On semigroups of transformations that preserve a double direction equivalence

Hui Chen; Xin Liu; Shoufeng Wang

doi:10.1515/math-2022-0606

Article Open Access

On semigroups of transformations that preserve a double direction equivalence

Hui Chen , Xin Liu and Shoufeng Wang

Published/Copyright: July 21, 2023

Published by

Become an author with De Gruyter Brill

Submit Manuscript Author Information Explore this Subject

$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

For a non-empty set X , denote the full transformation semigroup on X by T ( X ) and suppose that E is an equivalence relation on X . Evidently,

T E ∗ ( X ) = { α ∈ T ( X ) ∣ ( x , y ) ∈ E if and only if ( x α , y α ) ∈ E for all x , y ∈ X }

is a subsemigroup of T ( X ) . In this article, we investigate Green relations, Green ∗ -relations and Green ∼ -relations, various kinds of regularities, ℱ -abundant and G -abundant elements and left and right magnifying elements in T E ∗ ( X ) . More specifically, we first obtain the necessary and sufficient conditions under which ℒ (respectively, ℒ ∗ , ℒ ˜ , ℛ , ℛ ∗ , and ℛ ˜ ) is (left, right) compatible, ℛ = ℛ ∗ or ℒ = ℒ ˜ . Then, we give the sufficient and necessary conditions such that T E ∗ ( X ) is left regular (respectively, right regular, completely regular, intra-regular, and completely simple). Finally, we characterize the ℱ -abundant (respectively, G -abundant) and left (respectively, right) magnifying elements in T E ∗ ( X ) .

Keywords: Green (∼-, ∗-) relations; (ℱ-, 𝒢-) abundant; (left, right, intra-) regular; (left, right) magnifying

MSC 2010: 20M20; 20M10

1 Introduction

Let S be a semigroup. As usual, we use E ( S ) to denote the set of idempotents in S . It is well known that the Green relations on S are defined as follows: for all a , b ∈ S , a ℒ b (respectively, a ℛ b , a J b ) if and only if S 1 a = S 1 b (respectively, a S 1 = b S 1 , S 1 a S 1 = S 1 b S 1 ), and ℋ = L ∩ R and D is the smallest equivalence containing ℒ and ℛ . Moreover, ℒ is right compatible and ℛ is left compatible, respectively. An element a in S is called regular if there exists x ∈ S such that a = a x a . Moreover, if every element in S is regular, then S is called regular. Recall that a semigroup S is regular if and only if every ℒ -class (or every ℛ -class) of S contains idempotents. A regular semigroup S is called orthodox (respectively, completely regular) if E ( S ) forms a subsemigroup of S (respectively, if every ℋ -class of S contains an idempotent). A completely simple semigroup is a completely regular semigroup that has no proper ideal. On the other hand, from Clifford and Preston [1] (see Page 121), a semigroup S is called left (respectively, right, completely, intra-) regular if a ℒ a 2 (respectively, a ℛ a 2 , a ℋ a 2 , a J a 2 ) for all a ∈ S .

As generalizations of usual Green relations, Green ∗ -relations were introduced and investigated in the 1970s (e.g., see Fountain [2,3]). Let S be a semigroup. Then, the relations ℒ ∗ and ℛ ∗ on S are defined as follows:

ℒ ∗ = { ( a , b ) ∈ S × S ∣ a x = a y if and only if b x = b y for all x , y ∈ S 1 } , ℛ ∗ = { ( a , b ) ∈ S × S ∣ x a = y a if and only if x b = y b for all x , y ∈ S 1 } .

Let ℋ ∗ = L ∗ ∩ R ∗ . Obviously, ℒ ∗ and ℛ ∗ are equivalences, and ℒ ∗ is right compatible and ℛ ∗ is left compatible, respectively. In view of Fountain [3], an element a in S is called right abundant (respectively, left abundant) if the ℒ ∗ -class (respectively, ℛ ∗ -class) containing a has an idempotent. Moreover, if an element a in S is both left and right abundant, then a is called abundant. Further, a semigroup S is called left abundant (respectively, right abundant, abundant) if each element in S is left abundant (respectively, right abundant, abundant).

Now, let S be a semigroup. Recall from El-Qallali [4] (also see Lawson [5]) that the Green ∼ -relations ℒ ˜ and ℛ ˜ on S are given as follows:

ℒ ˜ = { ( a , b ) ∈ S × S ∣ a e = a if and only if b e = b for all e ∈ E ( S ) } , ℛ ˜ = { ( a , b ) ∈ S × S ∣ e a = a if and only if e b = b for all e ∈ E ( S ) } .

One can show that ℒ ˜ and ℛ ˜ are equivalences on S , and

(1.1) ℒ ⊆ ℒ ∗ ⊆ ℒ ˜ and ℛ ⊆ ℛ ∗ ⊆ ℛ ˜ .

However, ℒ ˜ (respectively, ℛ ˜ ) may not be right compatible (respectively, left compatible). For more facts on Green relations and their generalizations, the reader can consult the previous study [6]. Again, from El-Qallali [4] and Lawson [5], an element a in S is left semiabundant (respectively, right semiabundant) if the ℛ ˜ -class (respectively, ℒ ˜ -class) containing a has an idempotent. Moreover, a semiabundant element of S is an element that is both left and right semiabundant. If every element of a semigroup S is left semiabundant (respectively, right semiabundant, semiabundant), then S is said to be left semiabundant (respectively, right semiabundant, semiabundant). By (1.1), regular elements are always abundant, and left abundant (respectively, right abundant, abundant) elements are always left semiabundant (respectively, right semiabundant, semiabundant).

Recently, Stokes [7] introduced the equivalences ℱ and G on a semigroup S as follows:

ℱ = { ( a , b ) ∈ S × S ∣ x a = x if and only if x b = x for all x ∈ S } , G = { ( a , b ) ∈ S × S ∣ a x = x if and only if b x = x for all x ∈ S } .

According to Stokes [7], an element a in S is called ℱ -abundant (respectively, G -abundant) in S if the ℱ -class (respectively, G -class) containing a has an idempotent. Moreover, if every element of a semigroup S is ℱ -abundant (respectively, G -abundant), then we call S ℱ -abundant (respectively, G -abundant). Among other things, Stokes [7] investigated some properties of ℱ -abundant and G -abundant elements in semigroups. In particular, it is shown that the full transformation semigroup on a non-empty set is G -abundant, and the partial transformation semigroup and the symmetric inverse semigroup on a non-empty set are both ℱ -abundant and G -abundant.

We observe that the notions of left and right magnifying elements of semigroups were introduced by Ljapin [8]. An element a of a semigroup S is called left (respectively, right) magnifying if there exists a proper subset M of S such that S = a M (respectively, S = M a ). Some research of magnifying elements in general semigroups can be seen in [8–10].

As is well known, the theory of transformation semigroups plays an important role in the study of algebraic theory of semigroups as the Cayley theorem for semigroups says that every semigroup is isomorphic to some transformation semigroup. The transformation semigroups have been extensively investigated in the literature (see the monographs [1] and [11,12] and the articles [13–24] for example). In particular, left and right magnifying elements in some generalized partial transformation semigroups and left and right regular elements of the semigroups of transformations preserving an equivalence relation and a cross-section are investigated in the texts in [13,14] and [18], respectively.

Now, let X be a non-empty set, T ( X ) be the full transformation semigroup on X , and E be an equivalence on X . In 2010, Deng et al. [15] introduced the subsemigroup T E ∗ ( X ) of T ( X ) as follows:

T E ∗ ( X ) = { α ∈ T ( X ) ∣ ( x , y ) ∈ E if and only if ( x α , y α ) ∈ E for all x , y ∈ X } .

Obviously, the identity transformation ι X on X is the identity of T E ∗ ( X ) . Since then, the semigroup T E ∗ ( X ) has been investigated by several authors, and many useful and interesting results have been obtained (see the texts, [15–17] and [19,24] and their references therein). In fact, the articles [15] and [16] investigated the Green relations and regularity, and the Green ∗ -relations and abundance of T E ∗ ( X ) , respectively, Sun and Sun [19] described the natural partial order on T E ∗ ( X ) and determined its compatible and maximal (respectively, minimal) elements, and the text [11] studied the E -inversive elements and showed that E -inversive elements coincide with regular elements in T E ∗ ( X ) . In 2018, Yonthanthum [24] considered the variant semigroups of T E ∗ ( X ) .

The aim of this article is to study T E ∗ ( X ) along this direction. In Section 2, we first recall some results that are necessary in this article. In particular, we show that the set Reg ( T E ∗ ( X ) ) consisting of the regular elements of T E ∗ ( X ) forms a subsemigroup of T E ∗ ( X ) . In Section 3, we describe the Green ∼ -relation ℒ ˜ and investigate left and right compatibility and equality of Green relations, Green ∗ -relations, and Green ∼ -relations. More concretely, we obtain the necessary and sufficient conditions under which ℒ (respectively, ℒ ∗ , ℒ ˜ , ℛ , ℛ ∗ and ℛ ˜ ) is (left, right) compatible, ℛ = ℛ ∗ or ℒ = ℒ ˜ in T E ∗ ( X ) and Reg ( T E ∗ ( X ) ) . Section 4 is devoted to various kinds of regularities. In particular, we give the sufficient and necessary conditions such that T E ∗ ( X ) (or Reg ( T E ∗ ( X ) ) ) is a left regular (respectively, right regular, completely regular, intra-regular, and completely simple) semigroup. In Section 5, we characterize the ℱ -abundant and G -abundant elements in T E ∗ ( X ) and Reg ( T E ∗ ( X ) ) . Finally, we give the characterizations of left and right magnifying elements in T E ∗ ( X ) in the last section. The results of this article not only extend and enrich those of the related literatures, but also provide some examples and counter-examples for the study of Green’s relations and their generalizations in the theory of abstract semigroups.

2 Preliminaries

In the remainder of this article, we always fix a non-empty set X , an equivalence relation E on X , and denote the set of E -classes of X by X ∕ E . Write

X α = { x α ∣ x ∈ X } , ker α = { ( x , y ) ∈ X × X ∣ x α = y α } , X \ U = { x ∈ X ∣ x ∉ U }

and use ∣ U ∣ to represent the cardinality of U for all U ⊆ X and α ∈ T ( X ) . Furthermore, we denote the universal relation X × X on X by ω X and the identity relation { ( x , x ) ∣ x ∈ X } on X by ε X , respectively. In this section, among other things, we shall recall some known results on the semigroup T E ∗ ( X ) , which will be used in the next sections. We first present the characterizations of Green relations on this semigroup. Denote

Z ( α ) = { A ∈ X ∕ E ∣ A ∩ X α = ∅ }

for every α ∈ T E ∗ ( X ) .

Lemma 2.1

(Theorems 2.1–2.3 and 2.5 in [15]) Let α , β ∈ T E ∗ ( X ) .

( α , β ) ∈ ℒ if and only if X α = X β .
( α , β ) ∈ ℛ if and only if ker α = ker β and ∣ Z ( α ) ∣ = ∣ Z ( β ) ∣ .
( α , β ) ∈ ℋ if and only if ker α = ker β and X α = X β .
( α , β ) ∈ J if and only if ∣ X α ∣ = ∣ X β ∣ and there exist ρ , τ ∈ T E ∗ ( X ) , such that for any A ∈ X ∕ E , we have A α ⊆ B β ρ and A β ⊆ C α τ for some B , C ∈ X ∕ E .

The following lemma provides the characterizations of Green ∗ -relations on T E ∗ ( X ) .

Lemma 2.2

(Theorems 3.1 and 3.2 in [16]) Let α , β ∈ T E ∗ ( X ) .

( α , β ) ∈ ℒ ∗ if and only if X α = X β .
( α , β ) ∈ ℛ ∗ if and only if ker α = ker β .
( α , β ) ∈ ℋ ∗ if and only if X α = X β and ker α = ker β .

Regular (respectively, right abundant, abundant) elements in T E ∗ ( X ) can be described as follows. Recall that a cross-section of an equivalence E on X is a subset U of X such that ∣ U ∩ A ∣ = 1 for all A ∈ X ∕ E .

Lemma 2.3

(Theorem 3.1 in [15], Theorem 4.3 in [16], and Lemma 2.3 in [17]) Let α ∈ T E ∗ ( X ) . Then,

α i s r e g u l a r ⇔ α i s r i g h t a b u n d a n t ⇔ α i s a b u n d a n t

⇔ A ∩ X α ≠ ∅ f o r a l l A ∈ X ∕ E ⇔ Z ( α ) = ∅ ⇔ X α c o n t a i n s a c r o s s - s e c t i o n o f E .

In particular, if α is idempotent, then A α ⊆ A for all A ∈ X ∕ E .

Denote the set of the regular elements in T E ∗ ( X ) by Reg ( T E ∗ ( X ) ) . We shall show that Reg ( T E ∗ ( X ) ) is a subsemigroup of T E ∗ ( X ) . To this aim, we need the following result.

Lemma 2.4

Let α ∈ T E ∗ ( X ) and R be a cross-section of E. If α is regular, then R α = { r α ∣ r ∈ R } is also a cross-section of E.

Proof

Let r , s ∈ R with ( r , s ) ∉ E . Then, ( r α , s α ) ∉ E by the fact that α ∈ T E ∗ ( X ) . On the other hand, let x ∈ X . Then, by the regularity of α and Lemma 2.3, there exists y ∈ X such that ( x , y α ) ∈ E . Since R is a cross-section of E , it follows that ( y , r ) ∈ E for some r ∈ R , whence ( y α , r α ) ∈ E . This implies that ( x , r α ) ∈ E . Thus, R α is also a cross-section of E .□

Lemma 2.5

Reg ( T E ∗ ( X ) ) forms a subsemigroup of T E ∗ ( X ) .

Proof

Let α , β ∈ Reg ( T E ∗ ( X ) ) . By Lemma 2.3, X α contains a cross-section of E . This together with the fact that β ∈ Reg ( T E ∗ ( X ) ) gives that X α β also contains a cross-section by Lemma 2.4, and so α β ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3 again.□

The following lemma gives a necessary and sufficient condition under which T E ∗ ( X ) = Reg ( T E ∗ ( X ) ) .

Lemma 2.6

(Theorem 3.2 in [15] and Theorem 4.6 in [16]) For the semigroup T E ∗ ( X ) ,

T E ∗ ( X ) i s r e g u l a r ⇔ T E ∗ ( X ) i s r i g h t a b u n d a n t ,

⇔ T E ∗ ( X ) i s a b u n d a n t ⇔ X ∕ E i s f i n i t e .

However, the following lemma tells us that T E ∗ ( X ) is always left abundant.

Lemma 2.7

(Theorem 4.1 in [16]) Every ℛ ∗ -class of T E ∗ ( X ) contains an idempotent. That is, T E ∗ ( X ) is left abundant. And so T E ∗ ( X ) is left semiabundant by (1.1).

To compare the relations ℛ ∗ and ℛ ˜ in T E ∗ ( X ) , we quote the following known result.

Lemma 2.8

([4], also see Theorem 3.7 in [6]) Let S be a semigroup and a , b ∈ S . If a and b are regular, then

a ℒ b ⇔ a ℒ ∗ b ⇔ a ℒ ˜ b , a ℛ b ⇔ a ℛ ∗ b ⇔ a ℛ ˜ b .

If a and b are left abundant, then a ℛ ∗ b if and only if a ℛ ˜ b .

In view of Lemmas 2.1, 2.2, 2.7, and 2.8, we have the following corollary.

Corollary 2.9

In the semigroup T E ∗ ( X ) , we have ℒ = L ∗ , ℋ = H ∗ , and ℛ ∗ = ℛ ˜ .

To give the relationship between the relations ℒ and ℛ in T E ∗ ( X ) and those in Reg ( T E ∗ ( X ) ) , we need to recall a result presented in Howie [12].

Lemma 2.10

(Proposition 2.4.2 in [12]) Let U be a semigroup, S a regular subsemigroup of U, and a , b ∈ S . Then, a ℒ b (respectively, a ℛ b ) in S if and only if a ℒ b (respectively, a ℛ b ) in U .

Lemmas 2.1, 2.3, 2.5, and 2.10 give the following useful lemma.

Lemma 2.11

Let α , β ∈ Reg ( T E ∗ ( X ) ) .

α ℒ β in Reg ( T E ∗ ( X ) ) ⇔ α ℒ β in T E ∗ ( X ) ⇔ X α = X β .
α ℛ β in Reg ( T E ∗ ( X ) ) ⇔ α ℛ β in T E ∗ ( X ) ⇔ ker α = ker β .
α ℋ β in Reg ( T E ∗ ( X ) ) ⇔ α ℋ β in T E ∗ ( X ) ⇔ ker α = ker β and X α = X β .

To end this section, we state a result on regularities of semigroups, which is useful in our discussions later.

Lemma 2.12

(Theorem 4.4.3 in [1]) Let S be a regular semigroup. Then, the following conditions are equivalent:

S is left regular.
S is right regular.
S is completely regular.

3 On Green relations, Green ∗ -relations, and Green ∼ -relations

In this section, we shall give some results on left and right compatibility and equality of Green relations, Green ∗ -relations, and Green ∼ -relations. As mentioned in Section 1, in any semigroup, the relations ℒ and ℒ ∗ (respectively, ℛ and ℛ ∗ ) are always right (respectively, left) compatible. One can ask the following question naturally: When are the relations ℒ and ℒ ∗ (respectively, ℛ and ℛ ∗ ) in T E ∗ ( X ) and Reg ( T E ∗ ( X ) ) left (respectively, right) compatible? To answer this question, we first give the following simple results. For convenience, we denote

J ( X ) = { α ∈ T ( X ) ∣ α is injective } and S ( X ) = { α ∈ T ( X ) ∣ α is bijetive } .

Proposition 3.1

For the semigroup T E ∗ ( X ) , the following statements are true:

T E ∗ ( X ) = T ( X ) if and only if E = ω X . In this case, Reg ( T E ∗ ( X ) ) = T ( X ) .
T E ∗ ( X ) = J ( X ) if and only if E = ε X . In this case, Reg ( T E ∗ ( X ) ) = S ( X ) .

Proof

(1) The sufficiency is obvious. On the other hand, if E ≠ ω X , then X has at least two E -classes. Clearly, in this case, any constant transformation on X is not in T E ∗ ( X ) , and so T E ∗ ( X ) ≠ T ( X ) . The final assertion follows from the well-known fact that T ( X ) is regular.

(2) The sufficiency is obvious. If E ≠ ε X , then there exists an E -class A with ∣ A ∣ ≥ 2 . Take a ∈ A and define α ∈ T ( X ) as follows:

x α = a , x ∈ A x , otherwise.

Then, α ∈ T E ∗ ( X ) , but α ∉ J ( X ) , and so T E ∗ ( X ) ≠ J ( X ) . Now, let α ∈ T E ∗ ( X ) = J ( X ) . Then, by Lemma 2.3, α ∈ Reg ( T E ∗ ( X ) ) if and only if X α contains a cross-section of E . As E = ε X , the only cross-section of E is X itself, and so α ∈ Reg ( T E ∗ ( X ) ) if and only if X α = X . This gives that Reg ( T E ∗ ( X ) ) = S ( X ) .□

Proposition 3.2

The following statements are equivalent:

The Green relation ℒ in Reg ( T E ∗ ( X ) ) is left compatible.
The Green relation ℋ in Reg ( T E ∗ ( X ) ) is left compatible.
E = ε X .

Proof

(1) ⇒ (2). This follows from the facts that ℋ = L ∩ R and ℛ is always left compatible.

(2) ⇒ (3). If E ≠ ε X , then there exists A ∈ X ∕ E with ∣ A ∣ ≥ 2 . Take two distinct elements a and b in A and define α , β , γ ∈ T ( X ) as follows:

x α = b , x = a a , x ∈ A \ { a } x , otherwise , x β = a , x = a b , x ∈ A \ { a } x , otherwise , x γ = a , x ∈ A x , x ∈ X \ A .

Then, α , β , γ ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3 and X α = X β , ker α = ker β . This yields that ( α , β ) ∈ ℋ by Lemma 2.11. On the other hand, since X γ α = { b } ∪ ( X \ A ) and X γ β = { a } ∪ ( X \ A ) , we have X γ α ≠ X γ β , and so ( γ α , γ β ) ∉ ℋ by Lemma 2.11 again. This implies that ℋ is not left compatible.

(3) ⇒ (1). If E = ε X , then Reg ( T E ∗ ( X ) ) is the symmetric group S ( X ) by Proposition 3.1 (2), and so ℒ is the universal relation and certainly left compatible.□

Proposition 3.3

The following statements are equivalent:

The Green relation ℒ in T E ∗ ( X ) is left compatible.
The Green relation ℋ in T E ∗ ( X ) is left compatible.
X is finite and E = ε X .

If this is the case, we have T E ∗ ( X ) = Reg ( T E ∗ ( X ) ) = S ( X ) .

Proof

(1) ⇒ (2). This follows from the facts that ℋ = L ∩ R and ℛ is always left compatible.

(2) ⇒ (3). If the relation ℋ in T E ∗ ( X ) is left compatible, then ℋ in Reg ( T E ∗ ( X ) ) is also left compatible by Lemma 2.11. By Proposition 3.2, E = ε X . If X is infinite, then we can take a countable subset Y = { y i ∣ i ∈ N } . Define α , β ∈ T ( X ) as follows:

x α = y i + 1 , x = y i ∈ Y x , otherwise and x β = y 3 , x = y 0 y 2 , x = y 1 y 1 , x = y 2 y i + 1 , x = y i ∈ Y , i ≥ 3 x , otherwise .

Then, α , β ∈ T E ∗ ( X ) obviously and X α = X β , ker α = ker β . By Lemma 2.1, ( α , β ) ∈ ℋ . Observe that x α 2 = X \ { y 0 , y 1 } and x α β = X \ { y 0 , y 3 } , it follows that ( α 2 , α β ) ∉ ℋ by Lemma 2.1 again, and so ℋ is not left compatible, a contradiction. Thus, X is finite.

(3) ⇒ (1). If E = ε X and X is finite, then by Proposition 3.1 (2) and Lemma 2.6, we have T E ∗ ( X ) = Reg ( T E ∗ ( X ) ) = S ( X ) , and so ℒ is the universal relation and certainly left compatible.□

To give the next result, we need a lemma that is announced in the text [16]. For the completeness, we shall give a detailed proof for it. Recall that

Z ( α ) = { A ∈ X ∕ E ∣ A ∩ X α = ∅ }

for each α ∈ T E ∗ ( X ) .

Lemma 3.4

Let α , β ∈ T E ∗ ( X ) . Then, ∣ Z ( α β ) ∣ = ∣ Z ( α ) ∣ + ∣ Z ( β ) ∣ .

Proof

Denote

Z ( α ) = { A i ∣ i ∈ I } = { A ∈ X ∕ E ∣ A ∩ X α = ∅ } , Z ( β ) = { B j ∣ j ∈ J } = { B ∈ X ∕ E ∣ B ∩ X β = ∅ } .

Let A i ∈ Z ( α ) . Since β ∈ T E ∗ ( X ) , there exists a unique C i ∈ X ∕ E such that A i β ⊆ C i . So we can form the set U = { C i ∣ i ∈ I } . Observe that β ∈ T E ∗ ( X ) , it follows that

i = j ⇔ A i = A j ⇔ C i = C j .

So ∣ Z ( α ) ∣ = ∣ U ∣ . Let C i ∈ U . If C i ∩ X α β ≠ ∅ , then ( x 0 α ) β ∈ C i for some x 0 ∈ X . Observe that A i β ⊆ C i and β ∈ T E ∗ ( X ) , it follows that x 0 α ∈ A i . This contradicts to the fact that A i ∈ Z ( α ) . Thus, C i ∩ X α β = ∅ . That is, C i ∈ Z ( α β ) . On the other hand, if B j ∈ Z ( β ) , then B j ∩ X β = ∅ , and so B j ∩ X α β = ∅ . This implies that B j ∈ Z ( α β ) . As A i β ⊆ C i , we have C i ∩ X β ≠ ∅ for all i ∈ I . This yields that C i ≠ B j for all i ∈ I and j ∈ J . From the aforementioned discussions, we have

U ∪ Z ( β ) ⊆ Z ( α β ) , U ∩ Z ( β ) = ∅ , ∣ U ∣ = ∣ Z ( α ) ∣ ,

and so ∣ Z ( α β ) ∣ ≥ ∣ Z ( α ) ∣ + ∣ Z ( β ) ∣ .

Now, let D ∈ Z ( α β ) . Then, D ∩ X α β = ∅ . If D ∩ X β = ∅ , then D ∈ Z ( β ) . If D ∩ X β ≠ ∅ , then there exists x 0 ∈ X such that x 0 β ∈ D . Assume that A ∈ X ∕ E and x 0 ∈ A . As D ∩ X α β = ∅ and x 0 β ∈ D , we have ( x 0 β , x α β ) ∉ E , and so ( x 0 , x α ) ∉ E for all x ∈ X by the fact that β ∈ T E ∗ ( X ) . Observe that x 0 ∈ A , it follows that A ∩ X α = ∅ , and so A ∈ Z ( α ) . Suppose that A = A i for some i ∈ I . By the fact that x 0 ∈ A , we have x 0 β ∈ A i β ⊆ C i by the construction of C i . However, x 0 β ∈ D . Therefore, D ∩ C i ≠ ∅ , and so D = C i . This implies that Z ( α β ) ⊆ U ∪ Z ( β ) and U ∩ Z ( β ) = ∅ . Hence, ∣ Z ( α β ) ∣ ≤ ∣ U ∣ + ∣ Z ( β ) ∣ = ∣ Z ( α ) ∣ + ∣ Z ( β ) ∣ .□

Now, we are in a position to characterize the right compatibility of the relations ℛ , R ∗ , and ℋ .

Proposition 3.5

The following statements are equivalent:

The Green relation ℛ in Reg ( T E ∗ ( X ) ) is right compatible.
E = ε X , or there is a unique non-singleton E-class A and ∣ A ∣ = 2 .
The Green ∗ -relation ℛ ∗ in T E ∗ ( X ) is right compatible.
The Green relation ℛ in T E ∗ ( X ) is right compatible.
The Green relation ℋ in T E ∗ ( X ) is right compatible.
The Green relation ℋ in Reg ( T E ∗ ( X ) ) is right compatible.

Proof

(1) ⇒ (2) and (6) ⇒ (2). If (2) does not hold, then we have two cases.

Case 1: There are two non-singleton E -classes A 1 and A 2 . Take a 1 , b 1 ∈ A 1 and a 2 , b 2 ∈ A 2 with a 1 ≠ b 1 and a 2 ≠ b 2 , and define α , β , γ ∈ T ( X ) as follows:

x α = b 2 , x = a 1 a 2 , x ∈ A 1 \ { a 1 } b 1 , x = a 2 a 1 , x ∈ A 2 \ { a 2 } x , otherwise , x β = b 1 , x = a 1 a 1 , x ∈ A 1 \ { a 1 } b 2 , x = a 2 a 2 , x ∈ A 2 \ { a 2 } x , otherwise , x γ = a 2 , x ∈ A 1 b 1 , x = a 2 a 1 , x ∈ A 2 \ { a 2 } x , otherwise .

Then, α , β , γ ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3 and ker α = ker β , X α = X β , whence ( α , β ) ∈ ℋ in Reg ( T E ∗ ( X ) ) by Lemma 2.11. Observe that

a 1 β γ = b 1 γ = a 2 = a 1 γ = b 1 β γ and a 1 α γ = b 2 γ = a 1 ≠ b 1 = a 2 γ = b 1 α γ ,

it follows that ker β γ ≠ ker α γ , whence ( α γ , β γ ) ∉ ℛ in Reg ( T E ∗ ( X ) ) by Lemma 2.11 again. This implies that neither ℛ nor ℋ in Reg ( T E ∗ ( X ) ) is right compatible.

Case 2: There is an E -class A containing at least three elements. Take three distinct elements a , b , and c in A and define α , β , and γ ∈ T ( X ) as follows:

x α = b , x = a c , x = b a , x = c x , otherwise , x β = a , x = a c , x = b b , x = c x , otherwise , x γ = a , x = a , c c , x = b x , otherwise.

Then, α , β , γ ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3 and ker α = ker β , X α = X β . Moreover, Lemma 2.11 gives that ( α , β ) ∈ ℋ in Reg ( T E ∗ ( X ) ) . However, as

b α γ = c γ = a = a γ = c α γ , b β γ = c γ = a ≠ c = b γ = c β γ ,

we have ker α γ ≠ ker β γ , and so ( α γ , β γ ) ∉ ℛ in Reg ( T E ∗ ( X ) ) by Lemma 2.11 again. This also shows that neither ℛ nor ℋ in Reg ( T E ∗ ( X ) ) is right compatible.

(2) ⇒ (3). If E = ε X , then T E ∗ ( X ) = J ( X ) by Proposition 3.1 (2). This implies that ker α = ε X for every α ∈ T E ∗ ( X ) . By Lemma 2.2, ℛ ∗ is the universal relation in T E ∗ ( X ) and certainly right compatible. If A is the unique non-singleton E -class, ∣ A ∣ = 2 and α , β , γ ∈ T E ∗ ( X ) with ( α , β ) ∈ ℛ ∗ , then ker α = ker β by Lemma 2.2. Let ( x , y ) ∈ ker α γ . Then, x α γ = y α γ , and so ( x α , y α ) ∈ E , whence ( x , y ) ∈ E by the fact that γ , α ∈ T E ∗ ( X ) . By hypothesis, x = y or A = { x , y } . If x = y , then x β γ = y β γ certainly. Now assume that A = { x , y } . In the case that ( x , y ) ∈ ker α = ker β , we have x β = y β and so x β γ = y β γ . In the case that ( x , y ) ∉ ker α = ker β , we obtain that x α ≠ y α and x β ≠ y β . Since { x , y } = A and α , β ∈ T E ∗ ( X ) , we have ( x α , y α ) ∈ E , ( x β , y β ) ∈ E and so A = { x α , y α } = { x β , y β } . This together with the fact that x α γ = y α γ gives that x β γ = y β γ . Thus, we have x β γ = y β γ , and so ( x , y ) ∈ ker β γ in all cases. Therefore, ker α γ ⊆ ker β γ . This together with its dual gives that ker α γ = ker β γ . In view of Lemma 2.2, we have ( α γ , β γ ) ∈ ℛ ∗ . Thus, ℛ ∗ in T E ∗ ( X ) is right compatible.

(3) ⇒ (4). If α , β , γ ∈ T E ∗ ( X ) and α ℛ β , then ker α = ker β and ∣ Z ( α ) ∣ = ∣ Z ( β ) ∣ by Lemma 2.1. In view of Lemma 2.2, we have α ℛ ∗ β . By Item (3), we have α γ ℛ ∗ β γ , and so ker α γ = ker β γ by Lemma 2.2 again. Moreover, by Lemma 3.4,

∣ Z ( α γ ) = ∣ Z ( α ) ∣ + ∣ Z ( γ ) ∣ = ∣ Z ( β ) ∣ + ∣ Z ( γ ) ∣ = ∣ Z ( β γ ) ∣ .

Thus, ( α γ , β γ ) ∈ ℛ in T E ∗ ( X ) by Lemma 2.1 again.

(4) ⇒ (5). This follows from the facts that ℋ = L ∩ R and ℒ is always right compatible.

( 5 ) ⇒ ( 6 ) and ( 4 ) ⇒ ( 1 ) . This follows from Lemma 2.11.□

Now, we consider the relations ℒ ˜ and ℛ ˜ in T E ∗ ( X ) . By Corollary 2.9, ℛ ˜ = ℛ ∗ . So we only consider ℒ ˜ . We begin our discussion by giving a characterization of this relation.

Theorem 3.6

Let α , β ∈ T E ∗ ( X ) . Then, ( α , β ) ∈ ℒ ˜ if and only if X α = X β , or for all a ∈ ( X β \ X α ) ∪ ( X α \ X β ) , the E -class containing a is { a } .

Proof

Sufficiency. If X α = X β , then we have ( α , β ) ∈ ℒ ∗ by Lemma 2.2 and so ( α , β ) ∈ ℒ ˜ by (1.1). Now assume that X α ≠ X β . Take an arbitrary idempotent e in T E ∗ ( X ) and suppose that α e = α . Let x ∈ X . If x β ∈ X α , then there exists y ∈ X such that x β = y α . This implies that x β e = y α e = y α = x β . If x β ∈ X β X α , by hypothesis, the E -class containing x β is { x β } . Since e is idempotent, we have { x β } e ⊆ { x β } by Lemma 2.3, and so x β e = x β . We have shown that α e = α implies that β e = β . This together with its dual gives that ( α , β ) ∈ ℒ ˜ .

Necessity. Let ( α , β ) ∈ ℒ ˜ . Then, for every idempotent e in T E ∗ ( X ) , α e = α if and only if β e = β . Assume that X α ≠ X β and there exists a ∈ ( X β \ X α ) ∪ ( X α \ X β ) such that the E -class containing a has at least two elements. Without loss of generality, suppose that a ∈ X β \ X α , A ∈ X ∕ E and a , c ∈ A . As a ∈ X β , we have a = b β for some b ∈ X . Define e ∈ T ( X ) as follows: a e = c and x e = x for all X \ { a } . Then, e ∈ T E ∗ ( X ) and e 2 = e . Obviously, α e = α . However, b β e = a e = c ≠ a = b β , and so β ≠ β e , a contradiction.□

The following two propositions characterize the left and right compatibility of ℒ ˜ .

Proposition 3.7

In T E ∗ ( X ) , the relation ℒ ˜ is left compatible if and only if E = ε X .

Proof

If E = ε X , then all E -classes are singletons and so ℒ ˜ is the universal relation on T E ∗ ( X ) by Theorem 3.6, whence ℒ ˜ is left compatible certainly.

Conversely, if E ≠ ε X , then there exists a non-singleton E -class A. Take two distinct elements a , b ∈ A and define α , β , γ ∈ T ( X ) as in the proof of Proposition 3.2. Then, α , β , γ ∈ T E ∗ ( X ) and X α = X β . This yields that ( α , β ) ∈ ℒ ˜ by Theorem 3.6. On the other hand, observe that X γ α = { b } ∪ ( X \ A ) , X γ β = { a } ∪ ( X \ A ) and a , b ∈ A , it follows that ( γ α , γ β ) ∉ ℒ ˜ by Theorem 3.6 again, and so ℒ ˜ is not left compatible.□

Proposition 3.8

In T E ∗ ( X ) , the relation ℒ ˜ is right compatible if and only if one of the following conditions holds:

E = ε X .
X ∕ E is finite.
X ∕ E is infinite and every E-class contains at least two elements.

Proof

Sufficiency. First, if (1) holds, then by the proof of Proposition 3.7, ℒ ˜ is the universal relation on T E ∗ ( X ) , and so ℒ ˜ is right compatible. Second, if (2) holds, then by Lemma 2.6, T E ∗ ( X ) is regular, and so ℒ ˜ = ℒ ∗ = ℒ by Lemma 2.8, whence ℒ ˜ is right compatible. Finally, if (3) holds, then by Lemma 2.2 and Theorem 3.6, we have ℒ ˜ = ℒ ∗ , whence ℒ ˜ is also right compatible.

Necessity. Assume that the given conditions do not hold. Then, X ∕ E is infinite, and there are a non-singleton E -class A and a singleton E -class B = { b } . Since X ∕ E is infinite, X ∕ E and ( X ∕ E ) \ { A , B } have the same cardinality, and so there exists a bijection φ from X ∕ E onto ( X ∕ E ) \ { A , B } . Fix an element u M in φ ( M ) for each M ∈ X ∕ E and form the set U = { u M ∣ M ∈ X ∕ E } . Obviously, there also exists a bijection ψ from X ∕ E onto U ∪ B . Define α , β ∈ T ( X ) by the rule that x α = u M and x β = M ψ if x ∈ M and M ∈ X ∕ E . Then, one can easily see that α , β ∈ T E ∗ ( X ) and

X α = U , X β = U ∪ B , A ∩ X α = ∅ = B ∩ X α .

By Theorem 3.6, we have ( α , β ) ∈ ℒ ˜ as ( X β \ X α ) = B = { b } and ( X α \ X β ) = ∅ . Take an element a in A , and define γ ∈ T ( X ) as follows:

x γ = a , x ∈ B = { b } b , x ∈ A x , otherwise.

Then, γ ∈ T E ∗ ( X ) . Observe that

X α γ = X α , X β γ = X α ∪ { a } , a ∈ X β γ \ X α γ , a ∈ A ,

and A is not a singleton E -class, it follows that ( α γ , β γ ) ∉ ℒ ˜ by Theorem 3.6. Thus, ℒ ˜ is not right compatible.□

By Corollary 2.9, we have ℒ = L ∗ , ℋ = H ∗ and ℛ ∗ = ℛ ˜ . However, we have the following results.

Proposition 3.9

In T E ∗ ( X ) , ℛ = R ∗ if and only if X ∕ E is finite.

Proof

If X ∕ E is finite, then T E ∗ ( X ) is regular by Lemma 2.6 and so ℛ = ℛ ∗ by Lemma 2.8. Conversely, if ℛ = ℛ ∗ , then by Lemma 2.7, every ℛ ∗ -class contains an idempotent, and so every ℛ -class has an idempotent. This implies that T E ∗ ( X ) is regular, and so X ∕ E is finite by Lemma 2.6 again.□

Proposition 3.10

In T E ∗ ( X ) , ℒ ∗ = L ˜ if and only if X ∕ E is finite or X ∕ E is infinite and each E-class contains at least two elements.

Proof

The sufficiency follows from the proof of the corresponding part of Proposition 3.8. To see the necessity, let ℒ ∗ = ℒ ˜ . Then, ℒ ˜ is right compatible as ℒ ∗ is always right compatible. By Proposition 3.8, we only need to prove that it is impossible that both E = ε X and X is infinite. Otherwise, we fix an element a in X . Since X is infinite, X and X \ { a } have the same cardinality and so there exists a bijection φ from X onto X \ { a } . As E = ε X , T E ∗ ( X ) is exactly the set of injections on X by Proposition 3.1. Thus, both the identity transformation ι X and φ above are in T E ∗ ( X ) with X φ = X \ { a } and X ι X = X . By Lemma 2.2, Theorem 3.6, and the fact that { a } is an E -class, we have ( φ , ι X ) ∉ ℒ ∗ and ( φ , ι X ) ∈ ℒ ˜ , which gives that ℒ ∗ ≠ ℒ ˜ , a contradiction.□

At the end of this section, we consider the right semiabundant elements in T E ∗ ( X ) (i.e., elements that are ℒ ˜ -related to some idempotent).

Theorem 3.11

Let α ∈ T E ∗ ( X ) . Then, α is right semiabundant in T E ∗ ( X ) if and only if Z ( α ) = ∅ or ∣ A ∣ = 1 for all A ∈ Z ( α ) .

Proof

Assume that α is right semiabundant in T E ∗ ( X ) . Then, there exists an idempotent e ∈ T E ∗ ( X ) such that ( α , e ) ∈ ℒ ˜ . If the given condition is not valid, then there exists A ∈ Z ( α ) with ∣ A ∣ ≥ 2 , and so A ∩ X α = ∅ . Since e is an idempotent, we have A e ⊆ A by Lemma 2.3. This implies that X α ≠ X e . Take a ∈ A e . Then, a ∈ A and a ∈ X e \ X α . Observe that ∣ A ∣ ≥ 2 , it follows that ( α , e ) ∉ ℒ ˜ by Theorem 3.6, a contradiction.

Conversely, assume that ∣ Z ( α ) ∣ = 0 or ∣ A ∣ = 1 for all A ∈ Z ( α ) . If ∣ Z ( α ) ∣ = 0 , then α is regular and certainly is right semiabundant by Lemma 2.3. Suppose that ∣ A ∣ = 1 for all A ∈ Z ( α ) . We can define e ∈ T ( X ) as follows: for each A ∈ X ∕ E , if A ∩ X α = ∅ (i.e., A ∈ Z ( α ) ), we let x e = x for all x ∈ A ; if A ∩ X α ≠ ∅ , we fix an element u A ∈ X α ∩ A , in the case that x ∈ A ∩ X α , we let x e = x , in the case that x ∈ A \ X α , we let x e = u A . One can check routinely that e ∈ T E ∗ ( X ) and e 2 = e . By the definition of e , we obtain that X e = X α ∪ ( ⋃ A ∈ Z ( α ) A ) . As ∣ A ∣ = 1 for all A ∈ Z ( α ) , it follows that ( α , e ) ∈ ℒ ˜ by Theorem 3.6. This gives that α is right semiabundant in T E ∗ ( X ) .□

Proposition 3.12

The semigroup T E ∗ ( X ) is right semiabundant if and only if X ∕ E is finite or E = ε X .

Proof

We first consider the sufficiency. If X ∕ E is finite, then T E ∗ ( X ) is regular by Lemma 2.6, and so Z ( α ) = ∅ for every α ∈ T E ∗ ( X ) by Lemma 2.3. By Theorem 3.11, T E ∗ ( X ) is right semiabundant. If E = ε X , then each E -class contains exactly one element, and so T E ∗ ( X ) is right semiabundant by Theorem 3.11 again.

Conversely, assume that the given condition does not hold. Then, X ∕ E is infinite and there exists an E -class A with ∣ A ∣ ≥ 2 . Since X ∕ E is infinite, X ∕ E and ( X ∕ E ) \ { A } have the same cardinality. So there exists a bijection φ : X ∕ E → ( X ∕ E ) \ { A } . For each M ∈ X ∕ E , fix u M ∈ φ ( M ) and form the set U = { u M ∣ M ∈ X ∕ E } . Then, U ∩ A = ∅ . Obviously, the map ψ : X ∕ E → U , M ↦ u M is a bijection. Define α ∈ T ( X ) by the rule that x α = M ψ = u M for every M ∈ X ∕ E and x ∈ M . Then, α ∈ T E ∗ ( X ) and X α = U . This together with the fact that U ∩ A = ∅ implies that A ∈ Z ( α ) . Observe that ∣ A ∣ ≥ 2 , and it follows that α is not right semiabundant by Theorem 3.11. Thus, T E ∗ ( X ) is not right semiabundant.□

By Lemma 2.5, the set of regular elements in T E ∗ ( X ) always forms a subsemigroup of T E ∗ ( X ) . However, we have the following different result for right semiabundant elements.

Proposition 3.13

The set of right semiabundant elements of the semigroup T E ∗ ( X ) forms a subsemigroup if and only if one of the following conditions holds:

X ∕ E is finite.
E = ε X .
X ∕ E is infinite, and each E-class has at least two elements.

Proof

We first show the sufficiency. If X ∕ E is finite or E = ε X , then T E ∗ ( X ) itself is right semiabundant by Proposition 3.12. If ( 3 ) holds, α , β ∈ T E ∗ ( X ) , and α and β are right semiabundant, then we have ∣ Z ( α ) ∣ = 0 = ∣ Z ( β ) ∣ by Theorem 3.11. This implies that α and β are regular by Lemma 2.3. In view of Lemma 2.5, we obtain that α β is also regular and certainly is right semiabundant.

Conversely, if the given conditions do not hold, then X ∕ E is infinite and there exists a non-singleton E -class A and a singleton E -class B = { b } . Since X ∕ E is infinite, we can choose a countable subset { C i ∈ X ∕ E ∣ i ∈ N } of X ∕ E . Take a 1 , a 2 ∈ A and c i ∈ C i for each i ∈ N . Define α and β in T ( X ) as follows:

x α = a 1 , x ∈ A c 0 , x ∈ B c i + 1 , x ∈ C i x , otherwise and x β = b , x ∈ A a 1 , x ∈ B c i , x ∈ C i x , otherwise .

It is easy to see that α , β ∈ T E ∗ ( X ) , Z ( α ) = { B } and ∣ Z ( β ) ∣ = 0 . By Theorem 3.11, α and β are right semiabundant. Observe that

x α β = b , x ∈ A c 0 , x ∈ B c i + 1 , x ∈ C i x , otherwise ,

it follows that α β ∈ T E ∗ ( X ) and A ∈ Z ( α β ) . Since ∣ A ∣ ≥ 2 , we obtain that α β is not right semiabundant by Theorem 3.11 again. Thus, the set of right semiabundant elements of T E ∗ ( X ) does not form a subsemigroup.□

4 Various kinds of regularities

In this section, we consider several kinds of regularities of T E ∗ ( X ) . By Lemma 2.5, we have known that Reg ( T E ∗ ( X ) ) is a regular subsemigroup of T E ∗ ( X ) . The following result gives a sufficient and necessary condition under which Reg ( T E ∗ ( X ) ) is left (or equivalently, right, completely by Lemma 2.12) regular. Recall that a semigroup S is called left (respectively, right, completely, intra-) regular if a ℒ a 2 (respectively, a ℛ a 2 , a ℋ a 2 , a J a 2 ) for all a ∈ S .

Proposition 4.1

The following conditions are equivalent:

Reg ( T E ∗ ( X ) ) is left (or equivalently, right, completely) regular.
Reg ( T E ∗ ( X ) ) is intra-regular.
E = ε X , or there is a unique non-singleton E-class A and ∣ A ∣ = 2 .

Proof

(1) ⇒ (2). Trivial.

(2) ⇒ (3). Assume that ( 3 ) is not true. Then, either X has two distinct non-singleton E -classes or X has an E -class containing at least three elements.

In the former case, take two distinct E -classes A 1 and A 2 , and fix two elements a 1 and b 1 in A 1 and a 2 and b 2 in A 2 , respectively. Further, for any A ∈ ( X ∕ E ) \ { A 1 , A 2 } , fix an element x A in A . Define α ∈ T ( X ) as follows:

x α = a 2 , x = a 1 b 2 , x ∈ A 1 \ { a 1 } a 1 , x ∈ A 2 x A , x ∈ A ∈ ( X ∕ E ) \ { A 1 , A 2 } .

Then, α ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3, A 1 α = { a 2 , b 2 } and

x α 2 = a 1 , x ∈ A 1 a 2 , x ∈ A 2 x A , x ∈ A ∈ ( X ∕ E ) \ { A 1 , A 2 } .

This implies that ∣ B α 2 ∣ = 1 for all B ∈ X ∕ E . Therefore, for any B in X ∕ E and ρ ∈ T E ∗ ( X ) , it is impossible that { a 2 , b 2 } = A 1 α ⊆ B α 2 ρ . By Lemma 2.1, ( α , α 2 ) ∉ J in T E ∗ ( X ) and certainly ( α , α 2 ) ∉ J in Reg ( T E ∗ ( X ) ) . So neither T E ∗ ( X ) nor Reg ( T E ∗ ( X ) ) is intra-regular.

In the latter case, fix an E -class A having at least three elements and take three distinct elements a , b , and c in A . Further, for any B ∈ ( X ∕ E ) \ { A } , fix an element x B in B . Define α ∈ T ( X ) as follows:

x α = b , x = a c , x ∈ A \ { a } x B , x ∈ B ∈ ( X ∕ E ) \ { A } .

Then, α ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3, A α = { b , c } and

x α 2 = c , x ∈ A x B , x ∈ B ∈ ( X ∕ E ) \ { A } .

Thus, A α = { b , c } and ∣ B α 2 ∣ = 1 for all B ∈ X ∕ E . This implies that B α 2 ρ does not contain A α for all B ∈ X ∕ E and ρ ∈ T E ∗ ( X ) , whence ( α , α 2 ) ∉ J in T E ∗ ( X ) by Lemma 2.1 and certainly ( α , α 2 ) ∉ J in Reg ( T E ∗ ( X ) ) . This also yields that neither T E ∗ ( X ) nor Reg ( T E ∗ ( X ) ) is intra-regular.

(3) ⇒ (1). In the case that E = ε X , by Proposition 3.1 (2), Reg ( T E ∗ ( X ) ) is equal to the symmetric group S ( X ) , and so ℒ is the universal relation, whence Reg ( T E ∗ ( X ) ) is left regular clearly.

In the case that A = { a , b } is the only non-singleton E -class, for each α ∈ Reg ( T E ∗ ( X ) ) , we obtain that α 2 ∈ Reg ( T E ∗ ( X ) ) , X α 2 ⊆ X α , and X α 2 contains a cross-section of E by Lemma 2.3. Observe that the only cross-sections of E are

( X \ A ) ∪ { a } = X \ { b } and ( X \ A ) ∪ { b } = X \ { a } .

Without loss of generality, we assume that X α 2 contains X \ { b } . Suppose that X α ≠ X α 2 . Then, we have X α 2 = X \ { b } and X α = X , and so we can let x α = b and y α 2 = a for some x , y in X . Since { a , b } = A and α ∈ T E ∗ ( X ) , it follows that ( x , y α ) ∈ E . If x = y α , then b = x α = ( y α ) α = y α 2 = a , a contradiction. Thus, x ≠ y α . By hypothesis, we obtain { x , y α } = { a , b } . If b = x , then b = x α = b α = x α 2 = b α 2 , and so b ∈ X α 2 . Otherwise, b = y α . But this yields that a = x and

b = x α = a α = y α 3 = ( y α ) α 2 = b α 2 ,

and so b ∈ X α 2 . However, X α 2 = X \ { b } . Thus, both the aforementioned two cases are impossible. Therefore, X α = X α 2 , and so ( α , α 2 ) ∈ ℒ by Lemma 2.11 (1), so that ( 1 ) holds.□

Now, we consider the regularities of T E ∗ ( X ) .

Proposition 4.2

The following conditions are equivalent:

T E ∗ ( X ) is left regular.
T E ∗ ( X ) is right regular.
T E ∗ ( X ) is completely regular.
T E ∗ ( X ) is intra-regular.
X ∕ E is finite, and E = ε X or there exists a unique non-singleton E-class A and ∣ A ∣ = 2 .

Proof

If ( 5 ) holds, then by Lemma 2.6, Reg ( T E ∗ ( X ) ) = T E ∗ ( X ) , and so Items (1)–(4) hold by Proposition 4.1.

On the other hand, assume that one of Items (1)–(4) holds. Then, T E ∗ ( X ) is intra-regular. In view of the proof of “(2) ⇒ (3)” in the proof of Proposition 4.1, to see ( 5 ) , we only need to show that X ∕ E is finite. In fact, if X ∕ E is infinite, then X is infinite. We shall consider the following two cases.

In the case that E = ε X , we choose a countable subset U = { u i ∣ i ∈ N } of X and define α ∈ T ( X ) by the rule that

x α = u i + 1 , x = u i , i ∈ N x , otherwise .

Then, α ∈ T E ∗ ( X ) . If ( α , α 2 ) ∈ J in T E ∗ ( X ) , then by Lemma 2.1, there exists ρ ∈ T E ∗ ( X ) such that

X \ { u 0 } = X α ⊆ X α 2 ρ = ( X \ { u 0 , u 1 } ) ρ .

Since ρ is injective by Proposition 3.1, we obtain that neither u 0 ρ nor u 1 ρ is in ( X \ { u 0 , u 1 } ) ρ , and so u 0 ρ , u 1 ρ ∉ ( X \ { u 0 } ) . That is to say, u 0 ρ = u 1 ρ = u 0 . This contradicts the fact that ρ is injective. Thus, ( α , α 2 ) ∉ J , and so T E ∗ ( X ) is not intra-regular.

In the case that X has a unique non-singleton E -class A and ∣ A ∣ = 2 , we also choose a countable subset U = { u i ∣ i ∈ N } of X \ A and define α ∈ T ( X ) as follows:

x α = u 0 , x ∈ A u i + 1 , x = u i , i ∈ N x , otherwise .

Then, α ∈ T E ∗ ( X ) . If ( α , α 2 ) ∈ J in T E ∗ ( X ) , then by Lemma 2.1, there exists ρ ∈ T E ∗ ( X ) such that

X \ A = X α ⊆ X α 2 ρ = ( X \ ( A ∪ { u 0 } ) ) ρ .

If u 0 ρ ∈ X \ A , then u 0 ρ ∈ ( X \ ( A ∪ { u 0 } ) ) ρ , and so there is x 0 ∈ X \ ( A ∪ { u 0 } ) such that u 0 ρ = x 0 ρ , and so u 0 = x 0 by hypothesis. This is impossible by the choice of x 0 . Thus, u 0 ρ ∉ X \ A and so u 0 ρ ∈ A . Take a ∈ A . If a ρ ∈ X \ A , then a ρ ∈ ( X \ ( A ∪ { u 0 } ) ) ρ , and so a ρ = y ρ for some y in X \ ( A ∪ { u 0 } ) . Since ρ ∈ T E ∗ ( X ) , we have ( a , y ) ∈ E and so y ∈ A . This is a contradiction. Thus, a ρ ∈ A . From the aforementioned statements, we have u 0 ρ ∈ A and a ρ ∈ A . That yields that ( u 0 ρ , a ρ ) ∈ E and so ( u 0 , a ) ∈ E , whence u 0 ∈ A . This leads to a contradiction by the choice of u 0 . Thus, ( α , α 2 ) ∉ J and so T E ∗ ( X ) is not intra-regular.□

Recall from Howie [12, Theorem 3.3.1] that completely simple semigroups are exactly Rees matrix semigroups over groups, and completely simple semigroups with identity are exactly groups.

Proposition 4.3

The following conditions are equivalent:

Reg ( T E ∗ ( X ) ) is completely simple.
Reg ( T E ∗ ( X ) ) is a group.
E = ε X .

Proof

Since Reg ( T E ∗ ( X ) ) contains ι X as its identity, it is immediate that (1) is equivalent to (2) by the construction of completely simple semigroups.

If (3) holds, then (2) holds by Proposition 3.1. Conversely, assume that (2) is true. If (3) does not hold, then we take a non-singleton E -class A and fix an element a in A . Define α ∈ T ( X ) as follows:

x α = a , x ∈ A x , otherwise .

Then, α is an idempotent in Reg ( T E ∗ ( X ) ) and α ≠ ι X . This implies that Reg ( T E ∗ ( X ) ) is not a group as any group contains exactly one idempotent, a contradiction.□

The following proposition gives the sufficient and necessary condition under which Reg ( T E ∗ ( X ) ) forms an orthodox semigroup.

Proposition 4.4

Reg ( T E ∗ ( X ) ) is orthodox if and only if each E-class of X contains at most two elements.

Proof

Assume that X has an E -class A that contains at least three elements. Fix three distinct elements a , b , and c in A and define e , f ∈ T ( X ) as follows:

x e = b , x = a x , x ≠ a , x f = a , x = c x , x ≠ c .

Then, it is easy to see that e and f are idempotents in T E ∗ ( X ) . However, c ( e f ) = a ≠ b = c ( e f ) 2 . This gives that e f ≠ ( e f ) 2 , and so Reg ( T E ∗ ( X ) ) is not orthodox.

Conversely, if the given condition holds and e and f are idempotents in Reg ( T E ∗ ( X ) ) , then by Lemma 2.3, for all A ∈ X ∕ E , we have A e ⊆ A and A f ⊆ A . Let x ∈ X . If the E -class containing x is { x } , then x e = x = x f , and so x e f e f = x = x e f . If the E -class containing x is not { x } , then there exists a unique y ∈ X such that the E -class containing x is { x , y } . This yields that { x , y } e ⊆ { x , y } . Observe that e is idempotent, it follows that there are three cases:

x e = x , y e = y or x e = x = y e or x e = y = y e .

Similarly, we have

x f = x , y f = y or x f = x = y f or x f = y = y f .

Thus, we have the following nine cases:

x e = x , y e = y , x f = x , y f = y . In this case, x e f = x f = x and x e f e f = x e f = x .
x e = x , y e = y , x f = x = y f . In this case, x e f = x f = x and x e f e f = x e f = x .
x e = x , y e = y , x f = y = y f . In this case, x e f = x f = y and x e f e f = y e f = y .
x e = x = y e , x f = x , y f = y . In this case, x e f = x f = x and x e f e f = x e f = x .
x e = x = y e , x f = x = y f . In this case, x e f = x f = x and x e f e f = x e f = x .
x e = x = y e , x f = y = y f . In this case, x e f = x f = y and x e f e f = y e f = y .
x e = y = y e , x f = x , y f = y . In this case, x e f = y f = y and x e f e f = y e f = y .
x e = y = y e , x f = x = y f . In this case, x e f = y f = x and x e f e f = x e f = x .
x e = y = y e , x f = y = y f . In this case, x e f = y f = y and x e f e f = y e f = y .

This shows that ( e f ) 2 = e f . Thus, Reg ( T E ∗ ( X ) ) is orthodox.□

5 ℱ -abundant and G -abundant elements

In this section, we shall determine the ℱ -abundant and G -abundant elements in T E ∗ ( X ) , Reg ( T E ∗ ( X ) ) and ( T E ∗ ( X ) ) 0 , ( Reg ( T E ∗ ( X ) ) ) 0 , respectively. As a consequence, [7, Proposition 3.4] obtained by Stokes is generalized and enriched.

Theorem 5.1

Let α ∈ T E ∗ ( X ) . Then, α is ℱ -abundant if and only if for all A ∈ X ∕ E , there exists u ∈ A such that u α = u .

Proof

Assume that α is ℱ -abundant and there exists B ∈ X ∕ E such that x α ≠ x for all x ∈ B . Then, there exists an idempotent e in T E ∗ ( X ) such that ( α , e ) ∈ ℱ and B e ⊆ B by Lemma 2.3. Take b ∈ B , then b e ∈ B , and so b ( e α ) = ( b e ) α ≠ b e by hypothesis, whence e α ≠ e . However, e e = e . This implies that ( α , e ) ∉ ℱ , a contradiction.

Conversely, assume that the given condition holds. For each A ∈ X ∕ E , denote A ′ = { u ∈ A ∣ u α = u } and fix an element u A in A ′ . Define e ∈ T ( X ) as follows: for each A ∈ X ∕ E and x ∈ A ,

x e = x , x ∈ A ′ u A , x ∈ A \ A ′ .

Then, e ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3 and e 2 = e . Moreover, it is easy to see that x α = x if and only if x e = x for all x ∈ X . This yields that β α = β if and only if β e = β for all β ∈ T E ∗ ( X ) . Thus, ( α , e ) ∈ ℱ , and so α is ℱ -abundant.□

Proposition 5.2

T E ∗ ( X ) is ℱ -abundant if and only if ∣ X ∣ = 1 .

Proof

If ∣ X ∣ = 1 , then T E ∗ ( X ) is ℱ -abundant obviously. On the other hand, if ∣ X ∣ ≥ 2 , then either there is an E -class A with ∣ A ∣ ≥ 2 or there are two distinct E -classes B and C . In the former case, we consider an element α in T ( X ) such that α ∣ A is a bijective transformation on A without fixed elements (this is possible by the theory of groups) and α ∣ X \ A is the identity transformation on X \ A . Then, α ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3 and x α ≠ x for all x ∈ A . By Theorem 5.1, α is not ℱ -abundant. In the latter case, fix b 0 ∈ B and c 0 ∈ C , and define β ∈ T ( X ) as follows:

x β = c 0 , x ∈ B b 0 , x ∈ C x , otherwise .

Then, β ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3 and x β ≠ x for all x ∈ B . By Theorem 5.1, β is not ℱ -abundant. Thus, T E ∗ ( X ) is not ℱ -abundant in either case.□

By the proofs of Theorem 5.1 and Proposition 5.2, we can also see that, for each α ∈ Reg ( T E ∗ ( X ) ) , α is ℱ -abundant in Reg ( T E ∗ ( X ) ) if and only if for all A ∈ X ∕ E , there exists u ∈ A such that u α = u , and Reg ( T E ∗ ( X ) ) is ℱ -abundant if and only if ∣ X ∣ = 1 . However, we have the following result.

Proposition 5.3

( Reg ( T E ∗ ( X ) ) ) 0 is ℱ -abundant.

Proof

Let S = ( Reg ( T E ∗ ( X ) ) ) 0 and suppose α ∈ S . If α = 0 , then α is ℱ -abundant obviously. If α ≠ 0 and α is ℱ -abundant in T E ∗ ( X ) , then α is also ℱ -abundant in S by the definition of ℱ . Now, let α ≠ 0 such that α is not ℱ -abundant in T E ∗ ( X ) . Then, by Theorem 5.1, there exists B ∈ X ∕ E such that x α ≠ x for all x ∈ B . We shall show that ( α , 0 ) ∈ ℱ in S in this case. To see this, let β ∈ S . If β 0 = β , then β = 0 , and so β α = 0 α = 0 = β . On the other hand, assume that β α = β . We assert that β = 0 . Otherwise, there exists u ∈ X such that u β ∈ B by Lemma 2.3 as β ∈ Reg ( T E ∗ ( X ) ) . Observe that β α = β , it follows that u β ∈ B and ( u β ) α = u ( β α ) = u β , a contradiction. Thus, β = 0 , and so β 0 = 0 = β . From the aforementioned discussions, we obtain that ( α , 0 ) ∈ ℱ and so α is ℱ -abundant in S .□

Remark 5.4

( T E ∗ ( X ) ) 0 is not ℱ -abundant in general. In fact, let E = ε X and X be infinite. Choose a countable subset U = { x i ∣ i ∈ N } of X and define α , β ∈ T ( X ) as follows:

x α = x 1 , x = x 0 x 0 , x = x 1 x , otherwise , x β = x i + 2 , x = x i , i ∈ N x , otherwise .

Then, α , β ∈ T E ∗ ( X ) . Since E = ε X , we have T E ∗ ( X ) = J ( X ) by Proposition 3.1 (2) and so the only idempotents in ( T E ∗ ( X ) ) 0 are ι X and 0. Since ι X α = α ≠ ι X = ι X ι X and β 0 = 0 ≠ β = β α , we obtain ( α , ι X ) ∉ ℱ and ( α , 0 ) ∉ ℱ , and so α is not ℱ -abundant in ( T E ∗ ( X ) ) 0 .

Proposition 5.5

The set of ℱ -abundant elements in T E ∗ ( X ) forms a subsemigroup of T E ∗ ( X ) if and only if each E-class of X contains at most two elements.

Proof

Necessity. Let A ∈ X ∕ E and ∣ A ∣ ≥ 3 . Take a , b , c ∈ A and define α , β ∈ T ( X ) as follows:

x α = a , x ∈ A \ { b , c } c , x = b b , x = c x , otherwise and x β = b , x ∈ A \ { c } a , x = c x , otherwise .

Then, it is easy to see that α , β ∈ T E ∗ ( X ) . By Theorem 5.1, both α and β are ℱ -abundant in T E ∗ ( X ) . However, observe that

x α β = b , x ∈ A \ { b } a , x = b x , otherwise ,

it follows that α β is not ℱ -abundant by Theorem 5.1 again.

Sufficiency. Assume that each E -class of X contains at most two elements. Let α and β be ℱ -abundant in T E ∗ ( X ) . Then, by Theorem 5.1, for every A ∈ X ∕ E , there exist u , v ∈ A , such that u α = u and v β = v . Let A ∈ X ∕ E . If A = { x } , then we have x α = x = x β , and so x α β = x β = x .

If A = { x , y } , then x α = x = y α , or x α = y = y α , or x α = x , y α = y . Similarly, we have x β = x = y β , or x β = y = y β , or x β = x , y β = y . This gives the following nine cases:

x α = x = y α , x β = x = y β . In this case, x α β = x β = x .
x α = x = y α , x β = y = y β . In this case, y α β = x β = y .
x α = x = y α , x β = x , y β = y . In this case, x α β = x β = x .
x α = y = y α , x β = x = y β . In this case, x α β = y β = x .
x α = y = y α , x β = y = y β . In this case, y α β = y β = y .
x α = y = y α , x β = x , y β = y . In this case, y α β = y β = y .
x α = x , y α = y , x β = x = y β . In this case, x α β = x β = x .
x α = x , y α = y , x β = y = y β . In this case, y α β = y β = y .
x α = x , y α = y , x β = x , y β = y . In this case, x α β = x β = x .

This shows that there exists u ∈ A such that u α β = u in either case. Thus, α β is ℱ -abundant by Theorem 5.1 again.□

Now, we consider the G -abundance of T E ∗ ( X ) and Reg ( T E ∗ ( X ) ) .

Theorem 5.6

Let α ∈ T E ∗ ( X ) . Then, α is G -abundant if and only if A α ⊆ A for all A ∈ X ∕ E .

Proof

Necessity. Let e ∈ T E ∗ ( X ) , e 2 = e and ( α , e ) ∈ G . Then, α e = e as e 2 = e . Thus, for every A ∈ X ∕ E , we have ( A α ) e = A ( α e ) = A e ⊆ A by Lemma 2.3, and so A α ⊆ A as e ∈ T E ∗ ( X ) .

Sufficiency. Assume that A α ⊆ A for all A ∈ X ∕ E . Then, the restriction α ∣ A of α to A is a transformation on A for all A ∈ X ∕ E . For each A ∈ X ∕ E , we consider the equivalence σ A on A , which is generated by { ( x α , x ) ∣ x ∈ A } and denote the set of the σ A -classes of A by Ω A = { A i ∣ i ∈ I } . Fix an element a i in A i for each A i ∈ Ω A and define e A ∈ T ( A ) by the rule that x e A = a i for all x ∈ A i and i ∈ I . Then, e A 2 = e A , ker e A = σ A and so x ( α e A ) = ( x α ) e A = x e A for all x ∈ A . Now, define e ∈ T ( X ) as follows: for each A ∈ X ∕ E and x ∈ A , x e = x e A . Obviously, e ∈ T E ∗ ( X ) , e 2 = e , and α e = e .

Now let β ∈ T E ∗ ( X ) . If e β = β , then α β = α e β = e β = β . Conversely, if α β = β , then for all A ∈ X ∕ E and x ∈ A , we have ( x α ) β = x ( α β ) = x β and so ( x α , x ) ∈ ker β ∩ ( A × A ) . This implies that ker e A = σ A ⊆ ker β ∩ ( A × A ) . Thus, ker e ⊆ ker β by the construction of e . Since e 2 = e , for all x ∈ X , we have ( x e ) e = x e 2 = x e , and so ( x e , x ) ∈ ker e ⊆ ker β , whence x ( e β ) = ( x e ) β = x β . Thus, e β = β and we have shown that ( α , e ) ∈ G and e 2 = e . Therefore, α is G -abundant.□

Proposition 5.7

T E ∗ ( X ) is G -abundant if and only if E = ω X . In this case, T E ∗ ( X ) = T ( X ) .

Proof

If E = ω X , then the only E -class is X , and for each α ∈ T E ∗ ( X ) , we have X α ⊆ X . By Theorem 5.6, all elements in T E ∗ ( X ) are G -abundant. If E ≠ ω X , then we can pick two distinct E -classes A and B in X ∕ E . Fix a ∈ A and b ∈ B , and define α ∈ T ( X ) as follows:

x α = b , x ∈ A a , x ∈ B x , otherwise .

Then, α ∈ Reg ( T E ∗ ( X ) ) by Lemma 2.3. Furthermore, α is not G -abundant by Theorem 5.6 as A α is not contained in A . The final assertion follows from Proposition 3.1 (1).□

By the proofs of Theorem 5.6 and Proposition 5.7, we can also see that, for each α ∈ Reg ( T E ∗ ( X ) ) , α is G -abundant if and only if A α ⊆ A for all A ∈ X ∕ E , and Reg ( T E ∗ ( X ) ) is G -abundant if and only if E = ω X . However, we can prove that both ( T E ∗ ( X ) ) 0 and ( Reg ( T E ∗ ( X ) ) ) 0 are G -abundant.

Proposition 5.8

Both ( T E ∗ ( X ) ) 0 and ( Reg ( T E ∗ ( X ) ) ) 0 are G -abundant.

Proof

Let S = ( T E ∗ ( X ) ) 0 and suppose α ∈ S . If α = 0 , then α is G -abundant obviously. If α ≠ 0 and α is G -abundant in T E ∗ ( X ) , then α is also G -abundant in S by the definition of G .

Now, let α ≠ 0 be such that α is not G -abundant in T E ∗ ( X ) . Then, by Theorem 5.6, there exists A ∈ X ∕ E such that A α is not contained in A . We shall show that ( α , 0 ) ∈ G in this case. Let β ∈ S . If 0 β = β , then β = 0 and so α β = α 0 = 0 = β . On the other hand, assume that α β = β . Since A α is not contained in A and α ∈ T E ∗ ( X ) , we have A α ⊆ B for some B ∈ X ∕ E with B ≠ A . If β ≠ 0 , then β ∈ T E ∗ ( X ) , and so A β = A ( α β ) = ( A α ) β ⊆ B β by the fact that α β = β . This is impossible as B ≠ A and β ∈ T E ∗ ( X ) . Thus, β = 0 , and so 0 β = 0 = β . We have shown that ( α , 0 ) ∈ G , whence α is G -abundant in S . By similar arguments, we can prove that ( Reg ( T E ∗ ( X ) ) ) 0 is also G -abundant.□

Proposition 5.9

The set of G -abundant elements in T E ∗ ( X ) forms a subsemigroup of T E ∗ ( X ) .

Proof

Let α and β be G -abundant elements in T E ∗ ( X ) . Assume that A ∈ X ∕ E . By Theorem 5.6, we have A α ⊆ A and A β ⊆ A , and so A α β ⊆ A β ⊆ A . Thus, α β is G -abundant by Theorem 5.6 again.□

Remark 5.10

In view of Theorems 5.1, 5.6, and 3.11, Lemmas 2.3, 2.6, and 2.7, and Propositions 5.5, 5.7, and 5.9, in the semigroup T E ∗ ( X ) , in general we have

{ ℱ -abundant elements } ⊂ { G -abundant elements } ⊂ { regular elements } = { abundant elements } = { right abundant elements } ⊂ { semiabundant elements } = { right semiabundant elements } ⊂ T E ∗ ( X ) = { left abundant elements } = { left semiabundant elements } .

In the end of this section, we summary some results obtained in the previous three sections. First, by Propositions 3.1 (2), 3.2, 3.7, and 4.3, we have the following theorem.