1. Introduction and Preliminaries
Let
S be a semigroup. An element
x of
S is called left regular if
for some
. A right regular element is defined dually. Denote by
and
the sets of all left regular elements and right regular elements of
S, respectively. Left and right regular elements are important in semigroup theory because they play a key role in the study of regular semigroups, which are semigroups in which every element is both left and right regular. Regular semigroups have many interesting properties and are used in various areas of mathematics, including algebra, topology and theoretical computer science. An element
x of
S is called left (right) magnifying if there is a proper subset
M of
S satisfying
(
). In 1963, Ljapin [
1] studied the notion of right and left magnifying elements of a semigroup. Several years later, Migliorini introduced the concepts of the minimal subset related to a magnifying element of
S in [
2,
3]. Gutan [
4] researched semigroups with strong and nonstrong magnifying elements in 1996. Later, he proved that every semigroup with magnifying elements is factorizable in [
5].
Let
be the total transformation semigroup on a nonempty set
X. It is well known that
is a regular semigroup. Moreover, every semigroup is isomorphic to a subsemigroup of some total transformation semigroups. The most basic mathematical structures are transformation semigroups. In 1952, Malcev [
6] characterized ideals of
. Later, Miller and Doss [
7] studied its group
-classes and its Green’s relations. The generalization of these studies is the focus of this paper.
In 1975, Symons [
8] considered a subsemigroup of
defined by
where
Y is a nonempty subset of
X. He determined all the automorphisms of
. In 2005, Nenthein et al. [
9] described regular elements in
and counted the number of all regular elements of
when
X is a finite set. They described such a number in terms of
and their related Stirling numbers. A few years later, Sanwong and Sommanee [
10] studied regularity and Green’s relations for the semigroup
. They determined when
becomes a regular semigroup. Moreover, they gave a class of maximal inverse subsemigroups of
in 2008. After that, they proved that the set
is the largest regular subsemigroup of
and determined its Green’s relations. In [
11], Sanwong described Green’s relations and found all maximal regular subsemigroups of
. In 2009, maximal and minimal congruences on
were considered. Sanwong et al. [
12] found that
has only one maximal congruence if
X is a finite set. They generalized [
13] Theorem 3.4 for
Y being infinite. Furthermore, characterizations of all minimal congruences on
were given. In the same year, Sun [
14] proved that while the semigroup
is not left abundant, it is right abundant. Later in 2016, Lei Sun and Junling Sun [
15] investigated the natural partial order on
. Moreover, they determined the maximal elements and the minimal elements of
.
Consider the semigroup
of transformations that leave
Y invariant. In 1966, Magill [
16] constructed and discussed the semigroup
. In fact, if
, then
. Later in [
8], automorphism groups of a semigroup
were given by Symons. In 2005, Nenthein et al. [
9] characterized regularity for
. In addition, they found the number of regular elements in
for a finite set
X. Honyam and Sanwong [
17] studied its ideals, group
-classes and Green’s relations on
. Furthermore, they described when
is isomorphic to
for some set
A. A few years later, the left, right regular and intra-regular elements of a semigroup
were discussed by Choomanee, Honyam and Sanwong [
18]. Moreover, when
X is finite, they calculated the number of left regular elements in
. In [
19], natural partial orders on the semigroup
were considered by Sun and Wang. Moreover, they investigated left and right compatible elements with respect to this partial oder. Finally, they described the abundance of
. In [
20], all elements in the semigroup
that are left compatible with the natural partial order were studied. Left and right magnifying elements of
were given by Chiram and Baupradist in [
21]. In a recent study, Punkumkerd and Honyam [
22] provided a characterization of left and right magnifying elements on the semigroup
.
denotes the set of all partial transformations
from a subset of
X to
X and
, where
is the domain of
. Their results have shown to be more general than the previous findings from [
21].
In
Section 2, we consider left and right magnifying elements of
and
. We prove that each left magnifying element in
is not regular. Furthermore, we show that every left and right magnifying element in
is regular. In
Section 3 and
Section 4, we focus on left and right regularity on
and
. We show that every left magnifying element in
is a right regular element. Every right magnifying element is a left regular element. As [
10] determined when
becomes a regular semigroup, we also characterize whenever
and
is a left (right) regular semigroup.
Note that throughout this paper, we will write mappings from the right, rather than and compose that the left to the right, rather than where and . For each , we denote the set by and is the set for a subset . Then, it is obvious that is a partition of X.
2. Magnifying Elements
In this section, we focus on characterizations of left magnifying elements and right magnifying elements in and . The relationships between magnifying elements and regular elements are given.
Theorem 1 ([
23])
. Let . Then, α is right magnifying if and only if α is surjective but not injective and is such that for all and for some . Theorem 2 ([
23])
. A mapping is left magnifying if and only if α is injective but not surjective. Theorem 3 ([
23])
. Let . If and , then α is left magnifying if and only if α is injective. Remark 1. For each , we note from Theorem 1 that α is right magnifying if and only if and is not injective.
Theorem 4. Let and . Then, α is left magnifying in a semigroup if and only if α is injective.
Proof. Assume that
is left magnifying. Suppose that
M is a proper subset of
satisfying
. Let
be such that
. If
, then
contains exactly one element. Thus,
has no proper subset
M such that
. This is a contradiction. Therefore,
. Let
. Define
by
It is verifiable that
. From
, there exists
such that
. Suppose that
. Then,
, which is a contradiction. Hence,
and so
is injective.
Suppose that
is injective. Then, we choose
and let
M be the set
. From
, we have
. It follows from our assumption that every
, there is a unique
satisfying
. Let
. We define
by
It is verifiable that
. Now, let
. Thus,
. From
being injective and
, we obtain
. Therefore,
. Hence,
and so
. This implies that
is left magnifying. □
The set of natural numbers is represented by the letter . Additionally, we denote the set of even natural numbers and the set of all odd natural numbers greater than 3 by and , respectively.
Example 1. Let and . Define by for all . Then, . Clearly, α is injective. From Theorem 4, we obtain that α is left magnifying. We will show that α is not a regular element. Suppose that α is a regular element in . Thus, there exists such that . Consider . Thus, . Hence, , which is a contradiction. So α is not regular element.
From the above example, we will verify that in , each left magnifying element is not a regular element.
Theorem 5. If , then every left magnifying element of is not regular.
Proof. Assume that . Let be a left magnifying element. From Theorem 4, we obtain that is injective. Suppose that is a regular element in . Then, there exists such that . Since , we choose . Therefore, and then . Since is injective, we have . This is a contradiction with . So is not regular. □
Theorem 6. Every right magnifying element of a semigroup is regular.
Proof. Let be a right magnifying element of . By Remark 1, we obtain that . It follows that . From , we have . This means that and so is a regular element of . □
The following example shows that there exists an element in some which is regular but it is not right magnifying.
Example 2. Let and . Define byNote that . It is easy to see that and . Thus, α is regular but it is not right magnifying. In the rest of this section, we consider magnifying elements in .
Lemma 1 ([
21])
. Let . Then, α is a right magnifying element if and only if α is surjective but not injective such that for all . Lemma 2 ([
21])
. Let . Then, α is a left magnifying element if and only if α is injective but not surjective such that for all . Lemma 3 ([
9])
. Let . Then, α is a regular element if and only if . Theorem 7. Every left magnifying element of a semigroup is regular.
Proof. Suppose that is left magnifying. We will show that . Clearly, . Let . Then, there exists such that . Thus, by Lemma 2. This implies that . From Lemma 3, we obtain that is regular. □
Theorem 8. Every right magnifying element of a semigroup is regular.
Proof. Suppose that is right magnifying. We will show that . Clearly, . Let . By Lemma 1, we have . Thus, there exists . Hence, . Therefore, . From Lemma 3, we obtain is regular. □
Example 3. Let α be defined in Example 2. It is clear that and α is neither injective nor surjective. Since , this means that . Hence, α is regular, while it is neither a left nor right magnifying element in .
3. Left Regular and Right Regular Elements in
Now, we start with the characterizations of left regular and right regular elements in . Moreover, we determine whenever becomes a left regular semigroup and a right regular semigroup, respectively.
Theorem 9. Let . Then, the following statements are equivalent.
- (1)
α is left regular.
- (2)
.
- (3)
for any , .
Proof.
Suppose that is left regular. Thus, there exists satisfying . This implies that . Thus, .
Suppose that and let . Thus, there is satisfying . By assumption, we have . Thus, there exists such that ; that is, . It follows that . Therefore, .
Assume that (3) holds. For , there is a unique satisfying . From assumption, we have . So there exists such that . Define by for all . It is obvious that . Let . This implies that . Hence, and so is left regular. □
If we replace Y with X in Theorem 9, we have the following corollary.
Corollary 1. Let . Then, α is left regular if and only if for each .
Proof. By taking , we obtain and . By Theorem 9(3), we obtain that is regular if and only if for each . □
Theorem 10. Let . Then, α is right regular if and only if is injective.
Proof. Assume that
is right regular. Then, there exists
such that
. We will show that
is injective. Let
be such that
. Thus, there exist
such that
and
. We obtain that
Therefore,
is injective.
Conversely, suppose
is injective. Let
. Then, there exists a unique
such that
. We choose
. Define
by
From
, we obtain that
. Let
. Note that
and
where
. Since
is injective, we have
. So
. Hence,
and so
is right regular. □
Corollary 2. Let . Then, α is right regular if and only if is injective.
From Theorems 4 and 10, we obtain the following corollary immediately.
Corollary 3. For , every left magnifying element of a semigroup is right regular.
Corollary 4. Every right magnifying element of a semigroup is left regular.
Proof. Let be a right magnifying element. By Remark 1, we have . It follows that . Hence, and so is left regular. □
Example 4. Let and . Define byThen, . Clearly, α is not injective. We see that . This means that is an injection. This means that α is right regular but it is not left magnifying. Example 5. Let and . Define byClearly, . We see that and . Hence, α is a left regular element but it is not right magnifying. Notice that for , we obtain that is left and right regular. Now, we consider the other case.
Theorem 11. Let be such that . Then, is a right regular semigroup if and only if .
Proof. If
, then
and
is a right regular semigroup. Assume that
is a right regular semigroup and suppose
. Let
be distinct elements and
. Define
by
Then,
. However,
is not injective. Thus,
is not right regular, which is a contradiction. Hence,
. □
Theorem 12. Let be such that . Then, is a left regular semigroup if and only if .
Proof. If
, then
and
is a left regular semigroup. Assume that
is a left regular semigroup and
. Let
be distinct elements and
. Define
by
Then,
. Note that
and
. So
. Therefore,
is not left regular. This is a contradiction. Hence,
. □
Corollary 5. Every left regular element of a semigroup is regular.
Proof. We first note that is the largest regular subsemigroup of . Let be a left regular element of . It follows from Theorem 9(2) that . Thus, and so . Therefore, is a regular element of . □
Example 6. Let and . Define byClearly, . Consider and . Then, α is regular but it is not left regular. Example 7. Let and . Define byIt is verifiable that . We obtain that is injective and . Thus, α is right regular but it is not regular. Finally, we consider the set of all left regular elements and the set of all right regular elements of . We begin with the following example.
Example 8. Let and . We consider the mappings and . We note that . Moreover, and are injective; that is . Clearly, . This implies that is not injective and so . In this case, we obtain that is not a semigroup.
Theorem 13. Let . Then, is a semigroup if and only if .
Proof. Suppose that
. Let
be distict elements. Define
by
We see that
and
is injective. Define
by
Then,
and
is injective. Since
and
, we obtain
. From
, we conclude that
is not injective. Therefore,
and
is not closed.
Assume that . If , then is a right regular semigroup. Therefore, is a semigroup. Suppose that . Let and . We will show that is injective. Let be such that . If or is a constant mapping, then is a constant mapping. Thus, is injective. Suppose that and are not constant mappings. Then, . We observe that . If , then since is injective. Note that . Then, since is injective. This is a contradition. Hence, . So is injective. Therefore, is closed. □
Theorem 14. Let . Then, is a semigroup if and only if .
Proof. Suppose that
. Let
be distinct elements. Define
by
And we define
by
Then,
and
. Thus,
. It is easy to verify that
such that
and
. Clearly,
. Hence,
. So
is not left regular.
Conversely, suppose . If , then is a left regular semigroup. We have is a semigroup. Assume that . Let . If or , then is a constant mapping. Suppose that . Then, and so it is a left regular element. From being a left regular element, we obtain and . This implies that and . Similarly, and . It is easy to verify that and . Hence, and . Therefore, is a left regular element. □
Theorem 15. If Y is finite, then .
Proof. Assume that Y is finite. Let be a left regular element of . Then, . It follows that ; that is, . From , we obtain is surjective. Since and Y is a finite set, we obtain is an injection. So is right regular.
Assume that is right regular. Thus, is injective and also is surjective since is finite. This means that . We see that . Hence, and so . Therefore, is a left regular element of . □
Next, the cardinality of right regular elements in the semigroup are investigated when X is finite.
Theorem 16. Let and . Then,where . Proof. By Theorem 15, we have
. This implies that
. Let
and
. From
Y being finite, we have
is bijective for all
by Theorem 10. Notice that the number of image sets in
Y of cardinality
k is equal to
. Since there are
ways of partitioning the remaining
elements into
k subsets, we obtain
. Therefore,
□
4. Left Regular and Right Regular Elements in
Theorem 17 ([
18])
. Let . Then, α is left regular if and only if and . Theorem 18 ([
18])
. Let . Then, α is right regular if and only if and where . Although the left and right regular elements of
were characterized in [
18], in this section we obtain the different results; see the following theorems.
Theorem 19. Let . Then, α is a left regular element if and only if for every , and for every , .
Proof. Assume that is a left regular element. Thus, for some . Let and let . Then, and . We see that and . Therefore, . Let and let . Then, , and . We note that and . Hence, .
Conversely, suppose the conditions hold. Let . Since is a partition of X, there is a unique satisfying . If , then and by our assumption. So, we choose satisfying . If , then since , we choose satisfying . Define by for all . Then, is well-defined and . Let . Then, there is a unique such that . Thus, . Therefore, ; that is, . So is left regular. □
Theorem 20. Let . Then, α is a right regular element if and only if is injective and .
Proof. Assume that is right regular. Thus, is also a right regular element in . From Corollary 2, we have is injective. Next, we will show that . Let . Then, for some . Thus, for some since is a right regular element in . If , then , which is a contradiction. Hence, and so .
Assume that is injective and . Let be defined in the converse part of Theorem 10; we note that . It is enough to verify that . Let . If , then by the definition of , we have . Assume that . There is a unique satisfying . If , then . By assumption, we have which is a contradiction. This means that . Hence, and so . □
Example 9. Let and . Define byThen, and . So . Moreover, we obtain and . It is clear that for every and for all . From Theorem 19, α is left regular. Note that . By Theorem 3, α is also not regular. Example 10. Let α be defined in Example 2. Then, and also . Thus, and α is regular. Note that and . Hence, α is not a left regular element of .
Example 11. Let α be defined in Example 5. Then, and so . Consider and is not injective. Hence, α is regular but not right regular in .
Example 12. Recall α from Example 4. Then, and so . We see that is injective and . From Theorem 20, we obtain α is right regular. Consider and . Hence, . From Theorem 3, we obtain α is not regular.
Theorem 21. The following statements are equivalent.
- (1)
.
- (2)
.
- (3)
.
Proof.
Assume that
. We will show that
. Suppose that
. Let
be distinct elements and
. Define
by
Claim that
. If
, then
. Moreover, if
, then
. Thus,
when
or
. Note that
and
. Clearly,
. Hence,
is not left regular. For the case
and
, we define
by
Then,
,
and
. It follows that
but
. We conclude that
, which is a contradiction. So
.
Conversely, assume that . Then, it is easy to verify that is a left regular semigroup. Hence, .
Assume that . We will show that . Suppose that . We consider from condition Then . Since and is not injective, we have is not right regular. Similarly, and is not injective. So is not right regular, which is a contradition. Hence, .
Conversely, suppose that . Then, it is clear that is a right regular semigroup. Hence, . □
Theorem 22. The following statements are equivalent.
- (1)
is a semigroup.
- (2)
is a semigroup.
- (3)
.
Proof.
. Suppose that . Let be distinct elements and . It is enough to consider only two cases.
Case 1: . Recall from Theorem 14, we have . Note that and . Clearly, and . Thus, is left regular; that is, . Similarly, . Consider and . Therefore, and so is not left regular; that is, . Hence, is not a semigroup.
Case 2:
. Assume that
. Define
by
Then,
. So
. Note that
,
and
. Clearly,
,
and
. Therefore,
is a left regular element of
. Define
by
If
, then
. Thus,
. If
, then
. So
. We can show that
. Then, we note that
and
. Clearly,
. Hence,
and so
is not a semigroup.
Conversely, suppose . Then, we have is a semigroup from Theorem 21(1).
. Suppose that . Let be distinct elements and . Recall and from the proof of . It is enough to show that and are right regular elements of . Clearly, and are injective. Consider and .
Then, and are right regular elements of ; that is, . Note that is not injective. We conclude that is not right regular. Thus, and so is not a semigroup.
Conversely, suppose . Then, we have is a semigroup from Theorem 21(2). □