EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6172 ISSN 1307-5543 – ejpam.com Published by New York Business Global Finite Γ-AG-Groupoids With Left Identities and Left Zeros Cholathorn Chanoi1, Apichaya Kauppamung1, Panuwat Luangchaisri1, Thawhat Changphas1,∗ 1Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand Abstract. Let Γ be a nonempty set. A nonempty set A is called a Γ-AG-groupoid if there is a function f from A × Γ × A into A, customary denoted aγb for f(a, γ, b), satisfying the identity (aγb)βc = (cγb)βa for any a, b, c ∈ A and γ, β ∈ Γ. For each γ ∈ Γ, an operation on A associated to γ is given by ab = aγb. Suppose further that A is finite, contains a left identity and a left zero a0. The objective of this paper is to provide sufficient conditions under which the set A \ {a0} is a commutative group under the operation on A determined by γ for all γ ∈ Γ. 2020 Mathematics Subject Classifications: 20N02 Key Words and Phrases: AG-groupoid, Γ-AG-groupoid, semigroup, Γ-semigroup, group. 1. Introduction An Abel-Grassmann’s groupoid, abreviated by AG-groupoid, is a groupoid A such that the identity (ab)c = (cb)a holds for any a, b, c ∈ A. An AG-groupoid is also called a left almost semigroup (it is abreviated by LA-semigroup), a left invertive groupoid, or a right modular groupoid (cf. [1], [2], [3], [4]). An AG-groupoid is closely related to a commutative semigroup, because if an AG- groupoid contains a right identity, then it becomes a commutative monoid. Moreover, if an AG-groupoid A with a left identity and a left zero a0 is finite, then (under certain conditions) A \ {a0} is a commutative group (cf. [5] Theorem 2.2). The purpose of this paper is to extend this result to Γ-AG-groupoids. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6172 Email addresses: thacha@kku.ac.th (T. Changphas) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 2 of 9 2. Preliminaries Let Γ be a nonempty set. A nonempty set A is called a Γ-groupoid if there is a function f of A× Γ× A into A, we often write aγb for f(a, γ, b). Suppose A is a Γ-groupoid. For each γ ∈ Γ, an operation on A determined by γ is defined by for any a, b ∈ A, ab = aγb. A Γ-groupoid A is called a Γ-semigroup if the identity (aγb)βc = aγ(bβc) holds for all a, b, c ∈ A and γ, β ∈ Γ. A Γ-semigroup A is said to be commutative if aγb = bγa for all a, b ∈ A and γ ∈ Γ. The notion of Γ-semigroup was introduced and studied by M. K. Sen (cf. [6], [7]). Suppose A is a semigroup. For a nonempty set Γ, define aγb = ab for any a, b ∈ A and γ ∈ Γ; then A is a Γ-semigroup. If A is a Γ-semigroup, then for any γ ∈ Γ, A is a semigroup under the operation determined by γ. Example 1. Let A and Γ be the set of all nonpositive integers and the set of all nonpositive even integers, respectively. For a, b ∈ A and γ ∈ Γ, define aγb to be the usual multiplication of integers; then A is a Γ-semigroup. Example 2. Let Mat2×3(Q) denote the set of all 2×3 matrices over Q, the set of rational numbers. And, let Γ denote the set of all 3× 2 matrices over Q. For A,B ∈ Mat2×3(Q) and γ ∈ Γ, define AγB to be the usual matrix product. Then Mat2×3(Q) is a Γ-semigroup. Note that Mat2×3(Q) is not a semigroup under the usual product of matrices. We need the following theorem proved in [8] (also, in [9]). Theorem 1. Suppose A is a Γ-semigroup. If A is a group under the operation defined by γ for some γ ∈ Γ, then A is a group under the operation determined by γ for all γ ∈ Γ. Let Γ be a nonempty set. A nonempty set A is called a Γ-AG-groupoid if there is a function f from A× Γ×A into A, it is customary to write aγb for f(a, γ, b), such that (aγb)βc = (cγb)βa for all a, b, c ∈ A and γ, β ∈ Γ. Suppose A is an AG-groupoid and Γ is a nonempty set. Then A is a Γ-AG-groupoid under the function defined by aγb = ab for all a, b ∈ A and γ ∈ Γ. If A is a Γ-AG-groupoid, then for any γ ∈ Γ, A is an AG-groupoid under the operation determined by γ. Example 3. Let Γ = {1, 2, . . . , n}. Define a function from Z × Γ × Z into Z, the set of all integers, by aγb = b− γ − a for all a, b ∈ Z and γ ∈ Γ, where − is a usual subtraction of integers. Then Z is a Γ-AG-groupoid. C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 3 of 9 Example 4. Let A = Γ = {0, i,−i}. Define a function from A×Γ×A into A by aγb for all (a, γ, b) ∈ A × Γ × A; here aγb is the usual multiplication of complex numbers. Then A is a Γ-AG-groupoid, where as A is not an AG-groupoid. The following theorem is in [10] (see also in [11]). Theorem 2. Any Γ-AG-groupoid satisfies the medial law. That is, if A is a Γ-AG- groupoid, then (aγb)β(cαd) = (aγc)β(bαd) for any a, b, c, d ∈ A and γ, β, α ∈ Γ. An element e of a Γ-AG-groupoid A is said to be a left identity if for all a ∈ A and γ ∈ Γ, eγa = a. An element a0 of a Γ-AG-groupoid A is said to be a left zero if for all a ∈ A and γ ∈ Γ, a0γa = a0. A Γ-AG-groupoid A is said to be cancellative if for all a, b, c ∈ A and γ ∈ Γ, (aγc = bγc or cγa = cγb) imply a = b. Example 5. Consider an AG-groupoid A = {1, 2, 3, 4} with the operation defined as follows: · 1 2 3 5 1 1 2 3 5 2 3 3 3 3 3 3 3 3 3 5 2 3 3 3 Let Γ be a nonempty set. For a, b ∈ A and γ ∈ Γ, define aγb = a · b. We have that A is a finite Γ-AG-groupoid with left identity 1, and left zero 3. 3. Results We begin this section with the following theorem. Theorem 3. If A is a Γ-AG-groupoid satisfying the identity aγ(bβc) = (cγb)βa for all a, b, c ∈ A and γ, β ∈ Γ, then A is a Γ-semigroup. Proof. Assume the condition holds. Then, for a, b, c ∈ A and γ, β ∈ Γ, we have (aγb)βc = (cγb)βa = aγ(bβc). Thus A is a Γ-semigroup. C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 4 of 9 Theorem 4. If A is a cancellative Γ-AG-groupoid satisfying the identity aγ(bβc) = (cγb)βa for all a, b, c ∈ A and β, γ ∈ Γ, then for any γ ∈ Γ, A is a commutative semigroup under the operation determined by γ. Proof. Assume the condition holds. By Theorem 3, for any γ ∈ Γ, A is a semigroup under the operation determined by γ. Let a, b ∈ A and γ ∈ Γ. Consider: (aγ(aγb))γa = ((aγa)γb)γa = (aγa)γ(bγa) = (aγb)γ(aγa) = ((aγb)γa)γa. So (aγ(aγb))γa = ((aγb)γa)γa. By cancellative law, aγ(aγb) = (aγb)γa. By assumption, aγ(aγb) = aγ(bγa). Using cancellative law, aγb = bγa. Hence A is a commutative semi- group under the operation determined by γ. An AG-groupoid A is said to be cancellative if for all a, b, c ∈ A, ac = bc or ca = cb imply a = b. We specifically have the following corollary: Corollary 1. If A is a cancellative AG-groupoid satisfying the identity a(bc) = (cb)a for all a, b, c ∈ A, then A is a commutative semigroup. Now, we present the main result. Theorem 5. Let A be a finite Γ-AG-groupoid containing at least two elements (|A| > 1). Suppose A contains a left identity e and a left zero a0, and A satisfies the identity aγ(bβc) = (cγb)βa for all a, b, c ∈ A and γ, β ∈ Γ. Suppose further that there exist γ0 ∈ Γ and an operation ∗ of A×A into A, write a ∗ b for ∗(a, b), such that (i)-(v) hold: (i) A is an AG-groupoid under ∗. (ii) For any a ∈ A there exists b ∈ A such that b ∗ a = a0. (iii) a0 ∗ a = a for all a ∈ A. (iv) (a ∗ b)γ0c = (aγ0c) ∗ (bγ0c) for all a, b, c ∈ A. C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 5 of 9 (v) For any a, b ∈ A, if aγ0b = a0 then a = a0 or b = a0. Then A \ {a0} is a commutative group under the operation determined by γ for all γ ∈ Γ. Proof. Let A = {a0, a1, . . . , am}, where m ≥ 1. Claim 1: A \ {a0} is an AG-groupoid under the operation determined by γ0. Since m ≥ 1, A \ {a0} ≠ ∅. Suppose aiγ0aj = a0 for some ai, aj ∈ A \ {a0}. By (v), ai = a0 or aj = a0. This is a contradiction. Thus aiγ0aj ∈ A \ {a0} for all ai, aj ∈ A \ {a0}; so A \ {a0} is a groupoid under the operation determined by γ0. From A \ {a0} ⊆ A, it follows that (aiγ0aj)γ0ak = (akγ0aj)γ0ai for all ai, aj , ak ∈ A \ {a0}. Claim 2: e ̸= a0. Suppose not. If ai ∈ A then ai = eγ0ai = a0γ0ai = a0. Thus A = {a0}, this is a contradiction. So e ̸= a0. Claim 3: a0γ0ai = aiγ0a0 for all ai ∈ A. Let ai ∈ A. Consider: (aiγ0a0)γ0e = (eγ0a0)γ0ai = a0γ0ai = a0. By (v), aiγ0a0 = a0 or e = a0. By Claim 2, aiγ0a0 = a0. Hence a0γ0a = a0 = aiγ0a0. Claim 4: For each ak ∈ A \ {a0} there exists a−1 k ∈ A \ {a0} such that akγ0a −1 k = e = a−1 k γ0ak. Let ak ∈ A \ {a0}. Consider: Hk,γ0 = {akγ0a1, akγ0a2, . . . , akγ0am}. To show that |Hk,γ0 | = m, suppose akγ0ar = akγ0as for some r ̸= s. Consider: arγ0ak = (eγ0ar)γ0ak = (akγ0ar)γ0e = (akγ0as)γ0e = (eγ0as)γ0ak = asγ0ak. By (ii), there exists a−1 r ∈ A such that a−1 r ∗ ar = a0. Consider (Using (i), (iii), (iv)): (as ∗ a−1 r )γ0ak = (asγ0ak) ∗ (a−1 r γ0ak) = (arγ0ak) ∗ (a−1 r γ0ak) = (ar ∗ a−1 r )γ0ak = (a0 ∗ (ar ∗ a−1 r ))γ0ak C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 6 of 9 = ((a0 ∗ a0) ∗ (ar ∗ a−1 r ))γ0ak = (((ar ∗ a−1 r ) ∗ a0) ∗ a0)γ0ak = (((a0 ∗ a−1 r ) ∗ ar) ∗ a0)γ0ak = ((a−1 r ∗ ar) ∗ a0)γ0ak = (a0 ∗ a0)γ0ak = a0γ0ak = a0. By (v), as ∗ a−1 r = a0 or ak = a0. Since ak ̸= a0, as ∗ a−1 r = a0. Consider: ar = a0 ∗ ar = (as ∗ a−1 r ) ∗ ar = (ar ∗ a−1 r ) ∗ as = (a0 ∗ (ar ∗ a−1 r )) ∗ ak = ((a0 ∗ a0) ∗ (ar ∗ a−1 r )) ∗ ak = (((ar ∗ a−1 r ) ∗ a0) ∗ a0) ∗ ak = (((a0 ∗ a−1 r ) ∗ ar) ∗ a0) ∗ ak = ((a−1 r ∗ ar) ∗ a0) ∗ ak = (a0 ∗ a0) ∗ ak = a0 ∗ as = as. Then ar = as; this is a contradiction. Hence |Hk,γ0 | = m. Let akγ0aj ∈ Hk,γ0 . Suppose akγ0aj = a0. By (v), ak = a0 or aj = a0. This is a contradiction. Then akγ0aj ̸= a0, and akγ0aj ∈ A \ {a0}. So Hk,γ0 ⊆ A \ {a0}. From |Hk,γ0 | = m = |A \ {a0}|, it follows that Hk,γ0 = A \ {a0}. Since e ∈ Hk,γ0 , e = akγ0ai for some ai ∈ A \ {a0}. Moreover, aiγ0ak = eγ0(aiγ0ak) = (akγ0ai)γ0e = eγ0e = e. Setting a−1 k = ai, we then have akγ0a −1 k = e = a−1 k γ0ak. By Claim 4, A \ {a0} is a group under the operation determined by γ0. And, by Theorem 1, A \ {a0} is a group under the operation determined by γ for all γ ∈ Γ. Finally, by Theorem 4, we conclude that A \ {a0} is a commutative group under the operation determined by γ for all γ ∈ Γ. This completes the proof. An element e of an AG-groupoid A is said to be a left identity if for all a ∈ A, ea = a. An element a0 of A is said to be a left zero if for all a ∈ A, a0a = a0. The following corollary is particularly true. C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 7 of 9 Corollary 2. Let (A, ·) be a finite AG-groupoid with |A| > 1, a left identity e, a left zero a0, and a · (b · c) = (c · b) · a for all a, b, c ∈ A. Suppose there exists an operation ∗ on A such that (i)-(v) hold: (i) (A, ∗) is an AG-groupoid. (ii) For any a ∈ A there exists b ∈ A such that b ∗ a = a0. (iii) a0 ∗ a = a for all a ∈ A. (iv) (a ∗ b) · c = (a · c) ∗ (b · c) for all a, b, c ∈ A. (v) For any a, b ∈ A, a · b = a0 implies a = a0 or b = a0. Then (A \ {a0}, ·) is a commutative group. The following proposition demonstrates the necessity of the identity aγ(bβc) = (cγb)βa for all a, b, c ∈ A and γ, β ∈ Γ as stated in Theorem 5. Proposition 1. Let A be a finite Γ-AG-groupoid containing at least two elements (|A| > 1). Suppose A contains a left identity e and a left zero a0. Suppose further that there exist γ0 ∈ Γ and an operation ∗ of A×A into A, write a ∗ b for ∗(a, b), such that (i)-(v) hold: (i) A is an AG-groupoid under ∗. (ii) For any a ∈ A there exists b ∈ A such that b ∗ a = a0. (iii) a0 ∗ a = a for all a ∈ A. (iv) (a ∗ b)γ0c = (aγ0c) ∗ (bγ0c) for all a, b, c ∈ A. (v) For any a, b ∈ A, if aγ0b = a0 then a = a0 or b = a0. Then, under the operation determined by γ0, A \ {a0} is a cancellative AG-groupoid with left identity and inverses (i.e., for each ak ∈ A\{a0} there exists a−1 k ∈ A\{a0} such that akγ0a −1 k = e = a−1 k γ0ak). Proof. As the proof of Theorem 5, under the operation determined by γ0, we have A \ {a0} is an AG-groupoid with left identity, and for any ak ∈ A \ {a0}, e = akγ0ai for some ai ∈ A \ {a0}. Consider (using Theorem 2): aiγ0ak = eγ0(aiγ0ak) = (eγ0e)γ0(aiγ0ak) = (eγ0ai)γ0(eγ0ak) = ((eγ0ak)γ0ai)γ0e = (akγ0ai)γ0e C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 8 of 9 = eγ0e = e. Then aiγ0ak = e. Finally, let ai, aj , ak ∈ A \ {a0} be such that aiγ0ak = ajγ0ak. Moreover, as the proof of Theorem 5, there exists a−1 k ∈ A \ {a0} such that a−1 k γ0ak = e = akγ0a −1 k . Consider: ai = eγ0ai = (a−1 k γ0ak)γ0ai = (aiγ0ak)γ0a −1 k = (ajγ0ak)γ0a −1 k = (a−1 k γ0ak)γ0aj = eγ0aj = aj . Similarly, if ai, aj , ak ∈ A \ {a0} such that akγ0ai = akγ0aj then ai = aj . Hence the proof is complete. Acknowledgements The Research on ”Finite Γ-AG-groupoids with left identities and left zeros” is sup- ported by Research, Innovation and Academic Services Fund, Faculty of Science, Khon Kaen University. References [1] W. A. Dudek and R. S. Gigoń. Completely inverse AG**-groupoids. Semigroup Forum, 87(1):201–229, 2013. [2] M. A. Kazim and M. Naseeruddin. On almost semigroups. Portugaliae mathematica, 36(1):41–47, 1977. [3] A. D. Keedwell and J. Dénes. Latin squares and their applications. 1974. [4] P. V. Protic and N. Stevanovic. On Abel-Grassmann’s groupoids. Proc. Math. Conf. Priötina, page 31–38, 1994. [5] Q. Mushtaq and M. S. Kamran. Finite AG-groupoid with left identity and left zero. International Journal of Mathematics and Mathematical Sciences, 27(6):3873–389, 2000. [6] M. K. Sen. On Γ-semigroups, algebra and its applications. Algebra and its applica- tions, page 301–308, 1981. C. Chanoi et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6172 9 of 9 [7] M. K. Sen and N. Saha. K: On Γ-semigroup I. Bulletin of Calcutta Mathematical Society, 78:181–186, 1986. [8] M. K. Sen and S. Chattopadhyay. Wreath product of a semigroup and a Γ-semigroup. Discussiones Mathematicae General Algebra and Applications, 28:161–178, 2008. [9] K. Wattanatripop and T. Changphas. On left and right A-ideals of a Γ-semigroup. Thai Journal of Mathematics, page 87–96, 2018. [10] T. Shah and I. Rehman. On Γ-ideals and Γ-bi-ideals in Γ-AG-groupoids. International Journal of Algebra, 4:267–276, 2010. [11] M. Khan, S. Anis, and F. Faisal. On fuzzy-Γ-ideals of Γ-Abel-Grassmann’s groupoids. Research Journal of Applied Sciences, Engineering and Technology, 6:1326–1334, 2013.