# On characterizations of (0-)minimal and maximal ordered left (right) ideals in ordered ternary semigroups.

1. Introduction and Preliminaries

In 1995, Dixit and Dewan [3] introduced and studied properties of quasi-ideals and bi-ideals in ternary semigroups. In 2000, Cao and Xu [2] characterized minimal and maximal left ideals in ordered semigroups and gave some characterizations of minimal and maximal left ideals in ordered semigroups. In 2002, Arslanov and Kehayopulu [1] characterized minimal and maximal ideals in ordered semigroups. They proved two theorems as follow: In an ordered semigroup S, for which there exists an element a [member of] S such that the ideal of S generated by a is S, there is at most one maximal ideal which is the union of all proper ideals of S. In an ordered semigroup S containing unit, there is at most one maximal ideal which is the union of all proper ideals of S.

The concept of minimal and maximal left ideals is the really interested and important thing about ordered semigroups. Now we also characterize (0-)minimal and maximal ordered left ideals in ordered ternary semigroups and give some characterizations of (0-)minimal and maximal ordered left ideals in ordered ternary semigroups. Similar results hold if we replace the word "left" by "right".

Our purpose in this paper is fivefold.

(1) To give the definition of an ordered ternary semigroup.

(2) To introduce the concept of left simple and left 0-simple ordered ternary semigroups.

(3) To characterize properties of ordered left ideals in ordered ternary semigroups.

(4) To characterize the relationship between (0-)minimal ordered left ideals and left (0-)simple ordered ternary semigroups.

(5) To characterize the relationship between maximal ordered left ideals and left simple and left 0-simple ordered ternary semigroups.

To present the main theorems we first recall the definition of a ternary semigroup which is important here.

A nonempty set T is called a ternary semigroup [4] if there exists a ternary operation T x T x T [right arrow] T, written as ([x.sub.1], [x.sub.2], [x.sub.3]) [right arrow] [[X.sub.1][X.sub.2][X.sub.3]], satisfying the following identity for any [x.sub.1], [x.sub.2], [x.sub.3], [x.sub.4], [x.sub.5] [member of] T,

[[[x.sub.1][x.sub.2][x.sub.3]][x.sub.4][x.sub.5]] = [[x.sub.1][[x.sub.2][x.sub.3][x.sub.4]][x.sub.5]] = [[x.sub.1][x.sub.2][[x.sub.3][x.sub.4][x.sub.5]]].

For nonempty subsets A,B and C of a ternary semigroup T, let

[ABC] := {[abc] : a [member of] A, b [member of] B and c [member of] C}.

If A = {a}, then we also write [{a}BC] as [aBC], and similarly if B = {b} or C = {c} or A = {a} and B = {b} or A = {a} and C = {c} or B = {b} and C = {c}. A nonempty subset S of a ternary semigroup T is called a ternary subsemigroup [3] of T if [SSS] [subset or equal to] S.

Example 1.1. [3] Let T = {-i, 0, i}. Then T is a ternary semigroup under the multiplication over complex number while T is not a semigroup under complex number multiplication.

Example 1.2.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

Then T = {O, I,[A.sub.1],[A.sub.2],[A.sub.3],[A.sub.4]} is a ternary semigroup under matrix multiplication.

A partially ordered ternary semigroup T is called an ordered ternary semigroup if for any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.].

If (T, *, [less than or equal to]) is an ordered ternary semigroup and S is a ternary subsemigroup of T, then (S, *, [less than or equal to]) is an ordered ternary semigroup. For a subset H of an ordered ternary semigroup T, we denote (H] := {t [member of] T : t [less than or equal to] h for some h [member of] H} and H [union] a := H [union] {a} for all a:= If H [union] {a}, we also write ({a}] as (a]. We see that H [subset or equal to] (H], ((H ]] = (H] and for any subsets A and B of an ordered ternary semigroup T, we have (A] [subset or equal to] (B], if A [subset or equal to] B and (A [union] B] = (A] [union] (B]. A nonempty subset L (R) of an ordered ternary semigroup T is called a left ideal (right ideal) of T if [TTL] [subset or equal to] L ([RTT] [subset or equal to] R). A left ideal (right ideal) L (R) of an ordered ternary semigroup T is called an ordered left ideal (ordered right ideal) of T if for any b [member of] T and a [member of] L (a [member of] R), b [less than or equal to] a implies b [member of] L (b [member of] R). The intersection of all ordered left ideals of a ternary subsemigroup S of an ordered ternary semigroup T containing a nonempty subset A of S is the ordered left ideal of S generated by A. For A = {a}, let [L.sub.S] (a) denote the ordered left ideal of S generated by {a}. If S = T, then we also write [L.sub.T] (a) as L(a). An element a of an ordered ternary semigroup T with at least two elements is called a zero element of T if [a[t.sup.1][t.sup.2] = [[t.sup.1]a[t.sup.2]] = [[t.sup.1][t.sup.2a] = a for all [t.sup.1], [t.sup.2]. T and a = t for all t. T and we denote it by 0. If T is an ordered ternary semigroup with zero, then every ordered left ideal of T contains a zero element. An ordered ternary semigroup T without zero is called left simple if it has no proper ordered left ideals. An ordered ternary semigroup T with zero is called left 0-simple if it has no nonzero proper ordered left ideals and [T T T ] [not equal to] = {0}. An ordered left ideal L of an ordered ternary semigroup T without zero is called a minimal ordered left ideal of T if there is no an ordered left ideal A of T such that A [subset] L. Equivalently, if for any ordered left ideal A of T such that A [subset or equal to] L, we have A [subset] L. A nonzero ordered left ideal L of an ordered ternary semigroup T with zero is called a 0-minimal ordered left ideal of T if there is no a nonzero ordered left ideal A of T such that A [subset] L. Equivalently, if for any nonzero ordered left ideal A of T such that A [subset or equal to] L, we have A = L. Equivalently, if for any ordered left ideal A of T such that A [subset] L, we have A = {0}. A proper ordered left ideal L of an ordered ternary semigroup T is called a maximal ordered left ideal of T if for any ordered left ideal A of T such that L [subset] A, we have A = T. Equivalently, if for any proper ordered left ideal A of T such that L [subset or equal to] A, we have A = L.

We shall assume throughout this paper that T stands for an ordered ternary semigroup.

The following two lemmas are also necessary for our considerations and easy to verify.

Lemma 1.3. For any nonempty subsetA of T, ([TTA] [union] A] is the smallest ordered left ideal of T containing A. Furthermore, for any a [member of] T,

L(a) = ([T T a] [union]a].

Lemma 1.4. For any nonempty subset A of T, ([TTA]] is an ordered left ideal of T.

Lemma 1.5. If T has no a zero element, then the following statements are equivalent.

(a) T is left simple.

(b) ([T T a]] = T for all a [member of] T .

(c) L(a) = T for all a [member of] T .

Proof. By Lemma 1.4 and T is left simple, we have ([T T a]] = T for all a. T. Therefore (a) implies (b). By Lemma 1.3, L(a) = ([T T a] . a] = ([T T a]]. (a] = T. (a] = T. Thus (b) implies (c). Now let L be an ordered left ideal of T and let a [member of] L. Then T = L(a) [subset or equal to] L [subset or equal to] T, so L = T. Hence T is left simple, we have that (c) implies (a).

Hence the proof is completed.

Lemma 1.6. If T has a zero element, then the following statements hold.

(a) If T is left 0-simple, then L(a) = T for all a [member of] T {0}.

(b) If L(a) = T for all a [member of] T \ {0}, then either [T T T] = {0} or T is left 0-simple.

Proof. (a) Assume that T is left 0-simple. Then L(a) is a nonzero ordered left ideal of T for all a [member of] T \ {0}. Hence L(a) = T for all a [member of] T \ {0}.

(b) Assume that L(a) = T for all a [member of] T \ {0} and let [T T T ] = {0}. Now let L be a nonzero ordered left ideal of T and put a [member of] L \ {0}. Then T = L(a) [subset or equal to] L [subset or equal to] T, so L = T. Therefore T is left 0-simple.

The next lemma is easy to verify.

Lemma 1.7. Let {L[gamma] : [gamma] [member of] [GAMMA]} be a family of ordered left ideals of T. Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.] is an ordered left ideal of T and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

is also an ordered left ideal of T if [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

Lemma 1.8. If L is an ordered left ideal of T and S is a ternary subsemigroup of T, then the following statements hold.

(a) If S is left simple such that S [intersection] L [not equal to] then S [subset or equal to] L.

(b) If S is left 0-simple such that S \ {0} [intersection] L [not equal to] then S [intersection] L.

Proof. (a) Assume that S is left simple such that S n L. Then let a S [intersection] L. By Lemma 1.4, we have ([SSa]] n S is an ordered left ideal of S. This implies that ([SSa]] [intersection] S = [subset or equal to] Hence S [subset or equal to] ([SSa]] [subset or equal to] ([TTL]] [subset or equal to] (L] [subset or equal to] L, so S [subset or equal to] L.

(b) Assume that S is left 0-simple such that S [intersection L [not equal to] 0. Then let a [member of] S [intersection] L. By Lemmas 1.3 and 1.6 (a), we have S = [L.sub.S](a) = ([SSa] [union] a] [intersection] S [subset or equal to] ([SSa] [union] a] [union] ([T T a] [union] a] = L(a) [subset or equal to] L. Therefore S [??] L.

Hence the proof of the lemma is completed.

2. (0-)Minimal Ordered Left Ideals

In this section, we characterize the relationship between minimal and 0-minimal ordered left ideals and left simple and left 0-simple ordered ternary semigroups.

Theorem 2.1. If T has no a zero element and L is an ordered left ideal of T, then the following statements hold.

(a) L is a minimal ordered left ideal without zero of T if and only if L is left simple.

(b) If L is a minimal ordered left ideal with zero of T, then L is left 0-simple.

Proof. (a) Assume that L is a minimal ordered left ideal without zero of T. Now let A be an ordered left ideal of L. Then [LLA] [subset or equal to] A. Define H := {h [member of] A : h [less than or equal to] [[l.sub.1][l.sub.2a] for some l1, [l.sup.2]. L and a. A}. Then [not equal to] = H [subset or equal to] A [subset or equal to] L. To show that H is an ordered left ideal of T, let [t.sup.1], [t.sup.2] [member of] T and h [member of] H. Then h = [[l.sub.1][l.sup.2a] for some [l.sub.1], [l.sup.2] [member of] L and a [member of] A, so [[t.sup.1][t.sup.2]h] = [[t.sup.1][t.sup.2][[l.sup.1][l.sup.2]a]] = [[[t.sup.1][t.sup.2][l.sup.1]][l.sup.2]a]. Since L is an ordered left ideal of T, we have [[t.sup.1][t.sup.2]h], [[t.sup.1][t.sup.2][l.sup.1]]. L. Thus [[[t.sup.1][t.sup.2][l.sup.1]][l.sup.2]a], [[t.sup.1][t.sup.2]h]. A because A is an ordered left ideal of L. Hence [[t.sup.1][t.sup.2]h]. H, so [TTH]. H. Next let t [member of] T and h [member of] H be such that t [less than or equal to] h. Then h = [[l.sup.1][l.sup.2]a] for some [l.sup.1], [l.sup.2]. L and a. A, so t = [[l.sup.1][l.sup.2a] [member of] [LLA] [subset or equal to] A [subset or equal to] L. Since L is an ordered left ideal of T, we get t [member of] L. Thus t. A because A is an ordered left ideal of L. Hence t [member of] H, so H is an ordered left ideal of T. Since L is a minimal ordered left ideal of T, we have H = L. Hence A = L, so we conclude that L is left simple.

Conversely, assume that L is left simple. Now let A be an ordered left ideal of T such that A [subset or equal to] L. Then A [union] L [not equal to] [empty set], it follows from Lemma 1.8 (a) that L [subset or equal to] A. Hence A = L, so L is a minimal ordered left ideal of T .

(b) It is similar to the proof of necessary condition of statement (a). Therefore we complete the proof of the theorem.

Using the similar proof of Theorem 2.1 (a) and the Lemma 1.8 (b), we have Theorem 2.2.

Theorem 2.2. If T has a zero element and L is a nonzero ordered left ideal of T, then the following statements hold.

(a) If L is a 0-minimal ordered left ideal of T, then either [LLA] = {0} for some nonzero ordered left ideal A of L or L is left 0-simple.

(b) If L is left 0-simple, then L is a 0-minimal ordered left ideal of T.

Theorem 2.3. If T has no a zero element but it has proper ordered left ideals, then every proper ordered left ideal of T is minimal if and only if T contains exactly one proper ordered left ideal or T contains exactly two proper ordered left ideals [L.sub.1] and [L.sub.2], [L.sup.1] [union] [L.sup.2] = T and [L.sup.1] [intersection] [L.sup.2] = [empty set].

Proof. Assume that every proper ordered left ideal of T is minimal. Now let L be a proper ordered left ideal of T. Then L is a minimal ordered left ideal of T. We consider the following two cases:

Case 1: T = L(a) for all a. T \ L.

If K is also a proper ordered left ideal of T and K _= L, then K [not equal to] L [not equal to] [empty set] because L is a minimal ordered left ideal of T. Thus there exists a [member of] K \ L [subset or equal to] T \ L. Hence T = L(a) [subset or equal to] K [subset or equal to] T, so K = T. It is impossible, so K = L. In this case, L is the unique proper ordered left ideal of T .

Case 2: There exists a [member of] T \ L such that T [not equal to] L(a).

Then L(a) [not equal to] L and L(a) is a minimal ordered left ideal of T. By Lemma 1.7, L(a) [union] L is an ordered left ideal of T. By hypothesis and L [subset] L(a) [union] L, we get L(a) [union] L = T. Since L(a) [intersection] L [ L(a) [subset] and L(a) is a minimal ordered left ideal of T, L(a) [intersection] L = [empty set] Now let K be an arbitrary proper ordered left ideal of T. Then K is a minimal ordered left ideal of T. We observe that K = K n T = K n (L(a). L) = (K nL(a)).(K nL). If K nL = [not equal to], then K = L because K and L are minimal ordered left ideals of T. If K n L(a) _= [not equal to], then K = L(a) because K and L(a) are minimal ordered left ideals of T. In this case, T contains exactly two proper ordered left ideals L and L(a), L(a). L = T and L(a) n L = [not equal to].

The converse is obvious.

Using the same proof of Theorem 2.3, we have Theorem 2.4.

Theorem 2.4. If T has a zero element and nonzero proper ordered left ideals, then every nonzero proper ordered left ideal of T is 0-minimal if and only if T contains exactly one nonzero proper ordered left ideal or T contains exactly two nonzero proper ordered left ideals [L.sup.1] and [L.sup.2], [L.sup.1]. [L.sup.2] = T and [L.sup.1] n [L.sup.2] = {0}.

3. Maximal Ordered Left Ideals

In this section, we characterize the relationship between maximal ordered left ideals and the union U of all (nonzero) proper ordered left ideals in ordered ternary semigroups.

Theorem 3.1. If T has no a zero element but it has proper ordered left ideals, then every proper ordered left ideal of T is maximal if and only if T contains exactly one proper ordered left ideal or T contains exactly two proper ordered left ideals [L.sup.1] and [L.sup.2], [L.sup.1] [union] [L.sup.2] = T and [L.sup.1] [intersection] [L.sup.2] = [empty set].

Proof. Assume that every proper ordered left ideal of T is maximal. Now let L be a proper ordered left ideal of T. Then L is a maximal ordered left ideal of T. We consider the following two cases:

Case 1: T = L(a) for all a [member of] T \ L.

If K is also a proper ordered left ideal of T and K [not eqaul to] = L, then K is a maximal ordered left ideal of T. This implies that K L [not equal to] [empty set], so there exists a [member of] K \ L [subset or equal to] T \ L. Thus T = L(a) [subset or equal to] K [subset or equal to] T, so K = T. It is impossible, so K = L. In this case, L is the unique proper ordered left ideal of T .

Case 2: There exists a. T \ L such that T _= L(a).

Then L(a) [not equal to] = L and L(a) is a maximal ordered left ideal of T. By Lemma 1.7, L(a) [union] L is an ordered left ideal of T. Since L [subset] L(a) [union] L and L is a maximal ordered left ideal of T, L(a)[union]L = T. By hypothesis and L(a)[intersection] L [union] L(a), we get L(a) [intersection] L = [empty set]. Now let K be an arbitrary proper ordered left ideal of T. Then K is a maximal ordered left ideal of T. We observe that K = K [intersection] T = K [intersection] (L(a) [union] L) = (K [intersection] L (a))[union](K [intersection] L). If K [intersection] L [not equal to] [empty set], then K = L because K n L and L are maximal ordered left ideals of T . If K [intersection] L [not equal to] [empty set], then K = L(a) because K [intersection] L(a) and L(a) are maximal ordered left ideals of T. In this case, T contains exactly two proper ordered left ideals L and L(a), L(a) [union] L = T and L(a) [intersection] L = [empty set].

The converse is obvious.

Using the same proof of Theorem 3.1, we have Theorem 3.2.

Theorem 3.2. If T has a zero element and nonzero proper ordered left ideals, then every nonzero proper ordered left ideal of T is maximal if and only if T contains exactly one nonzero proper ordered left ideal or T contains exactly two nonzero proper ordered left ideals [L.sup.1] and [L.sup.2], [L.sup.1]. [L.sup.2] = T and [L.sup.1] [intersection] [L.sup.2] = {0}.

Theorem 3.3. A proper ordered left ideal L of T is maximal if and only if

(a) T \ L = {a} and ([aT a]] [subset or equal to] L for some a [member of] T or

(b) T \ L [subset or equal to] ([T T a]] for all a [member of] T L.

Proof. Assume that L is a maximal ordered left ideal of T. Then we consider the following two cases:

Case 1: There exists a [member of] T \ L such that ([T T a]] [subset or equal to] L.

Then ([aT a]] [subset or equal to] ([T T a]] [subset or equal to] L. By Lemma ??, we have L [union] (a] = (L [union] ([T T a]]) [union] (a] = L [union] (([T T a]] [union] (a]) = L [union] ([T T a] [union] a] = L [union] L(a). Thus L [union] (a] is an ordered left ideal of T because L [union] L(a) is an ordered left ideal of T. Since L is a maximal ordered left ideal of T and L [union] L [union] (a], we have L [union] (a] = T. Hence T \ L [subset or equal to] (a]. To show that T L = {a}, let x [member of] T \ L. Then x [less than or equal to] a, so ([T T x]] [subset or equal to] ([T T a]] [subset or equal to] L. From ([T T x]] [subset or equal to] L and x [member of] T \ L, a similar argument shows that T \ L [subset or equal to] (x]. Consequently a [less than or equal to] x, so x = a. Hence T \ L = {a}. In this case, the condition (a) is satisfied.

Case 2: ([T T a]] [??] L for all a [member of] T \ L.

If a [member of] T \ L, then ([T T a]] [??] L and ([T T a]] is an ordered left ideal of T by Lemma 1.4. By Lemma 1.7, we have L [union] ([T T a]] is an ordered left ideal of T and L [subset] L [union] ([T T a]]. Since L is a maximal ordered left ideal of T, L [union] ([T T a]] = T. Hence we conclude that T \ L. ([T T a]] for all a [member of] T \ L. In this case, the condition (b) is satisfied.

Conversely, let J be an ordered left ideal of T such that L [subset] J. Then J \ L [not equal to] [empty set]. If T \ L = {a} and ([aT a]] [subset or equal to] L for some a [member of] T, then J \ L [subset or equal to] T \ L = {a}. Thus J \ L = {a}, so J = L [union] a = T. Hence L is a maximal ordered left ideal of T. If T \ L [union] ([T T a]] for all a [member of] T \ L, then T \ L [subset or equal to] ([T T x]] [subset or equal to] ([T T J]] [subset or equal to] (J] [subset or equal to] J for all x [member of] J \ L. Hence T = (T \ L) [subset or equal to] L [subset or equal to] J [subset or equal to] J = J [subset or equal to] T, so J = T. Therefore L is a maximal ordered left ideal of T .

Hence the theorem is now completed.

For an ordered ternary semigroup T, let U denote the union of all nonzero proper ordered left ideals of T if T has a zero element and let U denote the union of all proper ordered left ideals of T if T has no a zero element. Then it is easy to verify Lemma 3.4.

Lemma 3.4. U = T if and only if L(a) [not equal to] T for all a [member of] T .

As a consequence of Theorem 3.3 and Lemma 3.4, we obtain

Theorem 3.5. If T has no a zero element, then one and only one of the following four conditions is satisfied.

(a) T is left simple.

(b) L(a) [not equal to] T for all a [member of] T .

(c) There exists a [member of] T such that L(a) = T, a [??] ([T T a]], ([aT a]] [subset or equal to] U = T \ {a} and U is the unique maximal ordered left ideal of T.

(d) T \U = {x. T : ([T T x]] = T } and U is the unique maximal ordered left ideal of T.

Proof. Assume that T is not left simple. Then there exists a proper ordered left ideal L of T, so U is an ordered left ideal of T. We consider the following two cases:

Case 1: U = T .

By Lemma 3.4, we have L(a) [not equal to] T for all a [member of] T. In this case, the condition (b) is satisfied.

Case 2: U [not equal to] T .

Then U is a maximal ordered left ideal of T. Nowassume that L is a maximal ordered left ideal of T. Then L [subset or equal to] U [subset] T because L is a proper ordered left ideal of T. Since L is a maximal ordered left ideal of T, we have L = U. Hence U is the unique maximal ordered left ideal of T. By Theorem 3.3, we get

(a) T \ U = {a} and ([aT a]] [subset or equal to] U for some a [member of] T or

(b) T \ U [subset or equal to] ([T T a]] for all a [member of] T U.

Suppose that T \ U = {a} and ([aT a]] [subset or equal to] U for some a [member of] T. Then ([aT a]] [subset or equal to] U = T \ {a}. Since a [??] U, we have L(a) = T. If a [member of] ([T T a]], then (a] [subset or equal to] ([T T a]]. By Lemma 1.3, we have T = L(a) = ([T T a] [union] a] = ([T T a]] [union] (a] = ([T T a]]. Thus a = [[t.sup.1][t.sup.2a] and [t.sup.1] = [t.sub.3][t.sub.4a] for some [t.sup.1], [t.sup.2], [t.sub.3], [t.sub.4] [member of] T. Hence a [less than or equal to] [[t.sup.1][t.sup.2a]] [less than or equal to] [[t.sub.3][t.sub.4a][t.sup.2a] = [[t.sub.3][t.sub.4][at.sup.2a]]. Since [at.sup.2a] [member of] U and U is an ordered left ideal of T, we get a [member of] U. It is impossible, so a [??] ([T T a]]. In this case, the condition (c) is satisfied.

Now suppose that T \ U [subset or equal to] ([T T a]] for all a [member of] T \ U. To show that T \ U = {x. T : ([T T x]] = T }, let x [member of] T \ U. Then x [member of] ([T T x]], so (x] [subset or equal to] ([T T x]]. By Lemma 1.3, we have L(x) = ([T T x] [union] x] = ([T T x]] [union] (x] = ([TT x]]. Since x [??] U, we have L(x) = T. Hence T = L(x) = ([T T x]]. Conversely, let x [member of] T be such that T = ([T T x]]. If x. U, then L(x) [subset or equal to] U [subset] T. By Lemma 1.3, we have L(x) = ([T T x] [union] x] = ([T T x]]. (x] = T. (x] = T. It is impossible, so x [member of] T \ U. Hence T \ U = {x [member of] T : ([T T x]] = T }. In this case, the condition (d) is satisfied.

Therefore the proof of the theorem is completed.

Using the same proof of Theorem 3.5, we have Theorem 3.6.

Theorem 3.6. If T has a zero element and [T T T ] [not equal to] {0}, then one and only one of the following four conditions is satisfied.

(a) T is left 0-simple.

(b) L(a) [not equal to] T for all a [member of] T .

(c) There exists a [member of] T such that L(a) = T, a [??] ([T T a]], ([aT a]] [subset or equal to] U = T \ {a} and U is the unique maximal ordered left ideal of T.

(d) T \U = {x [member of] T : ([T T x]] = T } and U is the unique maximal ordered left ideal of T.

Acknowledgement

The author would like to thank the referee for the useful comments and suggestions given in an earlier version of this paper.

References

[1] M.Arslanov and N. Kehayopulu,A note on minimal and maximal ideals of ordered semigroups, Lobachevskii Journal of Mathematics, 11:3-6, 1995.

[2] Y. Cao and X. Xu, On minimal and maximal left ideals in ordered semigroups, Semigroup Forum, 60:202-207, 2000.

[3] V.N. Dixit and S. Dewan, A note on quasi and bi-ideals in ternary semigroups, International Journal of Mathematics and Mathematical Sciences, 18:501-508, 1995.

[4] D.H. Lehmer, A ternary analogue of abelian groups, American Journal of Mathematics, 54:329-338, 1932.

Aiyared Iampan

Department of Mathematics, School of Science and Technology, Naresuan University Phayao, Phayao 56000, Thailand E-mail: aiyaredi@nu.ac.th
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