EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5976 ISSN 1307-5543 – ejpam.com Published by New York Business Global Quasi-Ideals and H-Classes on the Direct Product of Two Semigroups Panuwat Luangchaisri1, Ontima Pankoon1, Thawhat Changphas1,∗ 1 Department of Mathematics, Faculty of Science Khon Kaen University, Khon Kaen 40002, Thailand Abstract. Let S be a semigroup and x ∈ S. The principal quasi-ideal of S containing x is denoted by Q(x). An H-class of S containing x is denoted by Hx. Let S1, S2 be semigroups. The direct product S1×S2 is defined as the Cartesian product of S1 and S2 equipped with the componentwise binary operation. Let (a, b) ∈ S1×S2. The direct product of Q(a)×Q(b) need not to be Q((a, b)). In this paper, we provide necessary and sufficient conditions when Q(a)×Q(b) = Q((a, b)) and the conditions when H(a,b) = Ha ×Hb. 2020 Mathematics Subject Classifications: 20M12 Key Words and Phrases: Semigroups, direct product, quasi-ideal, H-class, maximal H-class 1. Introduction Let S be a semigroup. A nonempty subset Q of S is called a quasi-ideal of S if QS ∩ SQ ⊆ Q. The concept of quasi-ideals was introduced by Steinfeld [1]. Quasi- ideals have been studied extensively in semigroup theory and have been found to have many important applications and connections with other algebraic structures. One such connection is with congruences on semigroups. It was proved by Steinfeld [2] that for any a, b ∈ S, aHb if and only if Q(a) = Q(b). This result shows that a congruence H on a semigroup can be described in terms of the associated quasi-ideal. Given semigroups S1 and S2, the direct product S1 × S2 is defined as the Cartesian product of S1 and S2, equipped with the componentwise binary operation. That is, for any (s1, s2), (t1, t2) ∈ S1 × S2, we define (s1, s2)(t1, t2) = (s1t1, s2t2). Let (a, b) ∈ S1 × S2. Fabrici investigated the structure of L-classes in the direct product of S1 and S2 [3]. The author showed that L(a) × L(b) need not be equal to L((a, b)) and give the conditions when L(a) × L(b) = L((a, b)). Fabrici also investigated the conditions under which the direct product of L-classes La × Lb is an L-class L(a,b). Conditions for the direct product of J -classes Ja × Jb = J(a,b) were investigated as well [4]. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5976 Email addresses: panulu@kku.ac.th (P. Luangchaisri), ontimapa@kkumail.com (O. Pankoon), thacha@kku.ac.th (T. Changphas) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 2 of 14 Let m,n be nonnegative integers. A subsemigroup A of S is called an (m,n)-ideal of S if AmSAn ⊆ A [5]. Here, A0S = SA0 = S. The smallest (m,n)-ideal of S containing a is denoted by [a](m,n). Luangchaisri and Changphas provided necessary and sufficient con- ditions for [a](m,n) × [b](m,n) = [(a, b)](m,n) [6]. Moreover, they determined an equivalence class on a semigroup S by for any x ∈ S, J(m,n),x = {y ∈ S | [x](m,n) = [y](m,n)} and provided the condition for J(m,n),a × J(m,n),b = J(m,n),(a,b). In this paper, we show that the direct product Q(a) ×Q(b) need not to be Q((a, b)). In addition, we provide necessary and sufficient conditions when Q(a)×Q(b) = Q((a, b)) and the condition when Ha ×Hb = H(a,b). 2. Main Results Let S be a semigroup. For nonempty subsets A,B of S, the set product of A and B, denoted by AB, is defined by AB = {ab | a ∈ A and b ∈ B}. In particular, we write Ab instead of A{b} and write aB instead of {a}B. Definition 1. A nonempty subset Q of a semigroup S is called a quasi-ideal of S if QS ∩ SQ ⊆ Q. Definition 2. Let S be a semigroup and a ∈ S. The smallest quasi-ideal of S containing a, denoted by Q(a), is called the principal quasi-ideal of S generated by a. Remark 1. Let a ∈ S. Then Q(a) = {a} ∪ (aS ∩ Sa). Theorem 1. If Q1 and Q2 are quasi-ideals of semigroups S1 and S2, respectively, then Q1 ×Q2 is a quasi-ideal of S1 × S2. Proof. Assume that Q1 and Q2 be quasi-ideals of S1 and S2, respectively. Then (Q1 ×Q2)(S1 × S2) ∩ (S1 × S2)(Q1 ×Q2) = (Q1S1 ×Q2S2) ∩ (S1Q1 × S2Q2) = (Q1S1 ∩ S1Q1)× (Q2S2 ∩ S2Q2) ⊆ Q1 ×Q2. Therefore, Q1 ×Q2 is a quasi-ideal of S1 × S2. The above theorem states that the direct product of two quasi-ideals is a quasi-ideal. However, it is important to note that the direct product of two principal quasi-ideals does not necessarily result in a principal quasi-ideal. This can be seen in the following example: Example 1. ([7]) Consider two semigroups (S1, ∗) and (S2, ·), where S1 = {a1, a2, a3, a4} and S2 = {b1, b2, b3, b4}. The binary operation ∗ on S1 and the binary operation · on S2 are defined as follows: P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 3 of 14 ∗ a1 a2 a3 a4 a1 a1 a1 a1 a1 a2 a1 a2 a2 a4 a3 a1 a2 a2 a4 a4 a1 a4 a4 a2 · b1 b2 b3 b4 b1 b1 b1 b1 b4 b2 b1 b2 b2 b4 b3 b1 b3 b3 b4 b4 b4 b4 b4 b1 We observe that (a3, b4) ∈ Q(a3) × Q(b3), but (a3, b4) /∈ Q((a3, b3)). Thus, we have Q(a3) × Q(b3) ̸= Q((a3, b3)). Furthermore, for any (u, v) ∈ S1 × S2 such that (u, v) ̸= (a3, b4), we have (a3, b4) ∈ Q(a3)×Q(b3), but (a3, b4) /∈ {(u, v)}∪((uS1∩S1u)×(vS2∩S2v)). Therefore, we obtain Q(a3) ×Q(b3) ̸= Q((u, v)), which implies that Q(a3) ×Q(b3) is not the principal quasi-ideal of S1 × S2. From Example 1, we see that Q(a)×Q(b) need not to be Q((a, b)). Next, we present the inclusion of Q(a) × Q(b) and Q((a, b)) and give necessary and sufficient conditions when Q(a)×Q(b) = Q((a, b)). Theorem 2. Let (a, b) ∈ S1 × S2. Then Q((a, b)) ⊆ Q(a)×Q(b). Proof. Assume that (a, b) ∈ S1 × S2. Then Q((a, b)) = {(a, b)} ∪ ((a, b)(S1 × S2) ∩ (S1 × S2)(a, b)) = {(a, b)} ∪ ((aS1 × bS2) ∩ (S1a× S2b)) = {(a, b)} ∪ ((aS1 ∩ S1a)× (bS2 ∩ S2b)) ⊆ ({a} ∪ (aS1 ∩ S1a))× ({b} ∪ (bS2 ∩ S2b)) = Q(a)×Q(b). Therefore, Q((a, b)) ⊆ Q(a)×Q(b). Theorem 3. Let (a, b) ∈ S1 × S2. Then Q((a, b)) = Q(a) × Q(b) if and only if at least one of the following conditions holds: (1) aS1 ∩ S1a = {a}; (2) bS2 ∩ S2b = {b}; (3) a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b. Proof. Assume that (1), (2), and (3) do not hold. Then we have two cases to consider: (i) a /∈ aS1 ∩ S1a and bS2 ∩ S2b ̸= {b}; (ii) b /∈ bS2 ∩ S2b and aS1 ∩ S1a ̸= {a}. If (i) holds, then there exists v ∈ bS2 ∩ S2b such that v ̸= b. Then (a, v) ̸= (a, b) and (a, v) /∈ (aS1 ∩ S1a)× (bS2 ∩ S2b). Thus, (a, v) /∈ Q((a, b)). Since (a, v) ∈ Q(a)×Q(b), it follows that Q((a, b)) ̸= Q(a)×Q(b). We can prove similarly that Q((a, b)) ̸= Q(a)×Q(b) when (ii) holds. P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 4 of 14 Conversely, assume that (1) holds. Then we get Q(a)×Q(b) = ({a} ∪ (aS1 ∩ S1a))× ({b} ∪ (bS2 ∩ S2b)) = {a} × ({b} ∪ (bS2 ∩ S2b)) = {(a, b)} ∪ ({a} × (bS2 ∩ S2b)) = {(a, b)} ∪ ((aS1 ∩ S1a)× (bS2 ∩ S2b)) = Q((a, b)). On the same way, we have Q(a) × Q(b) = Q((a, b)) when the condition (2) is satisfied. Assume that the condition (3) occurs. Then we have Q(a)×Q(b) = ({a} ∪ (aS1 ∩ S1a))× ({b} ∪ (bS2 ∩ S2b)) = (aS1 ∩ S1a)× (bS2 ∩ S2b) = Q((a, b)). Therefore, Q(a)×Q(b) = Q((a, b)). The following theorem shows that if Q(a)×Q(b) ̸= Q((a, b)), then Q(a)×Q(b) is not the principal quasi ideal. Theorem 4. If Q(a) × Q(b) ̸= Q((a, b)), then Q(a) × Q(b) ̸= Q((u, v)) for all (u, v) ∈ S1 × S2. Proof. Assume that Q(a) × Q(b) ̸= Q((a, b)) and Q(a) × Q(b) = Q((u, v)) for some (u, v) ∈ S1 × S2. Then we obtain the following: (a, b) ∈ Q(a)×Q(b) = Q((u, v)) ⊆ Q(u)×Q(v) (u, v) ∈ Q((u, v)) = Q(a)×Q(b). This implies that (aS1 ∩ S1a)× (bS2 ∩ S2b) ⊆ (Q(u)S1 ∩ S1Q(u))× (Q(v)S2 ∩ S2Q(v)) = (uS1 ∩ S1u)× (vS2 ∩ S2v) ⊆ (Q(a)S1 ∩ S1Q(a))× (Q(b)S2 ∩ S2Q(b)) = (aS1 ∩ S1a)× (bS2 ∩ S2b). Since (a, b) ̸= (u, v) and (a, b) ∈ Q(a)×Q(b) = Q((u, v)), it follows that (a, b) ∈ (uS1 ∩ S1u)× (vS2 ∩ S2v) = (aS1 ∩ S1a)× (bS2 ∩ S2b). By Theorem 3, we have Q(a)×Q(b) = Q((a, b)), which contradicts the hypothesis. P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 5 of 14 The concept of Green’s relations were introduced by Green [8]. It can be found in many textbooks on semigroup theory (e.g. [9], [10]). Equivalence relations L and R on a semigroup S are defined by aLb ⇔ L(a) = L(b) aRb ⇔ R(a) = R(b), where L(u) (respectively, R(u)) is the principal left (respectively, right) ideal of S con- taining u. An equivalence relation H on S is define by H = L ∩R. For any a ∈ S, let Ha denote an H-class of S containing a, that is, Ha = {b ∈ S | aHb}. Lemma 1. ([2]) Let a, b ∈ S. Then aHb if and only if Q(a) = Q(b). This implies that Ha = Hb if and only if Q(a) = Q(b). We give an example to show that the direct product of two H-classes need not to be an H-class as the following illustrative example. Example 2. ([7]) Let (S1, ∗) and (S2, ·) be semigroups where S1 = {a1, a2, a3, a4} and S2 = {b1, b2, b3, b4} together with ∗ : S1 × S1 → S1 and · : S2 × S2 → S2 defined by ∗ a1 a2 a3 a4 a1 a1 a1 a1 a1 a2 a1 a2 a3 a4 a3 a1 a3 a2 a4 a4 a4 a4 a4 a4 · b1 b2 b3 b4 b1 b1 b1 b3 b4 b2 b1 b1 b3 b4 b3 b3 b3 b4 b1 b4 b4 b4 b1 b3 Since (a3, b2) ∈ Ha2×Hb2 and (a3, b2) /∈ H(a2,b2), we have that Ha2×Hb2 ⊈ H(a2,b2). Thus, Ha2 ×Hb2 is not an H-class of S1 × S2. Theorem 5. Let (a, b) ∈ S1 × S2. Then (1) H(a,b) ⊆ Ha ×Hb; (2) If H(a,b) ⊊ Ha ×Hb, then Ha ×Hb is a union of at least two H-classes. Proof. (1). Let (u, v) ∈ H(a,b). Then, Q((u, v)) = Q((a, b)). We obtain that Q(u) = Q(a) and Q(v) = Q(b). Indeed, (u, v) ∈ Q((u, v)) = Q((a, b)) ⊆ Q(a)×Q(b) P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 6 of 14 and (a, b) ∈ Q((a, b)) = Q((u, v)) ⊆ Q(u)×Q(v). By Lemma 1, we have (u, v) ∈ Ha ×Hb. (2). Assume that H(a,b) ⊊ Ha ×Hb. There is (u, v) ∈ S1 × S2 such that (u, v) ∈ Ha ×Hb and (u, v) /∈ H(a,b). By (1), we have that H(u,v) ⊆ Hu ×Hv = Ha ×Hb. Therefore, H(a,b) and H(u,v) are difference classes contained in Ha ×Hb. Theorem 6. Let (a, b) ∈ S1 × S2. Then H(a,b) = Ha × Hb if and only if at least one of the following conditions holds: (1) Ha = {a} and Hb = {b}; (2) a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b. Proof. Assume that H(a,b) = Ha ×Hb. If H(a,b) = {(a, b)}, then we have Ha ×Hb = H(a,b) = {(a, b)} = {a} × {b}. Hence, Ha = {a} and Hb = {b}. If H(a,b) ̸= {(a, b)}, then there exists (u, v) ∈ S1 × S2 such that (u, v) ̸= (a, b) and Q((u, v)) = Q((a, b)). Since (u, v) ∈ H(a,b) = Ha ×Hb, we have Q(u) = Q(a) and Q(v) = Q(b). Then (uS1 ∩ S1u)× (vS2 ∩ S2v) = (aS1 ∩ S1a)× (bS2 ∩ S2b). The assumptions (a, b) ∈ Q((a, b)) = Q((u, v)) and (a, b) ̸= (u, v) lead to (a, b) ∈ (uS1 ∩ S1u)× (vS2 ∩ S2v) = (aS1 ∩ S1a)× (bS2 ∩ S2b). Thus, a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b. Conversely, let (u, v) ∈ Ha ×Hb. If (1) holds, then (u, v) = (a, b) ∈ H(a,b). Suppose that (2) holds. By Theorem 3, we have Q((a, b)) = Q(a) × Q(b). Since (u, v) ∈ Ha ×Hb, it follows that Q(a) = Q(u) and Q(b) = Q(v). P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 7 of 14 Thus, Q((u, v)) ⊆ Q(u)×Q(v) = Q(a)×Q(b) = Q((a, b)). On the other hand, Q(a, b) = Q(a)×Q(b) = (aS1 ∩ S1a)× (bS2 ∩ S2b) = (Q(a)S1 ∩ S1Q(a))× (Q(b)S2 ∩ S2Q(b)) = (Q(u)S1 ∩ S1Q(u))× (Q(v)S2 ∩ S2Q(v)) = (uS1 ∩ S1u)× (vS2 ∩ S2v) = (u, v)(S1 × S2) ∩ (S1 × S2)(u, v) ⊆ Q((u, v)). Thus, Q((a, b)) = Q((u, v)). This means (u, v) ∈ H(a,b). By these two cases, we conclude that Ha ×Hb ⊆ H(a,b). By Theorem 5, we have H(a,b) ⊆ Ha ×Hb. Therefore, Ha ×Hb = H(a,b). Theorem 7. Let (a, b) ∈ S1 × S2. If Q((a, b)) = Q(a)×Q(b), then H(a,b) = Ha ×Hb. Proof. Assume that Q((a, b)) = Q(a) × Q(b). By Theorem 3, we have one of the following conditions holds: (i) aS1 ∩ S1a = {a}; (ii) bS2 ∩ S2b = {b}; (iii) a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b. Assume that aS1 ∩ S1a = {a}. Then Ha = {a}. If b ∈ bS2 ∩ S2b, we have that H(a,b) = Ha × Hb by Theorem 6 (2). If b /∈ bS2 ∩ S2b, then we obtain that Hb = {b}. Hence, H(a,b) = Ha × Hb by Theorem 6 (1). The case bS2 ∩ S2b = {b} can be proved similarly. The case a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b is obtained directly from Theorem 6 (2). Therefore, H(a,b) = Ha ×Hb. However, the converse of Theorem 7 is not true in general. That is, H(a,b) = Ha ×Hb does not implies that Q((a, b)) = Q(a)×Q(b). Example 3. ([7]) Let (S1, ∗) and (S2, ·) be semigroups where S1 = {a1, a2, a3, a4} and S2 = {b1, b2, b3, b4}. The binary operations ∗ : S1 × S1 −→ S1 and · : S2 × S2 −→ S2 are defined as: ∗ a1 a2 a3 a4 a1 a1 a1 a1 a1 a2 a1 a1 a1 a1 a3 a1 a1 a1 a2 a4 a1 a1 a2 a2 · b1 b2 b3 b4 b1 b1 b1 b1 b1 b2 b1 b1 b1 b1 b3 b1 b1 b3 b3 b4 b1 b1 b3 b3 We have Ha3 = {a3} and Hb4 = {b4}. Hence H(a3,b4) = Ha3 × Hb4 by Theorem 6. Since (a3, b1) ∈ Q(a3) × Q(b4) but (a3, b1) /∈ Q((a3, b4)), it follows that Q((a3, b4)) ̸= Q(a3)×Q(b4). P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 8 of 14 Theorem 8. Let (a, b) ∈ S1 × S2. If | Ha |> 1 and | Hb |> 1, then (1) a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b; (2) H(a,b) = Ha ×Hb. Proof. (1) Assume that | Ha |> 1 and | Hb |> 1. Then there exist u ∈ Ha and v ∈ Hb such that u ̸= a and v ̸= b. It can be observed that (a, b) ∈ (uS1 ∩ S1u)× (vS2 ∩ S2v) and (u, v) ∈ (aS1 ∩ S1a)× (bS2 ∩ S2b). Thus, (a, b) ∈ (uS1 ∩ S1u)× (vS2 ∩ S2v) = (Q(u)S1 ∩ S1Q(u))× (Q(v)S2 ∩ S2Q(v)) = (Q(a)S1 ∩ S1Q(a))× (Q(b)S2 ∩ S2Q(b)) = (aS1 ∩ S1a)× (bS2 ∩ S2b). Therefore, a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b. (2) It is obtained immediately by Theorem 6. We obtain that the reverse of Theorem 8 is not true as the following example. Example 4. [7] Let (S1, ∗) and (S2, ·) be semigroups where S1 = {a1, a2, a3, a4} and S2 = {b1, b2, b3, b4}. The binary operations ∗ : S1 × S1 −→ S1 and · : S2 × S2 −→ S2 are defined as: ∗ a1 a2 a3 a4 a1 a1 a1 a1 a1 a2 a1 a2 a2 a2 a3 a1 a2 a2 a2 a4 a1 a4 a4 a4 · b1 b2 b3 b4 b1 b1 b1 b1 b1 b2 b1 b2 b2 b2 b3 b1 b2 b2 b2 b4 b1 b2 b2 b2 We have that H(a1,b2) = {(a1, b2)} = {a1} × {b2} = Ha1 ×Hb2 but | Ha1 |=| Hb2 |= 1. Corollary 1. Let (a, b) ∈ S1 × S2. If Ha ×Hb is a union of at least two H-classes, then (1) | Ha |> 1 and Hb = {b} or (2) Ha = {a} and | Hb |> 1. Theorem 9. Let (a, b) ∈ S1 × S2. Then Ha ×Hb is a union of at least two H-classes if and only if (1) | Ha |> 1, Hb = {b} and b /∈ bS2 ∩ S2b or (2) a /∈ aS1 ∩ S1a,Ha = {a} and | Hb |> 1. P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 9 of 14 Proof. Assume that Ha ×Hb is a union of at least two H-classes. By Corollary 1, we have either | Ha |> 1 and Hb = {b} or Ha = {a} and | Hb |> 1. Case 1: | Ha |> 1 and Hb = {b}. We have that a ∈ aS1 ∩ S1a. Suppose that b ∈ bS2∩S2b. Form Theorem 3, we haveHa×Hb = H(a,b) which contradicts to the assumption. Therefore, b /∈ bS2 ∩ S2b. Case 2: Ha = {a} and | Hb |> 1. We can prove similarly Case 1. Conversely, we assume that (1) or (2) holds. If (1) holds, then H(a,b) ̸= Ha × Hb by Theorem 6. Hence, H(a,b) ⊊ Ha ×Hb. By Theorem 5, we obtain that Ha ×Hb is a union of at least two H-classes. We can proceed analogously in another case. Let u ∈ S. An L-class (respectively, R-class) of S containing u is denoted by Lu (respectively, Ru). Since relations L and R on S are defined in terms of ideals, the inclusion order among these ideals induces a partial order among equivalence classes. For any a, b ∈ S, La ≤ Lb ⇔ L(a) ⊆ L(b) Ra ≤ Rb ⇔ R(a) ⊆ R(b). These partial orders induce a partial order among H-classes as follows: Ha ≤ Hb ⇔ La ≤ Lb and Ra ≤ Rb. We say that Ha is maximal if there is no u ∈ S such that Ha ≤ Hu and Ha ̸= Hu. Equivalently, Ha is maximal if and only if there is no u ∈ S such that Ha ≤ Hu and Q(a) ̸= Q(u). A characterization of a maximal H-class is indicated as the following theorem. Theorem 10. Let a ∈ S. Then Ha is a maximal if and only if there is no b ∈ S such that Q(a) ⊊ Q(b). Proof. Suppose that there exists b ∈ S such that Q(a) ⊊ Q(b),which implies that Ha ̸= Hb. Since a ∈ Q(a) ⊊ Q(b) = L(b) ∩R(b), we have a ∈ L(b) and a ∈ R(b). Hence, L(a) ⊆ L(b) and R(a) ⊆ R(b). It follows that La ⩽ Lb and Ra ⩽ Rb. Thus, Ha is not maximal. Conversely, assume thatHa is not maximal. Then there exists b ∈ S such thatHa ⩽ Hb and Ha ̸= Hb. Thus, La ⩽ Lb and Ra ⩽ Rb. These imply L(a) ⊆ L(b) R(a) ⊆ R(b) Q(a) ̸= Q(b). P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 10 of 14 Hence, Q(a) = L(a) ∩R(a) ⊆ L(b) ∩R(b) = Q(b). Therefore, Q(a) ⊊ Q(b). Theorem 11. Let (a, b) ∈ S1 × S2 such that (a, b) ∈ (aS1 ∩ S1a) × (bS2 ∩ S2b). Then H(a,b) is maximal if and only if Ha and Hb are maximal. Proof. Assume that H(a,b) is not maximal. Then there exists (u, v) ∈ S1×S2 such that Q((a, b)) ⊊ Q((u, v)). By Theorem 3, we have Q(a)×Q(b) = Q((a, b)) ⊊ Q((u, v)) ⊆ Q(u)×Q(v). This means that Q(a) ⊊ Q(u) or Q(b) ⊊ Q(v). Thus, Ha or Hb is not maximal. Conversely, suppose that Ha is not maximal. Then there exists u ∈ S1 such that Q(a) ⊊ Q(u). This implies that Q(a) ⊆ uS1 ∩ S1u and u /∈ Q(a). Case 1: u ∈ uS1 ∩ S1u. Then Q((a, b)) = Q(a)×Q(b) ⊊ Q(u)×Q(b) = Q((u, b)). Case 2: u /∈ uS1 ∩ S1u. Then Q((a, b)) = Q(a)×Q(b) ⊊ {(u, b)} ∪ (Q(a)×Q(b)) = {(u, b)} ∪ ((aS1 ∩ S1a)× (bS2 ∩ S2b)) ⊆ {(u, b)} ∪ ((uS1 ∩ S1u)× (bS2 ∩ S2b)) = Q((u, b)). Therefore, H(a,b) is not maximal. We obtain the same conclusion when Hb is not maximal. Lemma 2. Let (a, b) ∈ S1 × S2 such that (a, b) /∈ (aS1 ∩ S1a) × (bS2 ∩ S2b). If Ha and Hb are maximal, then H(a,b) is maximal. P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 11 of 14 Proof. Assume that H(a,b) is not maximal. Then there exists (u, v) ∈ S1×S2 such that Q((a, b)) ⊊ Q((u, v)). It follows that (a, b) ̸= (u, v) and (a, b) ∈ (uS1∩S1u)×(vS2∩S2v). We obtain Q(a) ⊆ Q(u) and Q(b) ⊆ Q(v). If Q(a) = Q(u) and Q(b) = Q(v), then we have (a, b) ∈ (uS1 ∩ S1u)× (vS2 ∩ S2v) = (Q(u)S1 ∩ S1Q(u))× (Q(v)S2 ∩ S2Q(v)) = (Q(a)S1 ∩ S1Q(a))× (Q(b)S2 ∩ S2Q(b)) = (aS1 ∩ S1a)× (bS2 ∩ S2b). This contradicts to (a, b) /∈ (aS1 ∩ S1a)× (bS2 ∩ S2b). We therefore conclude that Q(a) ⊊ Q(u) or Q(b) ⊊ Q(v). Thus, Ha or Hb is not maximal. Lemma 3. Let H(a,b) and H(u,v) be difference H-classes contained in Ha ×Hb. If (a, b) /∈ (aS1 ∩ S1a)× (bS2 ∩ S2b), then (u, v) /∈ (uS1 ∩ S1u)× (vS2 ∩ S2v). Proof. Assume that (u, v) ∈ (uS1 ∩ S1u)× (vS2 ∩ S2v). Since H(u,v) ⊆ Ha × Hb, we have Hu × Hv = Ha × Hb. This implies Q(u) = Q(a) and Q(v) = Q(b). Thus, (a, b) ∈ Q(a)×Q(b) = Q(u)×Q(v) = (uS1 ∩ S1u)× (vS2 ∩ S2v) = (Q(u)S1 ∩ S1Q(u))× (Q(v)S2 ∩ S2Q(v)) = (Q(a)S1 ∩ S1Q(a))× (Q(b)S2 ∩ S2Q(b)) = (aS1 ∩ S1a)× (bS2 ∩ S2b). Lemma 4. Let Ha and Hb be maximal H-classes of S1 and S2, respectively. (1) If Ha = {a} and Hb = {b}, then Ha ×Hb = H(a,b) is maximal. (2) If | Ha |> 1 and | Hb |> 1, then Ha ×Hb = H(a,b) is maximal. Proof. (1) Assume that Ha = {a} and Hb = {b}. Let (u, v) ∈ S1 × S2 such that Q((a, b)) ⊆ Q((u, v)). If (u, v) /∈ Q((a, b)), then we have (u, v) ̸= (a, b), that is, u ̸= a or v ̸= b. Assume that u ̸= a. Then we have two cases to consider: P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 12 of 14 (i) If u /∈ aS1 ∩ S1a, then u /∈ Q(a). This implies that Q(a) ⊊ Q(u) which contradicts to Ha is maximal. (ii) If u ∈ aS1 ∩ S1a, then Q(a) = Q(u). Thus, u ∈ Ha which contradicts to Ha = {a}. The case v ̸= b can be proved similarly. Therefore, (u, v) ∈ Q((a, b)). Thus, Q((u, v)) = Q((a, b)). This means that H(a,b) is maximal. (2) Assume that | Ha |> 1 and | Hb |> 1. Then a ∈ aS1 ∩ S1a and b ∈ bS2 ∩ S2b. These imply that Ha ×Hb = H(a,b). By Theorem 11, the direct product Ha ×Hb is maximal. Theorem 12. Let (a, b) ∈ S1 × S2. If Ha and Hb are maximal, then one of the following conditions holds: (1) Ha ×Hb is maximal; (2) Ha ×Hb is the union of at least two maximal H-classes in S1 × S2. Proof. Assume that Ha and Hb are maximal. By Lemma 4, Ha × Hb = H(a,b) is maximal when one of the following conditions holds: (i) | Ha |= 1 and | Hb |= 1 or (ii) | Ha |> 1 and | Hb |> 1. Assume that | Ha |> 1 and | Hb |= 1. If (a, b) ∈ (aS1 ∩ S1a)× (bS2 ∩ S2b), then we obtain that Ha ×Hb = H(a,b) by Theorem 6. By Theorem 11, Ha ×Hb is maximal. On the other hand, assume that (a, b) /∈ (aS1 ∩ S1a)× (bS2 ∩ S2b). Then H(a,b) ⊊ Ha ×Hb. By Theorem 5, the direct product Ha ×Hb contains at least two H-classes in S1 × S2. Let H(u,v) be arbitrary H-class of S1 × S2 contained in Ha × Hb. By Lemma 3, we have (u, v) /∈ (us1 ∩ S1u)× (vS2 ∩ S2v). Suppose that H(u,v) is not maximal. By Lemma 2, Ha = Hu is not maximal or Hb = Hv is not maximal. This contradicts to assumption. Therefore, H(u,v) is maximal. The case | Ha |= 1 and | Hb |> 1 can be proved similarly. Definition 3. Let S be a semigroup. An element a ∈ S is decomposable if there are u, v ∈ S such that a = uv. We say that b ∈ S is indecomposable if b is not decomposable, equivalently, b ∈ S \ S2. Theorem 13. Let (a, b) ∈ S1 × S2. If (a, b) is indecomposable, then H(a,b) is a maximal H-class of S1 × S2. Proof. Assume that H(a,b) is not a maximal H-class. This means that there exists (u, v) ∈ S1 × S2 such that Q((a, b)) ⊊ Q((u, v)). P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 13 of 14 Since (a, b) ̸= (u, v) and (a, b) ∈ Q((u, v)), we have (a, b) ∈ (uS1 ∩ S1u)× (vS2 ∩ S2v). Therefore, (a, b) is decomposable. The following example shows that the reverse of the above theorem is not generally true. Example 5. [7] Let S = {a, b, c, d} be a semigroup with the binary operation ∗ on S defined by ∗ a b c d a a a a a b a a a a c a b c c d a b c c Then H(b,b) is maximal, but (b, b) is decomposable. 3. Conclusions In this paper, we studied quasi-ideals and H-classes in the direct product of two semi- groups S1 × S2 by focusing on being the principal quasi-ideal of Q(a) × Q(b) and being the H-class of Ha ×Hb in S1 × S2, where Q(a) and Q(b) are the principal quasi-ideal of S1 generated by a ∈ S1 and the principal quasi-ideal of S2 generated by b ∈ S2, respec- tively. Similarly, Ha and Hb are an H-class of S1 containing a ∈ S1 and an H-class of S2 containing b ∈ S2, respectively. First, we proved that Q(a) × Q(b) is a quasi-ideal in S1 × S2, and we provided an example to indicate that it is not guaranteed to be the principal quasi-ideal generated by (a, b). Furthermore, we ensured that if Q(a) × Q(b) is not the principal quasi-ideal generated by (a, b), then it is not the principal quasi-ideal generated by any other element in S1×S2. We then characterized when the direct product of two principal quasi-ideals is the principal quasi-ideal by presenting the sufficient and necessary conditions for Q(a) × Q(b) = Q((a, b)). In the context of the H-class, we gave an example to demonstrate that Ha × Hb is not guaranteed to be an H-class in S1 × S2. The relation between the H-class H(a,b) and the direct product Ha ×Hb is explained. After that, we provided the necessary and sufficient conditions for Ha × Hb = H(a,b). The connection between the two main points is as follows: If Q(a) × Q(b) is the principal quasi-ideal, we can ensure that Ha ×Hb is an H-class in S1 × S2. Finally, we also studied the maximal H-class. We proved under some conditions that the maximality of Ha and Hb implies the maximality of H(a,b). Furthermore, the converse of this statement also holds. In addition, we investigated the properties of the direct product of two maximal H-classes Ha × Hb and studied the maximal H-class of H(a,b) through the maximal H-class of Ha and Hb, including the element, namely, the indecomposable element. P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (2) (2025), 5976 14 of 14 Acknowledgements This work (Grant No. RGNS 65-054) was supported by Office of the Permanent Secretary, Ministry of Higher Education, Science, Research and Innovation (OPS MHESI), Thailand Science Research and Innovation (TSRI) and Khon Kaen University. References [1] O Steinfeld. Über die quasiideale von halbgruppend. Publ. Math. Debrecen, 4:262 – 275, 1956. [2] O Steinfeld. Quasi-ideal in rings and semigroups. Semigroup Forum, 19:371 – 372, 1980. [3] I Fabrici. One-sided principal ideals in the direct product of two semigroups. Math- ematica Bohemica, 118(4):337 – 342, 1993. [4] I Fabrici. 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