Kyungpook Mathematical Journal 2018; 58(4): 627-635
Regularity of a Particular Subsemigroup of the Semigroup of Transformations Preserving an Equivalence
Jittisak Rakbud
Department of Mathematics, Faculty of Science, Silpakorn University, Nakhon Pathom 73000, Thailand
e-mail : rakbud_j@su.ac.th
Received: July 26, 2018; Revised: October 19, 2018; Accepted: October 24, 2018; Published online: December 23, 2018.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

In this paper, we use the notion of characters of transformations provided in  by Purisang and Rakbud to define a notion of weak regularity of transformations on an arbitrarily fixed set X. The regularity of a semigroup of weakly regular transformations on a set X is also investigated.

Keywords: regularity, weak regularity, character, full transformation semigroup.
1. Introduction

For any semigroup , we call an element a of a regular element of if there exists an element b of such that aba = a. It is well-known that an element a of a semigroup is regular if and only if there is such that aca = a and cac = c. We denote the set of all regular elements of a semigroup by . A semigroup is said to be regular if every element of is regular, that is, if .

The notion of regularity plays an important role in semigroup theory. Over the years, there have been many people studying the regularity of subsemigroups of the regular semigroup T(X) of functions on a nonempty set X under the composition, called a full transformation semigroup. The following are two simple subsemigroups of T(X) which have widely been investigated or used as bases for building up some other subsemigroups of T(X):

$T(X,Y)={α∈T(X):Xα⊆Y}$

and

$T¯(X,Y)={α∈T(X):Yα⊆Y},$

where Y is a fixed nonempty subset of X (see [2, 3, 7, 9, 10, 11] for some references). Here are some results on the regularity of T(X, Y ) and (X, Y ) provided in  by Nenthein, Youngkhong and Kemprasit, in  by Honyam and Sanwong, and in  by Sanwong and Sommanee.

Theorem 1.1.([7, Theorem 2.1])

Let X be a nonempty set, and let Y be a nonempty subset of X. Then for any αT(X, Y ), the following statements are equivalent:

• αR(T(X, Y ));

• = Y α;

• Y ∩ ()α−1 ≠ ∅︀ for all xX;

• Y−1 ≠ ∅︀ for all xXα.

By Thorem 1.1, the following corollary was deduced.

Corollary 1.2.([7, Corollary 2.2])

Let X be a nonempty set, and let Y be a nonempty subset of X. Then T(X, Y ) is regular if and only if Y = X or |Y | = 1.

Remark 1.3

In the proof of the implication (4) ⇒ (1) in Theorem 1.1, the authors defined a function β in T(X, Y ) under the assumption that Y−1 ≠ ∅︀ for all x to make α regular by = yx if x and = c otherwise, where c is a fixed element of Y and for each x, yx is a fixed element of Y−1.

Theorem 1.4. ([7, Theorem 2.3])

Let X be a nonempty set, and let Y be a nonempty subset of X. Then for any α(X, Y ), the following statements are equivalent:

• αR((X, Y ));

• Y = Y α;

• Y ∩ ()α−1 ≠ ∅︀ for all xY α−1;

• Y−1 ≠ ∅︀ for all xY.

By Thorem 1.4, the following corollary was obtained.

Corollary 1.5. ([7, Corollary 2.4])

Let X be a nonempty set, and let Y be a nonempty subset of X. Then T̄(X, Y ) is regular if and only if Y = X or |Y | = 1.

Theorem 1.6. ([10, Theorem 2.4])

Let X be a nonempty set, and let Y be a nonempty subset of X. Then R(T(X, Y )) is the largest regular subsemigroup of T(X, Y ).

Theorem 1.7. ([3, Lemma 1])

Let X be a nonempty set, and let Y be a nonempty subset of X. Then R((X, Y )) is a subsemigroup of T̄(X, Y ) if and only if Y = X or |Y | = 1. In this trivial situation, R((X, Y )) = (X, Y ) is regular.

In this paper, by a partition of a nonempty set X, we mean a family ℱ = {Yi : iI} of nonempty subsets of X such that X = ∪iI Yi and YiYj for all i, jI with ij. Each of the two partitions {X} and {{x} : xX} is called a trivial partition of X.

Throughout the remainder of this paper, let X be a nonempty set, and let ℱ = {Yi : iI} be a partition of X, which are arbitrarily fixed. Let

$TF(X)={α∈T(X):∀i∈I∃j∈I,Yiα⊆Yj}.$

It is clear that T(X) is a subsemigroup of the full transformation semigroup T(X). Note that T(X) is exactly the semigroup of transformations preserving the equivalence ℰ induced by the partition ℱ (see  for more detials). There have been several works on the semigroup of transformations preserving an equivalence (see [1, 5, 6] for some references). For each αT(X), let χ(α) : II be defined by (α) = j if and only if YiαYj. By the defintion of a partition, we see that χ(α) is well-defined, that is, χ(α)T(I). For each αT(X), we call the function χ(α) the character of α with respect to ℱ. In addition to the set X and the partition ℱ of X, let J be an arbitrarily fixed nonempty subset of I. Let

$TF(J)(X)={α∈T(X):χ(α)∈T(I,J)}.$

It is clear that

$TF(J)(X)={α∈T(X):∀i∈I∃j∈J,Yiα⊆Yj}.$

The set $TF(J)(X)$, which is indeed a subsemigroup of T(X), as well as the notion of character were first introduced in  by Purisang and Rakbud. In that paper, the authors studied the regularity of the semigroup $TF(J)(X)$ and some other semigroups defined via the notion of character. We summarize some of their results as follows.

Proposition 1.8. ([8, Proposition 2.2])

Let Y = ∪jJ Yj. Then the following statementshold:

• $TF(J)(X)=T(X,Y)$if and only if |J| = 1 or ℱ = {{x} : xX}.

• $TF(J)(X)=T(X)$if and only if J = I oris trivial.

Lemma 1.9. ([8, Lemma 2.3])

For every α, $β∈TF(J)(X)$, χ(αβ) = χ(α)χ(β).

By using the notion of character, the authors defined two congruence relations χ and χ̃ on $TF(J)(X)$ as follows:

$(α,β)∈χ⇔χ(α)=χ(β),(α,β)∈χ˜⇔χ(α)∣J=χ(β)∣J.$

And then they studied the regularity of the quotient semigroups $TF(J)(X)/χ$ and $TF(J)(X)/χ˜$. The following are what they obtained.

Theorem 1.10. ([8, Theorem 2.4])

For each$α∈TF(J)(X)$, let [α] and$[α]˜$be the equivalence classes of α under the equivalence relations χ and χ̃ respectively. Then the following statements hold:

• $TF(J)(X)/χ≅T(I,J)$by the isomorphism [α] ↦ χ(α).

• $TF(J)(X)/χ˜≅T(J)$by the isomorphism$[α]˜↦χ(α)∣J$.

By Corollary 1.2 and Theorem 1.10, the following corollary was obtained.

Corollary 1.11. ([8, Corollary 2.5])

The following statements hold:

• The three statements (a), (b) and (c) below are all equivalent:

• the quotient semigroup$TF(J)(X)/χ$is regular;

• the semigroup T(I, J) is regular;

• J = I or |J| = 1.

• The quotient semigroup T(X)/χ, which is exactly$TF(I)(X)/χ$, is regular.

• The quotient semigroup$TF(J)(X)/χ˜$is regular.

In , the regularity of the semigroup $TF(J)(X)$ was obtained as follows.

Theorem 1.12. ([8, Theorem 2.6])

The semigroup$TF(J)(X)$is regular if and only if$∣TF(J)(X)∣=1$or$TF(J)(X)=T(X)$.

Note that, from Theorem 1.12, we immediately have that T(X) is regular if and only if ℱ is trivial. This can also be deduced from Proposition 2.4 of Huisheng .

It is clear that for each αT(X), the equivalence class [α] of α under the equivalence relation χ is a subsemigroup of T(X) if and only if χ(α) is an idempotent element of the full transformation semigroup T(I). The regularity of the semigroup [α], in the case where α is an idempotent element of T(I), was also studied in . In  as well, some other subsemigroups of T(X) were defined by using the notion of character as follows: Let I(X), S(X) and B(X) be the sets of all elements of T(X) whose characters are injective, surjective and bijective respectively. The regularity of each of these three semigroups was also studied.

Observe that the semigroups $TF(J)(X)$, [α] when χ(α) is idempotent, I(X), S(X) and B(X) can simultaneously be generalized by making use of the notion of character as follows: For every subsemigroup of T(I), let

$TF(S)(X)={α∈TF(X):χ(α)∈S}.$

By Lemma 1.9, we see that $TF(S)(X)$ is a subsemigroup of T(X). And, furthermore, Lemma 1.9 also implies that for every subsemigroup ℋ of T(X), ℋ is necessarily of the form $TF(S)(X)$ for some subsemigroup of T(I), in fact, . We state this pleasant result in the following theorem.

Theorem 1.13

For every ℋ ⊆ T(X), ℋ is a subsemigroup of T(X) if and only if there is a subsemigroup of T(I) such that$H=TF(S)(X)$. In this situation, .

Let be a subsemigroup of T(I). Then by considering the congruence relation χ on T(X) restricted to $TF(S)(X)$, we have the quotient semigroup $TF(S)(X)/χ$. It is clear that $TF(S)(X)/χ={[α]:α∈TF(S)(X)}$, and that $TF(S)(X)/χ$ is a subsemigroup of T(X)/χ. Analogously to Theorem 1.10(1), the following result is obtained.

Theorem 1.14

$TF(S)(X)/χ≅S$by the isomorpism [α] ↦ χ(α).

Immediately from Theorem 1.14, we have the following corollary.

Corollary 1.15

The quotient semigroup$TF(S)(X)/χ$is regular if and only if the semigroup is regular.

We note here that the regularity of may not imply $TF(S)(X)$ to be regular. For example, when is exactly the regular semigroup T(I), we have that $TF(S)(X)=TF(X)$ which is regular only when ℱ is trivial. By making use of the notion of character, we can define a subset of $TF(S)(X)$ as follows: Let

$Rw (TF(S)(X))={α∈TF(S)(X):χ(α)∈R(S)}.$

By Lemma 1.9, we have that $R (TF(S)(X))⊆Rw (TF(S)(X))$. And obviously, if $TF(S)(X)$ is regular, then $R (TF(S)(X))=TF(S)(X)=Rw (TF(S)(X))$. It is easy to see that the set $Rw (TF(S)(X))$ is a subsemigroup of $TF(S)(X)$ if and only if is a subsemigroup of . And in this situation, we have $Rw (TF(S)(X))=TF(R(S))(X)$.

From the inclusion $R (TF(S)(X))⊆Rw (TF(S)(X))$ and the defintion of the set $Rw (TF(S)(X))$, it makes perfect sense to call every $α∈Rw (TF(S)(X))$ a weakly regular transformation with respect to , or simply an -weakly-regular transformation. By Theorem 1.14, we have for any $α∈TF(S)(X)$ that α is an -weakly-regular transformation if and only if the equivalence class [α] of α under the congruence relation χ is a regular element of the quotient semigroup $TF(S)(X)/χ$. This yileds, in the case where is a subsemigroup of , that $Rw (TF(S)(X))/χ=R (TF(S)(X)/χ)$. Note that for any semigroup and a subsemigroup of , if , then . From this elementary fact, we immediately obtain that if is a subsemigroup of , then $R (Rw (TF(S)(X)))=R (TF(S)(X))$.

The aim of this paper is to investigate the regularity of $Rw (TF(S)(X))$ for a certain subsemigroup of T(I) with a subsemigroup of .

2. Regularity of a Semigroup of Weakly Regular Transformations

By virtue of Theorem 1.6, we have that the set $Rw (TF(J)(X))$ is a subsemigroup of $TF(J)(X)$. This section is devoted to studying the regualrity of this semigroup. From a result of Huisheng , the regularity of elements of the semigroup T(X) can immediately be deduced as follows.

Proposition 2.1

An element α of the semigroup T(X) is regualr if and only if for every iI, there exists jI such that YiYjα.

The result in Proposition 2.1 can straightforwardly be generalized to the semigroup $TF(J)(X)$ as follows.

Theorem 2.2

For every$α∈TF(J)(X)$, α is regualr if and only if for every iI, there exists jJ such that YiYjα.

Proof

Let $α∈TF(J)(X)$. We are now going to prove the necessity. Suppose that the condition holds. Let E = {iI : Yi ≠ ∅︀}. It is clear that E ≠ ∅︀. Let iE be arbitrarily fixed. By the assumption, we fix jiJ such that YiYjiα. For each xYi, by the inclusion YiYjiα, we fix $zx(i)∈Yji$ such that $zx(i)α=x$. Also, we fix ciYji, and then define a function βi : YiX by $xβi=zx(i)$ for all xYi and i = ci otherwise. It is clear that YiβiYji. Next, let β : XX be defined by β|Yi= βi for all iE and = a for all x ∈ ∪iIE Yi, where a is a fixed element in ∪jJ Yj. Then $β∈TF(J)(X)$ and αβα = α. To prove the sufficiency, suppose that α is regular. Then there is $β∈TF(J)(X)$ such that αβα = α. Let iI, and let j = (β). Then jJ. We will show that YiYjα. To see this, let xYi. Then xYi and there is zX such that = x. Since xYi and j = (β), we have Yj. Thus x = = zαβα = xβαYjα. Therefore, we obtain the inclusion YiYjα as desired.

Previously in Section 1, we have seen that $R (TF(J)(X))⊆Rw (TF(J)(X))$, and that the regulairty of the semigroup $TF(J)(X)$ suffices to obtain that $R (TF(J)(X))=Rw (TF(J)(X))$. Next, we give a necessary and sufficient condition for obtaining the equality $R (TF(J)(X))=Rw (TF(J)(X))$.

Theorem 2.3

$R (TF(J)(X))=Rw (TF(J)(X))$if and only if |Yj | = 1 for all jJ.

Proof

Suppose that Yj = {zj} for all jJ. We want to show that $Rw (TF(J)(X))⊆R (TF(J)(X))$. To see this, let $α∈Rw (TF(J)(X))$. Then the character χ(α) of α is a regular element of T(I, J). Thus, by Theorem 1.1, we have that (α) = (α). Let Y = ∪jJ Yj. Then Y α = {z(α) : jJ}. Since (α) = (α), we have for each iIJ that there is jiJ such that (α) = jiχ(α). This yields that Yiα = {zjiχ(α)} for all iIJ. Therefore, = Y α, which implies by Theorem 1.1 agian that α is a regular element of the semigroup T(X, Y), and that Y−1 ≠ ∅︀ for all x. As mentioned in Remark 1.3, we can define a function β : XX, under the condition that Y−1 ≠ ∅︀ for all x, which makes α regular in the semigroup T(X, Y ) as follows: Fix an element cY, and for each x, fix an element yxY−1. And then define β : XX by = yx for all x and = c otherwise. In our situation here, we can easily see from the way of defining the function β that $β∈TF(J)(X)$. Hence $α∈R (TF(J)(X))$; and thus $R (TF(J)(X))=Rw (TF(J)(X))$. Conversely, suppose that there is jJ such that |Yj| ≥ 2. We want to show that $Rw (TF(J)(X))⊈R (TF(J)(X))$. To see this, we fix two distinct elements a and b of Yj. And define α : XX by = a for all xYj and = b otherwise. Since Yjα = {a} and Yiα = {b} for all iI{j}, it follows that Yj = {a, b} ⊈ Yiα for all iJ. From this, we have by Theorem 2.2 that $α∉R (TF(J)(X))$. And since χ(α) is a constant function, we get that $α∈Rw (TF(J)(X))$. Thus $Rw (TF(J)(X))⊈R (TF(J)(X))$.

Since $Rw (TF(J)(X))$ is a subsemigroup of $TF(J)(X)$, we have $Rw (TF(J)(X))/χ$ is a subsemigroup of $TF(J)(X)/χ$. By Theorem 1.6 and Theorem 1.10(1), we obtain that $R (TF(J)(X)/χ)$ is a regular subsemigroup of $TF(J)(X)/χ$. Hecne the quotient semigroup $Rw (TF(J)(X))/χ=R (TF(J)(X)/χ)$ is regular. Next, we investigate the regularity of the semigroup $Rw (TF(J)(X))$ itself. By Theorem 2.3 and the fact that $R (Rw (TF(J)(X)))=R (TF(J)(X))$, the following result on the regularity of $Rw (TF(J)(X))$ is immediately obtained.

Corollary 2.4

The semigroup$Rw (TF(J)(X))$is regular if and only if |Yj | = 1 for all jJ.

By the definition of $TF(J)(X):=TF(T(I,J))(X)$, defined relatively to the semigroup T(I, J), and the regularity of R(T(I, J)), we expect to have that $R (TF(J)(X))$ is a regular subsemigroup of $TF(J)(X)$. But we find that, in general, the set $R (TF(J)(X))$ is not a subsemigroup of $TF(J)(X)$. This is affirmed by the following proposition.

Proposition 2.5

If |J| > 1 and there is jJ such that |Yj | > 1, then the set$R (TF(J)(X))$is not a subsemigroup of$TF(J)(X)$.

Proof

Suppose that |J| > 1 and there is jJ such that |Yj | > 1. Fix two distinct elements a and b in Yj, and define α : XX by = a for all xYj and = b otherwise. Clearly, $α∈TF(J)(X)$. And, as explained in the proof of Theorem 2.3, we have that α is not regular. Next, let kJ be different from j, and fix an element cYk. Let β : XX be defined by = c for all xYj and = b otherwise. And let γ : XX be defined by = b and = a for all xX{b}. Then β and α belong to $TF(J)(X)$. Since Yj = {b} = Ykβ, Yk = {c} = Yjβ and Yi = ∅︀ for all iI{j, k}, it follows from Theorem 2.2 that β is regular. Similarly, since Yj = {a, b} = Yjγ and Yi = ∅︀ for all iI{j}, we have that γ is regualr as well. Form the defintions of β and γ, it is easy to see that xβγ = a for all xYj and xβγ = b otherwise. Hence βγ = α. This yileds that the set $R (TF(J)(X))$ is not a subsemigroup of $TF(J)(X)$.

According to Proposition 2.5, the set $R (TF(J)(X))$ is not necessarily a subsemigroup of $TF(J)(X)$. The following result tells us when $R (TF(J)(X))$ is a subsemigroup of $TF(J)(X)$. It is immediately obtained from Theorem 1.6, Theorem 2.3 and Propostion 2.5.

Corollary 2.6

The set$R (TF(J)(X))$is a subsemigroup of$TF(J)(X)$if and only if |J| = 1 or |Yj | = 1 for all jJ. In this circumstance, $R (TF(J)(X))=R(T(X,Yj))$if |J| = 1 with J = {j}, and$R (TF(J)(X))=Rw (TF(J)(X))$if |Yj | = 1 for all jJ. Furthermore, we have in each case that the semigroup$R (TF(J)(X))$is regular.

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