On Congruences Induced by Certain Relations on “Semigroups”

K. V. R. Srinivas

Department of Mathematics, Regency Institute of Technology, Yanam, India.

DOI: 10.4236/apm.2015.59054                

Abstract

In his paper “On quasi-separative ‘semigroup’s’”, Krasilnikova, Yu. I. and Novikov, B. V. have studied congruences induced by certain relations on a “semigroup”. They further showed that if the “semigroup” is quasi separative then the induced congruence is a semilattice congruence. In this paper we continue the study of these relations and the induced congruences i.e., the congruences induced by certain relations on ‘‘semigroup’s”. In this paper mainly it is observed that if S is a quasi-separative and regular “semigroup” then the necessary and sufficient condition for to be the smallest semilattice congruence η is obtained.

Keywords

Cancellative “Semigroup”, Quasi-Separative ‘‘Semigroup’s”, Weakly Cancellative ‘‘Semigroup’s”, Weakly Balanced “Semigroup”

1. Introduction

In this paper “On quasi-separative ‘semigroup’s’”, Krasilnikova Yu. I. and Novikov B.V. have studied congruences induced by certain relations on a “semigroup”. They further showed that if the “semigroup” is quasi-se- parative then the induced congruence is a semilattice congruence. In this paper we continue the study of these relations and the induced congruences. In theorem 2, we have proved that the family of all relatios which satisfy the conditions from (1) to (3) in Def. 1 of this paper is a complete lattice. In theorem 3, we have also obtained that the family of all congruences which are induced by the relations in is a complete lattice. If S is a quasi-separative and regular “semigroup” then the necessary and sufficient condition for to be the smallest semilattice congruence which is denoted by h (throughout this chapter) is obtained, from which as a corollary that if S is a commutative regular “semigroup” then the congruence induced by the S × S is the smallest semilattice congruence [1] . The authors have remarked that a semilattice of weakly cancellative “semigroup’s” is weakly balanced, it is not known that “whether semilattice of weakly cancellative ‘semigroup’s’ [2] is weakly balanced”, show that the result is not true. It is also observed that every semilattice of weakly cancellative “semigroup’s”, need not be weakly balanced, for this an example is obtained.

2. Main Content

The following definition is due to Krasilnikova Yu. I. and Novikov B. V. (see [3] ).

Def 1: Let S be a “semigroup” and Ω be a relation on S satisfying conditions.

(1)

(2)

(3)

where and

Define a relation on S corresponding to Ω by if and only if. It is also equivalent to if and only if, this relation is a congruence on S.

Lemma 2: Let be the family of all relations on S which satisfy the conditions from (1) to (3) then is a complete lattice.

Proof: Let. Then clearly. Let be a subset of. Then both and are in. Therefore is a complete lattice.

Lemma 3: Let then is a complete lattice.

Proof: Since . Therefore. Also is the greatest element in. Let be a subset of. Then because.

Theorem 4: Let S be quasi-separative and regular “semigroup”. Then if and only if for any,.

Proof: Suppose S is quasi-separative and regular and Suppose. Then, so that it satisfies (1). Therefore if and only if for any. Conversely suppose that. Then. Since S is quasi-separative is semilattice congruence and hence

. Let so that and hence if and ony if for any, since S is regular there exists such that and. Put and then we have and so that hence. Therefore and similarly we have. Therefore if F is any filter in S then if and only if so that and hence.

Corollary 5: If S is a commutative regular “semigroup” then.

Corollary 6: If S is a completely regular and and if , for some then.

The following is an example of a completely regular “semigroup” in which.

Example 7: Let S be a left zero “semigroup” with at least two elements. If then then, which is a contradiction and hence.

Theorem 8: In a band S, if and only if S is a semilattice.

It is natural to ask whether every semilattice congruence on “semigroup” is of the form for some.

The following example shows that it is not true.

Example 9: Consider the non modular lattice in Figure 1 and let S be the “semigroup” is a semilattice. Clearly is a filter in L so that is a congruence on S. But for any.

Figure 1. a^c = b^c = 0, take c = f.

The following example shows that is a semi lattice congruence whenever then the “semigroup” need not be quasi-separative.

Example 10: Let be two element null “semigroup”. Clearly and which is a semilattice ongruence. But is not a quasi-separative (since and).

The following example shows that in non quasi-separative “semigroup’s” there exists such that is a semi-lattice congruence.

Example 11: Let S be a non quasi-separative “semigroup”, then 1_s is in, and is a semi-lattice congruence.

It is interesting to note that if S is a left or right zero “semigroup” then

.

In paper [1] they have remarked that it is not known that whether semilattice of weakly cancellative “semigroup’s” is quasi-separative and weakly balanced. In the following we are giving an example which shows that it is not true i.e. if a “semigroup’s” is isomorphic to a semilattice of weakly cancellative “semigroup’s” then S is a quasi-separative and weakly balanced.

Example 12: Consider the “semigroup” with multiplication table as follows:

Then h-classes are {a, b} and {c, d} which are right zero “semigroup’s” and hence S is a semilattice of weakly cancellative “semigroup’s”, but S is not weakly balanced since, , but.

The following is an example of quasi separative “semigroup”, which is not completely regular.

Example 13: Consider the “semigroup” on which “.” is defined by

where. Clearly S is quasi separative “semigroup”, and since the inverse of (m, n) is (n, m) and, S is not completely regular.

Thorem 14: Let S be a separative “semigroup”, and such that is a semilattice congruence then a Î.

Proof: Let S be a separative “semigroup” and a Î S such that E(a) is a semilattice congruence. Then for any, , so that . Now replace “y” by “b” and “x” by “a” then which implies. Since S is separative and we have so that. Since S is separative we have. Again since E(a) is a semilattice congruence, so that and hence.

Acknowledgements

We are very much thankful to the referees for their valuable suggestions.

Conflicts of Interest

The authors declare no conflicts of interest.

References

[1]	Burmistrovich, I.E. (1965) Commutative Bands of Cancellative “Semigroup’s’”. Sib.Mat.Zh., 6, 284-299 (Russian).
[2]	Petritch, M. (1973) Introduction to “‘Semigroup’ Theory”. Merill Books, Columbus.
[3]	Krasilnikova, Yu.I. and Novikov, B.V. (2005) On Quasi-separative “semigroup’s”. “Semigroup” Forum, 70, 347-355.