EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 7061
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Left Ideals and L-classes in the Finite Direct Product of
Semigroups

Panuwat Luangchaisri1, Ontima Pankoon1, Thawhat Changphas1,∗

1 Department of Mathematics, Faculty of Science Khon Kaen University, Khon Kaen 40002,
Thailand

Abstract. Let Si be a semigroup for all i ∈ {1, 2, . . . , n}. Then the Cartesian product of
S1, S2, . . . , Sn becomes a semigroup under componentwise multiplication. Let (s1, s2, . . . , sn) ∈
S1×S2 · · ·×Sn. In this paper, we give necessary and sufficient condition when the Cartesian prod-
uct of principal left ideals L(s1)× L(s2)× · · · × L(sn) is the principal left ideal L((s1, s2, . . . , sn))
and the Cartesian product of L-classes Ls1×Ls2×· · ·×Lsn is an L-class L(s1,s2,...,sn) in a semigroup
S1 × S2 × · · · × Sn.

2020 Mathematics Subject Classifications: 20M12

Key Words and Phrases: semigroup, direct product, principal left ideal, L-class

1. Introduction

Let S and T be semigroups. The Cartesian product S×T becomes a semigroup under
a binary operation on S × T defined by

(s, t)(s′, t′) = (ss′, tt′)

for all (s, t), (s′, t′) ∈ S × T . This semigroup is referred to as the direct product of S and
T . A nonempty subset A of S is a left ideal of S if SA ⊆ A. For any a ∈ S, the principal
left ideal of S generated by a, denoted by L(a), is the smallest left ideal of S containing
a. It is well-known that

L(a) = a ∪ Sa.

A relation L on S is then defined by the rule that aLb if and only if L(a) = L(b), i.e., if
and only if a ∪ Sa = b ∪ Sb. It is one of Green’s equivalence relations. Then the L-class
of S containing element a will be written by La. In [1], Fabrici considered a principal left
ideal and a relation L on the direct product of two semigroups S × T . Let (s, t) ∈ S × T .

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.7061

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 (4) (2025), 7061 2 of 10

Necessary and sufficient condition when L(s)× L(t) = L((s, t)) were provided. Moreover,
the author showed necessary and sufficient condition when L(s,t) = Ls ×Lt in S × T . The
principal (two-sided) ideals on the direct product of two semigroups were considered in
the same way [2].

Let m, n be nonnegative integers. A subsemigroup A of S is called an (m,n)-ideal of
S if AmSAn ⊆ A [3]. Here, A0S = SA0 = S. This definition is a generalized form of left
ideals, right ideals, and bi-ideals. For any element a in S, the smallest (m,n)-ideal of S
containing a is denoted by [a](m,n). Luangchaisri and Changphas [4] provided necessary
and sufficient condition for [s](m,n) × [t](m,n) = [(s, t)](m,n). 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)}. Then they provided the conditions for J(m,n),a × J(m,n),b = J(m,n),(a,b).

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 [5]. For each a ∈ S, the principal quasi-ideal
of S generated by a is denoted by Q(a). Luangchaisri et al. [6] considered necessary and
sufficient condition when Q(s)×Q(t) = Q((s, t)). Moreover, they characterized when the
Cartesian product of H-classes Hs ×Ht is an H-class of S × T .

According to the above examples, we can observe the research line to study various
kinds of ideals and equivalence relations on the direct product of two semigroups. In this
paper, we consider these concepts and extend to the finite direct product of semigroups.
The principal left ideals and L-classes are investigated. Moreover, we give an example
to show that the Cartesian product of principal left ideals need not be the principal left
ideal. In addition, an example for L-classes is also provided.

2. Main Results

Let {Si | i ∈ I} be a family of semigroups indexed by the set I = {1, 2, . . . , n}. Then
S1 × S2 × · · · × Sn becomes a semigroup under a componentwise multiplication, which is
defined by

(s1, s2, . . . , sn)(s
′
1, s

′
2, . . . , s

′
n) = (s1s

′
1, s2s

′
2, . . . , sns

′
n)

for all (s1, s2, . . . , sn), (s
′
1, s

′
2, . . . , s

′
n) ∈ S1 × S2 × · · · × Sn. This semigroup is called the

direct product of {Si | i ∈ I}. Note that the direct product of S1 is trivially the semigroup
S1. Therefore, we assume throughout that the indexed set I is not a singleton.

If Li is a left ideal of Si for all i ∈ I, then the Cartesian product L1 ×L2 × · · · ×Ln is
a left ideal of S1 × S2 × · · · × Sn. However, the Cartesian product of principal left ideals
need not be the principal left ideal. This is clarified by the following example:

Example 1. Let S = {s1, s2, s3, s4} be a semigroup under the following binary operation:

∗ s1 s2 s3 s4
s1 s1 s1 s1 s1
s2 s1 s2 s2 s4
s3 s1 s2 s2 s4
s4 s1 s4 s4 s2



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 3 of 10

This semigroup is applied from [7]. Since (s3, s1) ∈ L(s3)×L(s3) and (s3, s1) /∈ L((s3, s3)),
this shows that L(s3)×L(s3) ̸= L((s3, s3)). In addition, since (s3, s3) ∈ L(s3)×L(s3) and
(s3, s3) /∈ L((x, y)) for all (x, y) ∈ S×S\{(s3, s3)}, it follows that L(s3)×L(s3) ̸= L((x, y)
for all (x, y) ∈ S × S. Hence, L(s3) × L(s3) is not a principal left ideal of a semigroup
S × S.

We begin with Lemma 1 to mention about the inclusion of L((s1, s2, . . . , sn)) and
L(s1) × L(s2) × · · · × L(sn). Then, in Theorem 1, we give a necessary and sufficient
condition when L((s1, s2, . . . , sn)) = L(s1)× L(s2)× · · · × L(sn).

Lemma 1. Let Si be a semigroup and let si ∈ Si where i ∈ {1, 2, . . . n}. Then

L((s1, s2, . . . , sn)) ⊆ L(s1)× L(s2)× · · · × L(sn)

Proof. Let x ∈ L((s1, s2, . . . , sn)). Then

x ∈ (s1, s2, . . . , sn) ∪ (S1s1 × S2s2 × · · · × Snsn)

⊆ (s1 ∪ S1s1)× (s2 ∪ S2s2)× · · · × (sn ∪ Snsn)

= L(s1)× L(s2)× · · · × L(sn).

Thus, L((s1, s2, . . . , sn)) ⊆ L(s1)× L(s2)× · · · × L(sn).

Theorem 1. Let Si be a semigroup and let si ∈ Si, where i ∈ {1, 2, . . . , n}. Then

L((s1, s2, . . . , sn)) = L(s1)× L(s2)× · · · × L(sn)

if and only if at least one of the following conditions is satisfied:

(i) si ∈ Sisi for all i ∈ {1, 2, . . . , n};

(ii) there exists i ∈ {1, 2, . . . , n} such that Sjsj = {sj} for all j ∈ {1, 2, . . . , n} \ {i}.

Proof. Suppose (i) and (ii) do not hold. Then there exists i ∈ {1, 2, . . . , n} such
that si /∈ Sisi. This implies that Sisi ̸= {si}. Since (ii) does not hold, there exists
j ∈ {1, 2, . . . , n} such that j ̸= i and Sjsj ̸= {sj}. Without loss of generality, we let

s = (s1, s2, . . . , si, . . . , s
′
j , . . . , sn)

where s′j ∈ Sjsj \{sj}. Then s ∈ L(s1)×L(s2)×· · ·×L(sn). Since s /∈ S1s1×S2s2×· · ·×
Snsn and s ̸= (s1, s2, . . . , sn), we have s /∈ L((s1, s2, . . . , sn)). Thus, L((s1, s2, . . . , sn)) ̸=
L(s1)× L(s2)× · · · × L(sn).

Conversely, assume that (i) or (ii) holds. If (i) holds, then we have

L(s1)× L(s2)× · · · × L(sn) = S1s1 × S2s2 × · · · × Snsn

⊆ L((s1, s2, . . . , sn))

⊆ L(s1)× L(s2)× · · · × L(sn).



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 4 of 10

Thus, L(s1) × L(s2) × · · · × L(sn) = L((s1, s2, . . . , sn)). Meanwhile, if (ii) holds, let i ∈
{1, 2, . . . , n} be the index such that Sjsj = {sj} for all j ∈ {1, 2, . . . , n} \ {i}. Then

L(s1)× L(s2)× · · · × L(sn) = {s1} × {s2} × · · · × ({si} ∪ Sisi)× · · · × {sn}
= (s1, s2, . . . , sn) ∪ (S1s1 × S2s2 × · · · × Sisi × · · · × Snsn)

= L((s1, s2, . . . , sn)).

By these two cases, we conclude that L((s1, s2, . . . , sn)) = L(s1)× L(s2)× · · · × L(sn).

Remark 1. Let S be a semigroup defined as in Example 1. We have s3 /∈ {s1, s2, s4} =
Ss3. Thus, we immediately obtain from Theorem 1 that L(s3)× L(s3) ̸= L((s3, s3)).

In Theorem 1, we establish the sufficient and necessary condition when L((s1, s2, . . . , sn)) =
L(s1)×L(s2)× · · ·×L(sn). However, the negation of such condition does not ensure that
L(s1)×L(s2)×· · ·×L(sn) is not a principal left ideal. This assumption can be confirmed
by the following theorem.

Theorem 2. Let Si be a semigroup and let si ∈ Si, where i = 1, 2, . . . , n. If L((s1, s2, . . . , sn)) ̸=
L(s1)× L(s2)× · · · × L(sn), then L(s1)× L(s2)× · · · × L(sn) is not a principal left ideal.

Proof. Assume that L((s1, s2, . . . , sn)) ̸= L(s1) × L(s2) × · · · × L(sn). Suppose that
L(s1)× L(s2)× · · · × L(sn) is a principal left ideal of S1 × S2 × · · · × Sn. Then

L(s1)× L(s2)× · · · × L(sn) = L((t1, t2, . . . , tn))

for some (t1, t2, . . . , tn) ∈ S1 × S2 × · · · × Sn. We observe that

(s1, s2, . . . , sn) ∈ L(s1)× L(s2)× · · · × L(sn)

= L((t1, t2, . . . , tn))

⊆ L(t1)× L(t2)× · · · × L(tn).

On the same way, we also obtain (t1, t2, . . . , tn) ∈ L(s1)×L(s2)×· · ·×L(sn). These imply
that

S1s1 × S2s2 × · · · × Snsn = (S1 × S2 × · · · × Sn)(s1, s2, . . . , sn)

⊆ (S1 × S2 × · · · × Sn)(L(t1)× L(t2)× · · · × L(tn))

= S1t1 × S2t2 × · · · × Sntn

= (S1 × S2 × · · · × Sn)(t1, t2, . . . , tn)

⊆ (S1 × S2 × · · · × Sn)(L(s1)× L(s2)× · · · × L(sn))

= S1s1 × S2s2 × · · · × Snsn.

Thus, S1s1 × S2s2 × · · · × Snsn = S1t1 × S2t2 × · · · × Sntn. Since (s1, s2, . . . , sn) ̸=
(t1, t2, . . . , tn) and (s1, s2, . . . , sn) ∈ L((t1, t2, . . . , tn)), we have that

(s1, s2, . . . , sn) ∈ S1t1 × S2t2 × · · · × Sntn = S1s1 × S2s2 × · · · × Snsn.



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 5 of 10

By Theorem 1(i), L((s1, s2, . . . , sn)) = L(s1) × L(s2) × · · · × L(sn). This contradicts to
assumption. Therefore, L(s1)× L(s2)× · · · × L(sn) is not a principal left ideal.

The following example shows that the Cartesian product of L-classes need not be an
L-class.

Example 2. From the definition of the semigroup S in Example 1, we have Ls2 × Ls3 ̸=
L(s2,s3). Indeed: we have that L(s2,s3) = {(s2, s3)} and Ls2 × Ls3 = {s2, s4} × {s3} =
{(s2, s3), (s4, s3)}. Since (s4, s3) ∈ Ls2 × Ls3 and (s4, s3) /∈ L(s2,s3), we get Ls2 × Ls3 ̸=
L(s2,s3). This shows that Ls2 × Ls3 is not an L-class.

Next, we present Theorem 3 to mention the conclusion of Ls1 × Ls2 × · · · × Lsn and
L(s1,s2,...,sn). Then we give a necessary and sufficient condition when Ls1×Ls2×· · ·×Lsn =
L(s1,s2,...,sn) in Theorem 4. Furthermore, we provide the relation between the Cartesian
product of principal left ideals and the Cartesian product of L-classes in Theorem 5.

Theorem 3. Let Si be a semigroup and let si ∈ Si where i = 1, 2, . . . , n. Then the
following statements hold:

(i) L(s1,s2,...,sn) ⊆ Ls1 × Ls2 × · · · × Lsn;

(ii) if L(s1,s2,...,sn) ̸= Ls1 × Ls2 × · · · × Lsn, then Ls1 × Ls2 × · · · × Lsn contains at least
two L-classes in S1 × S2 × · · · × Sn.

Proof. (i) Let (t1, t2, . . . , tn) ∈ L(s1,s2,...,sn). Then L((t1, t2, . . . , tn)) = L((s1, s2, . . . , sn)).
This implies that

(s1, s2, . . . , sn) ∈ L((t1, t2, . . . , tn)) ⊆ L(t1)× L(t2)× · · · × L(tn)

and
(t1, t2, . . . , tn) ∈ L((s1, s2, . . . , sn)) ⊆ L(s1)× L(s2)× · · · × L(sn).

Thus, si ∈ L(ti) and ti ∈ L(si) for all i = 1, 2, . . . , n. It follows that

L(si) = si ∪ Sisi ⊆ L(ti) ∪ SiL(ti) = L(ti).

Similarly, we obtain L(ti) ⊆ L(si). Thus, L(si) = L(ti) for all i = 1, 2, . . . , n. Therefore,

(t1, t2, . . . , tn) ∈ Ls1 × Ls2 × · · · × Lsn .

(ii) Assume that L(s1,s2,...,sn) ̸= Ls1×Ls2×· · ·×Lsn . By (i), there exists (t1, t2, . . . , tn) ∈
Ls1 × Ls2 × · · · × Lsn such that (t1, t2, . . . , tn) /∈ L(s1,s2,...,sn). Then Lti = Lsi for all
i ∈ {1, 2, . . . , n}. Thus,

L(t1,t2,...,tn) ⊆ Lt1 × Lt2 × · · · × Ltn = Ls1 × Ls2 × · · · × Lsn .

Since L(s1,s2,...,sn) and L(t1,t2,...,tn) are difference, we obtain that Ls1 × Ls2 × · · · × Lsn

contains at least two L-classes of S1 × S2 × . . .× Sn.



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 6 of 10

Theorem 4. Let Si be a semigroup and let si ∈ Si where i = 1, 2, . . . , n. Then

L(s1,s2,...,sn) = Ls1 × Ls2 × · · · × Lsn

if and only if at least one of the following conditions is satisfied:

(i) Lsi = {si} for all i ∈ {1, 2, . . . , n};

(ii) si ∈ Sisi for all i ∈ {1, 2, . . . , n}.

Proof. Assume that L(s1,s2,...,sn) = Ls1×Ls2×· · ·×Lsn . If L(s1,s2,...,sn) = {(s1, s2, . . . , sn)},
then by assumption, we have

Ls1 × Ls2 × · · · × Lsn = {(s1, s2, . . . , sn)}.

Thus, Lsi = {si} for all i ∈ {1, 2, . . . , n}. Suppose that there exists (t1, t2, . . . , tn) ∈
L(s1,s2,...,sn) such that (t1, t2, . . . , tn) ̸= (s1, s2, . . . , sn). Then L((t1, t2, . . . , tn)) = L((s1, s2, . . . , sn)).
Since (t1, t2, . . . , tn) ∈ L((s1, s2, . . . , sn)) and (t1, t2, . . . , tn) ̸= (s1, s2, . . . , sn), we have that

(t1, t2, . . . , tn) ∈ S1s1 × S2s2 × · · · × Snsn.

Similarly, we get (s1, s2, . . . , sn) ∈ S1t1 × S2t2 × · · · × Sntn. Thus,

(s1, s2, . . . , sn) ∈ S1t1 × S2t2 × · · · × Sntn

= (S1 × S2 × · · · × Sn)(t1, t2, . . . , tn)

⊆ (S1 × S2 × · · · × Sn)(S1s1 × S2s2 × · · · × Snsn)

= S1S1s1 × S2S2s2 × · · · × SnSnsn

⊆ S1s1 × S2s2 × · · · × Snsn.

Therefore, si ∈ Sisi for all i ∈ {1, 2, . . . , n}.
Conversely, assume that (i) or (ii) holds. If (i) holds, then we get

Ls1 × Ls2 × · · · × Lsn = {(s1, s2, . . . , sn)}
⊆ L(s1,s2,...,sn)

⊆ Ls1 × Ls2 × · · · × Lsn .

Thus, L(s1,s2,...,sn) = Ls1 × Ls2 × · · · × Lsn . Assume that (ii) holds. Then we have

L((s1, s2, . . . , sn)) = S1s1 × S2s2 × · · · × Snsn.

Let (t1, t2, . . . , tn) ∈ Ls1 × Ls2 × · · · × Lsn . For each i ∈ {1, 2, . . . , n}, we have

Siti ⊆ L(ti) = L(si) = Sisi ⊆ Si(L(ti)) = Siti.

Thus, Sisi = Siti. That is ti ∈ L(ti) = L(si) = Sisi = Siti for all i ∈ {1, 2, . . . , n}. This
implies

L((t1, t2, . . . , tn)) = (t1, t2, . . . , tn) ∪ (S1t1 × S2t2 × · · · × Sntn)



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 7 of 10

= S1t1 × S2t2 × · · · × Sntn

= S1s1 × S2t2 × · · · × Snsn

= L((s1, s2, . . . , sn)).

Thus, (t1, t2, . . . , tn) ∈ L(s1,s2,...,sn). Hence Ls1×Ls2×· · ·×Lsn ⊆ L(s1,s2,...,sn). The opposite
inclusion is obtained by Theorem 3(i). Therefore, L(s1,s2,...,sn) = Ls1 × Ls2 × · · · × Lsn .

Theorem 5. Let Si be a semigroup and let si ∈ Si where i ∈ {1, 2, . . . , n}. If L((s1, s2, . . . , sn)) =
L(s1)× L(s2)× · · · × L(sn), then L(s1,s2,...,sn) = Ls1 × Ls2 × · · · × Lsn.

Proof. Assume that L((s1, s2, . . . , sn)) = L(s1)× L(s2)× · · · × L(sn). By Theorem 1,
there are two possible cases, as follows.
Case 1: si ∈ Sisi for all i ∈ {1, 2, . . . , n}. We obtain from Theorem 4(ii) that L(s1,s2,...,sn) =
Ls1 × Ls2 × · · · × Lsn .
Case 2: There exists i ∈ {1, 2, . . . , n} such that for each j ∈ {1, 2, . . . , n}\{i}, Sjsj = {sj},
which yields L(sj) = {sj} and Lsj = {sj}. By focusing on the index i, if si ∈ Sisi, we
obtain from Theorem 4(ii) that L(s1,s2,...,sn) = Ls1 × Ls2 × · · · × Lsn . On the other hand,
we suppose that si /∈ Sisi. We will prove that Lsi = {si} by a contradiction. Suppose that
there exists ti ∈ Si such that ti ̸= si and L(ti) = L(si), which implies Siti = Sisi. Since
si ∈ L(si) = L(ti) and si ̸= ti, we get si ∈ Siti = Sisi, which is a contradiction. Therefore,
Lsi = {si}. We thus conclude by Theorem 4(i) that L(s1,s2,...,sn) = Ls1 × Ls2 × · · · × Lsn

as required.

Let S be a semigroup. Since the relation L on S is defined in terms of left ideals of S,
the order among these left ideals induces a partial order among their equivalence classes,
that is,

La ≤ Lb if L(a) ⊆ L(b)

for all a, b ∈ S. Then an L-class Ls of S is maximal if there is no u ∈ S such that
L(s) ⊊ L(u).

Lemma 2. Let S be a semigroup and let s, u ∈ S. If L(s) ⊊ L(u), then the following
statements hold:

(i) u /∈ L(s);

(ii) L(s) ⊆ Su.

Proof. Assume that L(s) ⊊ L(u). To prove (i), suppose u ∈ L(s). Then

L(u) ⊆ L(s) ⊊ L(u).

This contradiction implies u /∈ L(s). To prove (ii), let t ∈ L(s). It follows that t ∈ L(u) =
u ∪ Su. If t = u, then L(s) ⊊ L(u) = L(t). By (i), we get that t /∈ L(s), which is a
contradiction. Thus, t ∈ Su.



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 8 of 10

Theorem 6. Let Si be a semigroup and let si ∈ Si where i ∈ {1, 2, . . . , n}.
If (s1, s2, . . . , sn) ∈ S1s1×S2s2× · · · ×Snsn, then L(s1,s2,...,sn) is a maximal L-class if and
only if Lsi is a maximal L-class for all i ∈ {1, 2, . . . , n}.

Proof. Assume that (s1, s2, . . . , sn) ∈ S1s1 × S2s2 × · · · × Snsn. By Theorem 1(i), we
have

L((s1, s2, . . . , sn)) = L(s1)× L(s2)× · · · × L(sn).

Suppose that Lsi is not a maximal L-class for some i ∈ {1, 2, . . . , n}. Then there exists
ui ∈ Si such that

L(si) ⊊ L(ui).

By Lemma 2(i), ui /∈ Sisi and Sisi ⊆ Siui. Thus,

L((s1, s2, . . . , si, . . . , sn)) = L(s1)× L(s2)× · · · × L(si)× · · · × L(sn)

= S1s1 × S2s2 × · · · × Sisi × · · · × Snsn

⊊ (s1, s2, . . . , ui, . . . , sn) ∪ (S1s1 × S2s2 × · · · × Siui × · · · × Snsn)

= L((s1, s2, . . . , ui, . . . , sn)).

Therefore, L(s1,s2,...,sn) is not a maximal L-class.
Conversely, assume that L(s1,s2,...,sn) is not a maximal L-class. Then there exists

(u1, u2, . . . , un) ∈ S1 × S2 × · · · × Sn such that

L((s1, s2, . . . , sn)) ⊊ L((u1, u2, . . . , un)).

By Lemma 2(i), we obtain

(u1, u2, . . . , un) /∈ L((s1, s2, . . . , sn))

= (s1, s2, . . . , sn) ∪ (S1s1 × S2s2 × · · · × Snsn)

= S1s1 × S2s2 × · · · × Snsn.

This implies that ui /∈ Sisi for some i ∈ {1, 2, . . . , n}. It follows by assumption that
ui ̸= si. Therefore, L(si) ⊆ L(ui) and ui /∈ si ∪ Sisi = L(si). These imply L(si) ⊊ L(ui).
Thus, Lsi is not a maximal L-class.

Corollary 1. Let Si be a semigroup and let si ∈ Si where i ∈ {1, 2, . . . , n}.
If (s1, s2, . . . , sn) ∈ S1s1×S2s2×· · ·×Snsn, then Ls1×Ls2×· · ·×Lsi is a maximal L-class
if and only if Lsi is a maximal L-class for all i ∈ {1, 2, . . . , n}.

Proof. The proof is obtained directly from Theorem 4 and Theorem 6

Definition 1. Let S be a semigroup. An element s ∈ S is decomposable if s ∈ S2. If an
element s ∈ S is not decomposable, we say that s is indecomposable.



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 9 of 10

In [1], the author remarked that if an element s of a semigroup S is indecompos-
able, then Ls = {s}. In the direct product of {Si | i ∈ {1, 2, . . . , n}} we have that if
(s1, s2, . . . , sn) ∈ S1×S2×· · ·×Sn is indecomposable, then L(s1,s2,...,sn) = {(s1, s2, . . . , sn)}
and si is indecomposable for some i ∈ {1, 2, . . . , n}. On the other hand, if there exists
i ∈ {1, 2, . . . , n} such that si is indecomposable, then we get that (s1, s2, . . . , sn) is in-
decomposable. Next, we consider relationships between indecomposable elements and
maximal L-classes.

Theorem 7. Let Si be a semigroup and let si ∈ Si where i ∈ {1, 2, . . . , n}.

(i) If (s1, s2, . . . , sn) is indecomposable, then L(s1,s2,...,sn) is a maximal L-class.

(ii) If L(s1,s2,...,sn) is a maximal L-class of S1 × S2 × · · · × Sn and (s1, s2, . . . , sn) /∈
S1s1 × S2s2 × · · · × Snsn, then (s1, s2, . . . , sn) is indecomposable.

Proof. (i) Assume that (s1, s2, . . . , sn) is indecomposable. Suppose that L(s1,s2,...,sn) is
not a maximal L-class. Then there exists (u1, u2, . . . , un) ∈ S1 × S2 × · · · × Sn such that

L((s1, s2, . . . , sn)) ⊊ L((u1, u2, . . . , un)).

By Lemma 2(ii), we obtain that

(s1, s2, . . . , sn) ∈ L((s1, s2, . . . , sn)) ⊆ S1u1 × S2u2 × · · · × Snun ⊆ S2
1 × S2

2 × · · · × S2
n.

This contradicts to an assumption. Therefore, L(s1,s2,...,sn) is a maximal L-class.
(ii) Assume that L(s1,s2,...,sn) is a maximal L-class and (s1, s2, . . . , sn) /∈ S1s1 × S2s2 ×

· · ·×Snsn. Suppose that (s1, s2, . . . , sn) is decomposable. Then there exists (t1, t2, . . . , tn)
such that (s1, s2, . . . , sn) ∈ (S1 × S2 × · · · × Sn)(t1, t2, . . . , tn). This implies that

L((s1, s2, . . . , sn)) ⊆ L((t1, t2, . . . , tn)).

We observe that (t1, t2, . . . , tn) ̸= (s1, s2, . . . , sn). Suppose that (t1, t2, . . . , tn) ∈ (S1×S2×
· · · × Sn)(s1, s2, . . . , sn), it follows that

(s1, s2, . . . , sn) ∈ S1t1 × S2t2 × · · · × Sntn = S1s1 × S2s2 × · · · × Snsn.

This contradicts to our assumption. Thus, (t1, t2, . . . , tn) /∈ L((s1, s2, . . . , sn)). Therefore,
L((s1, s2, . . . , sn)) ⊊ L((t1, t2, . . . , tn)). This contradicts to maximality of L(s1,s2,...,sn).
Hence, (s1, s2, . . . , sn) is indecomposable.

According to the observation of indecomposable elements on a direct product of semi-
groups together with Theorem 7, we have the following corollary.

Corollary 2. Let Si be a semigroup and let si ∈ Si where i ∈ {1, 2, . . . , n}.
If (s1, s2, . . . , sn) ∈ S1s1 × S2s2 × · · · × Snsn, then L(s1,s2,...,sn) is a maximal L-class in
S1 × S2 × · · · × Sn if and only if si ∈ Si is indecomposable for some i ∈ {1, 2, . . . , n}.



P. Luangchaisri, O. Pankoon, T. Changphas / Eur. J. Pure Appl. Math, 18 (4) (2025), 7061 10 of 10

3. Conclusions

In this paper, we studied the Cartesian product of principal left ideals and the Carte-
sian product of L-classes in the direct product of n semigroups (n ≥ 2). We gave an
explicit counterexample showing that the Cartesian product of principal left ideals is not
necessarily a principal left ideal. Similarly, an explicit counterexample for the Cartesian
product of L-classes was provided. We established the two main results, consisting of a
necessary and sufficient condition for the Cartesian product of principal left ideals to be a
principal left ideal, and a necessary and sufficient condition for the Cartesian product of
L-classes to be an L-class in the direct product of n semigroups. In addition, the condition
when the Cartesian product of maximal L-classes is maximal was also investigated.

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.

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