseteq fefef=\mathrm{dom}fef={\left(ef\right)}^{-1}(\mathrm{dom}f\cap \mathrm{ran}ef).$$

If $\left|X\right|=3$, then we have two cases to consider:

Case 1. $f\left(x\right)=x$. Then $e\left(x\right)=y$, $e\left(y\right)=y$ and $z\in \mathrm{dom}e$.

If $e\left(z\right)=y$, then $fe=\varnothing =fef$, which is a contradiction.

If $e\left(z\right)=z$, then $fe=\left(\begin{array}{c}z\\ z\end{array}\right)=fef$, which is a contradiction. Thus, this case is impossible.

Case 2. $f\left(x\right)=z$. Then $f\left(z\right)=z$ since f is an idempotent. In addition, $e\left(z\right)=y$. Now, either $e\left(x\right)=x$ or $e\left(x\right)=y$.

If $e\left(x\right)=x$ then $fe=\left(\begin{array}{c}x\\ z\end{array}\right)$ contradicting $fe$ is an idempotent.

If $e\left(x\right)=y$, then $fe=\varnothing =fef$, which is a contradiction. Thus, this case is impossible.

Therefore, $\left|X\right|⩾4$.    □

Lemma 3.

If $P\left(X\right)$ contains a right regular skew pair of idempotents $(e,f)$ such that $\mathrm{dom}{\left(ef\right)}^2=\mathrm{dom}ef$, then $\left|X\right|⩾4$.

Proof. 

Let $x\in \mathrm{dom}ef=\mathrm{dom}{\left(ef\right)}^2$ be such that

$$y=ef\left(x\right)\ne {\left(ef\right)}^2\left(x\right)=z.$$

Then, $x\ne y$ and $x\ne z$. Otherwise, we obtain the following:

$$y=ef\left(x\right)=ef\left(y\right)=efef\left(x\right)=z$$

or

$$y=ef\left(x\right)=ef\left(z\right)=efefef\left(x\right)=efef\left(x\right)=z.$$

If $\left|X\right|=3$, then the following hold:

$$\begin{array}{cc}\hfill e\left(y\right)& =y\hfill \\ \hfill e\left(z\right)& =z\hfill \\ \hfill f\left(z\right)& =z\hfill \\ \hfill ef\left(y\right)& =z\hfill \\ \hfill ef\left(z\right)& =z\hfill \\ \hfill {\left(ef\right)}^2\left(y\right)& =z\hfill \\ \hfill {\left(ef\right)}^2\left(z\right)& =z\hfill \\ \hfill fef\left(y\right)& =z\hfill \\ \hfill fef\left(z\right)& =z\hfill \\ \hfill fe\left(z\right)& =z\hfill \\ \hfill fe\left(y\right)& =z\hfill \\ \hfill efe\left(z\right)& =z\hfill \\ \hfill efe\left(y\right)& =z\hfill \\ \hfill f\left(y\right)& =z\hfill \\ \hfill f\left(x\right)& =x\hfill \\ \hfill e\left(x\right)& =y\hfill \\ \hfill fe\left(x\right)& =z\hfill \end{array}$$

Thus, $fe=\left(\begin{array}{ccc}x& y& z\\ z& z& z\end{array}\right)=fef$ is a contradiction. Therefore, $\left|X\right|⩾4$.    □

Theorem 7.

$P\left(X\right)$ contains a right regular skew pair of idempotents if and only if $\left|X\right|⩾4$.

Proof. 

Assume that $P\left(X\right)$ contains a right regular skew pair of idempotents. By Lemma 2 and 3, we have $\left|X\right|⩾4$. Conversely, if $\left|X\right|⩾4$, we let:

$$e=\left(\begin{array}{cccc}x& y& z& w\\ x& z& z& x\end{array}\right),f=\left(\begin{array}{cc}x& y\\ x& y\end{array}\right).$$

We can prove that $(e,f)$ is a skew pair of idempotents of $P\left(X\right)$.    □

If $P\left(X\right)$ contains a skew pair of idempotents $(e,f)$, then we have immediately that $\mathrm{dom}fe=\mathrm{dom}efe$. Therefore, the condition $fe\ne efe$ implies that there exists $x\in \mathrm{dom}fe$ such that $fe\left(x\right)\ne efe\left(x\right)$.

Lemma 4.

If $P\left(X\right)$ contains a discrete skew pair of idempotents $(e,f)$ such that $\mathrm{dom}{\left(ef\right)}^2\subset \mathrm{dom}ef$, then $\left|X\right|⩾5$.

Proof. 

Assume that $(e,f)$ is a discrete skew pair of idempotents of $P\left(X\right)$ such that $dom{\left(ef\right)}^2\subset \mathrm{dom}ef$. Then, there exists $x\in X$ such that:

$$a=fe\left(x\right)\ne efe\left(x\right)=b.$$

It is clear that

$$e\left(a\right)=e\left(b\right)=b\mathrm{and}f\left(a\right)=f\left(b\right)=a.$$

By assumption, there exist $c\in \mathrm{dom}ef\backslash \mathrm{dom}{\left(ef\right)}^2=\mathrm{dom}ef\backslash \mathrm{dom}fef$, which are different from a and b. It is clear that $f\left(c\right)\ne a,f\left(c\right)\ne b,e\left(c\right)\ne a,e\left(c\right)\ne b$. Assume that $\left|X\right|=4$. Then, $f\left(c\right)=c$ or $f\left(c\right)=d$. The first case implies $fe=fef$, and the second case implies $\mathrm{dom}f=X$, which is a contradiction. Thus, $\left|X\right|⩾5$.    □

Lemma 5.

If $P\left(X\right)$ contains a discrete skew pair of idempotents $(e,f)$ such that $\mathrm{dom}{\left(ef\right)}^2=\mathrm{dom}ef$, then $\left|X\right|⩾5$.

Proof. 

Let $x\in X$ such that:

$$a=fe\left(x\right)\ne efe\left(x\right)=b$$

By assumption, we require another element of X, say c, such that $ef\left(x\right)\ne efef\left(c\right)$. If $\left|X\right|=4$, X contains another element d, then $f\left(c\right)=c$ or $f\left(c\right)=d$.

Case 1. $f\left(c\right)=c$. Then, $e\left(c\right)=d$ and $e\left(d\right)=d$.

$\left(\alpha \right)$

$d\notin \mathrm{dom}f$ or $f\left(d\right)=c$ implies $efe=efef$;

$\left(\beta \right)$

$f\left(d\right)=a$ or $f\left(d\right)=d$ implies $fe=fef$.

Thus, this case is impossible.

Case 2. $f\left(c\right)=d$. Then, $f\left(d\right)=d$. Now we have the following equation:

$$f=\left(\begin{array}{cccc}a& b& c& d\\ a& a& d& d\end{array}\right),$$

but we cannot define the value of $e\left(d\right)$ satisfying our conditions. Thus, this case is also impossible.

Therefore, $\left|X\right|⩾5$.    □

Theorem 8.

$P\left(X\right)$ contains a discrete skew pair of idempotents if and only if $\left|X\right|⩾5$.

Proof. 

Assume that $P\left(X\right)$ contains a discrete skew pair of idempotents. Based on Lemmas 4 and 5, it follows that $\left|X\right|⩾5$. Conversely, if $X=\{x,y,z,w,v\}$, then we define:

$$e=\left(\begin{array}{ccccc}x& y& z& w& v\\ z& w& z& w& z\end{array}\right)\mathrm{and}f=\left(\begin{array}{ccc}x& y& z\\ x& y& x\end{array}\right).$$

Then,

$ef=\left(\begin{array}{ccc}x& y& z\\ z& w& z\end{array}\right)$,	$fe=\left(\begin{array}{ccc}x& z& v\\ x& x& x\end{array}\right)$,	${\left(fe\right)}^2=\left(\begin{array}{ccc}x& z& v\\ x& x& x\end{array}\right)$
${\left(ef\right)}^2=\left(\begin{array}{cc}x& z\\ z& z\end{array}\right)$,	$efe=\left(\begin{array}{ccc}x& z& v\\ z& z& z\end{array}\right)$,	$fef=\left(\begin{array}{cc}x& z\\ x& x\end{array}\right)$.

Thus, $(e,f)$ is a discrete skew pair of idempotents of $P\left(X\right)$.    □

3. Conclusions

A skew pair of idempotents on a regular semigroup is defined to be a pair $(e,f)$ such that $fe$ is an idempotent but $ef$ is not. It was proved that there are exactly four distinct types of skew pairs of idempotents, namely: strong, left regular, right regular, and discrete. In this paper, we considered the existence of skew pairs of idempotents on partial transformation semigroups using the cardinality of its based set. As in Theorem 4, we conclude that a partial transformation semigroup on a nonempty set X contains skew pairs of idempotents that are: strong if and only if $\left|X\right|\ge 2$; left regular if and only if $\left|X\right|\ge 4$; right regular if and only if $\left|X\right|\ge 4$; discrete if and only if $\left|X\right|\ge 5$.