EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 4, 2022, 2116-2126 ISSN 1307-5543 – ejpam.com Published by New York Business Global Regularity on variants of transformation semigroups that preserve an equivalence relation Piyaporn Tantong1, Nares Sawatraksa1,∗ 1 Division of Mathematics and Statistics, Faculty of Science and Technology, Nakhon Sawan Rajabhat University, Nakhon Sawan, Thailand Abstract. The variant of a semigroup S with respect to an element a ∈ S, is the semigroup with underlying set S and a new binary operation ∗ defined by x ∗ y = xay for x, y ∈ S. Let T (X) be the full transformation semigroup on a nonempty set X. For an arbitrary equivalence E on X, let TE(X) = {α ∈ T (X) : ∀a, b ∈ X, (a, b) ∈ E ⇒ (aα, bα) ∈ E}. Then TE(X) is a subsemigroup of T (X). In this paper, we investigate regular, left regular and right regular elements for the variant of some subsemigroups of the semigroup TE(X). 2020 Mathematics Subject Classifications: 20M20, 20M17 Key Words and Phrases: Transformation semigroup, Equivalence relation, Left regular, Right regular, Completely regular 1. Introduction and preliminaries Let S be a semigroup and a belong to S. We define a new binary operation ∗ on S by putting x ∗ y = xay for all x, y ∈ S. The operation ∗ is clearly associative. Hence (S, ∗) is a semigroup and it is called a variant of S. We usually write (S, a) rather than (S, ∗) to make the element explicit. Variants of abstract semigroups were first studied by Hickey [5]. Although concrete semigroups of relations had earlier been considered by Magill [10]. The study of semigroup variants goes back to the 1960 monograph of Lyapin [9] and a 1967 paper by Magill and Subbiah [6] that considers semigroups of functions X → Y under an operation defined by f · g = f ◦ θ ◦ g, where θ is some fixed function Y → X. In the case that X = Y , this provides an alternative product on the full transformation semigroup T (X) (consisting of all functions X → X). For an element a of a semigroup S, a is called regular if there exists x ∈ S such that a = axa. A semigroup S is regular semigroup if every element of S is regular. Regular ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4596 Email addresses: piyaporn.ta@nsru.ac.th (P. Tantong), nares.sa@nsru.ac.th (N. Sawatraksa) https://www.ejpam.com 2116 © 2022 EJPAM All rights reserved. P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2117 semigroups were introduced by Green [4] in his influential 1951 paper “On the structure of semigroups”. The concept of regularity in a semigroup was adapted from an analogous condition for rings, already considered by Neumann [12]. It was Green’s study of regular semigroups that led him to define his celebrated relations. According to a footnote in Green 1951, the suggestion that the notion of regularity be applied to semigroups was first made by Rees [15, 16]. This property of regular elements was first observed by Thierrin [17] in 1952. Another important kind of the regularity was introduced by Clifford [1] in 1941, who studied elements a of a semigroup S having the property that there exists x ∈ S such that a = axa and ax = xa, which we now call a completely regular element, and semigroups whose any element is completely regular, are called completely regular semigroups. The complete regularity was also investigated by Croisot [2] in 1953, who also studied elements a of a semigroup S for which a ∈ Sa2 (resp. a ∈ a2S), called left regular (resp. right regular) elements, and semigroups whose every element is left regular (resp. right regular), called left regular (resp. right regular) semigroups. In [13], Pei has introduced a family of subsemigroup of T (X) defined by TE(X) = {α ∈ T (X) : ∀a, b ∈ X, (a, b) ∈ E ⇒ (aα, bα) ∈ E} where E is an arbitrary equivalence relation on X. In [13], the author investigated reg- ularity and Green’s relations for TE(X). In [11], Namnak and Laysirikul investigated a necessary and sufficient condition when elements of TE(X) to be left regular, right regular and completely regular. For a fixed element θ in TE(X), the variant semigroup of TE(X) with the sandwich function θ will be denoted simply by TE(X, θ). Green’s equivalences for elements in the sandwich semigroup TE(X, θ) were characterized by Pei alone [13]. Deng, Zeng and Xu [7] introduced a subsemigroup of T (X) defined by TE∗(X) =: {α ∈ T (X) : ∀x, y ∈ X, (x, y) ∈ σ if and only if (xα, yα) ∈ σ}, the so-called semigroups of transformations that preserve double direction equivalence on X. They investigated the regularity and Green’s relations on TE∗(X). Later, Laysilikul and Namnak [8] investigated a necessary and sufficient condition for the left regularity, the right regularity and the completely regularity of elements in TE∗(X). Deng [3] dis- cussed the Green’s ∗-relations, certain ∗-ideal and certain Rees quotient semigroup for the semigroup TE∗(X) and proved that regular and abundant in the semigroup TE∗(X) coincided. The variant semigroup of TE∗(X) with the sandwich function θ, and denoted by (TE∗(X), θ). Yonthanthum [18] investigated a necessary and sufficient condition for an element of (TE∗(X), θ) to be regular and determined when (TE∗(X), θ) is a regular semigroup. However, it easy to see that the notations TE(X) (TE∗(X)) and TE(X, θ) (TE∗(X, θ)) have the same elements. Hence, these are the same set but need not be the same semi- group. If θ is the identity transformation, then these semigroups are coincided. Therefore semigroups TE(X, θ) (TE∗(X, θ)) is a generalization of TE(X) (TE∗(X)) and TE(X, θ). P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2118 In this paper, for a fixed element θ ∈ TE(X) (θ ∈ TE∗(X)), the variant semigroup of TE(X) (TE∗(X)) with the sandwich function θ will be denoted by TE(X, θ) (TE∗(X, θ)). This paper aims to characterize the regular, the left regular, the right regular and the completely regular for elements of TE(X, θ) and TE∗(X, θ). Moreover, we give a necessary and sufficient condition for the identity transformation idX of the semigroup TE(X) to be regular, left regular, right regular and completely regular elements in TE(X, θ) and TE∗(X, θ). In this introductory section, we present many notations and lemma most of which will be indispensable for our research. For arbitrary semigroup S, letReg(S), LReg(S), RReg(S) and CReg(S), denote the set of all regular elements, the set of all left regular elements, the set of all right regular elements and the set of all completely regular elements of S, respectively. For a nonempty set X and α ∈ T (X), we denote by π(α) the partition of X induced by α, namely, π(α) = {yα−1 : y ∈ Xα}. Then π(α) = X/ ker(α) where ker(α) = {(x, y) ∈ X × X : xα = yα}. For A ⊆ X, we define πA(α) = {P ∈ π(α) : P ∩A ̸= ∅}. In the remainder, let E be an equivalence relation on a nonempty set X. Denote by X/E the quotient set. For x ∈ X, we write Ex as for the set of all elements of X that are equivalent to x, that is, Ex = {y ∈ X : (x, y) ∈ E}. The following lemma is needed. Lemma 1. [13] Let α ∈ T (X). Then α ∈ TE(X) if and only if for each A ∈ X/E, there exists B ∈ X/E such that Aα ⊆ B. 2. Regularity of variants of transformation semigroups that preserve an equivalence relation In this section, we characterize the regular, left regular, right regular and completely regular elements of the variant semigroup TE(X, θ). The identity transformation on X, namely, idX is the identity of semigroup TE(X) but need not to be the identity of the semigroup TE(X, θ). In addition, we give a necessary and sufficient condition for idX of the semigroup TE(X) to be regular, left regular, right regular and completely regular elements in TE(X, θ). Theorem 1. Let α ∈ TE(X, θ). Then α ∈ Reg(TE(X, θ)) if and only if (i) ker(α) = ker(αθ) and (ii) for every A ∈ X/E, there exists B ∈ X/E such that A ∩Xαθ ⊆ Bθαθ. Proof. Assume that α is regular of TE(X, θ). Then α = α∗β∗α for some β ∈ TE(X, θ). Thus α = αθβθα. Clearly, ker(α) ⊆ ker(αθ). For the converse conclusion, let x, y ∈ X be such that (x, y) ∈ ker(αθ). Then xαθ = yαθ and so xα = xαθ(βθα) = yαθ(βθα) = yα. P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2119 This means that (x, y) ∈ ker(α). Hence (1) holds. For each A ∈ X/E, by Lemma 1 there exists B ∈ X/E such that Aβ ⊆ B. Thus Aβθ ⊆ Bθ. If a ∈ A ∩ Xαθ, then a ∈ A and a = xαθ for some x ∈ X. Thus aβθ ∈ Bθ. This implies that a = xαθ = xαθβθαθ = aβθαθ ∈ Bθαθ. Hence A ∩Xαθ ⊆ Bθαθ. Conversely, suppose that the conditions (1) and (2) hold. For each A ∈ X/E, we choose A′ ∈ X/E such that A ∩Xαθ ⊆ A′θαθ. Let x ∈ X. If x ∈ Xαθ, then by (2), we choose and fix an element x′ ∈ (Ex) ′ such that x = x′θαθ. Otherwise, if x /∈ Xαθ, we choose and fix an element x′ ∈ (Ex) ′. Define β : X → X by xβ = x′ for all x ∈ X. Then β is a well-defined mapping. Let x, y ∈ X be such that (x, y) ∈ E. Then Ex = Ey and so Ex′ = Ey′ . This implies that β ∈ TE(X, θ). It remains to be verified that α ∗β ∗α = α. If x ∈ X, then xαθβθα = (xαθ)′θα with (xαθ)′θαθ = xαθ. Thus ((xαθ)′θ, x) ∈ ker(αθ) = ker(α). This implies that xαθβθα = (xαθ)′θα = xα and therefore α = α ∗ β ∗ α. Hence α is regular in TE(X, θ). Let E be an equivalence relation on X and Y be a subset of X. A mapping α : Y → X is called E-preserving if for all x, y ∈ Y, (x, y) ∈ E implies (xα, yα) ∈ E. If α satisfies the condition that (xα, yα) ∈ E if and only if (x, y) ∈ E, then α is called E∗-preserving . It is easy to see that every α ∈ TE(X) is E-preserving but need not be E∗-preserving . The following result is immediate from Theorem 1. Corollary 1. idX ∈ Reg(TE(X, θ)) if and only if θ is an E∗-preserving bijection. Proof. Suppose that idX ∈ Reg(TE(X, θ)). Let x, y ∈ X be such that xθ = yθ. Then (x, y) ∈ ker(θ). By Theorem 1, ker(idX) = ker(idXθ) = ker(θ), which implies that x = xidX = yidX = y. Hence θ is an injection. For each x ∈ X, there exists B ∈ X/E such that xθ ∈ Exθ ∩XidXθ ⊆ BθidXθ. Thus xθ = x′θ2 for some x′ ∈ B. Since θ is injective, we deduce that x = x′θ. Consequently, θ is a bijection. Finally, for any x, y ∈ X, if (xθ, yθ) ∈ E, then there exists B ∈ X/E such that xθ, yθ ∈ Exθ ∩ XidXθ ⊆ BθidXθ. Therefore xθ = x′θ2 and yθ = y′θ2 where x′, y′ ∈ B. It follows from θ is injective that x = x′θ and y = y′θ. Since (x′, y′) ∈ E and θ ∈ TE(X), (x, y) = (x′θ, y′θ) ∈ E. We conclude that θ is an E∗-preserving bijection. Conversely, if θ is an E∗-preserving bijection, then θ−1 ∈ TE(X, θ) and idX = idXθ(θ−1θ−1)θidX = idX ∗ (θ−1θ−1) ∗ idX . Hence idX ∈ Reg(TE(X, θ)). In what follows we investigate when an element in TE(X, θ) is left regular. Theorem 2. Let α ∈ TE(X, θ). Then α ∈ LReg(TE(X, θ)) if and only if for every A ∈ X/E, there exists B ∈ X/E such that for each P ∈ πA(α), xθαθ ∈ P for some x ∈ B. P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2120 Proof. Assume that α ∈ LReg(TE(X, θ)). Then α = β ∗ α ∗ α for some β ∈ TE(X, θ) and so α = βθαθα. Let A ∈ X/E. By Lemma 1, there exists B ∈ X/E such that Aβ ⊆ B. Suppose that P ∈ πA(α) and let x ∈ P ∩ A. Hence xβ ∈ B and xα = xβθαθα which means xβθαθ ∈ (xα)α−1 = P . Conversely, for each A ∈ X/E, we choose A′ ∈ X/E such that for every P ∈ πA(α), xθαθ ∈ P for some x ∈ A′. Let x ∈ X. Since X/E and π(α) are partitions of X, there exist A ∈ X/E and P ∈ π(α) such that x ∈ A and x ∈ P . Hence P ∈ πA(α). By assumption, we choose and fix an element x′ ∈ A′ such that x′θαθ ∈ P and A′ ∈ X/E. We also have that x′θαθα = xα. Define β : X → X by xβ = x′ for all x ∈ X. Let x, y ∈ X be such that (x, y) ∈ E. Then Ex = Ey and thus Ex′ = Ey′ . This implies that β ∈ TE(X, θ). If x ∈ X, then xβθαθα = x′θαθα = xα which implies that α = βθαθα. Therefore α = β ∗ α ∗ α and hence α is left regular, as required. Corollary 2. idX ∈ LReg(TE(X, θ)) if and only if θ is a surjection. Proof. Suppose that idX ∈ LReg(TE(X, θ)). Let x ∈ X. Since {x} ∈ πEx(idX) and by Theorem 2, there exists B ∈ X/E such that bθidXθ ∈ {x} for some b ∈ B. Therefore x = bθidXθ = bθθ. Hence θ is a surjection on X. Conversely, assume that θ is a surjection. Thus θ2 is also surjective. Let A ∈ X/E. Note that πA(idX) = {{a} : a ∈ A}. Let P ∈ πA(idX), then P = {a} where a ∈ A and so a = xθθ for some x ∈ X. Choose B = Ex and so xθidXθ = xθθ = a ∈ {a} = P . Hence by Theorem 2, we conclude that idX ∈ LReg(TE(X, θ)). Theorem 3. Let α ∈ TE(X, θ). Then α ∈ RReg(TE(X, θ)) if and only if (θαθ)|Xα is an E∗-preserving injection. Proof. Suppose that α ∈ RReg(TE(X, θ)). Then there is β ∈ TE(X, θ) such that α = α ∗ α ∗ β and so α = αθαθβ. Since θαθ ∈ TE(X), (θαθ)|Xα is E-preserving. Let x, y ∈ Xα be such that x = x′α and y = y′α where x′, y′ ∈ X. If xθαθ = yθαθ, then x = x′α = x′αθαθβ = xθαθβ = yθαθβ = y′αθαθβ = y′α = y. It follows that (θαθ)|Xα is an injection. If (xθαθ, yθαθ) ∈ E, then since β ∈ TE(X, θ), we have that (xθαθβ, yθαθβ) ∈ E. Moreover, (x, y) = (x′α, y′α) = (x′αθαθβ, y′αθαθβ) = (xθαθβ, yθαθβ) ∈ E which implies that (θαθ)|Xα is an E∗-preserving injection. Conversely, assume that (θαθ)|Xα is an E∗-preserving injection. Let A ∈ X/E be such that A ∩ Xαθαθ ̸= ∅. We choose and fix an element xA ∈ A ∩ Xαθαθ. For each x ∈ A ∩ Xαθαθ, there exists a unique element x′ ∈ Xα such that x = x′θαθ by the condition (θαθ)|Xα is injective. We observe that (x′θαθ, x′Aθαθ) = (x, xA) ∈ E. It follows from (θαθ)|Xα is an E∗-preserving mapping that (x′, x′A) ∈ E. Define βA : A → Ex′ A by xβA = { x′ if x ∈ Xαθαθ, x′A otherwise. P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2121 Then we define the map β : X → X by β|A = { βA if A ∩Xαθαθ ̸= ∅, idA otherwise, for all A ∈ X/E. Since X/E is a partition of X, β is well-defined. Let x, y ∈ X be such that (x, y) ∈ E. Then x, y ∈ A for some A ∈ X/E. By the definition of β, we have (xβ, yβ) = (xβ|A, yβ|A). If A ∩ αβαβ = ∅, then (xβ, yβ) = (xidA, yidA) = (x, y) ∈ E. If A ∩ αβαβ ̸= ∅, then xβ, yβ ∈ Aβ = AβA ⊆ Ex′ A . Consequently, β ∈ TE(X, θ). Finally, to show that α = αθαθβ, let x ∈ X. Then xαθαθ ∈ Xαθαθ. Then there exists A ∈ X/E such that xαθαθ ∈ A. By the definition of βA, xαθαθβA = (xαθαθ)′ where (xαθαθ)′θαθ = xαθαθ = (xα)θαθ. Since (xαθαθ)′ is unique, we get that (xαθαθ)′ = xα. Thus xαθαθβ = xαθαθβA = xα. Therefore α = α ∗ α ∗ β. Hence α ∈ RReg(TE(X, θ)), as asserted. Corollary 3. idX ∈ RReg(TE(X, θ)) if and only if θ is an E∗-preserving injection. Proof. Assume that idX ∈ RReg(TE(X, θ)). By Theorem 3, we get that (θidXθ)|XidX is an E∗-preserving injection. This implies that θθ is an E∗-preserving injection. Hence θ is an E∗-preserving injection. This converse of corollary is clear. Final of this section, we give a characterization of completely regular elements in TE(X, θ). Recall that, an element a of a semigroup S is completely regular if and only if a is both left and right regular [14]. Hence, as an immediate consequence of Theorems 2 and 3, we have the following. Theorem 4. Let α ∈ TE(X, θ). Then α ∈ CReg(TE(X, θ)) if and only if (i) for every A ∈ X/E, there exists B ∈ X/E such that for each P ∈ πA(α), xθαθ ∈ P for some x ∈ B and (ii) (θαθ)|Xα is an E∗-preserving injection. As an immediate consequence of Corollaries 2 and 3. Corollary 4. idX ∈ CReg(TE(X, θ)) if and only if θ is an E∗-preserving bijection. Theorem 5. Let α ∈ TE(X, θ). If α ∈ CReg(TE(X, θ)), then every A ∈ X/E, there exists B ∈ X/E such that |P ∩Bθαθ| = |P ∩Xθαθ| = 1 for all P ∈ πA(α). Proof. Assume that α ∈ CReg(TE(X, θ)). Then α is regular, left regular and right regular. Let A ∈ X/E. By Theorem 2, there exists B ∈ X/E such that for each P ∈ πA(α), xθαθ ∈ P for some x ∈ B. For each P ∈ πA(α), we have xθαθ ∈ P for some x ∈ B. Therefore P ∩ Bθαθ ̸= ∅ and P ∩ Xθαθ ̸= ∅. Let y ∈ P ∩ Xθαθ. Then yα = xθαθα. By Theorem 3, we obtain that (θαθ)|Xα is injective. Claim that α|Xθαθ is also injective, let x1, x2 ∈ Xθαθ be such that x1α = x2α. Then x1 = x′1θαθ and P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2122 x2 = x′2θαθ for some x′1, x ′ 2 ∈ X. Thus x′1θαθα = x′2θαθα and so x′1θαθαθ = x′2θαθαθ. Since θαθ|Xα is injective, x′1θα = x′2θα which implies that x1 = x′1θαθ = x′2θαθ = x2. So, we have the claim. This implies that y = xθαθ and hence |P ∩Xθαθ| = 1. It follows from P ∩Bθαθ ⊆ P ∩Xθαθ that |P ∩Bθαθ| = |P ∩Xθαθ| = 1. 3. Regularity of variants of transformation semigroups that preserve double direction equivalence In this section, we characterize the regular, left regular, right regular and completely regular elements of the variant semigroup TE∗(X, θ). In addition, we give a necessary and sufficient condition for the identity transformation idX of the semigroup TE∗(X) to be regular, left regular, right regular and completely regular elements in TE∗(X, θ). Theorem 6. Let α ∈ TE∗(X, θ). Then α ∈ Reg(TE∗(X, θ)) if and only if (i) ker(α) = ker(αθ) and (ii) for every A ∈ X/E, there exists B ∈ X/E such that A ∩Xαθ = Bθαθ. Proof. Suppose that α ∈ Reg(TE∗(X, θ)). Then α ∈ Reg(TE(X, θ)). By Theorem 1(i), we have (i) holds. Let A ∈ X/E. Then by Theorem 1(ii), there exists B ∈ X/E such that A∩Xαθ ⊆ Bθαθ. Since θ, αθ ∈ TE(X) and by Lemma 1, there are C,D ∈ X/E such that Bθ ⊆ C and Cαθ ⊆ D. Therefore A ∩Xαθ ⊆ D. Since A and D are equivalence classes of X, A = D. This implies that A ∩ Xαθ ⊆ Bθαθ ⊆ Cαθ = Cαθ ∩ Xαθ ⊆ A ∩ Xαθ. Hence A ∩Xαθ = Bθαθ. Conversely, assume that conditions (i) and (ii) hold. Define β : X → X as in the proof of Theorem 1. Then β ∈ TE(X) and α = α ∗ β ∗ α. It remains to show that β ∈ TE∗(X). Let x, y ∈ X be such that (xβ, yβ) ∈ E. Then (x′, y′) ∈ E where Ex ∩Xαθ = Ex′θαθ and Ey ∩Xαθ = Ey′θαθ. Thus Ex′ = Ey′ and so Ex ∩Xαθ = Ex′θαθ = Ey′θαθ = Ey ∩Xαθ, which implies that Ex = Ey. Therefore (x, y) ∈ E and hence β ∈ TE∗(X). Corollary 5. idX ∈ Reg(TE∗(X, θ)) if and only if θ is a bijection. Now, we discuss a characterization of an element in the semigroup TE∗(X, θ) to be left regular. Theorem 7. Let α ∈ TE∗(X, θ). Then α ∈ LReg(TE∗(X, θ)) if and only if for every P ∈ π(α), P ∩Xθαθ ̸= ∅. Proof. Suppose that α ∈ LReg(TE∗(X, θ)). Then α ∈ LReg(TE(X, θ)). Let P ∈ π(α) and p ∈ P . Then P ∈ πEp(α). By Theorem 2, there exists B ∈ X/E such that xθαθ ∈ P for some x ∈ B. Hence P ∩Xθαθ ̸= ∅. Conversely, for every P ∈ π(α), P ∩Xθαθ ̸= ∅. We choose and fix an element xP θαθ ∈ P . For each x ∈ X, we let Px ∈ π(α) be such that x ∈ Px. Define β : X → X by xβ = xPx for all x ∈ X. Next, we will show that β ∈ TE∗(X, θ), let x, y ∈ X. If (x, y) ∈ E, then P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2123 (xα, yα) ∈ E and (xβ, yβ) = (xPx , xPy) where xPxθαθ ∈ Px and xPyθαθ ∈ Py. We then have (xPxθαθα, xPyθαθα) = (xα, yα) ∈ E. Since θαθα ∈ TE∗(X), (xβ, yβ) = (xPx , xPy) ∈ E. On the other hand, if (xβ, yβ) ∈ E, then (xPx , xPy) ∈ E where xPxθαθ ∈ Px and xPyθαθ ∈ Py and so (xα, yα) = (xPxθαθα, xPyθαθα) ∈ E. By α ∈ TE∗(X), it follows that (x, y) ∈ E. Hence β ∈ TE∗(X). Finally, to show that α = β ∗ α ∗ α. Let x ∈ X. Then xPxθαθ ∈ Px and hence xβθαθα = xPxθαθα = xα, as required. Corollary 6. idX ∈ LReg(TE∗(X, θ)) if and only if θ is a surjection. Proof. The necessity is clear from Corollary 2. To prove the sufficiency, we suppose that θ is a surjection. Then θθ is also a surjection. Since π(idX) = {{x} : x ∈ X}, P ∩ XθidXθ = P ∩ Xθθ ̸= ∅ for all P ∈ π(idX). Hence idX ∈ LReg(TE∗(X, θ)), by Theorem 7. Next, we characterize a right regular element of TE∗(X, θ). The following lemma is needed. Lemma 2. [8] Let α ∈ TE∗(X, θ) and A,B ∈ X/E. If Aα ⊆ B, then Bα−1 = A. Theorem 8. Let α ∈ TE∗(X, θ). Then α ∈ RReg(TE∗(X, θ)) if and only if (i) (θαθ)|Xα is an injection and (ii) if there exists A ∈ X/E such that A ∩ X(αθ)2 = ∅, then there exists an injection φ : {A ∈ X/E : A ∩X(αθ)2 = ∅} → {A ∈ X/E : A ∩Xα = ∅}. Proof. Assume that α ∈ RReg(TE∗(X, θ)). Then α ∈ RReg(TE(X, θ)). By Theorem 3, we then have (i) hold. Next, we prove that (ii) holds in the following. Suppose that {A ∈ X/E : A ∩ X(αθ)2 = ∅} ̸= ∅. Let A ∈ X/E be such that A ∩ X(αθ)2 = ∅. Since α ∈ RReg(TE∗(X, θ)), there exists β ∈ TE∗(X, θ) such that α = α ∗ α ∗ β and so α = αθαθβ. By Lemma 1, we let A′ ∈ X/E such that Aβ ⊆ A′. Claim that A′ ∩Xα = ∅, suppose not. Let x ∈ X be such that xα ∈ A′ and choose a ∈ A. Then aβ ∈ A′ and so (xαθαθβ, aβ) = (xα, aβ) ∈ E. Since β ∈ TE∗(X), we get (xαθαθ, a) ∈ E. Hence xαθαθ ∈ A which is a contradiction. Thus A′ ∩ Xα = ∅. Define φ : {A ∈ X/E : A ∩X(αθ)2 = ∅} → {A ∈ X/E : A ∩Xα = ∅} by Aφ = A′ for all A ∈ X/E and A ∩X(αθ)2 = ∅. To show that φ is injective, let A,B ∈ {A ∈ X/E : A ∩ X(αθ)2 = ∅} be such that Aφ = Bφ. By the definition of φ,Aφ = A′ and Bφ = B′ where Aβ ⊆ A′ and Bβ ⊆ B′ for some A′, B′ ∈ X/E. It follows from Lemma 2 that A = A′β−1 and B = B′β−1. Since A′ = B′, we deduce that A = B. Therefore φ is an injection. Hence (ii) holds. Conversely, suppose that the conditions (i) and (ii) hold. For each x ∈ X(αθ)2, we choose and fix an element x′ ∈ Xα such that x = x′θαθ. Let A ∈ X/E be such that A ∩X(αθ)2 ̸= ∅. Then we fix xA ∈ A ∩Xα and define βA : A → X by xβA = { x′ if x ∈ X(αθ)2, x′A otherwise. P. Tantong, N. Sawatraksa / Eur. J. Pure Appl. Math, 15 (4) (2022), 2116-2126 2124 Let A ∈ X/E be such that A ∩X(αθ)2 = ∅ and x ∈ A. By (ii), we fix x̃ ∈ Aφ and define βA : A → X by xβA = x̃ for all x ∈ A. Let β : X → X by β|A = βA for all A ∈ X/E. Since X/E is a partition of X,β is well-defined. Let x, y ∈ X be such that (x, y) ∈ E. Then x, y ∈ A for some A ∈ X/E. There are two cases to consider. Case 1. A ∩X(αθ)2 = ∅. Then (xβ, yβ) = (x̃, ỹ) ∈ E. Case 2. A ∩X(αθ)2 ̸= ∅. Without loss of generality, we assume that x, y ∈ X(αθ)2. Hence xβ = x′ and yβ = y′ where x = x′θαθ and y = y′θαθ, respectively. Since θαθ ∈ TE∗(X, θ) and (x′θαθ, y′θαθ) ∈ E, we conclude that (xβ, yβ) = (x′, y′) ∈ E. On the other hand, let x, y ∈ X be such that (xβ, yβ) ∈ E. Thus xβ, yβ ∈ B for some B ∈ X/E. If B ∩Xα = ∅, then by the definition of β, xβ, yβ ∈ B = Exφ = Eyφ. Since φ is an injection, Ex = Ey and hence (x, y) ∈ E. If B ∩Xα ̸= ∅, then by the definition of β, we may assume that xβ = x′, yβ = y′ for some x′, y′ ∈ Xα with x = x′θαθ and y = y′θαθ. Since (x′, y′) = (xβ, yβ) ∈ E and θαθ ∈ TE∗(X), we deduce that (x, y) = (x′θαθ, y′θαθ) ∈ E. It follows that β ∈ TE∗(X). Let x ∈ X, then x(αθ)2 ∈ X(αθ)2 and there exists (x(αθ)2)′ ∈ Xα such that (x(αθ)2)′θαθ = x(αθ)2 = (xα)θαθ. We note by (1) that (x(αθ)2)′ = xα. Therefore x(α ∗ α ∗ β) = xαθαθβ = x(αθ)2β = (x(αθ)2)′ = xα. Hence α is right regular, as required. Corollary 7. idX ∈ RReg(TE∗(X, θ)) if and only if (i) θ is an injection and (ii) A ∩Xθ ̸= ∅ for all A ∈ X/E. Proof. Assume that idX ∈ RReg(TE∗(X, θ)). By Corollary 3, we get that θ is injective. Suppose that A ∩ Xθ2 = ∅ for some A ∈ X/E. By (2) of Theorem 8, there exists an injection φ : {A ∈ X/E : A ∩X(idXθ)2 = ∅} → {A ∈ X/E : A ∩XidX = ∅}. This is a contradiction with {A ∈ X/E : A ∩ XidX = ∅} = {A ∈ X/E : A ∩ X = ∅} = ∅. Thus A∩Xθ2 ̸= ∅ for all A ∈ X/E. It follows from Xθ2 ⊆ Xθ that A∩Xθ ̸= ∅ for all A ∈ X/E. The converse of corollary follows from Theorem 8. The following result is obtained directly from Theorem 7 and Theorem 8. Theorem 9. Let α ∈ TE∗(X, θ). Then α ∈ CReg(TE∗(X, θ)) if and only if (i) for every P ∈ π(α), P ∩Xθαθ ̸= ∅, (ii) (θαθ)|Xα is an injection and (iii) if there exists A ∈ X/E such that A ∩ X(αθ)2 = ∅, then there exists an injection φ : {A ∈ X/E : A ∩X(αθ)2 = ∅} → {A ∈ X/E : A ∩Xα = ∅}. Corollary 8. idX ∈ CReg(TE∗(X, θ)) if and only if θ is a bijection. REFERENCES 2125 4. Conclusion and discussion In this work, we presented necessary and sufficient conditions when elements of the semigroups TE(X, θ) and TE∗(X, θ) to be regular, left regular, right regular and completely regular. 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