Regularity and abundance on semigroups of partial transformations with invariant set

Thapakorn Pantarak; Yanisa Chaiya

doi:10.1515/math-2023-0142

Article Open Access

Regularity and abundance on semigroups of partial transformations with invariant set

Thapakorn Pantarak and Yanisa Chaiya

Published/Copyright: November 11, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

Let P ( X ) be a partial transformation semigroup on a non-empty set X . For a fixed non-empty subset Y of X , let

P T ¯ ( X , Y ) = { α ∈ P ( X ) ∣ ( dom α ∩ Y ) α ⊆ Y } .

Then, P T ¯ ( X , Y ) consists of all the mapping in P ( X ) that leave Y ⊆ X as an invariant. It is a generalization of P ( X ) since P T ¯ ( X , X ) = P ( X ) . In this article, we present the necessary and sufficient conditions for elements of P T ¯ ( X , Y ) to be regular, left regular, and right regular. The results are used to describe the relationships between these elements and determine their number when X is a finite set. Moreover, we show that P T ¯ ( X , Y ) is always abundant.

Keywords: partial transformation semigroup; regularity; left regularity; right regularity; abundance

MSC 2010: 20M20

1 Introduction

An element a in a semigroup S is termed regular if there exists at least one solution to the equation a x a = a within S . The semigroup S is referred to as a regular semigroup if all of its elements are regular. The concept of regularity was initially introduced in ring theory by von Neumann in 1936 [1] and was formally presented in semigroup theory in 1955 by Munn and Penrose [2]. Subsequently, various groups of algebraists have undertaken extensive studies on regular semigroups and their structures [3–10].

Various concepts of regularity were explored by Croisot in 1953, and Croisot’s theory of decompositions of semigroup was presented in Section 4.1 in the study by Clifford and Preston [11]. In particular, the concepts of left and right regularities were important. We say that an element a in a semigroup S is considered left or right regular if there exists an element x in S such that a = x a 2 or a = a 2 x , respectively. Notably, if all elements in S are left or right regular, then S is referred to as a left or right regular semigroup, respectively. Furthermore, left and right regular semigroups and their structures have been extensively studied [12–15].

Since any semigroup can be realized as a transformation semigroup of an appropriate set, the latter family of semigroups is a significant topic of research. A crucial transformation semigroup is a full transformation semigroup T ( X ) , defined as a semigroup of all functions from a non-empty set X into itself under the operation of composition. In 1955, Doss [16] proved that T ( X ) is regular and characterized Green’s relations on this semigroup. This characterization can be directly applied to obtain characterizations for both left regularity and right regularity. Subsequently, the study of structural properties of the full transformation semigroup has garnered significant attention from researchers worldwide. In addition, the full transformation semigroup has been continuously generalized for expanding and recapturing known results. A famous generalization of T ( X ) is a semigroup T ¯ ( X , Y ) for ∅ ≠ Y ⊆ X , which is defined as follows:

T ¯ ( X , Y ) = { α ∈ T ( X ) ∣ Y α ⊆ Y } .

The semigroup T ¯ ( X , Y ) was first studied by Magill [17] in 1966. Subsequently, in 2005, Nenthein et al. [18] derived a characterization for the regularity of elements of T ¯ ( X , Y ) and demonstrated that T ¯ ( X , Y ) is regular if and only if X = Y or Y is a singleton set. In 2011, Honyam and Sanwong [19] characterized when T ¯ ( X , Y ) is isomorphic to T ( Z ) for some set Z , and fully described Green’s relations on T ¯ ( X , Y ) . Later in 2013, Choomanee et al. [20] characterized the left regular and right regular elements of T ¯ ( X , Y ) . Furthermore, they examined the relationship between these elements in terms of the finiteness of their images, revealing that left and right regular elements coincide when X α is a finite set. In addition, they enumerated the left regular elements in the case when X is finite. Moreover, in the same year, Sun and Wang [21] showed that T ¯ ( X , Y ) is abundant. Similarly, several other properties of transformation semigroups have been reported [22–26].

Given a non-empty set X , the super semigroup of all transformation semigroups on X is a partial transformation semigroup P ( X ) , which can be defined as follows:

P ( X ) = { α : A → X ∣ A ⊆ X } .

Evidently, T ( X ) and T ¯ ( X , Y ) are strictly contained in P ( X ) . This semigroup is also regular, and its Green’s relations were derived in the study by Howie [27]. In a similar vein, the partial transformation semigroups have gained popularity as a structure in research, and various other structural properties of semigroups have been established. For a fixed non-empty subset Y of X , in analogy with T ¯ ( X , Y ) , we can consider the following:

P T ¯ ( X , Y ) = { α ∈ P ( X ) ∣ ( dom α ∩ Y ) α ⊆ Y } .

Since P T ¯ ( X , X ) = P ( X ) , we can regard P T ¯ ( X , Y ) as a generalization of P ( X ) .

In this article, we determine the necessary and sufficient conditions for elements of P T ¯ ( X , Y ) to be regular, left regular, and right regular in Section 3. The results in this section recapture the known results on P ( X ) when we focus on Y = X . Moreover, they are used to deduce the results for T ¯ ( X , Y ) , when elements with X as their domain are considered. We also establish the relationship between different types of regularities and utilize the results to determine the number of elements of each type in Section 4. In Section 5, we show that P T ¯ ( X , Y ) is always abundant.

2 Preliminaries and notations

Some basic mathematical terminologies and relevant notations used in this article are described in this section. Further details were referred to and are discussed in the study by Howie [27].

Let X and Y be non-empty sets, with Y being a subset of X . We are considering a subsemigroup P T ¯ ( X , Y ) of P ( X ) defined as follows:

P T ¯ ( X , Y ) = { α ∈ P ( X ) ∣ ( dom α ∩ Y ) ⊆ Y } .

The identity map on X , denoted as i d X , belongs to P T ¯ ( X , Y ) . Throughout this article, functions are written on the right in the α β form, where α is applied first. For α ∈ P ( X ) , the notations dom α and im α denote the domain of α and the range of α , respectively. In addition, for x ∈ im α , x α − 1 denotes the set of inverse images of x under α , i.e., x α − 1 = { z ∈ dom α : z α = x } . Moreover, the following notation is used:

α = X t x t ,

where the script t belongs to some (unmentioned) index set T , the abbreviation { x t } denotes { x t : t ∈ T } and that im α = { x t } , x t α − 1 = X t ⊆ dom α , and ⋃ t X t = dom α . In particular, when α ∈ P T ¯ ( X , Y ) , we have Y α ⊆ Y . Thus, the domain of α can be divided into three parts, such that

α = A i B j C k a i b j c k ,

where A i ∩ Y ≠ ∅ , B j , C k ⊆ X \ Y , { a i } , { b j } ⊆ Y , and { c k } ⊆ X \ Y . Here, I , J , and K can be empty.

3 Regular, left regular, and right regular elements

In this section, we discuss regular, left regular, and right regular elements of P T ¯ ( X , Y ) in detail. For A ⊆ X , we denote ( dom α ∩ A ) α as A α , and if α , β , γ ∈ P T ¯ ( X , Y ) such that α = β γ , we readily obtain

(3.1) dom α = dom ( β γ ) = ( im β ∩ dom γ ) β − 1 ⊆ ( im β ) β − 1 = dom β ,

(3.2) im α = im ( β γ ) = ( im β ∩ dom γ ) γ ⊆ ( dom γ ) γ = im γ ,

(3.3) A α = ( dom ( β γ ) ∩ A ) β γ ⊆ ( dom β ∩ A ) β γ ⊆ ( dom γ ∩ A ) γ = A γ .

Moreover, an equivalence relation ker α can be defined as follows:

ker α = { ( a , b ) ∈ dom α × dom α ∣ a α = b α } .

Evidently, a ker α , the equivalence class of a ∈ dom α with respect to ker α , is given by ( a α ) α − 1 = { x ∈ dom α ∣ x α = a α } .

Theorem 3.1

Let α ∈ P T ¯ ( X , Y ) . The following statements are equivalent:

α is regular.
im α ∩ Y = Y α .
x ker α ∩ Y ≠ ∅ for all x ∈ dom α with x α ∈ Y .
x α − 1 ∩ Y ≠ ∅ for all x ∈ im α ∩ Y .

Proof

( 1 ) ⇒ ( 2 ) Assume α is regular. Then, there exists β ∈ P T ¯ ( X , Y ) such that α = α β α . Let x ∈ im α ∩ Y . Then, x = a α ∈ Y for some a ∈ dom α and a α ∈ dom β . Since β ∈ P T ¯ ( X , Y ) , we obtain Y β ⊆ Y and x = a α = a α β α ∈ Y β α ⊆ Y α . This implies that im α ∩ Y ⊆ Y α . Since Y α ⊆ im α ∩ Y , ( 2 ) follows.

( 2 ) ⇒ ( 3 ) Assume im α ∩ Y = Y α . Let x ∈ dom α be such that x α ∈ Y . Then, x α ∈ im α ∩ Y = Y α . Hence, there exists y ∈ dom α ∩ Y such that x α = y α , which yields y ∈ ( x α ) α − 1 = x ker α .

( 3 ) ⇒ ( 4 ) Assume ( 3 ) holds. Let x ∈ im α ∩ Y . Then, x = a α ∈ Y for some a ∈ dom α . Following equation ( 3 ) , x α − 1 ∩ Y = ( a α ) α − 1 ∩ Y = a ker α ∩ Y ≠ ∅ .

( 4 ) ⇒ ( 1 ) Assume x α − 1 ∩ Y ≠ ∅ for all x ∈ im α ∩ Y . Then, we can express

α = A i B j a i b j ,

where A i ∩ Y ≠ ∅ , B j ⊆ X \ Y , { a i } ⊆ Y , and { b j } ⊆ X \ Y . For each i and j , a ′ ∈ A i ∩ Y and b j ′ ∈ B j are chosen. Then, β : im α → X can be defined as follows:

β = a i b j a i ′ b j ′ .

Evidently, β ∈ P T ¯ ( X , Y ) and α = α β α .□

Corollary 3.2

The semigroup P T ¯ ( X , Y ) is regular if and only if X = Y .

Proof

If X = Y , then P T ¯ ( X , Y ) = P ( X ) , which is a regular semigroup. Conversely, we assume that Y is a proper subset of X . Then, there exists x ∈ X \ Y . Since Y ≠ ∅ , there exists y ∈ Y . Hence,

α = x y ∈ P T ¯ ( X , Y ) .

Obviously, im α ∩ Y = { y } ≠ ∅ = Y α . Thus, from Theorem 3.1, α is not regular.□

Lemma 3.3

Let α , β ∈ P T ¯ ( X , Y ) . Then, α = γ β for some γ ∈ P T ¯ ( X , Y ) if and only if im α ⊆ im β and Y α ⊆ Y β .

Proof

Assume α = γ β for some γ ∈ P T ¯ ( X , Y ) . From Corollaries 3.2 and 3.3, we obtain im α ⊆ im β and Y α ⊆ Y β .

Conversely, assume that im α ⊆ im β and Y α ⊆ Y β . Write

α = A i B j C k a i b j c k ,

where A i ∩ Y ≠ ∅ ; B j , C k ⊆ X \ Y ; { a i } , { b j } ⊆ Y ; and { c k } ⊆ X \ Y . From our assumptions, we can write

β = A i ′ B j ′ C k ′ D l a i b j c k d l ,

where A i ′ ∩ Y ≠ ∅ ; C k ′ ⊆ X \ Y ; B j ′ , D l , { d l } ⊆ X ; and possibly contain elements of Y . For each i , j , and k , we choose a i ′ ∈ A i ′ ∩ Y , b j ′ ∈ B j ′ , and c k ′ ∈ C k ′ , respectively. Let γ : dom α → X be defined as follows:

γ = A i B j C k a i ′ b j ′ c k ′ .

Evidently, γ ∈ P T ¯ ( X , Y ) and α = γ β .□

Since α 2 = α α , considering Equations (3.2) and (3.3), we obtain im α 2 ⊆ im α and Y α 2 ⊆ Y α , respectively. As a direct consequent of Lemma 3.3, the characterization of left regular elements on P T ¯ ( X , Y ) can be obtain as follows:

Theorem 3.4

Let α ∈ P T ¯ ( X , Y ) . Then, α is left regular if and only if im α = im α 2 and Y α = Y α 2 .

Corollary 3.5

The semigroup P T ¯ ( X , Y ) is left regular if and only if ∣ X ∣ = 1 .

Proof

Assume ∣ X ∣ > 1 . Let y ∈ Y and y ≠ x ∈ X . Then,

α = x y ∈ P T ¯ ( X , Y ) ,

where im α = { y } ≠ ∅ = im α 2 . Therefore, α is not left regular according to Theorem 3.4.

Conversely, on assuming ∣ X ∣ = 1 , we have P T ¯ ( X , Y ) = { ∅ , i d X } . Evidently, P T ¯ ( X , Y ) is a left regular semigroup.□

To derive a characterization of right regular elements of P T ¯ ( X , Y ) , we need the following lemma.

Lemma 3.6

Let α , β ∈ P T ¯ ( X , Y ) . Then, α = β γ for some γ ∈ P T ¯ ( X , Y ) if and only if all of the following conditions hold:

dom α ⊆ dom β .
ker β ∩ ( dom β × dom α ) ⊆ ker α .
If x ∈ ( dom α ∩ dom β ) \ Y and x β ∈ Y , then x α ∈ Y .

Proof

Assume α = β γ for some γ ∈ P T ¯ ( X , Y ) . From equation (3.1), we have dom α ⊆ dom β . Let ( x , x ′ ) ∈ ker β ∩ ( dom β × dom α ) . Then, x β = x ′ β , x ∈ dom β , and x ′ ∈ dom α = dom ( β γ ) = ( im β ∩ dom γ ) β − 1 . Hence, x β = x ′ β ∈ im β ∩ dom γ . Thus, x α = x ( β γ ) = ( x β ) γ = ( x ′ β ) γ = x ′ ( β γ ) = x ′ α , which yields ( x , x ′ ) ∈ ker α . Let z ∈ ( dom α ∩ dom β ) \ Y be such that z β ∈ Y . Then, z β ∈ dom γ and z α = z ( β γ ) = ( z β ) γ ∈ Y γ ⊆ Y .

Conversely, assume that all three aforementioned conditions hold. Consequently, for each x ∈ ( dom α ) β , there exists x ′ ∈ dom α such that x ′ β = x . Let γ : ( dom α ) β → X be defined by x γ = x ′ α for all x ∈ ( dom α ) β . Then, γ is well defined by (2). Moreover, we have ( dom α ) β β − 1 = dom α from (3), which yields dom α = dom ( β γ ) . To prove α = β γ , we assume that z ∈ dom α . Then, z β ∈ ( dom α ) β = dom γ and z ( β γ ) = ( z β ) γ = z α .□

Theorem 3.7

Let α ∈ P T ¯ ( X , Y ) . Then, α is right regular if and only if all of the following conditions hold:

ker α = ker α 2 .
For any x ∈ dom α 2 \ Y , if x α ∉ Y , then x α 2 ∉ Y .

Proof

Assume that α is right regular. Then, α = α 2 γ for some γ ∈ P T ¯ ( X , Y ) . By Lemma 3.6, we obtain dom α ⊆ dom α 2 and ker α 2 ∩ ( dom α 2 × dom α ) ⊆ ker α , and if x ∈ ( dom α ∩ dom α 2 ) \ Y and x α 2 ∈ Y , then x α ∈ Y . Since dom α 2 ⊆ dom α , we obtain dom α = dom α 2 , and so (2) holds. Since ker α ⊆ ker α 2 , we deduce that (1) holds.

Conversely, assume (1) and (2) hold. By (1), we obtain dom α = dom α 2 and so α and α 2 satisfy all conditions in Lemma 3.6. Hence, there exists γ ∈ P T ¯ ( X , Y ) such that α = α 2 γ . Therefore, α is right regular.□

To avoid difficulties arising from the calculation of ker α and ker α 2 , we rewrite the characterization of right regular elements in terms of α .

Lemma 3.8

Let α ∈ P T ¯ ( X , Y ) . Then, ker α = ker α 2 if and only if im α ⊆ dom α and α ∣ im α is injective.

Proof

Assume that ker α = ker α 2 . Let x α ∈ im α . Then, x ∈ dom α = dom α 2 = ( im α ∩ dom α ) α − 1 , and this implies x α ∈ dom α . Hence, im α ⊆ dom α . Let x 1 α , x 2 α ∈ im α be such that x 1 α 2 = x 2 α 2 . Then, ( x 1 , x 2 ) ∈ ker α 2 = ker α , which implies x 1 α = x 2 α . Hence, α ∣ im α is injective.

Conversely, assume that im α ⊆ dom α and α ∣ im α is injective. Let ( x 1 , x 2 ) ∈ ker α 2 . Then, x 1 α 2 = x 2 α 2 and x 1 , x 2 ∈ dom α 2 ⊆ dom α . Since ( x 1 α ) α = x 1 α 2 = x 2 α 2 = ( x 2 α ) α and α ∣ im α is injective, x 1 α = x 2 α , i.e., ( x 1 , x 2 ) ∈ ker α . This implies that ker α 2 ⊆ ker α . For the other containment, let ( x 3 , x 4 ) ∈ ker α . Then, x 3 α = x 4 α ∈ im α ⊆ dom α . Hence, x 3 α 2 = ( x 3 α ) α = ( x 4 α ) α = x 4 α 2 . Thus, ( x 3 , x 4 ) ∈ ker α 2 , and so ker α ⊆ ker α 2 . This completes the proof.□

Lemma 3.9

Let α ∈ P T ¯ ( X , Y ) . The following statements are equivalent:

For any x ∈ dom α 2 \ Y , if x α ∉ Y , then x α 2 ∉ Y .
im ( α ∣ im α \ Y ) ⊆ X \ Y .

Proof

( 1 ) ⇒ ( 2 ) Assume that (1) holds. Let x ∈ im ( α ∣ im α \ Y ) . Then, there exists x ′ ∈ ( im α \ Y ) ∩ dom α such that x ′ α = x . Since x ′ ∈ im α \ Y , there exists z ∈ dom α \ Y such that z α = x ′ ∈ dom α . So, z ∈ dom α 2 \ Y and z α ∉ Y . By (1), we obtain x = x ′ α = ( z α ) α = z α 2 ∉ Y . Hence, im ( α ∣ im α \ Y ) ⊆ X \ Y .

( 2 ) ⇒ ( 1 ) Assume that im ( α ∣ im α \ Y ) ⊆ X \ Y and let x ∈ dom α 2 \ Y be such that x α ∉ Y . Then, x α 2 = ( x α ) α ∈ im ( α ∣ im α \ Y ) ⊆ X \ Y .□

Combining Theorem 3.7 with Lemmas 3.8 and 3.9, we obtain the following theorem:

Theorem 3.10

Let α ∈ P T ¯ ( X , Y ) . Then, α is right regular if and only if all of the following statements hold:

im α ⊆ dom α .
α ∣ im α is injective.
im ( α ∣ im α \ Y ) ⊆ X \ Y .

If ∣ X ∣ > 1 , then α , as defined in Corollary 3.5, is not right regular. This yields the following corollary:

Corollary 3.11

The semigroup P T ¯ ( X , Y ) is right regular if and only if ∣ X ∣ = 1 .

Next, we establish that, in general, no relationship exists between regular, left regular, and right regular elements of P T ¯ ( X , Y ) .

Example 3.12

Let X be a set of all natural numbers and Y = { 1 , 2 , 3 , 4 , 5 } . Consider two elements α and β in P T ¯ ( X , Y ) be defined as follows:

α = { 1 , 5 , 6 } 2 3 n 3 1 2 n − 2 n ∈ { 7 , 8 , 9 , … } and β = 1 2 .

Evidently, im α = N \ { 4 } = im α 2 and Y α = { 1 , 2 , 3 } = Y α 2 . Then, according to Theorem 3.4, α is left regular. However, Y α ⊊ im α ∩ Y and α ∣ im α is not injective. Thus, α is neither regular nor right regular according to Theorems 3.1 and 3.10, respectively. On the other hand, we obtain im β ∩ Y = Y β , but im β ≠ im β 2 and im β ⊈ dom β . Hence, β is regular but not left and right regular.

Example 3.13

Let X be a set of all natural numbers and Y be a set of all positive even integers. Consider α ∈ P T ¯ ( X , Y ) be defined as follows:

α = 1 2 4 n 2 4 6 n + 2 n ∈ { 5 , 6 , 7 , … } .

Then, it is trivial that α is right regular. However, Y α ⊊ im α ∩ Y and Y α 2 ⊊ Y α , meaning that α is neither regular nor left regular.

Furthermore, we investigate some relationships between elements of varying regularities with respect to the finiteness of their images.

Lemma 3.14

Let ∅ ≠ α ∈ P T ¯ ( X , Y ) be such that im α is finite. If α is left or right regular, then ( X \ Y ) α ⊆ ( X \ Y ) ∪ Y α .

Proof

Assume that α is left regular or right regular. Write

α = A 1 … A r B 1 … B s C 1 … C t a 1 … a r b 1 … b s c 1 … c t ,

where A i ∩ Y ≠ ∅ , B j , C k ⊆ X \ Y , a i , b j ∈ Y , and c k ∈ X \ Y for all i ∈ I = { 1 , … , r } , j ∈ J = { 1 , … , s } , and k ∈ K = { 1 , … , t } .

Case 1: α is left regular. To prove that J = ∅ , we assume, to the contrary, that there exists j 0 ∈ J . Then, B j 0 α = b j 0 ∈ im α ∩ Y . Hence, b j 0 ∈ A i 0 for some i 0 ∈ I or b j 0 ∉ dom α . However, both cases yield ∣ im α 2 ∣ < ∣ im α ∣ , a contradiction. Therefore, J = ∅ , and consequently, ( X \ Y ) α ⊆ { a 1 , … , a r , c 1 , … , c t } ⊆ ( X \ Y ) ∪ Y α .

Case 2: α is right regular. To prove that J = ∅ , we assume, to the contrary, that there exists j 0 ∈ J . Then, a 1 , … , a r , b j 0 ∈ im α ⊆ dom α . Thus, a 1 , … , a r , b j 0 ∈ A i for some i ∈ I . According the pigeonhole principle, there exists some A i 0 that contains more than one element of { a 1 , … , a r , b j 0 } . Hence, the injectivity of α ∣ im α is contradictory. Therefore, J = ∅ , and consequently, ( X \ Y ) α ⊆ ( X \ Y ) ∪ Y α .□

Theorem 3.15

Let α ∈ P T ¯ ( X , Y ) be such that im α is finite. Then, the following statements hold:

If α is left regular, then α is regular.
If α is right regular, then α is regular.

Proof

(1) Assume that α is left regular. The assertion is trivial when α = ∅ . Considering α ≠ ∅ , we have, ( X \ Y ) α ⊆ ( X \ Y ) ∪ Y α according to Lemma 3.14. Hence,

im α ∩ Y = [ Y α ∪ ( X \ Y ) α ] ∩ Y ⊆ [ Y α ∪ ( X \ Y ) ] ∩ Y = Y α ⊆ im α ∩ Y .

Thus, im α ∩ Y = Y α and α is regular.

(2) The proof is similar to that of (1).□

In Example 3.12, β ∈ P T ¯ ( X , Y ) demonstrates that the converse of Theorem 3.15 does not hold.

Theorem 3.16

Let α ∈ P T ¯ ( X , Y ) be such that im α is finite. Then, α is left regular if and only if α is right regular.

Proof

The assertion is trivial when α = ∅ . Assume that α ≠ ∅ is left regular. By Lemma 3.14, we can express

α = A 1 … A r C 1 … C t a 1 … a r c 1 … c t ,

where A i ∩ Y ≠ ∅ , C k ⊆ X \ Y , a i ∈ Y , and c k ∈ X \ Y for all i ∈ I = { 1 , … , r } and k ∈ K = { 1 , … , t } . If ∣ A i ∩ { a 1 , … , a r } ∣ = 0 or ∣ A i ∩ { a 1 , … , a r } ∣ ≥ 2 for some i ∈ I , then ∣ Y α 2 ∣ < ∣ Y α ∣ , which contradicts Y α = Y α 2 . Thus, ∣ A i ∩ { a 1 , … , a r } ∣ = 1 for all i ∈ I . Furthermore, if ∣ C k ∩ { c 1 , … , c t } ∣ = 0 or ∣ C k ∩ { c 1 , … , c t } ∣ ≥ 2 for some k ∈ K , then ∣ im α 2 ∣ < ∣ im α ∣ , which contradicts im α = im α 2 . Thus, ∣ C k ∩ { c 1 , … , c t } ∣ = 1 for all k ∈ K . Hence, im α ⊆ dom α , α ∣ im α is injective, and im ( α ∣ im α \ Y ) ⊆ X \ Y . Therefore, α is right regular.

Conversely, assume that α ≠ ∅ is right regular. Again, we can write

α = A 1 … A r C 1 … C t a 1 … a r c 1 … c t ,

where A i ∩ Y ≠ ∅ , C k ⊆ X \ Y , a i ∈ Y , and c k ∈ X \ Y for all i ∈ I = { 1 , … , r } and k ∈ K = { 1 , … , t } . Since im α ⊆ dom α , a 1 , … , a r belong to A i for some i ∈ I . Since α ∣ im α is injective, ∣ A i ∩ { a 1 , … , a r } ∣ = 1 and c 1 , … , c t ∉ A i for all i ∈ I . Thus, c 1 , … , c t ∈ C k for some k ∈ K , which implies that ∣ C k ∩ { c 1 , … , c t } ∣ = 1 for all k ∈ K . Therefore, there exist permutations σ on I and φ on K , for which a i ∈ A i σ and c k ∈ C k φ for all i ∈ I and k ∈ K . Thus,

α 2 = A 1 … A r C 1 … C t a 1 σ … a r σ c 1 φ … c t φ ,

where Y α 2 = { a 1 σ , … , a σ } = { a 1 , … , a r } = Y α and im α 2 = { a 1 σ , … , a r σ , c 1 φ , … , c t φ } = { a 1 , … , a r , c 1 , … , c t } = im α . Therefore, α is left regular.□

Note that for a left regular element α in P T ¯ ( X , Y ) , where im α is finite, by Lemma 3.14, we can express this α as follows:

A 1 … A r C 1 … C t a 1 … a r c 1 … c t ,

where A i ∩ Y ≠ ∅ , C k ⊆ X \ Y , a i ∈ Y , and c k ∈ X \ Y for all i ∈ I = { 1 , … , r } and k ∈ K = { 1 , … , t } . Let Y ′ = im α ∩ Y = { a 1 , … , a r } and X ′ = im α \ Y α = { c 1 , … , c t } . By the proof of Theorem 3.16, ∣ A i ∩ Y ′ ∣ = 1 and ∣ C k ∩ X ′ ∣ = 1 for all i ∈ I and k ∈ K . Hence, we can express α ∣ Y ′ and α ∣ X ′ as follows:

α ∣ Y ′ = a 1 σ … a r σ a 1 … a r and α ∣ X ′ = c 1 φ … c t φ c 1 … c t ,

where σ and φ are bijections on I and K , respectively. Therefore, α ∣ Y ′ and α ∣ X ′ are permutations on Y ′ and X ′ , respectively.

4 Numbers of regular and left regular elements

Throughout this section, X is assumed to be a finite set with n elements, while ∅ ≠ Y ⊆ X is assumed to contain m elements. First, we determine the number of elements in P T ¯ ( X , Y ) in the following proposition.

Proposition 4.1

∣ P T ¯ ( X , Y ) ∣ = ∑ i = 0 m ∑ j = 0 n − m m i n − m j m i n j , where ∣ X ∣ = n and ∣ Y ∣ = m .

Proof

Consider α ∈ P T ¯ ( X , Y ) . Let Y ¯ = dom α ∩ Y be such that ∣ Y ¯ ∣ = i . Since Y α ⊆ Y , each element of Y ¯ can be matched with only one element of Y , i.e., there are m choices for an image of each element in Y . In particular, there are m i ways of constructing Y ¯ for 0 ≤ i ≤ m . Hence, α ∣ Y ¯ : Y ¯ → Y has ∑ i = 0 m m i m i different forms. Let A ¯ = dom α ∩ ( X \ Y ) be such that ∣ A ¯ ∣ = j . Then, there are n choices for the images of each element of A ¯ . In addition, the number of ways of choosing A ¯ equals n − m j for 0 ≤ j ≤ n − m . Hence, α ∣ A ¯ : A ¯ → X has ∑ j = 0 n − m n − m j n j different forms. Therefore,

∣ P T ¯ ( X , Y ) ∣ = ∑ i = 0 m m i m i ⋅ ∑ j = 0 n − m n − m j n j = ∑ i = 0 m ∑ j = 0 n − m m i n − m j m i n j .

This completes the proof.□

Next, we determine the number of regular elements in P T ¯ ( X , Y ) using the characterization given by Theorem 3.1. Recall that for 1 ≤ k ≤ r and r ∈ N , the Stirling number of the second kind S ( r , k ) is the number of partitions of an r -set into k -blocks, which satisfies the recurrence relation S ( r , k ) = S ( r − 1 , k − 1 ) + k S ( r − 1 , k ) with S ( r , 1 ) = S ( r , r ) = 1 . Moreover, it is known that S ( r , k ) = 1 k ! ∑ i = 0 k ( − 1 ) i k i ( k − i ) r .

Theorem 4.2

The number of regular elements in P T ¯ ( X , Y ) is

∑ k = 1 m ∑ i = k m ∑ j = 0 n − m m k m i n − m j k ! S ( i , k ) ( n − m + k ) j + ∑ j = 1 n − m n − m j ( n − m ) j + 1 ,

where ∣ X ∣ = n and ∣ Y ∣ = m .

Proof

Consider a regular element α ∈ P T ¯ ( X , Y ) .We first examine the case where im α ∩ Y = Y ′ ≠ ∅ and ∣ Y ′ ∣ = k . Let Y ¯ = dom α ∩ Y . Since im α ∩ Y = Y α , we obtain that ∣ Y ¯ ∣ = i , where k ≤ i ≤ m . Since the number of surjective functions from Y ¯ to Y ′ is k ! S ( i , k ) and the number of ways of choosing Y ′ and Y ¯ are m k and m i , respectively, we obtain that α ∣ Y ¯ : Y ¯ → onto Y ′ has ∑ i = k m m k m i k ! S ( i , k ) different forms. Let A ¯ = dom α ∩ ( X \ Y ) be such that ∣ A ¯ ∣ = j . Then, there are n − m + k choices for assigning an image to each element of A ¯ . The number of choice of A ¯ is equal to n − m j and 0 ≤ j ≤ n − m , α ∣ A ¯ : A ¯ → ( X \ Y ) ∪ Y ′ has ∑ j = 0 n − m n − m j ( n − m + k ) j different forms. Since 1 ≤ k ≤ m , the number of regular elements that contain some elements of Y in their ranges is

∑ k = 1 m ∑ i = k m ∑ j = 0 n − m m k m i n − m j k ! S ( i , k ) ( n − m + k ) j .

We now consider the case im α ∩ Y = ∅ . Then, dom α ⊆ X \ Y and α : dom α → X \ Y . If dom α = ∅ , then α = ∅ , which is regular. If dom α ≠ ∅ , then there are n − m choices for assigning an image to each element of dom α . The number of choice of dom α ⊆ X \ Y is equal to n − m j and 0 ≤ j ≤ n − m . So, in this case, α has

∑ j = 1 n − m n − m j ( n − m ) j + 1

different forms. Therefore, the number of regular elements in P T ¯ ( X , Y ) is

∑ k = 1 m ∑ i = k m ∑ j = 0 n − m m k m i n − m j k ! S ( i , k ) ( n − m + k ) j + ∑ j = 1 n − m n − m j ( n − m ) j + 1 ,

as required.□

Finally, we investigate the numbers of left regular and right elements in P T ¯ ( X , Y ) . Since X is a finite set, left regular and right regular elements coincide by Theorem 3.16. Thus, it suffices to enumerate the left regular elements.

Lemma 4.3

Let ∣ X ∣ = n , ∣ Y ∣ = m and α ∈ P T ¯ ( X , Y ) be left regular. If im α ∩ Y ≠ ∅ , then α ∣ Y : ( dom α ∩ Y ) → Y has ∑ r = 1 m ∑ s = 0 m − r m r m − r s r ! r s different forms.

Proof

Let im α ∩ Y = Y ′ = { a 1 , … , a r } . Then, we can write

α ∣ Y ′ = a 1 σ … a r σ a 1 … a r ,

where σ is a bijection on { 1 , … , r } and so α ∣ Y ′ is a permutation on Y ′ . Since the number of permutations on Y ′ is r ! and the number of ways of choosing Y ′ is equal to m r , we obtain that α ∣ Y ′ : Y ′ → Y ′ has m r r ! forms. Let Y * = ( dom α ∩ Y ) \ Y ′ be such that ∣ Y * ∣ = s . Then, Y * has ∑ s = 0 m − r m − r s different forms, and there are r choices of placing each element of Y * in A 1 , … , A r . Since 1 ≤ r ≤ m , we conclude that α ∣ Y has ∑ r = 1 m ∑ s = 0 m − r m r m − r s r ! r s different forms.□

Lemma 4.4

Let ∣ X ∣ = n , ∣ Y ∣ = m , α ∈ P T ¯ ( X , Y ) be left regular, and X ′ = im α \ Y α . Then, α ∣ X ′ : X ′ → X ′ has ∑ t = 0 n − m n − m t t ! different forms.

Proof

Suppose that ∣ X ′ ∣ = t . Then, α ∣ X ′ is a permutation on X ′ , which has t ! forms. Since the number of ways of choosing X ′ is n − m t and 0 ≤ t ≤ n − m , α ∣ X ′ has ∑ t = 0 n − m n − m t t ! different forms.□

Theorem 4.5

The number of left regular elements in P T ¯ ( X , Y ) is

∑ r = 1 m ∑ s = 0 m − r ∑ t = 0 n − m ∑ j = 0 n − m − t m r m − r s n − m t n − m − t j r ! t ! r s ( r + t ) j

+ ∑ t = 1 n − m ∑ j = 0 n − m − t n − m t n − m − t j t ! t j + 1 ,

where ∣ X ∣ = n and ∣ Y ∣ = m .

Proof

Let α ∈ P T ¯ ( X , Y ) be left regular. Assume that Y ′ = Y α = { a 1 , … , a r } and X ′ = im α \ Y α = { c 1 , … , c t } . When Y ′ ≠ ∅ , we can write

α = A 1 … A r C 1 … C t a 1 … a r c 1 … c t ,

where C k ⊆ X \ Y ∣ A i ∩ { a 1 , … , a r } ∣ = 1 , and ∣ C k ∩ { c 1 , … , c t } ∣ = 1 for all i = 1 , … , r and k = 1 , … , t . Let Y * = ( dom α ∩ Y ) \ Y ′ be such that ∣ Y * ∣ = s . By Lemma 4.3, α ∣ Y : ( dom α ∩ Y ) → Y has ∑ r = 1 m ∑ s = 0 m − r m r m − r s r ! r s different forms. By Lemma 4.4, α ∣ X ′ : X ′ → X ′ has ∑ t = 0 n − m n − m t t ! forms. Hence, α ∣ Y ∪ X ′ has ∑ r = 1 m ∑ s = 0 m − r ∑ t = 0 n − m m r m − r s n − m t r ! t ! r s different forms. Let X * = dom α \ ( Y ∪ X ′ ) be such that ∣ X * ∣ = j . Then, there are r + t ways of placing each element of X * in A 1 , … , A r and C 1 , … , C t , and X * has n − m − t j forms, where 0 ≤ j ≤ n − m − t . Hence, the number of left regular elements in this case is

∑ r = 1 m ∑ s = 0 m − r ∑ t = 0 n − m ∑ j = 0 n − m − t m r m − r s n − m t n − m − t j r ! t ! r s ( r + t ) j .

We now consider the case Y ′ = ∅ . If X ′ = ∅ , then α = ∅ , which is left regular. If X ′ ≠ ∅ , then by Lemma 4.4, α ∣ X ′ : X ′ → X ′ has ∑ t = 0 n − m n − m t t ! different forms. So, the number of left regular elements in this case is

∑ t = 1 n − m ∑ j = 0 n − m − t n − m t n − m − t j t ! t j .

Therefore, the number of left regular elements in P T ¯ ( X , Y ) is

∑ r = 1 m ∑ s = 0 m − r ∑ t = 0 n − m ∑ j = 0 n − m − t m r m − r s n − m t n − m − t j r ! t ! r s ( r + t ) j + ∑ t = 1 n − m ∑ j = 0 n − m − t n − m t n − m − t j t ! t j + 1 .

This completes the proof.□

5 Abundance of P T ¯ ( X , Y )

In a semigroup S , elements a and b belonging to S are ℒ * -related in S if and only if a and b are related by Green’s relation ℒ in some oversemigroup of S . The relation ℛ * is defined dually. A semigroup S is considered abundant if every ℒ * -class and every ℛ * -class contains an idempotent. It is known that regular semigroups are abundant, and ℒ = ℒ * and ℛ = ℛ * . In this section, we demonstrate that P T ¯ ( X , Y ) is abundant, even though it is not regular. Here, i d A denotes the identity map on the set A . The characterizations of the relations ℒ and ℛ on P ( X ) , and ℒ * and ℛ * on any semigroup S are discussed in Lemmas 5.1 and 5.2, respectively.

Lemma 5.1

[27] Let α , β ∈ P ( X ) . Then,

( α , β ) ∈ ℒ if and only if im α = im β ;
( α , β ) ∈ ℛ if and only if ker α = ker β .

Lemma 5.2

[28] Let S be a semigroup. Then,

ℒ * = { ( a , b ) ∈ S × S ∣ ( ∀ s , t ∈ S 1 ) a s = a t ⇔ b s = b t } , ℛ * = { ( a , b ) ∈ S × S ∣ ( ∀ s , t ∈ S 1 ) s a = t a ⇔ s b = t b } .

For the semigroup P T ¯ ( X , Y ) , the characterizations of the relations ℒ * and ℛ * can be demonstrated in the following two lemmas:

Lemma 5.3

Let α , β ∈ P T ¯ ( X , Y ) . Then, ( α , β ) ∈ ℒ * if and only if im α = im β .

Proof

Assume that im α = im β . Then, according to Lemma 5.1(1), α and β are ℒ -related in P ( X ) . Therefore, α and β are ℒ * -related in P T ¯ ( X , Y ) .

Conversely, assuming ( α , β ) ∈ ℒ * and defining γ = i d im α , we evidently have im γ = im α and α γ = α . Applying the characterization of the relation ℒ * from Lemma 5.2 (with α , β in the roles of a , b and γ and the identity in the roles of s and t , respectively), we can conclude that β γ = β and im β = im ( β γ ) = ( im β ∩ dom γ ) γ ⊆ im γ = im α . Similarly, im α ⊆ im β , from which follows im α = im β .□

Lemma 5.4

Let α , β ∈ P T ¯ ( X , Y ) . Then, ( α , β ) ∈ ℛ * if and only if ker α = ker β .

Proof

Assume that ker α = ker β . Then, according to Lemma 5.1(2), α and β are ℛ -related in P T ¯ ( X , Y ) . Therefore, α and β are ℛ * -related in P T ¯ ( X , Y ) .

Conversely, assume that ( α , β ) ∈ ℛ * . We begin by demonstrating that dom α = dom β . Since i d dom α α = α , according to Lemma 5.2, it follows that i d dom α β = β . Consequently, dom β = dom ( i d dom α β ) ⊆ dom ( i d dom α ) = dom α . Similarly, dom α ⊆ dom β , leading to the conclusion that dom α = dom β . Now, consider ( a , b ) ∈ ker α . This implies that a α = b α , and two cases arise.

Case 1: a , b ∈ Y or a , b ∈ X \ Y . Let Y \ { a , b } = { y i } , X \ ( Y ∪ { a , b } ) = { x j } , and define γ ∈ P T ¯ ( X , Y ) as follows:

γ = { a , b } y i x j a y i x j .

We observe that γ α = α , and then, by Lemma 5.2, γ β = β . Hence, b β = b γ β = a β , which implies ( a , b ) ∈ ker β .

Case 2: a ∈ Y and b ∈ X \ Y . Let Y \ a = y i , X \ ( Y ∪ b ) = x j , and define γ as explained in Case 1. Utilizing the same proof as presented in Case 1, we can establish that ( a , b ) ∈ ker ( β ) . Moreover, we similarly deduce ker β ⊆ ker α , thereby implying ker α = ker β , as needed.□

Using Lemmas 5.3 and 5.4, we obtain the following:

Theorem 5.5

The semigroup P T ¯ ( X , Y ) is an abundant semigroup.

Proof

Let α ∈ P T ¯ ( X , Y ) . Then, we have that i d im α is an idempotent in the ℒ * -class of α . Moreover, we can express

α = A i B j C k a i b j c k ,

where A i ∩ Y ≠ ∅ , B j , C k ⊆ X \ Y , a i , b j ∈ Y , and c k ∈ X \ Y , for all i , j , and k . For each i , j , and k , choose a i ′ ∈ A i ∩ Y , b j ′ ∈ B j , and c k ′ ∈ C k and let

γ = A i B j C k a i ′ b j ′ c k ′ .

Then, γ is an idempotent in P T ¯ ( X , Y ) with ker α = ker γ , i.e., γ is in the ℛ * -class of α . Therefore, P T ¯ ( X , Y ) is abundant.□

Funding information: This study was supported by Thammasat University Research Fund, Contract No. TUFT 3/2566.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The authors state that there are no conflicts of interest.

References

[1] J. von Neumann, On regular rings, Proc. Natl. Acad. Sci. USA 22 (1936), no. 12, 707–713. 10.1073/pnas.22.12.707Search in Google Scholar PubMed PubMed Central

[2] W. D. Munn and R. Penrose, A note on inverse semigroups, Math. Proc. Cambridge Philos. Soc. 51 (1955), 396–399. 10.1017/S030500410003036XSearch in Google Scholar

[3] T. E. Hall, On regular semigroups, J. Algebra 24 (1973), 1–24. 10.1016/0021-8693(73)90150-6Search in Google Scholar

[4] A. G. Pierre, The structure of regular semigroups, I: A representation, Semigroup Forum 8 (1974), 177–183. 10.1007/BF02194760Search in Google Scholar

[5] A. G. Pierre, The structure of regular semigroups, II: Cross-connections, Semigroup Forum 8 (1974), 254–259. 10.1007/BF02194766Search in Google Scholar

[6] A. G. Pierre, The structure of regular semigroups, III: The reduced case, Semigroup Forum 8 (1974), 260–265. 10.1007/BF02194767Search in Google Scholar

[7] K. S. S. Nambooripad, Structure of regular semigroups, I, Fundamental regular semigroups, Semigroup Forum 9 (1974), 354–363. 10.1007/BF02194864Search in Google Scholar

[8] R. Feigenbaum, Regular semigroup congruences, Semigroup Forum 17 (1979), 373–377. 10.1007/BF02194336Search in Google Scholar

[9] K. S. S. Nambooripad, The natural partial order on a regular semigroup, Proc. Edinb. Math. Soc. 23 (1980), 249–260. 10.1017/S0013091500003801Search in Google Scholar

[10] P. M. Edwards, Fundamental semigroups, Proc. Edinb. Math. Soc. 99 (1985), no. 3–4, 313–317. 10.1017/S0308210500014323Search in Google Scholar

[11] A. H. Clifford and G. B. Preston, The Algebraic Theory of Semigroups, Vol. I, Mathematical Surveys and Monographs, No. 7, American Mathematical Society, Providence, 1961. 10.1090/surv/007.1Search in Google Scholar

[12] I. Kiss, On a generalization of a right [left] regular element of a semigroup, Acta Math. Hungar. 23 (1972), 101–103. 10.1007/BF01889906Search in Google Scholar

[13] S. Bogdanović and M. Ćirić, A note on left regular semigroups, Publ. Math. Debrecen 48 (1996), 285–291. 10.5486/PMD.1996.1638Search in Google Scholar

[14] N. Kehayopulu and M. Tsingelis, On left regular ordered semigroups, Southeast Asian Bull. Math. 25 (2002), 609–615. 10.1007/s100120200005Search in Google Scholar

[15] N. Kehayopulu and M. Tsingelis, On left regular and intra-regular semigroups, Math. Slovaca 64 (2014), 1123–1134. 10.2478/s12175-014-0263-1Search in Google Scholar

[16] C. G. Doss, Certain equivalence relations in transformation semigroups, MSc thesis, University of Tennessee, Tennessee, 1955. Search in Google Scholar

[17] K. D. Magill Jr., Subsemigroups of S(X), Math. Japon. 11 (1966), 109–115. Search in Google Scholar

[18] S. Nenthein, P. Youngkhong, and Y. Kemprasit, Regular elements of some transformation semigroups, Pure Math. Appl. 16 (2005), no. 3, 307–314. Search in Google Scholar

[19] P. Honyam and J. Sanwong, Semigroups of transformations with invariant set, J. Korean Math. Soc. 48 (2011), no. 2, 289–300. 10.4134/JKMS.2011.48.2.289Search in Google Scholar

[20] W. Choomanee, P. Honyam, and J. Sanwong, Regularity in semigroups of transformations with invariant sets, Int. J. Pure Appl. Math. 87 (2013), 151–164. 10.12732/ijpam.v87i1.9Search in Google Scholar

[21] L. Sun and L. Wang, Natural partial order in semigroups of transformations with invariant set, Bull. Aust. Math. Soc. 87 (2013), 94–107. 10.1017/S0004972712000287Search in Google Scholar

[22] P. Honyam and J. Sanwong, Semigroups of linear transformations with invariant subspaces, Int. J. Algebra 6 (2012), no. 8, 375–386. Search in Google Scholar

[23] L. Sun and J. Sun, A note on naturally ordered semigroups of transformations with invariant set, Bull. Aust. Math. Soc. 91 (2015), 264–267. 10.1017/S0004972714000860Search in Google Scholar

[24] Y. Chaiya, Natural partial order and finiteness conditions on semigroups of linear transformations with invariant subspaces, Semigroup Forum 99 (2019), 579–590. 10.1007/s00233-019-10012-5Search in Google Scholar

[25] R. Chinram and S. Baupradist, Magnifying elements of semigroups of transformations with invariant set, Asian-Eur. J. Math. 14 (2019), no. 4, 1950056, DOI: https://doi.org/10.1142/S1793557119500566. 10.1142/S1793557119500566Search in Google Scholar

[26] M. Sarkar and S. N. Singh, On certain semigroups of transformations with an invariant set, Asian-Eur. J. Math. 15 (2022), no. 11, 2250198, DOI: https://doi.org/10.1142/S1793557122501984. 10.1142/S1793557122501984Search in Google Scholar

[27] J. M. Howie, Fundamentals of Semigroup Theory, London Mathematical Society Monographs, New Series, 12, Oxford Science Publications, The Clarendon Press, Oxford University Press, New York, 1995. Search in Google Scholar

[28] E. S. Lyapin, Semigroups, American Mathematical Society, Providence, 1963. Search in Google Scholar

Received: 2023-01-18

Revised: 2023-10-07

Accepted: 2023-10-13

Published Online: 2023-11-11

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https://doi.org/10.1515/math-2023-0142

Keywords for this article

partial transformation semigroup; regularity; left regularity; right regularity; abundance

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