Lobachevskii Journal of Mathematics Vol. 11, 2002, 19-21 \begin{matrix}c \\ ◯ \end{matrix}N. Kehayopulu, M. Tsingelis

N. Kehayopulu, M. Tsingelis

A NOTE ON PSEUDOCONGRUENCES IN SEMIGROUPS

(submitted by M. M. Arslanov)

Abstract. 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 this short note we introduce the concept of pseudocongruence in semigroups and we prove that each pseudocongruence on a semigroup S induces a congruence σ on S such that S/σ is an ordered semigroup.

If (S, ., ≤) is an ordered semigroup, then is there any congruence σ on S for which S/σ is an ordered semigroup? The answer to this question is important for studying the structure of ordered semigroups. The question led us to the concept of pseudoorder introduced in [1]. By a pseudoorder on S we mean a relation ρ on S such that:

1) ≤⊆ ρ;

2) (a, b) ∈ ρ, (b, c) ∈ ρ → (a, c) ∈ ρ;

3) (a, b) ∈ ρ, c ∈ S → (ac, bc) ∈ ρ and (ca, cb) ∈ ρ. In [1] we proved that if S is an ordered semigroup and ρ a pseudoorder on S, then the relation σ on S defined by

σ = {(a, b) ∈ S × S | (a, b) ∈ ρ&(b, a) ∈ ρ}

is a congruence on S and the set S/σ is an ordered semigroup. The following question is natural: if S is a semigroup, then is there any congruence /sigma on S such that S/σ is an ordered semigroup? The proposition below gives the answer. We first introduce the following definition.

Definition 0.1. Let (S, .) be a semigroup. A relation ρ on S is called pseudocongruence if

1) (a, a) ∈ ρ for all a ∈ S;

2) (a, b) ∈ ρ, (b, c) ∈ ρ → (a, c) ∈ ρ;

3) (a, b) ∈ ρ, c ∈ S → (ac, bc) ∈ ρ and (ca, cb) ∈ ρ.

Key words and phrases. Ordered semigroup, pseudocongruence in semigroup.

Proposition 0.2. Let (S, .) be a semigroup and ρ apseudocongruence on S. Then, there exists a congruence σ on S such that S/σ is an ordered semigroup.

Proof. 1) Let σ be the relation on S defined as follows:

σ = {(a, b) ∈ S × S | (a, b) ∈ ρ&(b, a) ∈ ρ.

σ is a congruence on S. In fact we have the following.

Let a ∈ S. Since (a, a) ∈ ρ, we have (a, a) ∈ σ. If (a, b) ∈ σ, then (b, a) ∈ ρ and (a, b) ∈ ρ, i. e. (b, a) ∈ σ.

If (a, b) ∈ σ, then (b, c) ∈ σ and (a, b) ∈ ρ, (b, a) ∈ ρ, (b, c) ∈ ρ, (c, b) ∈ ρ. It follows that (a, c) ∈ ρ, (c, a) ∈ ρ, i. e. (a.c) ∈ σ. Now let (a, b) ∈ σ, c ∈ S. Since (a, b) ∈ ρ(b, a) ∈ ρ, we have (ac, bc) ∈ ρ, (ca, cb) ∈ ρ, (bc, ac) ∈ ρ, (cb, ca) ∈ ρ. Then (ac, bc) ∈ σ, (ca, cb) ∈ σ.

2) We define an operator ”⊕” on S/σ as follows:

⊕ : S/σ × S/σ → S/σ | ((a)_{σ}, (b)_{σ}) → (a)_{σ} ⊕ (b)_{σ} := (ab)_{σ}.

Since σ is a congruence on S, the operation is well defined and (S/σ, ⊕) is a semigroup.

3) Now we define a relation ” ≤ ” on S/σ as follows:

≤:= {((a)_{σ}, (b)_{σ}) | ∃x ∈ (a)_{σ}∃y ∈ (b)_{σ} : (x, y) ∈ ρ.

It is easy to see that there are following equivalent definitions:

≤:= {((a)_{σ}, (b)_{σ}) | ∀x ∈ (a)_{σ}∀y ∈ (b)_{σ} : (x, y) ∈ ρ.
(a)_{σ} ≤ (b)_{σ} ⟺ (a, b) ∈ ρ.

To show that the definitions are equivalent suppose that a, b ∈ S. We consider the following three statements: α) ∃x ∈ (a)_{σ}∃y ∈ (b)_{σ} : (x, y) ∈ ρ. β) ∀x ∈ (a)_{σ}∀x ∈ (a)_{σ} : (x, y) ∈ ρ. γ) (a, b) ∈ ρ. α) → β). Let x ∈ (a)_{σ}, y ∈ (b)_{σ}, (x, y) ∈ ρ and let z ∈ (a)_{σ}, t ∈ (b)_{σ}.

Wwe have (x, a) ∈ σ, (y, b) ∈ σ, (x, y) ∈ ρ, (z, a) ∈ σ, (t, b) ∈ σ. Then we have (z, a) ∈ ρ, (a, x) ∈ ρ, (x, y) ∈ ρ, y, b) ∈ ρ, (b, t) ∈ ρ. This means that (z, t) ∈ ρ.

For β) → γ) and γ) → α) note that a ∈ (a)_{σ} and b ∈ (b)_{σ}.

To show that (S/σ, ⊕, ≤) is an ordered semigroup let (a)_{σ} ∈ S/σ. Since a ∈ S, we have (a, a) ∈ ρ. Then (a)_{σ} ≤ (a)_{σ}.

Let (a)_{σ} ≤ (b)_{σ}, (b)_{σ} ≤ (a)_{σ}. We have (a, b) ∈ ρ, (b, a) ∈ ρ. It follows that (a, b) ∈ σ and (a)_{σ} = (b)_{σ}.

Let now (a)_{σ} ≤ (b)_{σ}, (b)_{σ} ≤ (c)_{σ}. We have (a, b) ∈ ρ, (b, c) ∈ ρ. Therefore, (a, c) ∈ ρ and (a)_{σ} ≤ (c)_{σ}.

Let (a)_{σ} ≤ (b)_{σ}, (c)_{σ} ∈ S/σ. We have (a, b) ∈ ρ, c ∈ S. Then (ac, bc) ∈ ρ, (ca, cb) ∈ ρ. It follows that (ac)_{σ} ≤ (bc)_{σ}, (ca)_{σ} ≤ (cb)_{σ}. Hence (a)_{σ} ⊕ (c)_{σ} ≤ (b)_{σ} ⊕ (c)_{σ}, and (c)_{σ} ⊕ (a)_{σ} ≤ (c)_{σ} ⊕ (b)_{σ}.

Examples of pseudosemicongruences. For a semigroup S, a fuzzy subset f of S and an element a ∈ S, we consider

the extension of f by a, introduced by Xie Xiang-Yun in [2]. We denote it as ⟨a, f⟩, and this is the fuzzy subset of S defined by:

⟨a, f⟩ : S → [0, 1] | x → ⟨a, f⟩(x) := f(ax).

For fuzzy subsets f, g on S, we define f ⊆ g ⟺ f(a) ≤ g(a), for all a ∈ S. Let (S, .) be a semigroup. A fuzzy subset f of S is called prime if for all

a, b ∈ S f(ab) = max{f(a), f(b)}.

Example 1. Let S be a semigroup and f a prime fuzzy subset of S. Then the relation ρ = {(a, b) ∈ S × S | ⟨a, f⟩ ⊆ ⟨b, f⟩} is a pseudocongruence on S.

It is clear that ρ refexive and transitive. Let (a, b) ∈ ρ, c ∈ S. We need to show that (ac, bc) ∈ ρ. We have ⟨ac, f⟩ ⊆ ⟨bc, f⟩, i. e. ⟨ac, f⟩(x) ≤ ⟨bc, f⟩(x), for all x ∈ S. Indeed,

let x ∈ S. Then (a, b) ∈ ρ → ⟨a, f⟩ ⊆ ⟨b, f⟩ → ⟨a, f⟩(y) ≤ ⟨b, f⟩(y) for all y ∈ S → ⟨a, f⟩(cx) ≤ ⟨b, f⟩(cx) → f(acx) ≤ f(bcx) → ⟨ac, f⟩(x) ≤ ⟨bc, f⟩(x). Let now (a, b) ∈ rho, c ∈ S. We will prove that (ca, cb) ∈ ρ. Let x ∈ S.

To prove that ⟨ca, f⟩(x) ≤ ⟨cb, f⟩(x) consider (a, b) ∈ ρ. It follows that for all y ∈ S we have ⟨a, f⟩(y) ≤ ⟨b, f⟩(y) Then ⟨a, f⟩(x) ≤ ⟨b, f⟩(x). This means that f(ax) ≤ f(bx), and, therefore, max{f(c), f(ax)} ≤ max{f(c), f(bx)}. Since f is prime this means that ⟨ca, f⟩(x) ≤ ⟨cb, f⟩(x).

Example 2. Let S be a commutative semigroup and f a fuzzy subset of S. Then the relation ρ = {(a, b) ∈ S × S | ⟨a, f⟩ ⊆ ⟨b, f⟩} is a pseudocongruence on S. Indeed, it is clear that ρ is reflexive and transitive. If (a, b) ∈ ρ, c ∈ S, then (ac, bc) ∈ ρ also (exactly as in Example 1). Since S is commutative, we have (ac, bc) ∈ ρ.

References

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

[2] Xie Xiang-Yun, Fuzzy ideal extensions of semigroups, Preprint, to appear.

Niovi Kehayopulu, University of Athens, Department of Mathematics, Nikomidias 18, 161 22 Kesariani, Greece E-mail address: nkehayop@cc.uoa.gr

Received October, 12, 2002