Lobachevskii Journal of Mathematics Vol. 13, 2003, 51 – 56 \begin{matrix}c \\ ◯ \end{matrix}Niovi Kehayopulu and Michael Tsingelis

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 S/ρ 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 S/σ 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 S/σ 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 S/σ is an ordered semigroup induces a semi-pseudoorder on S.

For a given ordered semigroup (S, ., ≤) is essential to know if there exists a congruence ρ on S such that S/ρ 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 S/σ := {(a)_{σ} | a ∈ S} ((a)_{σ} is the σ-class of S containing a (a ∈ S)) is a semigroup and the operation on S/σ 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 S/σ an ordered semigroup ? A probable order on S/σ could be the relation ” ⪯ ” on S/σ defined by means of the order ” ≤ ” on S, that is

⪯: = {(t, z) ∈ S/σ × S/σ | ∃ (a, b) ∈≤ such that t = (a)_{σ}, z = (b)_{σ}}
= {((x)_{σ}, (y)_{σ}) | ∃ a ∈ (x)_{σ}, b = (y)_{σ} such that (a, b) ∈≤}.

Key words and phrases. Pseudoorder, pseudocongruence, semi-pseudoorder.

2000 Mathematical Subject Classification. 06F05, 20M10.

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 S/σ 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

S/σ̄ is an ordered semigroup. So according to [1],

each pseudoorder on an ordered semigroup S induces a congruence σ̄ on S such that S/σ̄ 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) ∈ ρ ∀ a ∈ 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 S/ρ 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 S/ρ 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 S/ρ such that the (S/ρ, ∗, ∇) 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 S/ρ 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 ∩ σ^{−1}.

The relation σ̄ is a congruence on S. Indeed: If a ∈ S, then (a, a) ∈ σ, then (a, a) ∈ σ^{−1}, so (a, a) ∈ σ ∩ σ^{−1} := σ̄. If (a, b) ∈ σ̄, then (a, b) ∈ σ and (a, b) ∈ σ^{−1}, then (b, a) ∈ σ^{−1} and (b, a) ∈ σ, so (b, a) ∈ σ^{−1} ∩ σ := σ̄. If (a, b) ∈ σ̄ and (b, c) ∈ σ̄, then (a, b) ∈ σ, (a, b) ∈ σ^{−1}, (b, c) ∈ σ, (b, c) ∈ σ^{−1}, then (a, c) ∈ σ and (a, c) ∈ σ^{−1}, thus (a, c) ∈ σ ∩ σ^{−1} := σ̄. Let (a, b) ∈ σ̄ and c ∈ S. We have (a, b) ∈ σ and (a, b) ∈ σ^{−1}. Since (a, b) ∈ σ, c ∈ S, we have (ac, bc) ∈ σ, (ca, cb) ∈ σ. Since (a, b) ∈ σ^{−1}, we have (b, a) ∈ σ, then (bc, ac) ∈ σ, (ca, cb) ∈ σ, hence (ac, bc) ∈ σ^{−1}, (ca, cb) ∈ σ^{−1}. Then we have (ac, bc) ∈ σ ∩ σ^{−1} := σ̄ and (ca, cb) ∈ σ ∩ σ^{−1} := 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 S/σ̄ with the operation ” ∗ ” on S/σ̄ 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 S/σ̄ as follows: (a)σ̄∇(b)σ̄ if and only if (a, b) ∈ σ. The relation ”∇” on S/σ̄ 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 ∩ σ^{−1} ⊆ σ^{−1}, then (c, a) ∈ σ. Since (b)σ̄ = (d)σ̄, we have (b, d) ∈ σ̄ := σ∩σ^{−1} ⊆ 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 (S/σ̄, ∗, ∇) is an ordered semigroup. Let ρ be a congruence on S and suppose there exists an order relation ” ⪯ ” on S/ρ such that (S/ρ̄, ∗, ∇) 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 S/ρ such that (S/ρ, ∗, ⪯) 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} ∈ S/ρ. Since (S/ρ, ∗, ⪯) 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, ., ≤) → (S/ρ̄, ∗, ∇) | 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)ρ̄∇f(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 semipseudoorder 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 = ρ^{−1} and ρ is a semi-pseudoorder on S. Thus ρ ∈ SP(S) and

f((ρ)_{≈}) := ρ̄:= ρ ∩ ρ^{−1} = 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 S/ρ such that (S/ρ, ∗, ∇) 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 semipseudoorder on S, then ρ̄ is a congruence on S. Then, by the Theorem, the set (S/ρ̄, ∗, ∇) 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.

SEMI-PSEUDOORDERS IN SEMIGROUPS 55

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.

Univ. of Athens, Dept. of Mathematics Home: Niovi Kehayopulu, Nikomidias 18, 161 22 KESARIANI, GREECE E-mail address: nkehayop@cc.uoa.gr

ReceivedSeptember 30, 2003