EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 3, 2023, 1772-1793 ISSN 1307-5543 – ejpam.com Published by New York Business Global On Γ-ideals, Γ-submonoids and Isomorphism Theorems of Γ-monoids via Γ-submonoids Hulsen T. Sarapuddin1,∗, Jocelyn P. Vilela1 1 Department of Mathematics and Statistics, College of Science and Mathematics, Center of Mathematical and Theoretical Physical Sciences-PRISM, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. This study introduces the concept of Γ-ideals and Γ-submonoids of Γ-monoids and investigates their relationships with the existing Γ-order-ideals. Moreover, quotient of Γ-monoids and isomorphism theorems via Γ-submonoids are proved. 2020 Mathematics Subject Classifications: 20M32 Key Words and Phrases: Γ-monoids, Γ-monoid Homomorphism, Γ-order-ideals, Γ-ideals, Γ- submonoids, Isomorphism Theorem 1. Introduction The talented monoid of a row-finite directed graph E = (E0, E1, r, s), denoted by TE , is the commutative monoid generated by {v(i) : v ∈ E0, i ∈ Z} such that v(i) =∑ e∈s−1(v) r(e)(i + 1) for every i ∈ Z and every v ∈ E0 that is not a sink. The additive group Z of integers acts on TE by monoid automorphisms by shifting indices: for each n, i ∈ Z and v ∈ E0, define nv(i) = v(i + n), which extends to an action of Z on TE [3]. Monoids with a group Γ acting (by monoid automorphisms) on it, called Γ-monoids, was first introduced in the paper of Hazrat and Li [1] as a tool in the study of talented monoids. In the same paper, Γ-order-ideals of Γ-monoids are also introduced. Sebandal and Vilela [5] prove some properties, including the isomorphism theorems for Γ-monoids and Γ-order-ideals are established. This paper extends the study of Γ-monoids by defining the concept of Γ-ideals and Γ-submonoids and establishing some of their properties. Moreover, this paper studies quotient of Γ-monoids via equivalence classes of Γ-submonoids and proves isomorphism theorems. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i3.4793 Email addresses: hulsen.sarapuddin@g.msuiit.edu.ph (H. T. Sarapuddin), jocelyn.vilela@g.msuiit.edu.ph (J. P. Vilela) https://www.ejpam.com 1772 © 2023 EJPAM All rights reserved. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1773 2. Preliminaries In this section, we present some basic concepts and known results that are useful in this study. Definition 1. [2] A semigroup is a nonempty set M together with a binary operation ∗ on M which is associative, that is, for all a, b, c ∈ M , a ∗ (b ∗ c) = (a ∗ b) ∗ c. Definition 2. [2] A monoid is a semigroupM which contains an identity element 1M ∈ M such that 1M ∗m = m ∗ 1M = m for all m ∈ M . For a monoid M with the binary operation ∗, we may also say that M is a monoid under ∗. A monoid M is said to be commutative if for all x, y ∈ M , x ∗ y = y ∗ x. If no confusion arises, by a monoid M , we shall mean a triple (M, 1M , ∗) unless other- wise specified. Definition 3. [6] Let (M, ∗) be a monoid. A submonoid is a subset S of M which is closed under the binary operation on M and contains the identity 1M of M . Definition 4. [6] Let (M, ∗) and (N, ·) be monoids. A monoid homomorphism is a mapping φ : M → N such that φ(a ∗ b) = φ(a) · φ(b) and φ(1M ) = 1N for all a, b ∈ M where 1M and 1N are the identities in M and N , respectively. Example 1. Consider the monoids M = (N,+) and N = (N, ·) and the mapping φ : M → N defined by φ(x) = bx, where b ∈ N \ {0}. For any x, y ∈ M , we have φ(x+ y) = bx+y = bx · by = φ(x) · φ(y) and φ(0) = b0 = 1. Therefore, φ is a monoid homomorphism. Definition 5. [6] A congruence on a monoid M is an equivalence relation ρ on M which satisfies the condition: For all u, v, x, y ∈ M , if xρy, then (u ∗ x ∗ v)ρ(u ∗ y ∗ v). Proposition 1. [6] Let ρ be a congruence on a monoid M . Then M/ρ is a monoid with binary operation ◦ given by ρ(x) ◦ ρ(y) = ρ(x ∗ y) for all x, y ∈ M . Definition 6. [4] Let M be a commutative monoid. For any submonoid H of M , we define a binary relation ρH in M by xρHy if and only if (x ∗H) ∩ (y ∗H) ̸= ∅. Remark 1. [4] For any submonoid H of a commutative monoid M , ρH is an equivalence relation on M . Definition 7. [2] An action of a group (G, ◦) in a set S is a function ϕ : G×S −→ S such that for all x ∈ S, and g1, g2 ∈ G: ϕ((1G, x)) = x and ϕ((g1 ◦ g2, x)) = ϕ((g1, ϕ((g2, x)))). When such an action is given, G is said to act on the set S. Example 2. Consider the group G = Z under the usual addition and the set S = R of real numbers and the mapping ϕ : G × S → S given by ϕ((g, x)) = 2gx. Let (g, x), (h, y) ∈ G × S such that (g, x) = (h, y). Then g = h and x = y. Thus, we have ϕ((g, x)) = 2gx = 2hy = ϕ((h, y)) and ϕ is well-defined. Now, for any g1, g2 ∈ G and x ∈ S, we have ϕ((0, x)) = 20x = x and ϕ((g1 + g2, x)) = 2g1+g2x = 2g12g2x = ϕ((g1, ϕ((g2, x)))). Therefore, ϕ is an action. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1774 Definition 8. [3] Let M be a monoid and Γ a group. M is said to be a Γ-monoid if there is an action ϕ : Γ×M → M of Γ on M via monoid automorphism, that is, ϕ is an action which satisfies: for all α ∈ Γ and x, y ∈ M , ϕ((α, x ∗ y)) = ϕ((α, x)) ∗ ϕ((α, y)). For α ∈ Γ and a ∈ M , the action of α on a shall be denoted by αa. Example 3. Consider Γ = Z a group of integers under the usual addition and the set M = R with the usual addition as its binary operation. Then, (M,+) is a monoid with identity 0. Consider the action ϕ : Γ ×M → M given by ϕ((α, x)) = 2αx in Example 2. Now, let α ∈ Γ and x, y ∈ M . Then we have ϕ((α, x + y)) = 2α(x + y) = 2αx + 2αy = ϕ((α, x)) + ϕ((α, y)). Therefore, M is a Γ-monoid. Example 4. Let Γ be a group of integers under addition and let T = M2(R) under matrix addition. Consider the mapping ϕ : Γ × T → T given by ( α, ( a b c d )) 7→ α ( a b c d ) =( 2αa 2αb 2αc 2αd ) . Let ( α, ( a b c d )) , ( β, ( e f g h )) ∈ Γ× T such that( α, ( a b c d )) = ( β, ( e f g h )) . Then α = β and ( a b c d ) = ( e f g h ) . Thus, ( 2αa 2αb 2αc 2αd ) =( 2βe 2βf 2βg 2βh ) and ϕ is well-defined. Now, for any α, β ∈ Γ and a, b, c, d ∈ R, we have ϕ (( 0, ( a b c d ))) = 0 ( a b c d ) = ( 20a 20b 20c 20d ) = ( a b c d ) and ϕ (( α+ β, ( a b c d ))) = α+β ( a b c d ) = ( 2α+βa 2α+βb 2α+βc 2α+βd ) = ( 2α2βa 2α3βb 2α4βc 2α5βd ) = ϕ (( α, ( 2βa 2βb 2βc 2βd ))) = ϕ (( α,φ (( β, ( a b c d ))))) . Thus, ϕ is an action. Now, let α ∈ Γ and ( a b c d ) , ( e f g h ) ∈ T. Then we have ϕ (( α, ( a b c d ) + ( e f g h ))) = ϕ (( α, ( a+ e b+ f c+ g d+ h ))) = ( 2α(a+ e) 2α(b+ f) 2α(c+ g) 2α(d+ h) ) H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1775 = ( 2αa+ 2αe 2αb+ 2αf 2αc+ 2αg 2αd+ 2αh ) = ( 2αa 2αb 2αc 2αd ) + ( 2αe 2αf 2αg 2αh ) = ϕ (( α, ( a b c d ))) + ϕ (( α, ( e f g h ))) . Therefore, T is a Γ-monoid. Example 5. Consider the set M = {1, a, b, c, d, e} and an operation ∗ given by ∗ 1 a b c d e 1 1 a b c d e a a a a a a a b b b b b b b c c c c c c c d d d d d d d e e e e e e e The operation ∗ is closed and associative since for all x, y ∈ M , x ∗ y = x holds for all x ̸= 1. Clearly, 1 is an identity in M . Thus, M is a monoid. With a group Γ acting trivially on M , we obtain that M is a Γ-monoid. Definition 9. [1] Let M , M1 and M2 be monoids and let Γ be a group acting on M , M1 and M2. (i) A Γ-monoid homomorphism is a monoid homomorphism ϕ : M1 −→ M2 that re- spects the action of Γ, this means ϕ(αa) = αϕ(a). (ii) A Γ-order-ideal of a monoid M is a subset I of M such that for any α, β ∈ Γ, αa ∗ βb ∈ I if and only if a, b ∈ I. Remark 2. [1] A Γ-order-ideal is a submonoid I of M which is closed under the action of Γ. Example 6. Let a group Γ acts trivially on both monoids M = (N,+) and N = (N, ·), that is, for all α ∈ Γ, we have ϕ((α,m)) = αm = m and ϕ((α, n)) = αn = n for all m ∈ M and n ∈ N . Now, let α ∈ Γ and x, y ∈ M . Then, ϕ((α, x + y)) = α(x+ y) = x + y = αx + αy = ϕ((α, x)) + ϕ((α, y)). Thus, M and N are Γ-monoids. Consider the monoid homomorphism φ : M → N defined by φ(x) = bx, where b ∈ N \ {0} in Example 1. For all α ∈ Γ and a ∈ M , we have φ(αa) = φ(a) = αφ(a). Thus, by Definition 9(ii), φ is a Γ-monoid homomorphism. Example 7. Consider the Γ-monoid M = R under the usual addition in Example 3 and the Γ-monoid T = M2(R) under matrix addition in Example 4. Define a mapping H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1776 ϕ : T → M by ϕ (( a b c d )) = 2(a + b + c + d). Let ( a b c d ) , ( e f g h ) ∈ T such that( a b c d ) = ( e f g h ) . Then a = e, b = f , c = g and d = h. Thus, 2(a + b + c + d) = 2(e + f + g + h) and ϕ is well-defined. Now, for any ( a b c d ) , ( e f g h ) ∈ T , we have ϕ (( 0 0 0 0 )) = 2(0 + 0 + 0 + 0) = 2(0) = 0 and ϕ (( a b c d ) + ( e f g h )) = ϕ (( a+ e b+ f c+ g d+ h )) = 2((a+ e) + (b+ f) + (c+ g) + (d+ h)) = 2((a+ b+ c+ d) + (e+ f + g + h)) = 2(a+ b+ c+ d) + 2(e+ f + g + h) = ϕ (( a b c d )) + ϕ (( e f g h )) . Thus, ϕ is a monoid homomorphism. Also, for all α ∈ Γ and ( a b c d ) ∈ T , we have ϕ ( α ( a b c d )) = ϕ (( 2αa 2αb 2αc 2αd )) = 2(2αa+ 2αb+ 2αc+ 2αd) = 2α2(a+ b+ c+ d) = αϕ (( a b c d )) . Hence, ϕ is a Γ-monoid homomorphism. Theorem 1. [4] Let M1 and M2 be commutative monoids and let f : M1 −→ M2 be a homomorphism. There exists a unique homomorphism φ : M1/ ker f −→ M2 such that the following diagram is commutative M1 M2 M1/ ker f rker f f φ that is, φ ◦ rker f = f , where rker f (x) := ρker f (x). Moreover, φ is onto and it has a trivial kernel, namely, kerφ = {ker f}. However, φ is an isomorphism if and only if ρf = ρker f . H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1777 3. Γ-ideals In this section, we discuss the properties of Γ-ideals of Γ-monoids. Let M be a Γ-monoid and x ∈ M . By Definition 8, for all α ∈ Γ, αx ∗ α1M = α(x ∗ 1M ) = αx and α1M ∗ αx = α(1M ∗ x) = αx. By uniqueness of the identity element in M , α1M = 1M . Remark 3. For a Γ-monoid M and α ∈ Γ, α1M = 1M . Definition 10. Let M be a Γ-monoid. A left Γ-ideal (respectively, right Γ-ideal) of M is a subset I of M such that for any α, β ∈ Γ, for all a ∈ I and m ∈ M , αm ∗ βa ∈ I (respectively, αa ∗ βm ∈ I). A Γ-ideal of M is a subset I of M such that I is both a left and right Γ-ideal of M . Let (M, ∗) be a Γ-monoid and A a Γ-ideal of M with a ∈ A. Then for all α, β ∈ Γ, we have αa = αa ∗ α1M ∈ A. Thus, we have the following remark. Remark 4. Let (M, ∗) be a Γ-monoid and A be a Γ-ideal of M . (i) M is a Γ-ideal. (ii) For all α ∈ Γ and for all a ∈ A, αa ∈ A. Lemma 1. Let A and B be Γ-ideals of a Γ-monoid M . Then A ∗B is a Γ-ideal of M . Proof. Let A and B be Γ-ideals of a Γ-monoid M . Clearly, A ∗B ⊆ M . Let x ∈ A ∗B and m ∈ M . Then x = a ∗ b for some a ∈ A and b ∈ B. Now, for all α, β ∈ Γ, αx∗βm = α(a ∗ b)∗βm = αa∗αb∗βm = αa∗(αb∗βm) ∈ A∗B by Remark 4(ii) and Defini- tion 10. Similarly, for all α, β ∈ Γ, αm∗βx ∈ A∗B. Therefore, A∗B is a Γ-ideal of M . The following example shows that a Γ-ideal is not necessarily a Γ-order-ideal. Example 8. Consider the set M = {1, n, h, s} and operation ∗ given by ∗ 1 n h s 1 1 n h s n n n h s h h h h s s s s s s Clearly, the operation is commutative. It can be verified that ∗ is associative. Since 1 ∗ 1 = 1, 1 ∗ n = n, 1 ∗ h = h and 1 ∗ s = s, it follows that 1 is the identity in M . Thus, M is a commutative monoid. Let Γ be a group and the mapping ϕ : Γ×M −→ M given by (α, a) 7→ αa = a. For any α, β ∈ Γ and a ∈ M , we have ϕ((0, a)) = 0a = a and ϕ((α+ β, a)) = α+βa = a = ϕ(β, a) = βa = ϕ((α, βa)) = ϕ((α, ϕ((β, a)))). H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1778 Thus, ϕ is an action. Now, let α ∈ Γ and a, b ∈ M . Then ϕ((α, a ∗ b)) = α(a ∗ b) = a ∗ b = αa ∗ αb = ϕ((α, a)) ∗ ϕ((α, b)). Hence, M is a Γ-monoid. Let C = {n, h, s}. Then for any α, β ∈ Γ, we have for all a ∈ C and m ∈ M , αa ∗ βm = αn ∗ β1 = n ∗ 1 = n ∈ C, αa ∗ βm = αn ∗ βh = n ∗ h = h ∈ C, αa ∗ βm = αh ∗ β1 = h ∗ 1 = h ∈ C, αa ∗ βm = αh ∗ βh = h ∗ h = h ∈ C, αa ∗ βm = αs ∗ β1 = s ∗ 1 = s ∈ C, αa ∗ βm = αs ∗ βh = s ∗ h = s ∈ C, αa ∗ βm = αn ∗ βn = n ∗ n = n ∈ C; αa ∗ βm = αn ∗ βs = n ∗ s = s ∈ C; αa ∗ βm = αh ∗ βn = h ∗ n = h ∈ C; αa ∗ βm = αh ∗ βs = h ∗ s = s ∈ C; αa ∗ βm = αs ∗ βn = s ∗ n = s ∈ C; αa ∗ βm = αs ∗ βs = s ∗ s = s ∈ C. Since M is commutative, βm ∗ αa = αa ∗ βm ∈ C. Thus, by Definition 10, C is a Γ-ideal. However, the identity 1 /∈ C. Thus, C is not a Γ-order-ideal of M . The following example shows that Γ-order-ideal is not necessarily a Γ-ideal. Example 9. Consider the Γ-monoid M = {1, n, h, s} in Example 8. Let A = {1, n, h}. Now, suppose that for all a, b ∈ M and for all α, β ∈ Γ, αa ∗ βb ∈ A. Then a ∗ b ∈ A. We consider the following three cases. Case 1. a ∗ b = 1. Then a = 1 and b = 1. Thus a, b ∈ A. Case 2. a ∗ b = n. Then a ∗ b = 1 ∗ n = n ∗ 1 = n ∗ n. Clearly, a, b ∈ A. Case 3. a ∗ b = h. Then a ∗ b = 1 ∗ h = n ∗ h = h ∗ 1 = h ∗ n. Clearly, a, b ∈ A. Thus, a, b ∈ A. Now, suppose that a, b ∈ A. Then, we have αa ∗ βb = α1 ∗ β1 = 1 ∗ 1 = 1 ∈ A; αa ∗ βb = α1 ∗ βn = 1 ∗ n = n ∈ A; αa ∗ βb = α1 ∗ βh = 1 ∗ h = h ∈ A; αa ∗ βb = αn ∗ βn = n ∗ n = n ∈ A; αa ∗ βb = αn ∗ βh = n ∗ h = h ∈ A; αa ∗ βb = αh ∗ βh = h ∗ h = h ∈ A. Thus, αa ∗ βb ∈ A. Hence, A is a Γ-order-ideal of M . Observe that there exist n ∈ A and s ∈ M such that for any α, β ∈ Γ, αn∗βs = n∗s = s /∈ A. Thus, by Definition 10, A is not a Γ-ideal. Remark 5. If I is a Γ-ideal, in general I is not necessarily a Γ-order-ideal. Similarly, if I is a Γ-order-ideal, in general I is not necessarily a Γ-ideal. Lemma 2. Let I be a Γ-ideal of a Γ-monoid M . Then the identity 1M ∈ I if and only if I = M . Proof. Let I is a Γ-ideal of M . Suppose that the identity 1M ∈ I and m ∈ M . Then for any α, β ∈ Γ, we have α1M ∗βm ∈ I. For α = β = 0, we have 01M ∗ 0m = 1M ∗m = m ∈ I. Thus, M ⊆ I. Consequently, I = M . Conversely, suppose that I = M . Thus, the identity 1M ∈ I. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1779 Theorems 2 and 3 imply that there exists no proper Γ-order-ideal which is also a Γ-ideal and vice versa. Theorem 2. Let I be a Γ-ideal of a Γ-monoid M . Then I is a Γ-order-ideal of M if and only if I = M . Proof. Let I be a Γ-ideal of M . Suppose that I is a Γ-order-ideal of M . Then the identity 1M ∈ I. By Lemma 2, I = M . Conversely, suppose that I = M . Thus, I is a Γ-order-ideal. Theorem 3. Let I be a Γ-order-ideal of a Γ-monoid M . Then I is a Γ-ideal of M if and only if I = M . Proof. Let I be a Γ-order-ideal of a Γ-monoid M . Then 1M ∈ I since I is also a submonoid. Suppose that I is a Γ-ideal of M . By Lemma 2, I = M . Conversely, suppose that I = M . Thus, by Remark 4(i), I is a Γ-ideal. Lemma 3. Let A and B be Γ-ideals of a Γ-monoid M . Then A∩B and A∪B are Γ-ideals of M . Proof. Let A and B be Γ-ideals of M . Let x ∈ A ∩ B and m ∈ M . Then x ∈ A and x ∈ B. Since A and B are Γ-ideals of M , for all α, β ∈ Γ, we have αx∗βm, αm∗βx ∈ A and αx ∗ βm, αm ∗ βx ∈ B. Hence, for all α, β ∈ Γ, αx ∗ βm, αm ∗ βx ∈ A∩B. Therefore, A∩B is a Γ-ideal of M . Now, let x ∈ A∪B and m ∈ M . Then x ∈ A or x ∈ B. Since A and B are Γ-ideals of M , for all α, β ∈ Γ, we have αx ∗ βm, αm ∗ βx ∈ A or αx ∗ βm, αm ∗ βx ∈ B. Hence, for all α, β ∈ Γ αx ∗ βm, αm ∗ βx ∈ A∪B. Therefore, A∪B is a Γ-ideal of M . Theorem 4. Let I be a Γ-order-ideal of a Γ-monoid M and J a Γ-ideal of M . (i) If J ∩ I ̸= ∅, then J ∩ I is a Γ-ideal of I. (ii) If M is commutative, then J ∪ I is a Γ-order-ideal of M . Proof. Let I be a Γ-order-ideal of M and J a Γ-ideal of M . (i) Let x ∈ J ∩ I and a ∈ I. Then x ∈ J and x ∈ I. Since J is a Γ-ideal of M , for all α, β ∈ Γ, αx ∗ βa, αa ∗ βx ∈ J . Also, since I is a Γ-order-ideal of M , for all α, β ∈ Γ, αx ∗ βa, αa ∗ βx ∈ I. Thus, for all α, β ∈ Γ, αx ∗ βa, αa ∗ βx ∈ J ∩ I. Therefore, J ∩ I is a Γ-ideal of I. (ii) Suppose that αx ∗ βa ∈ J ∪ I for all α, β ∈ Γ. Then, αx ∗ βa ∈ J or αx ∗ βa ∈ I. Since I is a Γ-order-ideal of M , it follows that x, a ∈ I ⊆ J ∪ I. Now, suppose that x, a ∈ J ∪ I. Consider the following cases. Case 1. x, a ∈ I. Then, since I is a Γ-order-ideal of M , for all α, β ∈ Γ, αx ∗ βa ∈ I ⊆ J ∪ I. Case 2. x ∈ I, a ∈ J . Then, since J is a Γ-ideal of M and M is commutative, for all α, β ∈ Γ, we have αx ∗ βa = βa ∗ αx ∈ J ⊆ J ∪ I. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1780 Case 3. x ∈ J, a ∈ I. Then, since J is a Γ-ideal of M , for all α, β ∈ Γ, we have αx ∗ βa ∈ J ⊆ J ∪ I. Case 4. x, a ∈ J . Then, since J is a Γ-ideal of M , for all α, β ∈ Γ, we have αx ∗ βa ∈ J ⊆ J ∪ I. Thus, J ∪ I is a Γ-order-ideal of M . Definition 11. Let (M, ∗) and (N, ·) be Γ-monoids and φ : M → N a Γ-monoid homo- morphism. The kernel of φ is denoted and defined by kerφ = {m ∈ M : φ(m) = 1N}. Proposition 2. Let (M, ∗) and (N, ·) be Γ-monoids and φ : M → N a Γ-monoid homo- morphism. (i) If φ is surjective and I is a Γ-ideal of M , then φ(I) is a Γ-ideal of N . (ii) If J is a Γ-ideal of N , then φ−1(J) is a Γ-ideal of M . Proof. Let φ : M → N be a Γ-monoid homomorphism. (i) Let x ∈ φ(I) and z ∈ N . Since φ is surjective, z = φ(n) for some n ∈ M and x = φ(y) for some y ∈ I. Then for all α, β ∈ Γ, αx ∗ βz = αφ(y) · βφ(n) = φ(αy) · φ(βn) = φ(αy ∗ βn). Since I is a Γ-ideal of M , αy ∗ βn ∈ I, so, αx ∗ βz ∈ φ(I). Similarly, for all α, β ∈ Γ, αz ∗ βx ∈ φ(I). Therefore, φ(I) is a Γ-ideal of N . (ii) Let y ∈ φ−1(J) and m ∈ M . Then φ(y) ∈ J and φ(m) ∈ N . Thus, for all α, β ∈ Γ, φ(αy ∗ βm) = φ(αy) · φ(βm) = αφ(y) · βφ(m) ∈ J , since J is a Γ-ideal of N . Hence, αy ∗ βm ∈ φ−1(J) for all α, β ∈ Γ. Similarly, for all α, β ∈ Γ, αm ∗ βy ∈ φ−1(J). Therefore, φ−1(J) is a Γ-ideal of M . Example 10. Consider the Γ-monoid homomorphism φ : M → N defined by φ(x) = bx, where b ̸= 0 in Example 6. Note that kerφ = {x ∈ M : φ(x) = 1} = {x ∈ M : bx = 1} = {x ∈ M : b = 1 or x = 0}. Take x = 0 ∈ kerφ, m = 2 ∈ M , and b = 2. Then for all α, β ∈ Γ, φ(αx + βm) = φ(α0 + β2) = φ(0 + 2) = φ(2) = 22 ̸= 1. This implies that αx + βm /∈ kerφ. By Definition 10, kerφ is not a Γ-ideal of M . Remark 6. For any Γ-monoids M and N , the kernel of a Γ-monoid homomorphism φ : M → N is not necessarily a Γ-ideal of M . Proposition 3. Let (M, ∗) and (N, ·) be Γ-monoids and φ : M → N a Γ-monoid homo- morphism. Then kerφ is a Γ-ideal of M if and only if kerφ = M. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1781 Proof. Let φ : M → N be a Γ-monoid homomorphism. Then 1M ∈ kerφ. Suppose that kerφ is a Γ-ideal ofM . Then by Lemma 2, kerφ = M . Now, suppose that kerφ = M . Then by Remark 4(i), kerφ is a Γ-ideal of M . By Proposition 3, kerφ is a Γ-ideal if and only if φ is a zero map. Thus, isomorphism theorems via Γ-ideals are irrelevant. 4. Γ-submonoids This section presents the discussions on Γ-submonoids of Γ-monoids. Definition 12. Let (M, ∗) be a Γ-monoid. A Γ-submonoid is a subset S of M such that the identity 1M ∈ S and, for all α, β ∈ Γ and for all s, t ∈ S, αs ∗ βt ∈ S. Let S be a Γ-submonoid of M . Then 1M ∈ S and for all α, β ∈ Γ and for all s, t ∈ S, we have αs ∗ βt ∈ S. Take α = β = 0. Thus, we have s ∗ t = 0s ∗ 0t ∈ S. Hence, S is a submonoid of M . Remark 7. Let S be a Γ-submonoid of a Γ-monoid M . (i) S is a submonoid of M , hence a monoid itself. (ii) For all s ∈ S and for all α ∈ Γ, αs ∈ S. (iii) M is a Γ-submonoid of M . Let S be a Γ-submonoid of a Γ-monoid M and let ϕ : Γ × M → M be the action (by monoid automorphism) of a group Γ on M . By Remark 7, S is a monoid. Moreover, by restricting the action ϕ to S, ϕ acts on S by monoid automorphism and hence, S is a Γ-monoid. Remark 8. A Γ-submonoid of a Γ-monoid is itself a Γ-monoid. Example 11. Consider the set M = {0, 1, x, y, z, s, b} and an operation + given by + 0 1 x y z s b 0 0 1 x y z s b 1 1 1 1 s s s b x x 1 1 s s s b y y s s y y s b z z s s y y s b s s s s s s s b b b b b b b b s H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1782 It was shown in [5] thatM is a commutative Γ-monoid with identity 0, where the trivial group Γ = {0} acts trivially on M . Let S = {0, y, s, b}, U = {0, 1, y, s, b}, V = {0, 1, x} and W = {0, y}. Note that the identity 0 is in S,U, V and W . Now, we have 00 + 00 = 0 + 0 = 0 ∈ S, 00 + 0s = 0 + s = s ∈ S, 0y + 0y = y + y = y ∈ S, 0y + 0b = y + b = b ∈ S, 0s+ 0b = s+ b = b ∈ S, 00 + 0y = 0 + y = y ∈ S; 00 + 0b = 0 + b = b ∈ S; 0y + 0s = y + s = s ∈ S; 0s+ 0s = s+ s = s ∈ S; 0b+ 0b = b+ b = b ∈ S. Thus, by Definition 12, S is Γ-submonoid of M . Similarly, U, V and W are Γ-submonoids of M . Consider the Γ-submonoid S = {0, y, s, b}. Now, take 0 ∈ S and z ∈ M . Then 0 ∗ z = z /∈ S. Thus, S is not a Γ-ideal of M . Remark 9. Let M be a Γ-monoid. A Γ-submonoid of M is not necessarily a Γ-ideal of M . Theorems 5 and 6 imply that there is no proper Γ-submonoid which is also a Γ-ideal and vice versa. Theorem 5. Let S be a Γ-submonoid of a Γ-monoid M . Then S is a Γ-ideal of M if and only if S = M . Proof. Let S be a Γ-submonoid of M . Suppose that S is a Γ-ideal of M . Since S is a Γ-submonoid, 1M ∈ S and thus, by Lemma 2, S = M . Conversely, suppose that S = M . Then, by Remark 4(i), S is a Γ-ideal of M . Theorem 6. Let I be a Γ-ideal of a Γ-monoid M . Then I is a Γ-submonoid of M if and only if I = M . Proof. Let I be a Γ-ideal of a Γ-monoid M . Suppose that I is a Γ-submonoid of M . Then 1M ∈ I and I = M . Conversely, suppose that I = M . By Remark 7(iii), I is a Γ-submonoid of M . Example 12. Consider the Γ-submonoid S = {0, y, s, b} in Example 11. Note that x ∗ z = s ∈ S. However, x, z /∈ S. Thus, S is not a Γ-order-ideal of M . Note that if S is a Γ-order-ideal of a Γ-monoid M , then by Remark 2, S is a submonoid and 1M ∈ S. Also, since S is a Γ-order-ideal, for all α, β ∈ Γ and for all s, t ∈ S, we have αs ∗ βt ∈ S. Thus, S is a Γ-submonoid of M and the following remark holds. Remark 10. Every Γ-order-ideal of a Γ-monoid M is a Γ-submonoid of M . However, a Γ-submonoid of M is not necessarily a Γ-order-ideal of M . The following example shows that a Γ-submonoid is not necessarily a normal sub- monoid. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1783 Example 13. Consider the Γ-submonoid U = {0, 1, y, s, b} in Example 11 which is also commutative. Observe that y, z ∈ M such that y, y ∗ z = y ∈ U . However, z /∈ U . Thus, U is not a normal submonoid of M . Remark 11. In general, a Γ-submonoid of a Γ-monoid M is not necessarily a normal submonoid of M . Theorem 7. Let S be a subset of a Γ-monoid M . Then S is a Γ-order-ideal if and only if S is a Γ-submonoid such that x ∗ y ∈ S implies x, y ∈ S. Proof. Let S be a subset of a Γ-monoid M . Suppose S is a Γ-order-ideal of M . Then by Remark 10, S is a Γ-submonoid and for α = β = 0, we have x ∗ y = 0x ∗ 0y ∈ S implies x, y ∈ S since S is a Γ-order-ideal. Now, suppose S is a Γ-submonoid such that x ∗ y ∈ S implies x, y ∈ S. Then for all α, β ∈ Γ and for all x, y ∈ S, αx ∗ βy ∈ S. Suppose for all α, β ∈ Γ, αx ∗ βy ∈ S. Take α = β = 0. Then x ∗ y = 0x ∗ 0y ∈ S which implies that x, y ∈ S. Therefore, S is a Γ-order-ideal. Lemma 4. Let A and B be Γ-submonoids of a Γ-monoid M . Then (i) A ∩B is a Γ-submonoid of M . (ii) If M is commutative and A, B are normal, then A∩B is a normal Γ-submonoid of M . Proof. Let A and B be Γ-submonoids of a Γ-monoid M . (i) Since A and B are Γ-submonoids of M , the identity 1M ∈ A and 1M ∈ B. Thus, 1M ∈ A∩B. Now, let a, b ∈ A∩B. Then, a, b ∈ A and a, b ∈ B. Since A and B are Γ-submonoids, for all α, β ∈ Γ, αa ∗ βb ∈ A and αa ∗ βb ∈ B. Hence, αa ∗ βb ∈ A∩B. Therefore, A ∩B is a Γ-submonoid of M . (ii) By (i), A∩B is a Γ-submonoid of M . It remains to show that A∩B is normal. Let x, x ∗ y ∈ A ∩ B. Then x, x ∗ y ∈ A and x, x ∗ y ∈ B. Since A and B are normal, y ∈ A and y ∈ B. Therefore, y ∈ A ∩ B and A ∩ B is a normal Γ-submonoid of M . Example 14. Consider the Γ-submonoids V = {0, 1, x} and W = {0, y} in Example 11. Then, V ∪W = {0, 1, x, y}. Now, for x, y ∈ V ∪W , we have x ∗ y = s /∈ V ∪W . Thus, V ∪W is not a Γ-submonoid of M . Remark 12. The union of two Γ-submonoids of a Γ-monoid M is not necessarily a Γ- submonoid of M . Theorem 8. Let (M, ∗) and (N, ·) be Γ-monoids and φ : M → N a Γ-monoid homomor- phism. (i) If S is a Γ-submonoid of M , then φ(S) is a Γ-submonoid of N . In particular, φ(M) is a Γ-submonoid of N . H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1784 (ii) If T is a Γ-submonoid of N , then φ−1(T ) is a Γ-submonoid of M . (iii) kerφ is a Γ-submonoid of M . (iv) If M is commutative, then kerφ is normal. Proof. Let φ : M → N be a Γ-monoid homomorphism. (i) Let S be a Γ-submonoid of M . Then 1M ∈ S and 1N = φ(1M ) ∈ φ(S). Let x, y ∈ φ(S). Then x = φ(a) and y = φ(b) for some a, b ∈ S. Since S is a Γ- submonoid, for all α, β ∈ Γ, αa ∗ βb ∈ S. Now, for all α, β ∈ Γ, we have αx · βy = αφ(a) · βφ(b) = φ(αa) · φ(βb) = φ(αa ∗ βb). Since αa ∗ βb ∈ S, it follows that αx · βy = φ(αa ∗ βb) ∈ φ(S). Thus, φ(S) is a Γ-submonoid of N . (ii) Let T be a Γ-submonoid of N . Then, φ(1M ) = 1N ∈ T and 1M ∈ φ−1(T ). Let x, y ∈ φ−1(T ). Then φ(x), φ(y) ∈ T . Now, for all α, β ∈ Γ, we have φ(αx ∗ βy) = φ(αx) ·φ(βy) = αφ(x) · βφ(y) ∈ T since T is a Γ-submonoid of N . This implies that for all α, β ∈ Γ, we have αx ∗ βy ∈ φ−1(T ). Therefore, φ−1(T ) is a Γ-submonoid of M . (iii) Since φ is a Γ-monoid homomorphism, φ(1M ) = 1N . Thus, 1M ∈ kerφ. Now, let x, y ∈ kerφ. Then φ(x) = 1N and φ(y) = 1N . Thus, by Remark 3, for all α, β ∈ Γ, φ(αx ∗ βy) = φ(αx) · φ(βy) = αφ(x) · βφ(y) = α1N · β1N = 1N · 1N = 1N . Hence, for all α, β ∈ Γ, αx ∗ βy ∈ kerφ. Therefore, kerφ is a Γ-submonoid of M . (iv) Let x, x ∗ y ∈ kerφ. Then φ(x) = 1N and φ(x ∗ y) = 1N . Thus, φ(y) = 1N · φ(y) = φ(x) · φ(y) = φ(x ∗ y) = 1N . This implies that y ∈ kerφ and thus, kerφ is normal. Theorem 9. Let J be a Γ-ideal and S a Γ-submonoid of a Γ-monoid M such that J ∩ S ̸= ∅. Then (i) J ∩ S is a Γ-ideal of S; (ii) J ∪ S is a Γ-submonoid of M . Proof. Let J be a Γ-ideal and S a Γ-submonoid of M such that J ∩ S ̸= ∅. (i) Let x ∈ J ∩S and s ∈ S. Then x ∈ J and x, s ∈ S. Since J is a Γ-ideal of M , for all α, β ∈ Γ, αx ∗ βs, αs ∗ βx ∈ J . Also, since S is a Γ-submonoid of M , for all α, β ∈ Γ, αx ∗ βs, αs ∗ βx ∈ S. Thus, for all α, β ∈ Γ, αx ∗ βs, αs ∗ βx ∈ J ∩ S and so, J ∩ S is a Γ-ideal of S. (ii) Let x, y ∈ J ∪ S. We consider the following cases. Case 1. x, y ∈ J . Since J is a Γ-ideal of M , for all α, β ∈ Γ, αx ∗ βy ∈ J ⊆ J ∪ S. Case 2. x ∈ J, y ∈ S. Since J is a Γ-ideal of M , for all α, β ∈ Γ, αx ∗ βy ∈ J ⊆ J ∪ S. Case 3. x, y ∈ S. Since S is a Γ-submonoid of M , for all α, β ∈ Γ, αx ∗ βy ∈ S ⊆ J ∪ S. Case 4. y ∈ J, x ∈ S. Since J is a Γ-ideal of M , for all α, β ∈ Γ, αx ∗ βy ∈ J ⊆ J ∪ S. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1785 Also, since S is a Γ-submonoid of M , 1M ∈ S ⊆ J ∪S. Therefore, J ∪S is a Γ-submonoid of M . Remark 13. Theorem 4(i) is also a consequence of Theorem9(i). Lemma 5. Let A and B be Γ-submonoids of a commutative Γ-monoid M . Then A ∗ B is a Γ-submonoid of M . Proof. Let x, y ∈ A∗B and α, β ∈ Γ. Then x = a1∗b1 and y = a2∗b2 for some a1, a2 ∈ A and b1, b2 ∈ B. Since A and B are Γ-submonoids, αa1 ∗ βa2 ∈ A and αb1 ∗ βb2 ∈ B. Note that 1M = 1M ∗ 1M ∈ A ∗B. Since M is commutative, αx ∗ βy = α(a1 ∗ b1) ∗ β(a2 ∗ b2) = (αa1 ∗ αb1) ∗ (βa2 ∗ βb2) = (αa1 ∗ βa2) ∗ (αb1 ∗ βb2). This implies that αx ∗ βy ∈ A ∗B. Therefore, A ∗B is a Γ-submonoid of M . Lemma 6. Let A and B be Γ-submonoids of a commutative Γ-monoid M . Then the map f : A → A ∗B defined by f(a) = a ∗ 1M is a Γ-monoid homomorphism. Proof. Let x, y ∈ A such that x = y. Then f(x) = x ∗ 1M = x = y = y ∗ 1M = f(y) and f is well-defined. Let x, y ∈ A. Then (i) f(x ∗ y) = x ∗ y ∗ 1M = x ∗ y = (x ∗ 1M ) ∗ (y ∗ 1M ) = f(x) ∗ f(y), (ii) f(1M ) = 1M ∗ 1M , the identity in A ∗B. Thus, f is a monoid homomorphism. Now, for all α ∈ Γ and x ∈ A, f(αx) = αx ∗ 1M = αx ∗ α1M = α(x ∗ 1M ) = αf(x). Thus, f is a Γ-monoid homomorphism. 5. Quotient Γ-monoids In [5], the quotient Γ-monoid M/S was established using the equivalence relation in Definition 6 such that the commutative Γ-monoidM and Γ-order-ideal S ofM were treated as commutative monoid and submonoid, respectively. Further, the third isomorphism theorem for Γ-monoids via Γ-order-ideals was proved. Here, we define an equivalence relation and construct quotient Γ-monoids via Γ- submonoids. Moreover, we prove the isomorphism theorems. Definition 13. Let M be a Γ-monoid. For any Γ-submonoid S of M and for all x, y ∈ M , we define a binary relation ρS in M by xρSy if and only if for all α ∈ Γ, (αx∗S)∩(αy∗S) ̸= ∅. The next example shows that if a Γ-submonoid S of a Γ-monoid M is not commutative, then ρS is not an equivalence relation. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1786 Example 15. Consider the Γ-monoid M = {1, a, b, c, d, e} in Example 5 with operation ∗ given by ∗ 1 a b c d e 1 1 a b c d e a a a a a a a b b b b b b b c c c c c c c d d d d d d d e e e e e e e Let S = {1, a, b}. Then, by routine calculation, S is a Γ-submonoid of M . Also, S is not commutative since a∗b = a ̸= b = b∗a. Now, for all α ∈ Γ, we have α1∗S = 1∗S = {1, a, b}, αa ∗ S = a ∗ S = {a} and αb ∗ S = b ∗ S = {b}. Thus, (αa ∗ S)∩ (α1 ∗ S) = {a} ≠ ∅ which implies that aρS1. Also, ( α1 ∗ S) ∩ (αb ∗ S) = {b} ≠ ∅ which implies that 1ρSb. However, (αa ∗S)∩ (βb ∗S) = ∅ which implies that a is not related to b under ρS , that is, ρS is not transitive, hence not an equivalence relation. The following result tells us that ρS is an equivalence relation for any commutative Γ-submonoid S of a Γ-monoid M . Further, if M is commutative, then ρS is a congruence relation on M . Theorem 10. Let S be a commutative Γ-submonoid of a Γ-monoid M . Then (i) ρS is an equivalence relation on M . (ii) If M is commutative, then ρS is a congruence relation on M . Proof. Let S be a commutative Γ-submonoid of a Γ-monoid M . (i) Let x ∈ M and S a Γ-submonoid of M . Then, for α ∈ Γ, we have (αx∗S)∩(αx∗S) = αx ∗ S ̸= ∅ since αx = αx ∗ 1M ∈ αx ∗ S. Thus, xρSx and ρS is reflexive. Let xρSy. Then, for all α ∈ Γ, (αx ∗ S) ∩ (αy ∗ S) ̸= ∅. Thus, (αy ∗ S) ∩ (αx ∗ S) = (αx ∗ S) ∩ (αy ∗ S) ̸= ∅. Hence, yρSx and ρS is symmetric. Now, let xρSy and yρSz. Then, for all α, β ∈ Γ, (αx ∗ S) ∩ (αy ∗ S) ̸= ∅ and (βy ∗ S) ∩ (βz ∗ S) ̸= ∅. Thus, we have αx ∗ s1 = αy ∗ s2 and βy ∗ s3 = βz ∗ s4 for some s1, s2, s3, s4 ∈ S. Hence, for all α ∈ Γ, αx ∗ s1 ∗ s3 = αy ∗ s2 ∗ s3 = αz ∗ s2 ∗ s4 and s1 ∗ s3, s2 ∗ s4 ∈ S since S is a Γ-submonoid. Hence, (αx ∗S)∩ (αz ∗S) ̸= ∅ and xρSz. Therefore, ρS is transitive. Consequently, ρS is an equivalence relation on M . (ii) Let M be a commutative Γ-monoid. Suppose that xρSy and u, v ∈ M . Then, we have for all α, β ∈ Γ, (αx ∗ S) ∩ (αy ∗ S) ̸= ∅ and thus, αx ∗ s1 = αy ∗ s2 for some s1, s2 ∈ S. Hence, (αx ∗ s1) ∗ α(u ∗ v) = (αy ∗ s2) ∗ α(u ∗ v). Since M is commutative, for all α ∈ Γ, α(u ∗ x ∗ v) ∗ s1 = α(u ∗ y ∗ v) ∗ s2 and (u ∗ x ∗ v)ρS(u ∗ y ∗ v). Thus, ρS is a congruence relation on M . H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1787 Definition 14. Let S be a commutative Γ-submonoid of a Γ-monoid M . Then for all x ∈ M , the equivalence class of x is denoted and defined by ρS(x) = {y ∈ M : xρSy}. Let S be a commutative Γ-submonoid of a Γ-monoid M and let m ∈ M . Then for all α ∈ Γ, (αm ∗ S) ∩ (αm ∗ S) = αm ∗ S ̸= ∅ since for α = 0, m = m ∗ 1M ∈ m ∗ S. Thus, m ∈ ρS(m). Hence, the following remark holds. Remark 14. Let S be a commutative Γ-submonoid of a Γ-monoidM and letm1,m2 ∈ M . (i) For all m ∈ M , m ∈ ρS(m). (ii) ρS(m1) = ρS(m2) if and only if (αm1 ∗ S) ∩ (αm2 ∗ S) ̸= ∅ for all α ∈ Γ. The quotient M/S using equivalence relation in Definition 6, where M is a monoid and S is a submonoid ofM is different fromM/S using the equivalence relation in Definition 13, where M is a Γ-monoid and S is a Γ-submonoid as shown in the following example. Example 16. Let Γ = Z the additive group of integers and M = Z8 = {0, 1, 2, 3, 4, 5, 6, 7} under addition modulo 8. Then M is a monoid with identity 0. Consider a mapping ϕ : Γ × M → M given by ϕ((α,m)) = 7αm. Let (α, x), (β, y) ∈ Γ × M such that (α, x) = (β, y). Then α = β and x = y. Thus, 7αx = 7βy and ϕ is well-defined. Now, let α, β ∈ Γ and m ∈ M . Observe that (i) ϕ((0,m)) = 70m = m; (ii) ϕ((α+ β,m)) = 7α+βm = 7α7βm = ϕ((α, ϕ((β,m)))). This implies that ϕ is an action. Now, let α ∈ Γ and x, y ∈ M . Then ϕ((α, x+8 y)) = ϕ((α, x+8 y)) = 7α(x+8 y) = 7αx+8 7αy = ϕ((α, x)) +8 ϕ((α, y)). Therefore, M is a Γ-monoid. Let S = {0, 4}. Observe that the identity 0 ∈ S and 0 +8 0 = 0, 0 +8 4 = 4 +8 0 = 4, 4 +8 4 = 0 ∈ S. This implies that S is a submonoid of M . Now, note that 0 +8 S = 0 +8 {0, 4} = {0, 4}, 4 +8 S = 4 +8 {0, 4} = {0, 4}; 1 +8 S = 1 +8 {0, 4} = {1, 5}, 5 +8 S = 5 +8 {0, 4} = {1, 5}; 2 +8 S = 2 +8 {0, 4} = {2, 6}, 6 +8 S = 6 +8 {0, 4} = {2, 6}; 4 +8 S = 4 +8 {0, 4} = {0, 4}, 7 +8 S = 7 +8 {0, 4} = {3, 7}. Moreover, ρS(0) = {0, 4}, ρS(1) = {1, 5}, ρS(2) = {2, 6}, ρS(3) = {3, 7}, ρS(4) = {0, 4}, ρS(5) = {1, 5}, ρS(6) = {2, 6}, and ρS(7) = {3, 7}. Thus, the quotient M/S = {ρS(0), ρS(1), ρS(2), ρS(3)} using the equivalence relation in Definition 6. Now, observe that for all α ∈ Γ, 7α = 1 or 7α = 7. Note that the identity 0 ∈ S and for all α, β ∈ Γ, α0 +8 β0 = 7α0 +8 7β0 = 0 +8 0 ∈ S; H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1788 α0 +8 β4 = 7α0 +8 7β4 = 7β4 = 4 ∈ S; α4 +8 β4 = 7α4 +8 7β4 = 0 or 4 ∈ S. This implies that S is a Γ-submonoid of M . Now, note that for all α ∈ Γ, α0 +8 S = α0 +8 {0, 4} = 7α0 +8 {0, 4} = {0, 4}; α1 +8 S = α1 +8 {0, 4} = 7α1 +8 {0, 4} = {1, 5} or {3, 7}; α2 +8 S = α2 +8 {0, 4} = 7α2 +8 {0, 4} = {2, 6}; α3 +8 S = α3 +8 {0, 4} = 7α3 +8 {0, 4} = {1, 5} or {3, 7}; α4 +8 S = α4 +8 {0, 4} = 7α4 +8 {0, 4} = {0, 4}; α5 +8 S = α5 +8 {0, 4} = 7α5 +8 {0, 4} = {1, 5} or {3, 7}; α6 +8 S = α6 +8 {0, 4} = 7α6 +8 {0, 4} = {2, 6}; α7 +8 S = α7 +8 {0, 4} = 7α7 +8 {0, 4} = {1, 5} or {3, 7}. Moreover, we have ρS(0) = ρS(4) = {0, 4}, ρS(2) = ρS(6) = {2, 6}, and ρS(1) = ρS(3) = ρS(5) = ρS(7) = {1, 3, 5, 7}. Thus, the quotient M/S = {ρS(0), ρS(1), ρS(2)} using the Definition 13. Observe that M/S yield is not equal to M/S above. Moreover, ρS(0) is the same with ρS(0) above, however, ρS(1)s are different. This implies that their equivalence classes are not equal. Hence, M/S via Γ-submonoid is different from M/S via submonoid, where M is a monoid. Theorem 11. If M is a commutative Γ-monoid and S a Γ-submonoid of M , then M/S is a Γ-monoid. Proof. Let M be a commutative Γ-monoid and S a Γ-submonoid of M . By Proposi- tion 1, since ρS is a congruence on M , we have M/ρS = M/S is a monoid with binary operation ◦ given by ρS(x) ◦ ρS(y) = ρS(x ∗ y) with identity ρS(1M ). Consider a map- ping ϕ : Γ × M/S −→ M/S given by (α, ρS(x)) 7→ αρS(x) = ρS( αx) for all α ∈ Γ and x ∈ M . Let (α, ρS(x)), (β, ρS(y)) ∈ Γ × M/S such that (α, ρS(x)) = (β, ρS(y)). Then α = β and ρS(x) = ρS(y). Thus, by Remark 14(ii), (α ′ x ∗ S) ∩ (α ′ y ∗ S) ̸= ∅ for all α′ ∈ Γ, which implies that α′ x ∗ s1 = α′ y ∗ s2 for some s1, s2 ∈ S. Accord- ingly, α(α ′ x ∗ s1) = α(α ′ x) ∗ αs1 = α(α ′ y ∗ s2) = α(α ′ y) ∗ αs2. Since S is a Γ-submonoid, αs1, αs2 ∈ S and (α+α′ x ∗ S) ∩ (α+α′ y ∗ S) ̸= ∅. This means that ϕ(α, ρS(x)) = αρS(x) = ρS( αx) = ρS( βy) = βρS(y) = ϕ(β, ρS(y)). Hence, ϕ is well-defined. Now, for any α, β ∈ Γ and x ∈ M , ϕ((0, ρS(x))) = 0ρS(x) = ρS( 0x) = ρS(x) and ϕ((α+ β, ρS(x))) = α+βρS(x) = α(βρS(x)) = ϕ((α, ϕ((β, ρS(x))))). Thus, ϕ is an action. Now, let α ∈ Γ and x, y ∈ M . Then ϕ((α, ρS(x) ◦ ρS(y))) = α(ρS(x) ◦ ρS(y)) = α(ρS(x ∗ y)) H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1789 = ρS( α(x ∗ y)) = ρS( αx ∗ αy) = ρS( αx) ◦ ρS(αy) = αρS(x) ◦ αρS(y) = ϕ((α, ρS(x))) ◦ ϕ((α, ρS(y))). Therefore, M/S is a Γ-monoid. Proposition 4. Let S be a normal Γ-submonoid of a commutative Γ-monoid M . Then ρS(h) = ρS(1M ) if and only if h ∈ S. Proof. Suppose h ∈ S. Let x ∈ ρS(h). Then, for all α ∈ Γ, (αx∗S)∩(αh∗S) ̸= ∅. This implies that there exist h1, h2 ∈ S such that αx ∗ h1 = αh ∗ h2 ∈ S. Since S is a normal Γ-submonoid and h1, αx ∗ h1 ∈ S, it follows that αx ∈ S for all α ∈ Γ. Accordingly, for all α ∈ Γ, αx∗α1M = α1M∗αx implies (αx∗S)∩(α1M∗S) ̸= ∅. Hence, xρS1M and x ∈ ρS(1M ). It follows that ρS(h) ⊆ ρS(1M ). Let x ∈ ρS(1M ). Then, (αx ∗ S) ∩ (α1M ∗ S) ̸= ∅ for all α ∈ Γ. Thus, there exist h1, h2 ∈ S such that for all α ∈ Γ, αx ∗ h1 = α1M ∗ h2 ∈ S. Since S is a normal Γ-submonoid and h1, αx∗h1 ∈ S, it follows that αx ∈ S. Observe that for all α ∈ Γ, αh = αh∗1M ∈ S since S is a Γ-submonoid. Accordingly, αx∗αh = αh∗αx implies (αx ∗ S) ∩ (αh ∗ S) ̸= ∅. Hence, xρSh and x ∈ ρS(h). Consequently, ρS(1M ) ⊆ ρS(h). Therefore, ρS(1M ) = ρS(h). Now, suppose ρS(1M ) = ρS(h). Then, by Remark 14(ii), (α1M ∗ S) ∩ (αh ∗ S) ̸= ∅ for all α ∈ Γ. Thus, there exist h1, h2 ∈ S such that αh ∗ h2 = α1M ∗ h1 ∈ S for all α ∈ Γ. Since S is a normal Γ-submonoid and h2, αh ∗ h2 ∈ S, it follows that αh ∈ S for all α ∈ Γ. Therefore, h ∈ S. Proposition 5. Let S be a normal Γ-submonoid of a commutative Γ-monoid M . Then M = S if and only if M/S = {ρS(1M )}. Proof. Suppose M = S. Let x ∈ M/S = M/M . Then x = ρM (y) for some y ∈ M . By Proposition 4, we have ρM (1M ) = ρM (y) = x. Hence, M/M = M/S = {ρM (1M )}. Conversely, suppose M/S = {ρS(1M )}. Let x ∈ M . Then ρS(x) ∈ M/S. Thus, ρS(x) = ρS(1M ). By Proposition 4, x ∈ S. Hence, M ⊆ S. Accordingly, M = S. Proposition 6. Let S be a normal Γ-submonoid of a commutative Γ-monoid M . Every Γ-submonoid of M/S is of the form R/S, where R is a Γ-submonoid of M containing S. Proof. Let H be a Γ-submonoid of M/S. Then H ⊆ M/S. Let R = {m ∈ M : ρS(m) ∈ H}. We show that R is a Γ-submonoid of M . Note that the identity in M/S is ρS(1M ) ∈ H and thus, 1M ∈ R. Now, let x, y ∈ R and α, β ∈ Γ. Then ρS(x), ρS(y) ∈ H and αρS(x) ∗ βρS(y) ∈ H since H is a Γ-submonoid. Accordingly, we have ρS( αx ∗ βy) = ρS( αx) ◦ ρS(βy) = αρS(x) ◦ βρS(y) ∈ H. It follows that αx ∗ βy ∈ R. Accordingly, R is a Γ-submonoid of M . Now, we show that S ⊆ R. Let x ∈ S. Then by Proposition 4, we have ρS(x) = ρS(1M ). Since ρS(1M ) is the identity in M/S and H is a Γ-submonoid of M/S, we must have ρS(x) = ρS(1M ) ∈ H. Thus, x ∈ R. Therefore, S ⊆ R. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1790 Theorem 12. Let M be a commutative Γ-monoid and S a normal Γ-submonoid of M . Then the mapping πS : M −→ M/S given by πS(x) = ρS(x) is a Γ-monoid epimorphism with kernel S. Proof. Let x, y ∈ M such that x = y. Then, πS(x) = ρS(x) = ρS(y) = π(y). Thus, πS is well-defined. Now, let x, y ∈ M . Then, we have πS(x ∗ y) = ρS(x ∗ y) = ρS(x) ◦ ρS(y) = πS(x) ◦ πS(y) and πS(1M ) = ρS(1M ). Thus, by Definition 4, πS is a monoid homomorphism. Since απS(x) = αρS(x) = ρS( αx) = πS( αx), by Definition, πS is a Γ-monoid homomorphism. Now, let b ∈ M/S. Then, b = ρS(a) for some a ∈ M . Thus, b = ρS(a) = πS(a) and so, π is surjective. Therefore, πS is an epimorphism. Now, since S is normal, by Proposition 4 we have kerπS = {m ∈ M : ρS(m) = ρS(1M )} = {m ∈ M : m ∈ S} = S ∩M = S as desired. The map πS in Theorem 12 is called the canonical epimorphism. Proposition 7. Let M be a Γ-monoid. Then for any A ⊆ M and S a commutative Γ-submonoid of M , π−1 S (πS(A)) = ⋃ x∈A ρS(x). Proof. Suppose y ∈ π−1 S (πS(A)). Then ρS(y) = πS(y) ∈ πS(A). Since πS is an epimorphism, there exists an x ∈ A such that πS(x) = ρS(y). Hence, ρS(x) = ρS(y). By Remark 14(ii), (αx ∗ S) ∩ (αy ∗ S) ̸= ∅ for all α ∈ Γ, that is, xρSy. This implies that y ∈ ρS(x) for some x ∈ A. It follows that y ∈ ⋃ x∈A ρS(x) so that π−1 S (πS(A)) ⊆⋃ x∈A ρS(x). Conversely, suppose y ∈ ⋃ x∈A ρS(x). Then y ∈ ρS(x) for some x ∈ A. This implies that yρSx, that is, (αy ∗ S) ∩ (αx ∗ S) ̸= ∅ for all α ∈ Γ. By Remark 14(ii), ρS(y) = ρS(x). Thus, πS(y) = πS(x). Since πS(x) ∈ πS(A), it follows that πS(y) ∈ πS(A) implying that y ∈ π−1 S (πS(A)). Hence, ⋃ x∈A ρS(x) ⊆ π−1 S (πS(A)). Therefore, π−1 S (πS(A)) = ⋃ x∈A ρS(x). 6. Isomorphism Theorems In [5], the isomorphism theorems for Γ-monoids via Γ-order-ideals are established. Here, we prove isomorphism theorems for Γ-monoids via Γ-submonoids. As shown already in Example 16, the quotient M/S in our discussion is not the same with the quotient discussed in [5]. Theorem 13. Let (M, ∗) and (N, ·) be commutative Γ-monoids and let f : M → N be a Γ- monoid homomorphism. There exists a unique Γ-monoid homomorphism φ : M/ ker f → N such that the following diagram is commutative M N M/ ker f πker f f φ H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1791 that is, φ ◦πker f = f , where πker f (x) := ρker f (x). Moreover, φ is onto and it has a trivial kernel, namely, kerφ = {ker f}. However, φ is a Γ-monoid isomorphism if and only if ρf = ρker f . Proof. Let (M, ∗) and (N, ·) be commutative Γ-monoids and let f : M → N be a Γ-monoid homomorphism. Since Γ-monoids are monoids and Γ-monoid homomorphism is a monoid homomorphism, by Theorem 1, there exists a unique monoid homomorphism φ : M/ ker f → N such that the following diagram is commutative M N M/ ker f πker f f φ that is, φ ◦ πker f = f , where πker f (x) := ρker f (x). Moreover, φ is onto and it has a trivial kernel, namely, kerφ = {ker f}. However, φ is an isomorphism if and only if ρf = ρker f . Thus, it remains to show that φ is a Γ-monoid homomorphism. Now, let ρker f (x) ∈ M/ ker f and α ∈ Γ. Since f is a Γ-monoid homomorphism, we have φ(αρker f (x)) = φ(ρker f ( αx)) = f(αx) = αf(x) = αφ(ρker f (x)). Hence, φ is a Γ-monoid homomorphism. Corollary 1. Let M and N be commutative Γ-monoids and f : M → N be a Γ-monoid homomorphism. Then f induces a Γ-monoid isomorphism M/ ker f ∼= Imf . Proof. Suppose f : M → N is a Γ-monoid homomorphism. Then, by Theorem 13, there exists a Γ-monoid homomorphism φ : M/ ker f → N . If we set N = Imf , then φ : M/ ker f → Imf is a Γ-monoid epimorphism. Thus, kerφ = {ρker f (x) : f(x) = 1N} = {ker f} implies that ρker f (x) = ker f and x ∈ ker f . Hence, by Proposition 4, ρker f (x) = ρker f (1M ) which implies that kerφ = {ρker f (1M )} and φ is injective. Accord- ingly, M/ ker f ∼= Imf . Corollary 2. Let K and L be normal Γ-submonoids of a commutative Γ-monoid M . Then K/(K ∩ L) ∼= (K ∗ L)/L. Proof. Consider the map f : K → K ∗ L defined by f(k) = k ∗ 1M and πL : K ∗ L → (K∗L)/L defined by πL(k∗l) = ρL(k∗l). Then φ : K → (K∗L)/L defined by φ(k) = ρL(k) is a Γ-monoid homomorphism. Let x ∈ (K ∗ L)/L. Then x = ρL(k ∗ l) for some k ∈ K and l ∈ L. Observe that x = ρL(k ∗ l) = ρL(k)◦ρL(l) = ρL(k)◦ρL(1M ) = ρL(k). So, there is a k ∈ K such that φ(k) = ρL(k) = x and φ is onto. Moreover, kerφ = {k ∈ K : ρL(k) = ρL(1M )} = {k ∈ K : k ∈ L} = K ∩ L. By Corollary 1, K/ kerφ ∼= Imφ = (K ∗ L)/L. The following theorem is the counterpart to the third isomorphism theorem of groups for Γ-monoids via Γ-submonoids. H. Sarapuddin, J. Vilela / Eur. J. Pure Appl. Math, 16 (3) (2023), 1772-1793 1792 Theorem 14. Let S and T be normal Γ-submonoids of a commutative Γ-monoid M with S ⊆ T . Then (M/S)/(T/S) ∼= M/T . Proof. Define f : M/S → M/T by f(ρS(h)) = ρT (h) for all ρS(h) ∈ M/S. Let ρS(h1), ρS(h2) ∈ M/S and suppose that ρS(h1) = ρS(h2). Then, ( αh1 ∗S)∩ (αh2 ∗S) ̸= ∅ for all α ∈ Γ. Thus, αh1 ∗w1 = αh2 ∗w2 for some w1, w2 ∈ S ⊆ T . Thus, (αh1 ∗T )∩ (αh2 ∗ T ) ̸= ∅ for all α ∈ Γ. By Remark 14(ii), ρT (h1) = ρT (h2). Thus, f(ρS(h1)) = f(ρS(h2)). Hence, f is well-defined. Let ρS(h1), ρS(h2) ∈ M/S. Then f(ρS(h1) ◦ ρS(h2)) = f(ρS(h1 ∗ h2)) = ρT (h1) ◦ ρT (h2) = f(ρS(h1)) ◦ f(ρS(h2)). Hence, f is a homomorphism. Let ρS(h) ∈ ker f . Then f(ρS(h)) = ρT (1M ), the identity in M/T . Thus, ρT (h) = ρT (1M ). By Proposition 4, h ∈ T . Hence, ρS(h) ∈ T/S. Thus, ker f ⊆ T/S. Let ρS(h) ∈ T/S. Then h ∈ T . By Proposition 4, ρT (h) = ρT (1M ). Thus, f(ρS(h)) = ρT (h) = ρT (1M ). Accordingly, ρS(h) ∈ ker f . Hence, T/S ⊆ ker f . So, T/S = ker f . For ρS(x), ρS(y) ∈ M/S and α ∈ Γ, recall that ρS(x)ρfρS(y) if and only if f(αρS(x)) = f(αρS(y)). We claim that ρf = ρker f . Let ρS(z) ∈ M/S. We show that ρf (ρS(z)) = ρker f (ρS(z)). Let ρS(w) ∈ ρker f (ρS(z)). Then (αρS(z) ◦ ker f) ∩ (αρS(w) ◦ ker f) ̸= ∅. Thus, there exist y1, y2 ∈ ker f such that αρS(z) ◦ y1 = αρS(w) ◦ y2. Hence, f(αρS(z)) = f(αρS(z)) ◦ ρT (1M ) = f(αρS(z)) ◦ f(y1) = f(αρS(z) ◦ y1) and f(αρS(w)) = f(αρS(w)) ◦ ρT (1M ) = f(αρS(w)) ◦ f(y2) = f(αρS(w) ◦ y2). So, by well- definedness of f , we have f(αρS(z)) = f(αρS(z) ◦ y1) = f(αρS(w) ◦ y2) = f(αρS(w)). Accordingly, ρS(w) ∈ ρf (ρS(z)). Thus, ρker f (ρS(z)) ⊆ ρf (ρS(z)). Now, let ρS(w) ∈ ρf (ρS(z)) and α ∈ Γ. Then f(αρS(z)) = f(αρS(w)), that is, αρT (z) = αρT (w). Thus, ρT ( αz) = ρT ( αw) implies (αw ∗ T ) ∩ (αz ∗ T ) ̸= ∅. Thus, there exist h1, h2 ∈ T such that αw∗h1 = αz∗h2. Hence, ρS(h1), ρS(h2) ∈ T/S = ker f . Consequently, ρS( αw) ◦ ρS(h1) = ρS( αw ∗ h1) = ρS( αz ∗ h2) = ρS( αz) ◦ ρS(h2) for all α ∈ Γ. This implies that (αρS(w) ◦ ker f) ∩ (αρS(z) ◦ ker f) ̸= ∅. Hence, ρS(w) ∈ ρker f (ρS(z)). Accordingly, ρf (ρS(z)) ⊆ ρker f (ρS(z)). Therefore, ρf (ρS(z)) = ρker f (ρS(z)) for all ρS(z) ∈ M/S, that is, ρf = ρker f . By Theorem 13, these all imply that (M/S)/(T/S) = (M/S)/ ker f ∼= M/T . 7. Conclusion : In this paper, we have shown that Γ-ideals and Γ-submonoids of a Γ-monoid M are not equivalent to the existing Γ-order-ideals of M . For any Γ-monoids M and N , we proved that the kernel of a Γ-monoid homomorphism φ : M → N is a Γ-submonoid of M . Also, for any Γ-submonoid S of a Γ-monoid M , ρS is a congruence relation if M is commutative and thus, M/S = M/ρS is defined for commutative Γ-monoid M . Moreover, isomorphism theorems for Γ-monoids via Γ-submonoids were proved. 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