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 ![](//html.scirp.org/file/7-5300865x13.png)
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
.
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 1s 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.