Lobachevskii Journal of Mathematics Vol. 13, 2003, 51 – 56

Niovi Kehayopulu and Michael Tsingelis
A NOTE ON SEMI-PSEUDOORDERS IN SEMIGROUPS
(submitted by M. M. Arslanov)

ABSTRACT. An important problem for studying the structure of an ordered semigroup S is to know conditions under which for a given congruence ρ on S the set Sρ is an ordered semigroup. In [1] we introduced the concept of pseudoorder in ordered semigroups and we proved that each pseudoorder on an ordered semigroup S induces a congruence σ on S such that Sσ is an ordered semigroup. In [3] we introduced the concept of semi-pseudoorder (also called pseudocongruence) in semigroups and we proved that each semi-pseudoorder on a semigroup S induces a congruence σ on S such that Sσ is an ordered semigroup. In this note we prove that the converse of the last statement also holds. That is each congruence σ on a semigroup (S,.) such that Sσ is an ordered semigroup induces a semi-pseudoorder on S.

 ________________ Key words and phrases. Pseudoorder, pseudocongruence, semi-pseudoorder. 2000 Mathematical Subject Classification. 06F05, 20M10.

For a given ordered semigroup (S,.,) is essential to know if there exists a congruence ρ on S such that Sρ be an ordered semigroup. This plays an important role for studying the structure of ordered semigroups. If S is a semigroup (resp. an ordered semigroup), by a congruence on S we mean an equivalence relation σ on S such that (a,b) σ implies (ac,bc) σ and (ca,cb) σ for all c S. If S is a semigroup and σ a congruence on S, then the set Sσ := {(a)σ a S} ( (a)σ is the σ-class of S containing a (a S)) is a semigroup and the operation on Sσ is defined via the operation on S. The following question is natural: If (S,.,) is an ordered semigroup and σ a congruence on S, then is the set Sσ an ordered semigroup ? A probable order on Sσ could be the relation on Sσ defined by means of the order on S, that is

: = {(t,z) Sσ × Sσ (a,b)  such that t = (a)σ,z = (b)σ} = {((x)σ, (y)σ) a (x)σ,b = (y)σ such that (a,b) }.

But this relation is not an order, in general. An example can be found in [1]. The following question arises: Is there a congruence σ on S for which Sσ is an ordered semigroup ? This let to the concept of pseudoorder introduced by the same authors in [1]. Let (S,.,) be an ordered semigroup. A relation ρ on S is called pseudoorder if

1. ρ
2. (a,b) ρ and (b,c) ρ imply (a,c) ρ.
3. (a,b) ρ implies (ac,bc) ρ and (ca,cb) ρ for each c S.

According to Lemma 1 in [1], if (S,.,) is an ordered semigroup and σ a pseudoorder on S, then the relation σ̄ on S defined by

σ̄ := {(a,b) S × S (a,b) σ and (b,a) σ}

is a congruence on S and the set

Sσ̄ is an ordered semigroup. So according to [1],

each pseudoorder on an ordered semigroup S induces a congruence σ̄ on S such that Sσ̄ is an ordered semigroup. For a further study of pseudoorders in ordered semigroups we refer to [2]. On the other hand, the concept of pseudocongruences in semigroups has been introduced by the same authors in [3]. If (S,.) is a semigroup, by a pseudocongruence on S we mean a relation ρ on S such that

1. (a,a) ρa S
2. (a,b) ρ and (b,c) ρ imply (a,c) ρ.
3. (a,b) ρ implies (ac,bc) ρ and (ca,cb) ρ for each c S.

If (S,.,) is an ordered semigroup, then each pseudoorder on

S is a pseudocongruence on S. Indeed, if ρ is a

pseudoorder on S and a S, then (a,a) ρ. Pseudocongruences can be also called semi-pseudoorders, and from now on we will keep that terminology of semi-pseudoorders. We have seen in [3], that each semi-pseudoorder on a semigroup S induces a congruence ρ on S such that Sρ is an ordered semigroup. In this paper we prove that the converse of this statement also holds. For a semigroup (S,.) we define a multiplication on on Sρ defined by (a)ρ (b)ρ := (ab)ρ. If (S,.) is a semigroup and ρ a congruence on S and if there exists an order relation on Sρ such that the (Sρ,,) is an ordered semigroup, then there exists a semi-pseudoorder σ on S such that ρ = σ̄. So each congruence ρ on a semigroup (S,.) such that Sρ is an ordered semigroup induces a semi-pseudoorder on S.

If (S,.) is a semigroup and σ a semi-pseudoorder on S, we define

σ̄ := σ σ1.

The relation σ̄ is a congruence on S. Indeed: If a S, then (a,a) σ, then (a,a) σ1, so (a,a) σ σ1 := σ̄. If (a, b) σ̄, then (a, b) σ and (a, b) σ1, then (b, a) σ1 and (b, a) σ, so (b, a) σ1 σ := σ̄. If (a, b) σ̄ and (b, c) σ̄, then (a, b) σ, (a, b) σ1, (b, c) σ, (b, c) σ1, then (a, c) σ and (a, c) σ1, thus (a, c) σ σ1 := σ̄. Let (a, b) σ̄ and c S. We have (a,b) σ and (a,b) σ1. Since (a,b) σ, c S, we have (ac,bc) σ, (ca,cb) σ. Since (a, b) σ1, we have (b,a) σ, then (bc,ac) σ, (ca,cb) σ, hence (ac,bc) σ1, (ca,cb) σ1. Then we have (ac,bc) σ σ1 := σ̄ and (ca,cb) σ σ1 := σ̄.

[It might be also noted that σ̄ = {(a,b) S × S (a,b) σ and (b,a) σ}. Hence σ̄ is a congruence on S (cf. [3])]. Since σ̄ is a congruence on S, the set Sσ̄ with the operation on Sσ̄ defined by (a)σ̄ (b)σ̄ := (ab)σ̄ is a semigroup (It is known).

If (S,.) is a semigroup and σ a

semi-pseudoorder on S, we define a relation on Sσ̄ as follows: (a)σ̄(b)σ̄ if and only if (a,b) σ. The relation on Sσ̄ is well defined. Indeed: Let (a)σ̄ = (c)σ̄, (b)σ̄ = (d)σ̄ and (a)σ̄ (b)σ̄. Since (a)σ̄(b)σ̄, we have (a, b) σ. Since (a)σ̄ = (c)σ̄, we have (a, c) σ̄ := σ σ1 σ1, then (c, a) σ. Since (b)σ̄ = (d)σ̄, we have (b,d) σ̄ := σ σ1 σ, then (b,d) σ. Then (c,d) σ, and (c)σ̄(d)σ̄. (Cf. also [3]).

Theorem. Let (S, .) be a semigroup. If σ is a semi-pseudoorder on S, then the set (Sσ̄,,) is an ordered semigroup. Let ρ be a congruence on S and suppose there exists an order relation on Sρ such that (Sρ̄,,) be an ordered semigroup. Then there exists a semi-pseudoorder σ on S such that

ρ = σ̄ and  = .

Proof. For the first part of the Theorem we refer to the Theorem in [3]. Let now ρ be a congruence on S and an order on Sρ such that (Sρ,,) be an ordered semigroup. Let σ be the relation on S defined by

σ := {(a,b) S × S (a)ρ (b)ρ}.

1) σ is a semi-pseudoorder on S. In fact:
Let a S. Since (a)ρ (a)ρ, we have (a,a) σ. Let (a,b) σ, (b, c) σ. Then (a)ρ (b)ρ, (b)ρ (c)ρ, then (a)ρ (c)ρ, and (a, c) σ. Let (a, b) σ and c S. Then (a)ρ (b)ρ and (c)ρ Sρ. Since (Sρ,,) is an ordered semigroup, we have (a)ρ (c)ρ (b)ρ (c)ρ, then (ac)ρ (bc)ρ, and (ac,bc) σ. Similarly (a,b) σ and c S, imply (ca,cb) σ.

2) ρ = σ̄. Indeed: We have

(a,b) ρ (a)ρ = (b)ρ (a)ρ (b)ρ and (b)ρ (a)ρ (a,b) σ and (b,a) σ (a,b) σ̄.

3) = . Indeed:
Let (a)ρ (b)ρ. Since (a,b) σ, we have (a)σ̄(b)σ̄. By 2), ρ = σ̄. So (a)σ̄ = (a)ρ and (b)σ̄ = (b)ρ. Then (a)ρ(b)ρ.
Let (a)ρ(b)ρ. Since ρ = σ̄, we have (a)ρ = (a)σ̄ and (b)ρ = (b)σ̄. Then (a)σ̄(b)σ̄, hence (a,b) σ, and (a)ρ (b)ρ.

Remark 1. If (S,.,) is an ordered semigroup and ρ a pseudoorder on S, then the mapping

f(S,.,) (Sρ̄,,) a (a)ρ̄

is a homomorphism. In fact, if a,b S, then

f(ab) := (ab)ρ̄ := (a)ρ̄ (b)ρ̄ = f(a) f(b).

Let now a b. Since (a,b) ρ, we have (a,b) ρ. Then, since ρ is a semipseudoorder on S, we have (a)ρ̄f(b), and f(a)f(b).

For a semigroup S, we denote by SP(S) the set of semi-pseudoorders on S and by C(S) the set of congruences on S. Let be the equivalence relation on S defined as follows:

ρ σ if and only if ρ̄ = σ̄.

Remark 2. If S is a semigroup and ρ a semi-pseudoorder on S, then the mapping

f : SP(S)C(S) (ρ) ρ̄

is (1-1) and onto. In fact: The mapping f is well defined: If ρ is a semi-pseudoorder on S, then ρ̄ is a congruence on S. Let ρ,σ SP(S) and (ρ) = (σ). Then we have ρ σ, and ρ̄ = σ̄.
f is (1-1): Let ρ,σ SP(S) such that ρ̄ = σ̄. Then ρ σ, and (ρ) = (σ).
f is onto: Let ρ C(S). Then ρ = ρ1 and ρ is a semi-pseudoorder on S. Thus ρ SP(S) and

f((ρ)) := ρ̄ := ρ ρ1 = ρ ρ = ρ.

For a semigroup S, we denote by OC(S) the set of all congruences ρ on S for which there exists an order relation on Sρ such that (Sρ,,) is an ordered semigroup.

Remark 3. If S is a semigroup, then the mapping

f : SP(S)OC(S) (ρ) ρ̄

is (1-1) and onto. In fact: The mapping f is well defined: If ρ is a semi-pseudoorder on S, then ρ ̄ is a congruence on S. Then, by the Theorem, the set (Sρ̄,,) is an ordered semigroup. Which means that ρ̄ OC(S).
Let ρ,σ SP(S) and (ρ) = (σ). Then we have ρ σ, and ρ̄ = σ̄.
f is (1-1): Let ρ,σ SP(S) and ρ̄ = σ̄. Then ρ σ, and (ρ) = (σ).
f is onto: Let ρ OC(S). By the Theorem, there exists a semi-pseudoorder σ on S such that ρ = σ̄. Then σ SP(S), and f((σ)) := σ̄ = ρ.

Acknowledgment. This research was supported by the Special Research Account of the University of Athens (Grant No. 70/4/5630).

### References

[1]   N. Kehayopulu, M. Tsingelis, On subdirectly irreducible ordered semigroups, Semigroup Forum 50 (1995), 161-177.

[2]   N. Kehayopulu, M. Tsingelis, Pseudoorder in ordered semigroups, Semigroup Forum 50 (1995), 389-392.

[3]   N. Kehayopulu, M. Tsingelis A note on pseudocongruences in semigroups, Lobachevskii J. Math. 11 (2002), 19-21.

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