EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 4, 2020, 987-994 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Left and Right Magnifying Elements in the Semigroup of all Binary Relations Watchara Teparos1, Soontorn Boonta1, Thitiya Theparod2,∗ 1 Department of General Science, Faculty of Science and Engineering, Kasetsart University, Chalermphrakiat Sakon Nakhon Province Campus, Muang, Sakon Nakhon, Thailand 2 Department of Mathematics, Faculty of Science, Mahasarakham University, Khamriang Sub-District, Kantarawichai District, Maha Sarakham, Thailand Abstract. An element a of a semigroup S is called left [right] magnifying if there exists a proper subset M of S such that S = aM [S = Ma]. Let X be a nonempty set and BX the semigroup of binary relations on X. In this paper, we give necessary and sufficient conditions for elements in BX to be left or right magnifying. 2020 Mathematics Subject Classifications: 20M20 Key Words and Phrases: Relations, functions, magnifying elements, the semigroup of all binary relations 1. Introduction and Preliminaries An element a of a semigroup S is called left [right] magnifying if there exists a proper subset M of S such that S = aM [S = Ma] in which the concept of such definition was first introduced by Ljapin [4]. In 1971, Migliorini [6] studies week left [right] and strong left [right] magnifying element of semigroups which specifically gives a definition of a proper subset M of S as a subsemigroup. In the following years, there are several studies on other properties of magnifying elements (see [1–3, 5, 7]). The semigroup of all binary relations are widely known and there are many research in the area of this type of semigroup ([8, 11, 12]). In 2018, Chinram, Petchkaew and Baupradist [9] give necessary and sufficient conditions for elements in some generalized linear transformation semigroups. Then in the next year, Baupradist, Panityakul and Chinram [10] give necessary and sufficient conditions for elements in semigroups of linear transformations with restricted range to be left or right magnifying. Our research is motivated by these studies. In this paper, we give necessary and sufficient conditions for elements in the semigroups of binary relations to be left and right magnifying. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3870 Email addresses: watchara.tha@ku.th (W. Teparos), soontorn.bo@ku.th (S. Boonta), thitiya.t@msu.ac.th (T. Theparod) https://www.ejpam.com 987 c© 2020 EJPAM All rights reserved. W. Teparos, S. Boonta, T. Theparod / Eur. J. Pure Appl. Math, 13 (4) (2020), 987-994 988 Throughout this paper, we let X be a nonempty set and |X| be the cardinality of X. It is well-known that the set of all binary relations on X form a semigroup, denoted by BX , under composition: α ◦ β ={(x, y) | ∃z ∈ X such that (x, z) ∈ α and (z, y) ∈ β} for all α, β ∈ BX . Let a, b ∈ X, A ⊆ X and α ∈ BX . The domain and range of α are denoted by domα = {x ∈ X | (x, y) ∈ α} and ranα = {y ∈ X | (x, y) ∈ α}, respectively. We use the following notions: α−1 = {(a, b) | (a, b) ∈ α}, (a)α = {y ∈ X | (a, y) ∈ α}, (a)α−1 = {x ∈ X | (x, a) ∈ α}, (A)α = {y ∈ X | y ∈ (a)α for some a ∈ A}, (A)α−1 = {x ∈ X | x ∈ (a)α−1 for some a ∈ A}, α|A = {(a, b) | a ∈ A and (a, b) ∈ α} Note that we can write (a)α instead of ({a})α. The universal relation X×X is denoted by ω and the identity relation on X is denoted by iX . Here, we will write a relation from the right, (a)α rather than α(a) and compose from the left to the right, (a)(αβ) rather than (β ◦ α)(a), for α, β ∈ BX . 2. Left Magnifying Elements of BX Lemma 1. Let α ∈ BX . |(y)α−1| = 1 for all y ∈ ranα if and only if for every a, b ∈ domα, (a)α ∩ (b)α 6= ∅ implies a = b. Proof. Assume that |(y)α−1| = 1 for all y ∈ ranα. Let a, b ∈ domα. Suppose that (a)α ∩ (b)α 6= ∅. Let c ∈ (a)α ∩ (b)α. Thus a ∈ (c)α−1 and b ∈ (c)α−1. Since a, b ∈ ranα, we have |(c)α−1| = 1, then a = b. Conversely, assume that for every a, b ∈ domα, (a)α∩ (b)α 6= ∅ implies a = b. Let y ∈ ranα. Suppose that |(y)α−1| > 1. Then there exist two distinct elements a, b ∈ domα such that a, b ∈ (y)α−1. Thus y ∈ (a)α and y ∈ (b)α. That is, (a)α∩ (b)α 6= ∅. As a result, a = b in which we obtain a contradiction. Therefore |(y)α−1| = 1. Lemma 2. Let α ∈ BX . |(y)α−1| = 1 for all y ∈ ranα if and only if (Aα)α−1 = A for every A ⊆ domα. Proof. Assume that |(y)α−1| = 1 for all y ∈ ranα. It is clear that A ⊆ (Aα)α−1. We want to show that (Aα)α−1 ⊆ A. Let x ∈ (Aα)α−1. Then there exists b ∈ Aα such that x ∈ (b)α−1. We have that b ∈ (x)α and therefore there exists a ∈ A such that b ∈ (a)α. That is, b ∈ (x)α ∩ (a)α. As a result, (x)α ∩ (a)α 6= ∅. By Lemma 1, x = a ∈ A. W. Teparos, S. Boonta, T. Theparod / Eur. J. Pure Appl. Math, 13 (4) (2020), 987-994 989 Conversely, assume that (Aα)α−1 = A for all A ⊆ domα. Let y ∈ ranα. We will show that |(y)α−1| = 1. Since y ∈ ranα, there exist a ∈ domα such that y ∈ (a)α. Clearly, a ∈ (y)α−1. Suppose that |(y)α−1| > 1. Then there exist b ∈ (y)α−1 such that b 6= a. Since {a} ⊆ domα and by assumption, we have (({a})α)α−1 = {a}. Therefore b ∈ (y)α−1 ⊆ ((a)α)α−1 = {a}. Thus b = a, a contradiction. As a result, |(y)α−1| = 1. Combining Lemmas 1 and 2 we have the following Lemma. Lemma 3. Let α ∈ BX . Then the following are equivalent: (i) for every a, b ∈ domα, (a)α ∩ (b)α 6= ∅ implies a = b; (ii) |(y)α−1| = 1 for all y ∈ ranα; (iii) (Aα)α−1 = A for every A ⊆ domα. Lemma 4. Let α ∈ BX . If α is a left magnifying element of BX , then (i) dom(α) = X, (ii) for any x, y ∈ dom(α), (a)α ∩ (b)α 6= ∅ implies a = b. Proof. Assume that α is a left magnifying element of BX . Then there exists a proper subset M of BX such that αM = BX . Since iX ∈ BX , there exists a relation β ∈M such that αβ = iX and since dom(iX) = X, we obtain domα = X. Let x, y ∈ domα. Suppose that (a)α ∩ (b)α 6= ∅. Then there exists an element z ∈ (x)α ∩ (y)α. We get (z)β ⊆ ((x)α)β = (x)(αβ) = (x)iX = {x}, and (z)β ⊆ ((y)α)β = (y)(αβ) = (y)iX = {y}. Thus (z)β ⊆ {x} and (z)β ⊆ {y}. We have that x = y. Therefore (x)α ∩ (y)α = ∅. Lemma 5. Let α ∈ BX . If for any x, y ∈ dom(α) such that (x)α ∩ (y)α 6= ∅, x = y and α is a function, then α is a one-to-one function. The proof of Lemma 5 is obvious and immediately obtained. Lemma 6. Let α ∈ BX . If α is a bijective function on X, then α is not left magnifying element of BX . Proof. We are going to proof this Lemma similarly to [9, Lemma 2.]. Suppose that α is a left magnifying element of BX . Then there exists a proper subset M of BX such that αM = BX . Since α is a bijective function, αM = αα−1αM = αα−1BX ⊆ αBX ⊆ BX = αM. Then αM = αBX . Hence M = α−1αM = α−1αBX = BX ; we achieve a contradiction. Therefore α is not a left magnifying element of BX . W. Teparos, S. Boonta, T. Theparod / Eur. J. Pure Appl. Math, 13 (4) (2020), 987-994 990 Lemma 7. Let α ∈ BX and domα = X. If α is not a bijective function on X and for every a, b ∈ domα, (a)α ∩ (b)α 6= ∅ implies a = b, then α is a left magnifying element of BX . Proof. Let α ∈ BX and domα = X. Assume that α is not a bijective function on X and for every a, b ∈ domα, (a)α ∩ (b)α 6= ∅ implies a = b. Let M = {h ∈ BX | domh 6= X or h is not a one-to-one function}. Then M is a proper subset of BX . We will show that αM = BX . Let β be a relation in BX . For each x ∈ (domβ)α. Then there exists a unique x′ ∈ domβ such that x ∈ (x′)α. Define a relation γ in BX by (x)γ = (x′)β where x′ ∈ domβ such that x ∈ (x′)α. If (domβ)α 6= X, then domγ 6= X. So γ ∈M . If β is not a function, by the definition of γ, clearly, γ is not a function. Consequently γ ∈M . Suppose that (domβ)α = X and β is a function. We will show that γ is not a one-to- one function. Let x ∈ X and c ∈ (x)α. Thus we have c ∈ X = (domβ)α. Then there exists x′ ∈ domβ such that c ∈ (x′)α. That is, (x)α ∩ (x′)α 6= ∅. Consequently, x = x′ ∈ domβ. Therefore domβ = X. Next, we will show that α is not a function from domβ = X onto X. Suppose the contrary that α is a function from X onto X. By our assumption and Lemma 5, α is a one-to-one function. Hence α is a bijective function on X. This is a contradiction. Therefore α is not a function from domβ = X onto X. Finally, we will show that γ is not a one-to-one function. Since α is not a function from domβ onto X, there exists x0 ∈ domβ such that |(x0)α| > 1 and that there exist two distinct elements u, v in (x0)α. By the definition of γ, we have (u)γ = (v)γ = (x0)β. Therefore γ can not be a one-to-one function. Hence γ ∈M . By Lemma 3(iii), we obtain dom(αγ) = (ranα ∩ domγ)α−1 = (ranα ∩ (domβ)α)α−1 = ((domβ)α)α−1 = domβ. Therefore dom(αγ) = domβ. For each x ∈ dom(αγ), we have (x)(αγ) = ((x)α)γ = (x)β. Then αγ = β, and as a result, αM = BX . Hence the proof is complete. Theorem 1. Let α ∈ BX . Then α is a left magnifying element of BX if and only if (i) domα = X, (ii) for every x, y ∈ domα, (x)α ∩ (y)α 6= ∅ implies x = y, and (iii) α is not a bijective function on X. W. Teparos, S. Boonta, T. Theparod / Eur. J. Pure Appl. Math, 13 (4) (2020), 987-994 991 Proof. It follows from Lemma 4, Lemma 6 and Lemma 7. Example 1. Let X = N and α ∈ BX defined by (x)α = {2x, 3x} for all x ∈ X. Following the assumption we have domα = X, for every x, y ∈ X such that x 6= y, (x)α ∩ (y)α = ∅ and α is not a function. So that α is not a bijective on X. Let M = {h ∈ BX | domh 6= X or h is not a one-to-one function}. Let β be any relation in BX . By Lemma 7, we can define a relation γ ∈ BX such that γ ∈ M and γα = β. For example, if β is a relation in BX such that (x)β = {2x} for all odd integer x. Thus domβ = {x ∈ N | x is odd}. So (domβ)α = {2x | x is odd} ∪ {3x | x is odd}. Define a relation γ in BX by (x)γ = (y)β if x = 2y or x = 3y for some odd integer y. Thus domγ = {2x | x is odd}∪{3x | x is odd} 6= X and domγ = {2x | x is odd}∪{3x | x is odd} ⊆ ranα. We have dom(αγ) = (ranα ∩ domγ)α−1 = (domγ)α−1 = ({2x | x is odd} ∪ {3x | x is odd})α−1 = {x ∈ N | x is odd} = domβ. That is, dom(αγ) = domβ. For each an odd integer x, (x)(αγ) = ((x)α)γ = ({2x, 3x}) γ = (x)β ∪ (x)β = (x)β. Therefore αγ = β. 3. Right Magnifying Elements of BX Lemma 8. If α is a right magnifying element of BX , then there exists a subset A of X such that α|A is an onto function from A to X. Proof. Assume that α is a right magnifying element of BX . Then there exists a proper subset M of BX such that Mα = BX . Since iX ∈ BX , then there exists γ ∈M such that γα = iX . It is clear that ranγ ∩domα 6= ∅. We put A := ranγ ∩domα. First, we will show that α|A is a function from A to X. Let x ∈ A. Then there exists u ∈ domγ such that x ∈ (u)γ. Therefore (x)α ⊆ ((u)γ)α = (u)(γα) = (u)iX = {u}. Thus |(x)α| = 1. As a result, α|A is a function from A to X. Next, we will show that α|A is onto. Let y ∈ X. Then y ∈ ran(iX) = ran(γα), giving that there exists v ∈ dom(γα) such that y ∈ (v)(γα) = ((v)γ)α. We have y ∈ ((v)γ)α when (v)γ ∈ (ranγ ∩ domα). This leads us to a conclusion that α|A is onto. The proof of the following Lemma is done similarly to [9, Lemma 5.]. W. Teparos, S. Boonta, T. Theparod / Eur. J. Pure Appl. Math, 13 (4) (2020), 987-994 992 Lemma 9. If α ∈ BX is bijective, then α is not a right magnifying element. Proof. Assume that α is a right magnifying element of BX . Then there exists a proper subset M of BX such that Mα = BX . Since α is a bijective function, Mα = Mαα−1α = BXα−1α ⊆ BXα ⊆ BX = Mα. Then Mα = BXα. Hence M = Mαα−1 = BXαα−1 = BX . This is a contradiction. Therefore α is not a right magnifying element of BX . Lemma 10. Let α ∈ BX . If there exists a subset A of X such that α|A is an onto function from A to X and α is not a bijective function on X, then α is a right magnifying element of BX . Proof. Assume that there exists a subset A of X such that α|A is an onto function from A to X and α is not a bijective function on X. Let M = {h ∈ BX | ranh 6= X}. Then M is a proper subset of BX . We will show that there exists a proper subset C of X such that α|C is an onto function from C to X. If A 6= X, then it is immediate by setting C := A. If A = X, then α is an onto function on X, but as α is not a bijective function on X, α is not a one-to-one function on X. Then there exists u ∈ X which α|X\{u} is an onto function from X \ {u} to X. By put C := X \ {u}, the proof is immediately obtained. Let β ∈ BX and x ∈ domβ. Then (x)β ⊆ X. Since α|C is onto, there exists a nonempty subset Cx of C such that (Cx)α = (x)β. From these we defined a relation γ ∈ BX by (x)γ = Cx for each x ∈ domβ. It is clearly that ranγ ⊆ C 6= X, then γ ∈M . Consider (x)β = (Cx)α = ((x)γ)α = (x)(γα), where x ∈ domβ, we have γα = β. Therefore BX = Mα. Theorem 2. Let α ∈ BX . Then α is a right magnifying element of BX if and only if there exists a subset A of X such that α is an onto function from A to X and α is not a bijective function on X. Proof. Trivial from Lemma 8, Lemma 9 and Lemma 10. Example 2. Let X = N and α be a relation in BX defined by (x)α = {{ x 2 } for all even x, {x, x+ 1} for all odd x. Then α|A is an onto function from A to N, where A = {x ∈ N | x is even}, and α is not a bijective function on N. We put M := {h ∈ BX | ranh 6= X}. Let β ∈ BX . By Lemma 3, there exists a relation γ in BX such that γα = β. W. Teparos, S. Boonta, T. Theparod / Eur. J. Pure Appl. Math, 13 (4) (2020), 987-994 993 For example, (a) if β ∈ BX such that (x)β = {x+ 3} for each x where x is odd. Then domβ = {x ∈ N | x is odd}. Define γ ∈ BX by (x)γ = 2(x+ 3) for all x that is odd. Thus domγ = domβ and ranγ ⊆ domα 6= X. We have γ ∈ M . For each odd integer x, then (x)(γα) = ((x)γ)α = (2(x+ 3))α = { 2(x+ 3) 2 } = {x+ 3} = (x)β. Therefore γα = β. (b) If β = {(1, 2), (1, 3), (2, 3)}, then β ∈ BX . Define γ ∈ BX by γ = {(1, 2), (1, 6), (2, 6)}. Thus (1)(γα) = ((1)γ)α = ({2, 6})α = {1, 3}, and (2)γα = ((1)γ)α = ({6})α = {3}. Therefore, γα = β. (c) If β ∈ BX such that (x)β = {x, x+ 2} for all x ∈ N. Define γ ∈ BX by (x)γ = {2x, 2(x+ 2)} for all x is an integer. Clearly that dom(γα) = domβ. Let x ∈ N. Then (x)(γα) = ((x)γ)α = ({2x, 2(x+ 2)})α = { 2x 2 , 2(x+ 2) 2 } = {x, x+ 2} = (x)β. Therefore (x)(γα) = (x)β for all x ∈ N. Acknowledgements The authors thank Faculty of Science and Engineering, Kasetsart University, Chalermphrakiat Sakon Nakhon Province Campus, for support. TT was financially supported by Faculty of Science, Mahasarakham University Grant Year 2018. 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