Regularity in the Semigroup of Transformations Preserving a Zig-Zag Order

Abstract

It is well known that the semigroup T(X) of all transformations on a set X is regular, but its subsemigroups do not need to be. Consider an ordered set $(X;\le )$ whose order forms a path with alternating orientation. The semigroup $\mathrm{OT}(X)$ of all order-preserving transformations on $(X;\le )$ is a subsemigroup of T(X). Some characterizations of regular semigroups $\mathrm{OT}(X)$ were given. For a non-empty subset S of X, we define the set

$$\begin{aligned} \mathrm{OT}_S(X)=\{\alpha \in \mathrm{OT}(X)\mid {{\,\mathrm{ran}\,}}\alpha =S \}. \end{aligned}$$

In this paper, we show that in general, $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X). A necessary and sufficient condition for $\mathrm{OT}_S(X)$ to be a regular subsemigroup of T(X) is given. Some regular subsemigroups of T(X) that are contained in $\mathrm{OT}_S(X)$ are studied. We investigate the left (right) regularity of $\mathrm{OT}(X)$. We obtain that left regular and right regular elements in $\mathrm{OT}(X)$ are identical. A characterization of left (right) elements in $\mathrm{OT}(X)$ is given. We classify all left (right) regular semigroups $\mathrm{OT}(X)$.

1 Introduction and Preliminaries

The regularity was first introduced in ring theory by von Neumann [21] (cf. [3]) in 1936. Later, Green [12] transmitted this concept in semigroup theory and became one of the most-studied topics.

An element a of a semigroup S is said to be regular if there exists an element $b\in S$ such that $a=aba$. A ${ regular}$ semigroup is a semigroup in which all elements are regular.

Some properties of semigroups can be deduced from their regularities. For example, Neumann [21] showed that any principle left (right) ideal of a regular semigroup has an idempotent generator and Hall [13] proved that a Green’s relation H is a congruence on a certain regular semigroup. For more information about regular semigroups, we refer to [3, 14, 16].

It is well known that any semigroup can be embedded in the semigroup T(X) of all transformations on a set X. So, T(X) and their subsemigroups play a significant role to study properties of semigroups. It is well known that T(X) is regular. However, a subsemigroup of T(X) does not need to be regular. A related question is: What is a regular subsemigroup of T(X)? The different types of regular subsemigroups were studied. In 2005, Huisheng [17] investigated the regularity for the subsemigroup

$$\begin{aligned} T_E(X)=\left\{ \alpha \in T(X)\mid \forall (x,y)\in E, (x\alpha ,y\alpha )\in E\right\} \end{aligned}$$

of T(X) where E is an equivalence relation on X. Sanwong and Sommanee [24] characterized the subsemigroups $T(X,Y)=\{\alpha \in T(X)\mid x\alpha \subseteq Y\}$ of T(X) which are regular.

A map $\alpha {:}\,X\rightarrow Y$ is order-preserving from an ordered set X to an ordered set Y if $a\alpha \le b\alpha$ in Y whenever $a\le b$ in X. We denote by $\mathrm{OT}(X)$ the semigroup of all order-preserving transformations on an ordered set X. Such a semigroup is a subsemigroup of T(X). In [11], Gluskin showed that if $\mathrm{OT}(X)$ is isomorphic to $\mathrm{OT}(Y)$, then the ordered sets X and Y are isomorphic or anti-isomorphic. Repnitskǐ and Vernitskǐ [22, Lemma 1.1] proved that every free semigroup can be represented by the semigroup $\mathrm{OT}(X)$ of a chain (or a totally ordered set). Later, Higgins et al. [15] considered the relative rank of T(X) modulo the semigroup $\mathrm{OT}(X)$ of a finite chain X. These results show that the semigroups $\mathrm{OT}(X)$ can be used to determine the structure of the initial one.

For a chain X, the regularity of $\mathrm{OT}(X)$ was widely investigated. It was shown that $\mathrm{OT}(X)$ is a regular subsemigroup of T(X) if X is finite (see [14, page 203, Exercise 6.1.9]). Kemprasit and Changphas [19] showed that if X is order-isomorphic to a subchain of ${\mathbb {Z}}$, then $\mathrm{OT}(X)$ is regular. In [9], Fernandes et al. described the largest regular subsemigroup of $\mathrm{OT}(X)$.

Our interest focuses on an important ordered set which is closed to chains and paths, respectively. A fenceX is an ordered set that the order forms a path with alternating orientation. Indeed, X is in which either

$$\begin{aligned} x_1\le x_2\ge x_3,\ldots ,x_{2m-1}\le x_{2m}\ge x_{2m+1},\ldots \end{aligned}$$

or

$$\begin{aligned} x_1\ge x_2 \le x_3,\ldots ,x_{2m-1}\ge x_{2m} \le x_{2m+1},\ldots \end{aligned}$$

are the only comparability relations where $X=\{x_1,x_2,\ldots ,x_n,\ldots \}$. Every element in X is minimal or maximal.

Properties of order-preserving transformations of a fence have been long considered (see for example: [5, 8, 23]). Recently, in 2015, Jendana and Srithus [18] gave a characterization for a finite fence X having $\mathrm{OT}(X)$ as a coregular semigroup and described coregular elements of $\mathrm{OT}(X)$. Tanyawong et al. [26] in 2016 described all regular semigroups $\mathrm{OT}(X)$ where X is a finite fence. In 2017, Dimitrova and Koppitz [6] considered the regularity and Green’s relations for the semigroup of all partial injection order-preserving transformations on a fence. Dimitrova et al. [7] determined the relative rank of the monoid $\mathrm{PF}_{{\mathbb {N}}}$ of all partial transformations on the set ${\mathbb {N}}$ of natural numbers preserving a zig-zag order on ${\mathbb {N}}$ modulo a set containing all idempotents and all surjections in $\mathrm{PF}_{{\mathbb {N}}}$. Particular maximal regular subsemigroups of the monoid $\mathrm{OP}_{{\mathbb {N}}}$ of all order-preserving partial transformations on ${\mathbb {N}}$ were determined by Lohapan and Koppitz [20] and they showed that $\mathrm{OP}_{{\mathbb {N}}}$ has infinitely many maximal regular subsemigroups. Later, in 2018, Fernandes et al. [10] studied the rank of the semigroup $\mathrm{OT}(X)$ when X is a fence. For the present, in Sect. 3, we focus on regular subsemigroups of T(X) that are contained in $\mathrm{OT}(X)$ when X is a finite fence.

Various types of regularity have been introduced by Croisot [4] (cf. [3]) in 1953 as Croisot’s theory. One of the central concepts in this theory is the left (right) regular. An element a of a semigroup S is said to be left regular (right regular) if there exists an element $b\in S$ such that $a=ba^2$ ($a=a^2b$). A left regular (right regular) semigroup is a semigroup in which all elements are left regular (right regular). Notice that the left (right) regularity is not a generalization of the regularity and in general, left regular, right regular and regular elements might not be related (see examples in [2, 25]).

Properties of semigroups can be deduced from their left (right) regularities. For example, if a is both left and right regular of a semigroup S, then a is in a subgroup of S and has an inverse which commutes with a [1]. Every left semigroup can decomposite into a union of left simple semigroups [3]. So, the left (right) regularity is a powerful tool in semigroup theory. Characterizations of left (regular) semigroups were given. Here, we show only some results for subsemigroups of T(X).

Sirasuntorn and Kemprasit [25] in 2010 characterized the left regular and right regular elements of the semigroup of all 1–1 transformations of a set and of all 1–1 transformations of a vector space. Later, in 2013, Choomanee et al. [2] described the left regular and right regular elements of the semigroup $S(X,Y)=\{\alpha \in T(X)\mid Y\alpha \subseteq Y\}$.

Throughout this paper, the word “fence” will mean a finite fence. Our main purpose is to investigate the left (right) regularity of the semigroup $\mathrm{OT}(X)$ where X is a fence. In Sect. 2, we give a necessary and sufficient condition for $\mathrm{OT}(X)$ to be left (right) regular. Left (right) regular elements in $\mathrm{OT}(X)$ are completely described. In Sect. 3, we define the set

$$\begin{aligned} \mathrm{OT}_S(X)=\{\alpha \in \mathrm{OT}(X) \mid {{\,\mathrm{ran}\,}}\alpha =S\} \end{aligned}$$

where $S\subseteq X$ and ${{\,\mathrm{ran}\,}}\alpha$ is the range of $\alpha$. In general, $\mathrm{OT}_S(X)$ does not need to be a subsemigroup of T(X). We characterize a subset S of a fence X having $\mathrm{OT}_S(X)$ as a subsemigroup of T(X). Finally, we show that if $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), then $\mathrm{OT}_S(X)$ is a regular subsemigroup of T(X). Some regular subsemigroups of T(X) that are contained in $\mathrm{OT}_S(X)$ are discussed.

2 The Left and Right Regularity for $\mathrm{OT}(X)$

In this section, we will be interested in the left and right regularity for $\mathrm{OT}(X)$ where X is a fence. A general question is whether the semigroup $\mathrm{OT}(X)$ is left (right) regular. For attacking this question, we start by giving basic knowledge involving order-preserving maps. An ordered set P is called connected if for all $a,b\in P$ there is a fence $F\subseteq P$ with endpoints a and b. It is well known that if P is connected and $\alpha {:}\,P\rightarrow Q$ is order-preserving from P to an ordered set Q, then $P\alpha$ is connected. Consequently, every order-preserving map $\alpha$ maps order-connected sets to order-connected sets. Because order-connected subsets of a fence X are precisely the subfences, an order-preserving map $\alpha {:}\,X\rightarrow X$ maps subfences to subfences.

Consider a fence X with $|X|\ge 3$. The following lemma is obtained.

Lemma 1

Let X be a fence with $|X|\ge 3$ and let a, b, c be distinct elements in X with $a<b>c$ or $a>b<c$. Define the map $\alpha {:}\,X \rightarrow X$ by

$$\begin{aligned} x\alpha = {\left\{ \begin{array}{ll} a,\quad x\,\text {is}\,\text {minimal}\,\text {and}\,x\not =a\\ b,\quad \text{ otherwise }. \end{array}\right. } or \, x\alpha = {\left\{ \begin{array}{ll} a,\quad x\,\text {is}\,\text {maximal}\,\text {and}\,x\not =a\\ b,\quad \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Then, the following statements hold:

• (i) The map $\alpha$ is an element of $\mathrm{OT}(X)$.

• (ii) The map $\alpha$ is regular in $\mathrm{OT}(X)$.

• (iii) The map $\alpha$ is not left regular in $\mathrm{OT}(X)$.

• (iv) The map $\alpha$ is not right regular in $\mathrm{OT}(X)$.

Proof

Consider the map $\alpha$ of the case $a<b>c$. We show that $\alpha$ satisfies the statements (i)–(iv).

• (i) Let $x,y\in X$ with $x\le y$. If x is minimal and $x\not =a$, then $x\alpha =a$. Since $y\alpha \in \{a,b\}$ and $a\le b$, we have $x\alpha \le y\alpha$. If x is maximal or $x=a$, then y is maximal or $y=a$ implying $x\alpha =b=y\alpha$. So, $\alpha$ is order-preserving.

• (ii) Define the map $\beta {:}\,X \rightarrow X$ by

  $$\begin{aligned} x\beta = {\left\{ \begin{array}{ll} c, &{} x~\text {is minimal}\\ b, &{} x~\text {is maximal}. \end{array}\right. } \end{aligned}$$

  Then, $\beta$ is order-preserving. Let $x\in X$. If x is minimal and $x\not = a$, then $x\alpha =a$ implies $x\alpha \beta \alpha =a\beta \alpha =c\alpha =a=x\alpha$. Similarly, $x\alpha \beta \alpha =x\alpha$ if x is maximal or $x=a$. So, $\alpha$ is regular in $\mathrm{OT}(X)$.

• (iii) Suppose that $\alpha$ is left regular in $\mathrm{OT}(X)$. Then there is a $\beta \in \mathrm{OT}(X)$ with $\alpha =\beta \alpha ^2$. Hence, $a=c\alpha =c\beta \alpha ^2=c\alpha ^2=a\alpha =b$ which is a contradiction.

• (iv) Suppose that $\alpha$ is right regular in $\mathrm{OT}(X)$. Then there is a $\beta \in \mathrm{OT}(X)$ with $\alpha =\alpha ^2\beta$. Hence, $a=c\alpha =c\alpha ^2\beta =a\alpha \beta =b\beta =b$ which is a contradiction.

  Similarly, the statements (i)–(iv) hold for the map $\alpha$ of the case $a>b<c$. $\square$

Lemma 1 shows that there is a map in $\mathrm{OT}(X)$ which is neither left nor right regular, but regular in $\mathrm{OT}(X)$ if $|X|\ge 3$. We have immediately the following proposition.

Proposition 1

Let X be a fence. If $|X|\ge 3$, then $\mathrm{OT}(X)$ is neither left regular nor right regular.

Observe that for an ordered set P, the identity map $id_P$ and a constant $C_a$ which maps all elements in P to $a\in P$ are order-preserving. Since $id_P^3=id_P$ and $C_a^3=C_a$, the both maps are left regular, right regular and regular in $\mathrm{OT}(P)$. It follows that every semigroup $\mathrm{OT}(P)$ has always a left regular and right regular element, even though $\mathrm{OT}(P)$ is not left or right regular. It is natural to ask when an element $\alpha$ in $\mathrm{OT}(X)$ is left and right regular, respectively. The answer is shown in the following theorem. To do so, we need some basic concepts. A map $\alpha {:}\,P\rightarrow Q$ is said to be order-embedding from an ordered set P to an ordered set Q if it satisfies the condition: for every $x,y\in P$,

$$\begin{aligned} x\le y~\text {in}~P~\text {if and only if}~ x\alpha \le y\alpha ~\text {in}~Q. \end{aligned}$$

If $\alpha$ is onto, then $\alpha$ is called an order-isomorphism.

In general, left regular, right regular and regular elements in a semigroup might not be related. Theorem 1 says that for a fence X, left regular and right regular elements in $\mathrm{OT}(X)$ are identical. Every left (right) regular element is regular in $\mathrm{OT}(X)$. Moreover, a characterization of all left (right) regular elements in $\mathrm{OT}(X)$ is given.

Theorem 1

Let X be a fence and $\alpha \in \mathrm{OT}(X)$ with $Y={{\,\mathrm{ran}\,}}\alpha$. Then, the following statements are equivalent : 

• (i) $\alpha$ is left regular in $\mathrm{OT}(X)$.

• (ii) $\alpha |_Y{:}\,Y\rightarrow Y$ is bijective.

• (iii) $\alpha$ is right regular in $\mathrm{OT}(X)$.

Moreover, if $\alpha$ is left regular (or right regular) in $\mathrm{OT}(X)$, then there is a map $\beta$ in $\mathrm{OT}(X)$ with ${{\,\mathrm{ran}\,}}\beta =Y={{\,\mathrm{ran}\,}}\alpha$ and $\alpha \beta \alpha =\alpha$ and therefore, $\alpha$ is regular in $\mathrm{OT}(X)$.

Proof

To begin with, let $\gamma :=\alpha |_Y$ and for each $y\in Y$, let ${\bar{y}}=\{x\in X{\setminus } Y\mid x\alpha =y\alpha \}$. We use the symbol $x\parallel y$ to mean that x and y are not comparable.

(i) $\Rightarrow$ (ii) Assume that $\alpha$ is left regular in $\mathrm{OT}(X)$. Suppose that $\gamma {:}\,Y\rightarrow Y$ is not bijective. Since Y is finite, $\gamma$ is neither injective nor surjective. It follows that there are elements $a\in Y={{\,\mathrm{ran}\,}}\alpha$ and $z\in X{\setminus } Y$ with $z\alpha =a$ and $x\alpha \not =a$ for all $x\in Y$. By the left regularity of $\alpha$, there is a map $\beta \in \mathrm{OT}(X)$ with $\alpha =\beta \alpha ^2$ implying $a=z\alpha =z\beta \alpha ^2=((z\beta )\alpha )\alpha$. Since $(z\beta )\alpha \in Y$, there is an element $y\in Y$ with $y\alpha =a$, a contradiction.

(ii) $\Rightarrow$ (i) Assume that $\gamma {:}\,Y\rightarrow Y$ is bijective. First, we show that $\gamma$ is order-embedding on Y. Let $x,y\in Y$ with $x\gamma \le y\gamma$. By the order-preserving property of $\alpha$, $x\le y$ or $x\parallel y$ since otherwise $x\gamma =x\alpha > y\alpha =y\gamma$. If $x\parallel y$, then there is a subfence F of Y having x and y as the endpoints. Since $x\parallel y$, the number of elements in F is greater than 2. By the injectivity of $\gamma$, $|F\alpha |\ge 3$ and $x\gamma \parallel y\gamma$ which is a contradiction. So, $x\le y$ and therefore, $\gamma$ is order-embedding. Because $\gamma$ is surjective, $\gamma$ is an order-isomorphism. It follows that $\gamma ^{-1}$ is an order-isomorphism. Define a map $\beta {:}\,X\rightarrow X$ by

$$\begin{aligned} x\beta = {\left\{ \begin{array}{ll} x\gamma ^{-1}, &{} x\in Y\\ y\gamma ^{-1}, &{} x\in {\bar{y}}. \end{array}\right. } \end{aligned}$$

Clearly, ${{\,\mathrm{ran}\,}}\beta =Y$. To show that $\beta$ is order-preserving, we let $a,b\in X$ with $a\le b$. Then, there are $y,z\in Y$ with $a\alpha =y\alpha =y\gamma$ and $b\alpha =z\alpha =z\gamma$. By the definition of $\beta$, we get $a\beta =y\gamma ^{-1}$ and $b\beta =z\gamma ^{-1}$. Since $\alpha$ is order-preserving, $y\gamma =a\alpha \le b\alpha =z\gamma$. It follows that $y=z$. So, $a\beta =y\gamma ^{-1}\le z\gamma ^{-1}=b\beta$.

Next, we show that $\beta \alpha ^2=\alpha$. Let $x\in X$. Then, there exists $y\in Y$ with $x\alpha =y\alpha$ and hence, $x\beta =y\beta$. By the definition of $\beta$, we have $x\beta \alpha ^2=y\beta \alpha ^2=y\gamma ^{-1}\gamma \alpha =y\alpha =x\alpha$. Therefore, $\alpha$ is left regular in $\mathrm{OT}(X)$.

(ii) $\Rightarrow$ (iii) Assume that $\gamma {:}\,Y\rightarrow Y$ is bijective. We show that the map $\beta$ as defined above satisfies the condition $\alpha ^2\beta =\alpha$. Let $x\in X$. Then, $x\alpha \in Y$ and by the definition of $\beta$, we have $x\alpha ^2\beta =x\alpha \gamma \gamma ^{-1}=x\alpha$. So, $\alpha$ is right regular in $\mathrm{OT}(X)$.

(iii) $\Rightarrow$ (ii) Assume that $\alpha$ is right regular in $\mathrm{OT}(X)$. We prove that $\gamma$ is bijective. Let $a,b\in Y$ with $a\gamma =b\gamma$. Then, there are $y,z\in Y$ with $y\gamma =y\alpha =a$ and $z\gamma =z\alpha =b$. So, $y\alpha ^2=a\gamma =b\gamma =y\alpha ^2$. Because $\alpha$ is right regular, there exists a map $\beta \in \mathrm{OT}(X)$ with $\alpha ^2\beta =\alpha$ and hence, $a=y\alpha =y\alpha ^2\beta =z\alpha ^2\beta =z\alpha =b$. So, $\gamma$ is injective which implies that $\gamma$ is bijective since Y is finite.

Finally, assume that $\alpha$ is left regular in $\mathrm{OT}(X)$. Consider the map $\beta$ as defined above. Let $x\in X$. Then, $x\alpha =y\alpha$ for some $y\in Y$ and so, $x\alpha \beta \alpha =y\alpha \beta \alpha =y\gamma \gamma ^{-1} \alpha =y\alpha =x\alpha$. Therefore, $\alpha$ is regular in $\mathrm{OT}(X)$. $\square$

To finish this section, it remains to describe a fence X having $\mathrm{OT}(X)$ to be left (right) regular.

Theorem 2

Let X be a fence. Then, $\mathrm{OT}(X)$ is left (right) regular if and only if $|X|\le 2$.

Proof

With the use of Proposition 1, the semigroup $\mathrm{OT}(X)$ is not left regular if $|X|> 2$. Conversely, assume that $|X|\le 2$. If $|X|=1$, then $\mathrm{OT}(X)$ is the set of a constant map. Hence, $\mathrm{OT}(X)$ is left regular. Let $|X|=2$ and $X=\{a,b\}$ with $a<b$. Then, $\mathrm{OT}(X)$ contains exactly 3 elements, namely 2 constant maps and the identity map. Hence, $\mathrm{OT}(X)$ is left regular. $\square$

3 Regular Subsemigroups of T(X)

Consider a fence X. In 2016, Tanyawong et al. [26] described all regular semigroups $\mathrm{OT}(X)$ of order-preserving transformations on X. Since $\mathrm{OT}(X)$ is a subsemigroup of T(X), every regular semigroup $\mathrm{OT}(X)$ is a regular subsemigroup of T(X). This means that we can classify all $\mathrm{OT}(X)$ which are regular subsemigroups of T(X). The aim of this section is to investigate regular subsemigroups of $\mathrm{OT}(X)$ when X is a fence.

We now focus on the set $\mathrm{OT}_S(X)$ of all order-preserving transformations on a fence X having S as their range, i.e.,

$$\begin{aligned} \mathrm{OT}_S(X)=\left\{ \alpha \in \mathrm{OT}(X)\mid {{\,\mathrm{ran}\,}}\alpha =S\right\} \end{aligned}$$

where $S\subseteq X$. Unfortunately, in general, $\mathrm{OT}_S(X)$ might not be a subsemigroup of T(X) as shown in the following example.

Example 1

Consider a 5-element fence X as shown in Fig. 1.

Fig. 1

The 5-element fence X

Define the maps $\alpha {:}\,X\rightarrow X$ and $\beta {:}\,X\rightarrow X$ by

$$\begin{aligned} \alpha := \begin{pmatrix} a&{}\quad b&{}\quad c&{}\quad d&{}\quad e\\ d&{}\quad d&{}\quad c&{}\quad d&{}\quad e \end{pmatrix}&\hbox {and}&\beta := \begin{pmatrix} a&{}\quad b&{}\quad c&{}\quad d&{}\quad e\\ c&{}\quad d&{}\quad e&{}\quad d&{}\quad d \end{pmatrix}, \end{aligned}$$

respectively. It is clear that $\alpha$ and $\beta$ are in $\mathrm{OT}_S(X)$ where $S=\{c,d,e\}$. But $\alpha \beta$ is not in $\mathrm{OT}_S(X)$ since ${{\,\mathrm{ran}\,}}\alpha \beta =\{d,e\}$. Hence, $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X).

Although $\mathrm{OT}_S(X)$ in Example 1 is not a subsemigroup of T(X), there is a subset S of X such that $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), for example, $S=\{a\}$ where $a\in X$. In what follows, we describe all ordered sets X having $\mathrm{OT}_S(X)$ as a subsemigroup of T(X).

Proposition 2

Let X be an ordered set and let S be a subset of X. Then, $\mathrm{OT}_S(X)$ is a subsemigroup of T(X) if and only if $S\alpha =S$ for all $\alpha \in \mathrm{OT}_S(X)$.

Proof

If $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), then $\alpha ^2\in \mathrm{OT}_S(X)$ for all $\alpha \in \mathrm{OT}_S(X)$. It follows that $S={{\,\mathrm{ran}\,}}\alpha ^2=(X\alpha )\alpha =S\alpha$ for all $\alpha \in \mathrm{OT}_S(X)$.

Conversely, assume that $S\alpha =S$ for all $\alpha \in \mathrm{OT}_S(X)$. Let $\alpha ,\beta \in \mathrm{OT}_S(X)$. Then, $X\alpha =S=X\beta$ implies ${{\,\mathrm{ran}\,}}\alpha \beta =(X\alpha )\beta =S\beta =S$. Because the composition of any two order-preserving maps is order-preserving, $\alpha \beta \in \mathrm{OT}_S(X)$. Hence, $\mathrm{OT}_S(X)$ is a subsemigroup of T(X). $\square$

Corollary 1

Let X be an ordered set and let S be a subset of X. If $S=X$ or $|S|=1$, then $\mathrm{OT}_S(X)$ is a subsemigroup of T(X).

A subset S of an ordered set X is called a subordered set of X if S forms an ordered set under the order of X. If S is a fence, then S is called a subfence of X and S is a non-trivial subfence of X if $|S|\ge 2$ and $S\not =X$.

It is known from Theorem 1 that for a fence X, a map $\alpha$ is left (right) regular in $\mathrm{OT}(X)$ if and only if $\alpha |_{{{\,\mathrm{ran}\,}}\alpha }$ is a bijection on ${{\,\mathrm{ran}\,}}\alpha$. Since $\mathrm{OT}_S(X)$ is a subsemigroup of T(X) if $S\alpha =S$ for all $\alpha \in \mathrm{OT}_S(X)$, i.e., $\alpha |_{S}$ is a bijection on S, we have that $\mathrm{OT}_S(X)$ is the set of left (right) regular elements in $\mathrm{OT}(X)$ if $\mathrm{OT}_S(X)$ is a subsemigroup of T(X). Again by Theorem 1, if $\alpha$ is left (right) regular in $\mathrm{OT}(X)$, there is a map $\beta \in \mathrm{OT}(X)$ with ${{\,\mathrm{ran}\,}}\beta ={{\,\mathrm{ran}\,}}\alpha$ and $\alpha \beta \alpha =\alpha$. If $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), then $\mathrm{OT}_S(X)$ is a regular semigroup and therefore, is a regular subsemigroup of T(X).

For the rest of this section, we restrict our study to subsets S of a fence X having $\mathrm{OT}_S(X)$ as a subsemigroup of T(X). Recall that if $\alpha$ is an order-preserving transformation on a fence X, then its range is a subfence of X. This means that the set S of $\mathrm{OT}_S(X)$ is a subfence of X. We start by proving a lemma.

Lemma 2

Let $S=\{a_1,a_2,\ldots ,a_n\}$ be a subfence of a fence X with $a_1<a_2>a_3<\cdots >(<)a_n$. Assume that

$$\begin{aligned} X=\left\{ A_1,A_2,\ldots ,A_k,a_1,a_2,\ldots ,a_n,B_1,B_2,\ldots ,B_l\right\} \end{aligned}$$

where $k,l\in {\mathbb {N}}$. If $k\ge 2$ or $l\ge 2$, then $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X).

Proof

We consider k and l in the following 2 cases.

Case 1$k\ge 2$. We define the map $\alpha {:}\,X\rightarrow X$ by

$$\begin{aligned} x\alpha = {\left\{ \begin{array}{ll} a_2, &{} x=a_1\\ x, &{} x\in S{\setminus } \{a_1\}\\ a_2, &{} x=A_k\\ a_1, &{} x\in \{A_1,\ldots ,A_{k-1}\}\\ a_n, &{} x\in \{B_1,\ldots ,B_l\}. \end{array}\right. } \end{aligned}$$

It is clear that $\alpha \in \mathrm{OT}_S(X)$ and $S\alpha =S{\setminus }\{a_1\}$.

Case 2$l\ge 2$. If $a_n$ is minimal, then $\alpha$ is defined similarly to Case 1. If $a_n$ is maximal, then the map $\alpha {:}\,X\rightarrow X$ defined by

$$\begin{aligned} x\alpha = {\left\{ \begin{array}{ll} x, &{} x\in S{\setminus } \{a_n\}\\ a_{n-1}, &{} x=a_n\\ a_n, &{} x=B_2\\ a_{n-1}, &{} x\in \{B_1,B_3,\ldots ,B_l\}\\ a_1, &{} x\in \{A_1,\ldots ,A_{k}\} \end{array}\right. } \end{aligned}$$

is an element in $\mathrm{OT}_S(X)$ with $S\alpha =S{\setminus }\{a_n\}$.

Proposition 2 implies that $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X). $\square$

The following propositions give useful properties of subfences S that are used later.

Proposition 3

Let S be a non-trivial subfence of a fence X. If $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), then $|S|\ge 3$, $|X{\setminus } S|\le 2$ and $X{\setminus } S$ is not a 2-element fence.

Proof

Assume that $S=\{a_1,a_2,\ldots ,a_n\}$ with $a_1<a_2>a_3<\cdots >(<)a_n$. Since $S\subset X$, the cardinal number $|X{\setminus } S|\ge 1$. Let

$$\begin{aligned} X=\left\{ A_1,A_2,\ldots ,A_k,a_1,a_2,\ldots ,a_n,B_1,B_2,\ldots ,B_l\right\} \end{aligned}$$

when $k\ge 1$ or $l\ge 1$. Then, $a_{2t+1}$ and $a_{2t}$ are minimal and maximal, respectively. Since $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), by Proposition 2, $S\alpha =S$ for all $\alpha \in \mathrm{OT}_S(X)$. If $|X{\setminus } S|> 2$ or $X{\setminus } S$ is a 2-element fence, then $k\ge 2$ or $l\ge 2$. Lemma 2 implies that $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X). Hence, $|X{\setminus } S|\le 2$ and $X{\setminus } S$ is not a 2-element fence.

To show that $|S|\not =2$, we prove by contradiction. Suppose that $|S|=2$. If $k\ge 2$ or $l\ge 2$, then $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X). So, $k\le 1$ and $l\le 1$. Since S is a proper subset of X, $k=1$ or $l=1$. Consider $k=1$. Define the map $\alpha {:}\,X\rightarrow X$ by

$$\begin{aligned} x\alpha = {\left\{ \begin{array}{ll} a_2, &{} x=A_1,\\ a_1, &{} \text {otherwise}.\\ \end{array}\right. } \end{aligned}$$

The map $\alpha$ is in $\mathrm{OT}_S(X)$ and $S\alpha =\{a_1\}\not =S$, a contradiction. If $k=0$, then $l=1$. The map $\alpha {:}\,X\rightarrow X$ which is defined by

$$\begin{aligned} x\alpha = {\left\{ \begin{array}{ll} a_1, &{} x=B_1,\\ a_2, &{} \text {otherwise}.\\ \end{array}\right. } \end{aligned}$$

is in $\mathrm{OT}_S(X)$ and $S\alpha =\{a_1\}\not =S$, a contradiction. Thus, $|S|\not =2$ and therefore, $|S|\ge 3$ since S is non-trivial. $\square$

Proposition 4

Let S be a non-trivial subfence of a fence X for which |S| is even. Then, $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X).

Proof

Since |S| is even, the endpoints of S are not both either minimal or maximal. We may assume that $S=\{a_1,a_2,\ldots ,a_n\}$ with $a_1<a_2>a_3<\cdots >a_{n-1}<a_n$. If $|X{\setminus } S|>2$ or $X{\setminus } S$ is a 2-element fence, then $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X) by Lemma 2. Consider $|X{\setminus } S|\le 2$ and $X{\setminus } S$ is not a 2-element fence. The fence X can be written as $A_1>a_1<a_2>\cdots <a_n$ or $a_1<a_2>\cdots <a_n>B_1$ or $A_1>a_1<a_2>\cdots <a_n>B_1$. Define the map $\alpha {:}\,X\rightarrow X$ as follows:

$$\begin{aligned} x\alpha = {\left\{ \begin{array}{ll} a_n, &{} x=A_1,\\ a_{n-i}, &{} x=a_i\in S{\setminus } \{a_n\},\\ a_1, &{} x=\{a_n,B_1\}, \end{array}\right. } \end{aligned}$$

when $A_1\in X$ and

$$\begin{aligned} x\alpha = {\left\{ \begin{array}{ll} a_1, &{} x=B_1,\\ a_{n-(i-2)}, &{} x=a_i\in S{\setminus } \{a_1\},\\ a_n, &{} x=a_1, \end{array}\right. } \end{aligned}$$

when $A_1\not \in X$. It is clear that $\alpha$ is order-preserving with ${{\,\mathrm{ran}\,}}\alpha =S$ and $S\alpha \subset S$. Proposition 2 implies that $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X). $\square$

In the theorem below, a necessary and sufficient condition for a subfence S of a fence X having $\mathrm{OT}_S(X)$ as a subsemigroup of T(X) is given.

Theorem 3

Let S be a non-trivial subfence of a fence X. Then, $\mathrm{OT}_S(X)$ is a subsemigroup of T(X) if and only if |S| is odd, $|X{\setminus } S|\le 2$ and $X{\setminus } S$ is not a 2-element fence.

Proof

Assume that $\mathrm{OT}_S(X)$ is a subsemigroup of $\mathrm{OT}(X)$. As a consequence of Propositions 3 and 4, |S| is odd, $|X{\setminus } S|\le 2$ and $X{\setminus } S$ is not a 2-element fence.

Conversely, assume that |S| is odd, $|X{\setminus } S|\le 2$ and $X{\setminus } S$ is not a 2-element fence. Then, $|S|\ge 3$ and the endpoints of S are both either minimal or maximal. Let $S=\{a_1,a_2,\ldots ,a_n\}$ with $a_1<a_2>a_3<\cdots >a_n$, that is, $a_{2m-1}$ and $a_{2m}$ are minimal and maximal, respectively. Then, X can be written as $A>a_1<a_2>a_3<\cdots >a_n$ or $a_1<a_2>a_3<\cdots >a_n<B$ or $A>a_1<a_2>a_3<\cdots >a_n<B$. Because of $|X{\setminus } S|\le 2$ and ${{\,\mathrm{ran}\,}}\alpha =S$, the cardinal number $|S{\setminus } S\alpha |\le 2$.

To show that $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), we prove by contradiction.

Suppose that $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X). Then, there is a map $\alpha \in \mathrm{OT}_S(X)$ with $S\alpha \not =S$. It follows that for every $y\in S{\setminus } S\alpha$, there is an element $x\in X{\setminus } S$ with $x\alpha =y$, that is, $A\alpha =y$ or $B\alpha =y$. $\square$

Claim 1

Every element in $S{\setminus } S\alpha$ is maximal.

Let $a\in S{\setminus } S\alpha$. Then, from ${{\,\mathrm{ran}\,}}\alpha =S$, we have $A\alpha =a$ or $B\alpha =a$. We may assume that $A\alpha =a$. Then, $a_1\alpha \le A\alpha =a$ since $a_1<A$. If a is minimal, then $a_1\alpha =a$ implies $a\in S\alpha$. Hence, a is maximal.

By Claim 1, every minimal element in S is in $S\alpha$.

If $|S|=3$, then $S{\setminus } S\alpha =\{a_2\}$ implies $a_2\alpha \not =a_2$ and $A\alpha =a_2$ or $B\alpha =a_2$. So, $a_2\alpha \in \{a_1,a_3\}$. Since $\alpha$ is order-preserving and $a_1<a_2>a_3$, we get $a_1\alpha =a_2\alpha =a_3\alpha$. Hence, $|\{a_1,a_2,a_3\}\alpha |=1$. If $|X{\setminus } S|=\{A\}$, then $|S|=|X\alpha |=|\{A,a_1,a_2,a_3\}\alpha |=2$. So, $|X{\setminus } S|=\{A,B\}$. If $A\alpha =a_2$, then from $a_3<B$, we have $a_3\alpha \le B\alpha$ and so, $B\alpha \in \{a_2,a_3\alpha \}$. Hence, $|S|=|X\alpha |=|\{A,a_1,a_2,a_3,B\}\alpha |=2$, a contradiction. Thus, $|S|\ge 5$ implies that there are at least 2 maximal elements in S.

Claim 2

Every element in $S{\setminus } S\alpha$ is mapped to a maximal element in S.

Suppose that there is an element $a_k\in S{\setminus } S\alpha$ such that $a_k\alpha =a_l$ for some minimal element $a_l$ in S. Then, $A\alpha =a_k$ or $B\alpha =a_k$. Assume that $A\alpha =a_k$. Because $a_k$ is maximal, $a_{k-1},a_{k+1}\in S$ implies $a_{k-1}\alpha =a_l=a_{k+1}\alpha$. So, $|\{A,a_{k-1},a_k,a_{k+1}\}\alpha |=2$. If $|X{\setminus } S|=\{A\}$, then $X\alpha \not =S$. Thus, $|X{\setminus }~S|=\{A,B\}$. If $B\alpha =a_{k-1}$ which is minimal, then from $a_n<B$, we have $a_n\alpha =B\alpha$. It follows that $|\{A,a_{k-1},a_k,a_{k+1},a_n,B\}\alpha |=3$. From $|X{\setminus } S|=2$ and $|S{\setminus } S\alpha |\le 2$, we have $X\alpha \not =S$, a contradiction. Hence, $B\alpha \not =a_{k-1}$ and similarly, $B\alpha \not =a_{k+1}$. Therefore, $X\alpha \not =S$ since $|\{a_{k-1},a_k,a_{k+1}\}\alpha |=1$, a contradiction.

From $|S{\setminus } S\alpha |\le 2$, we consider $S{\setminus } S\alpha$ in the following 2 cases.

Case 1$|S{\setminus } S\alpha |=1$, say $a_k\in S{\setminus } S\alpha$. Then, $A\alpha =a_k$ or $B\alpha =a_k$. We may assume that $A\alpha =a_k$. Then $a_1\alpha \in \{a_{k-1},a_{k+1}\}$. Suppose that $a_k=a_2$. Then $a_1\alpha \in \{a_1,a_3\}$. By Claim 2, $a_2\alpha =a_4$ since $a_1<a_2>a_3<a_4$ and so, $a_1\alpha =a_3$. From $a_1\in S={{\,\mathrm{ran}\,}}\alpha$, there is an element x in X with $x\alpha =a_1$. If x is maximal, there is a minimal element in S that is mapped to $a_1$. Hence, $a_l\alpha =a_1$ for some minimal element $a_l\in S{\setminus } \{a_1\}$. Since $a_2$ is the only element in S that is comparable to $a_1$ and $\alpha$ is order-preserving, $a_{l-1}\alpha =a_1=a_{l+1}\alpha$ if $a_l\not =a_n$ and $a_{n-1}\alpha =a_1$ if $a_l=a_n$. It follows that $s\alpha =a_1$ for all $s\in S{\setminus } \{a_1,a_2\}$. From $a_2\alpha =a_4$, we get $a_3\alpha \in \{a_3,a_4,a_5\}$ which is a contradiction. Thus $a_k\not =a_2$. If $a_1\alpha =a_{k-1}$, then $a_2\alpha =a_{k-2}$ since $a_k\not \in S\alpha$. Since $\alpha$ is order-preserving, there is no element x in X with $x\alpha =a_{k+1}$. So, $a_{k+1}\not \in {{\,\mathrm{ran}\,}}\alpha =S$ which is a contradiction. Similarly, we obtain a contradiction if $a_1\alpha =a_{k+1}$.

Case 2$|S{\setminus } S\alpha |=2$, say $a_k,a_l\in S{\setminus } S\alpha$. Let $A_1\alpha =a_k$ and $B_1\alpha =a_l$.

Case 2.1$k<l$. We consider $a_1$ and $a_n$ in the following cases.

If $a_1\alpha =a_{k-1}$ and $a_n\alpha =a_{l-1}$, then from $a_1$ and $a_n$ are not comparable, there is no element $x\in X$ with $x\alpha =a_{l+1}$ since $\alpha$ is order-preserving. So, $a_{l+1}\not \in {{\,\mathrm{ran}\,}}\alpha =S$. By a similar argument, we obtain the following:

If $a_1\alpha =a_{k+1}$ and $a_n\alpha =a_{l+1}$, then $a_{k-1}\not \in S$.

If $a_1\alpha =a_{k-1}$ and $a_n\alpha =a_{l+1}$, then $a_{k+1}\not \in S$.

If $a_1\alpha =a_{k+1}$ and $a_n\alpha =a_{l-1}$, then from $a_k,a_l\not \in S\alpha$, we have $a_{k-1}, a_{l+1} \not \in ~S$.

Thus, we get a contradiction.

Case 2.2$k>l$. By a similar argument as the case $k<l$, we get a contradiction.

Therefore, $\mathrm{OT}_S(X)$ is a subsemigroup of T(X). $\square$

We end with some results concerning the regular subsemigroup case. As we known, if $\mathrm{OT}_S(X)$ is a subsemigroup of T(X), then $\mathrm{OT}_S(X)$ is a regular subsemigroup of T(X). The following characterization is obtained immediately from Theorem 3.

Theorem 4

Let S be a non-trivial subfence of a fence X. Then, $\mathrm{OT}_S(X)$ is a regular subsemigroup of T(X) if and only if |S| is odd, $|X{\setminus } S|\le 2$ and $X{\setminus } S$ is not a 2-element fence.

For the case that $\mathrm{OT}_S(X)$ is not a subsemigroup of T(X), we are looking for other regular subsemigroups of T(X) that are contained in $\mathrm{OT}_S(X)$.

Proposition 5

Let S be a subfence of a fence X. Then, the set $\{\alpha \in \mathrm{OT}_S(X) \mid S\alpha =S\}$ is a regular subsemigroup of T(X).

Proof

Let $T=\{\alpha \in \mathrm{OT}_S(X) \mid S\alpha =S\}$. A similar argument as the proof of Proposition 2 shows that T is a subsemigroup of T(X). Theorem 1 implies that T is a regular subsemigroup of T(X). $\square$