On the Monoid of All Order Preserving Full Contractions with a Fixed Set ()
1. Introduction and Preliminaries
Denote
to be a finite
-chain
. A map that has both its domain and range subsets of
is said to be a transformation of the set
. A transformation that has its domain subset of
is said to be partial. The collection of all partial transformations on
is known as the semigroup of partial transformations and is usually denoted by
. A partial transformation whose domain is equal to
is said to be a full (or total) transformation. The collection of all full transformations on
is known as the semigroup of full transformations, which is usually denoted by
. A map
is said to be order preserving if (for all
)
implies
. The collection of all order preserving full transformations on
is known as the semigroup of order preserving full transformations and is usually denoted by
. The algebraic, combinatorial and the rank properties of the semigroup
have been extensively studied over the years, see for example [1]-[11].
Let
be a nonempty subset of
. Denote by
to be the collection of all
that fix elements of
that is,
. The set
is known as the semigroup of transformations with a fixed set under the usual composition of functions. This semigroup first appeared in [12], where its algebraic, combinatorial and rank properties were investigated. In particular, the authors in [12] investigated its Green’s relations, ideals, isomorphisms and rank properties. Later, Chaiya et al. [13] characterized all the maximal subsemigroups of
, while the natural partial order on
was investigated in [14]. There are various semigroups of transformations with fixed sets that have been studied by various authors, see for example [15] [16]. For basic concepts in semigroup theory, we refer the reader to the books [17]-[19].
A map
is said to be an isometry if for all
,
; and is said to be a contraction if for all
,
. Let
(1)
be the semigroup of all of full contractions on
and
(2)
be the semigroup of all order-preserving full contractions of
. A general study of these semigroups was initiated in 2013 by Umar and Al-Kharousi [20], in a research proposal to the Research Council of Oman. In this proposal, notations for these semigroups and their various subesmigroups were given. We are adopting the same notations for these semigroups in this paper.
The combinatorial properties of the semigroup
have been investigated by Adeshola and Umar [21], where they showed that the order of
is
. Ali et al. [22] characterized both the regular elements and Green’s relations of the semigroups
and
. Moreover, they proved that the collection of all regular elements of
(denoted by
) is a subsemigroup, and also showed that the Rees quotient of the two-sided ideal of
is an inverse semigroup. Recently, Toker [23] investigated the rank properties of
.
For a nonempty subset
of
, let
(3)
be the monoid of all order preserving full contraction mappings with a fixed set
. It is straightforward to show that
is a subsemigroup of
. Moreover, it is evident that
(where
denotes the identity map on
) for every nonempty subset
. Hence,
is a monoid. Notice that if
, then
(i.e.,
is the trivial group). Thus, for the rest of the paper, we will only consider case where
does not contain
. Furthermore, let
and
, and let
. Then, we shall refer to
as closure of
.
Now, for
, let
(4)
be the two-sided ideal of
, and let
(5)
be the collection of elements in
of height
. Evidently,
. Additionally, let
(6)
Furthermore, for
, let
(7)
be the Rees quotient semigroup of
, where the product of two elements
and
in
is zero if their product has height less than
, otherwise it is
.
The monoid
does not seem to have been studied in the literature. In this paper, we investigate the algebraic, combinatorial and rank properties of this monoid. Section 1 of this paper contains the definitions of basic terms. In Section 2, we give a characterization of regular elements as well as a necessary and sufficient condition for the monoid
to be regular. Moreover, we characterize all the Green’s relations and starred Green’s relations on the monoid
. In Section 3, we compute the cardinality and the rank of
. Finally, we study certain isomorphism properties on the monoid.
In line with [17], every transformation
can be written as
(8)
where
(also referred to as blocks) are equivalence classes under the relation
. The collection of all the equivalence classes of the relation
, is the partition of
usually denoted by
, i.e.,
(
). A subset
of
is said to be convex if
(
) and if there exists
such that
implies
. Let
be a partition of
, then
is called convex if any element
of
is convex. Moreover, a partition
is said to be ordered if for all
,
if and only if
for all
and
. Notice that if
is an ordered partition then its necessarily convex. A subset
of
is said to be a transversal of the partition
if
, and
.
For any transformation
, we shall denote
,
,
and
to mean the image set of
,
, the set of fixed points of
and the number of the fixed points of
(i.e.,
), respectively. For
, we shall write the composition of
and
as
for all
.
Let
be a semigroup, a subset
of
is said to be a generating set of
, if every element of
can be written as a product of elements of
. If
generates
, we simply write
. The rank of a semigroup
is defined and denoted by
An element
is said to be an idempotent if
. The collection of all idempotents of
is usually denoted by
. It is well known that an element
(where
is expressed as in (8)) in any transformation semigroup is an idempotent if and only if
is stationary in the sense that
for all
.
Before we begin our discussion, the following remark is worth noticing:
Remark 1.1.
1) For
and any nonempty subset
of
of order 1,
. Evidently, for all
and
, the element
is in
, but not in
.
Moreover, if
or
, the element
, but does not fix
and
, so
.
2) For
and any nonempty subset
of
of order
,
.
Notice that for any
with
and
, the element
but not in
since the element
is an element of
and is not a fixed point, and so,
.
The following Lemma from [21] is needed in our subsequent discussion.
Lemma 1.2. ([21] Lemma 1.2) Let
and let
. Then,
is convex.
We now have the following lemma.
Lemma 1.3. If
. Then,
for all
. In other words, if
, then
must fix
.
Proof. Let
and let
and
. Now, suppose by way of contradiction that there exists
such that
. Notice that
. Then, by the order preserving property of
, we must have
, i.e.,
. Now since
, then either
or
. Thus, we consider the two cases separately.
Case 1. Suppose
, i.e.,
. Then,
, contradicting the fact that
is a contraction.
Case 2. Suppose
i.e.,
. Then,
, contradicting the fact that
is a contraction. The result now follows. □
The elements in the monoid
have a general expression as in the lemma below.
Lemma 1.4. Let
and
for some
and for some
, so that
, where
. Then, every
of height
can be expressed as
(9)
Proof. Let
be of height
. Then, as in (8),
can be expressed as
Now since
is a contraction, then by Lemma 1.2,
is convex i.e.,
for some
. Thus,
can be expressed as
Now since
, then by Lemma 1.3,
must fix
. Notice that
for some
, and so,
for all
, where
and
. Thus,
can be expressed as
as required. □
We note the following remark.
Remark 1.5. It is worth noting that the block
(
and
) can be empty.
For the purpose of illustrations, take
, i.e.,
and
. Let
be
.
Notice that
and
. Moreover, the position of the first element in
is in
, i.e.,
, and so
. The block containing the maximum element of
is at the position
. Thus, the blocks containing the elements of
are
,
and
. Other blocks are
,
and
.
Now, consider
.
Clearly,
is the block containing the minimum element 4 while
contain the maximum element 6 in
, and so the blocks that contains the elements of
are
,
and
. It is worth noting that there is no any block before
and there is one block after
, i.e., the block
.
Moreover, consider
Notice that the blocks containing the elements in
are
,
and
. While other blocks are
and
. Obviously, there are two blocks before
and there is no block after
.
Furthermore, consider
as
.
is the block containing the minimum element of
while
contain the maximum element. Thus, there is not any block before
and after
.
The following is worth remarking.
Remark 1.6. Clearly if
, then
.
We now present the following lemma.
Lemma 1.7. For any
, .
Proof. Let
. Then, by Lemma 1.3,
fix
, and so .
On the other hand, if , then
fix
(since
) and therefore
. Thus, the result follows.
2. Regularity and Green’s Relations on
Let
be a semigroup without identity element and
be a monoid. The five equivalences classes on
known as Green’s relations were first introduced by J. A. Green in 1951. The primary aim of defining these relations is to study the structure of a semigroup
. These relations are defined as follows. For
,
if and only if
;
if and only if
;
if and if
. The relation
, while the relation
is a join of the relations
and
, i.e.,
. For more details on the properties of Green’s relations, we refer the reader to [3] [18] [19].
An element
in a semigroup
is regular if there exists
such that
. If every element of
is regular, then
is called a regular semigroup. If a semigroup with idempotent elements is not regular, then there is need to investigate the regular elements, so as to identify the
-classes and
-classes that contain idempotents. The semigroup
is not regular in general, but can be regular for certain
, as we are going to discuss below, in this section.
The Greens relations for the semigroup
and some of its subsemigroups have been investigated in [22]. Here, we also characterize these relations on the semigroup
. Throughout this section, we will consider
.
We begin our investigation by first noting the following well-known lemmas:
Lemma 2.1. ([22], Corollary 44) Let
be as expressed as in (8). Then,
if and only
and
.
Lemma 2.2. ([22], Corollary 45) Let
be expressed as in (8). Then,
if and only
.
Lemma 2.3. ([21], Lemma 1.1) Let
be such that
. Then,
. Equivalently,
is convex.
Lemma 2.4. If
such that
, then
.
Proof. Let
such that
. Suppose
is expressed as in (9), i.e.,
and let
Thus, by Lemma 1.2,
is convex, and therefore
. This implies
, and as such
, as required. □
From this point onward, we shall let
and
be of the form:
(10)
and
(11)
Next, we characterize the Green’s relations on
.
Theorem 2.5. Let
be expressed as in and (11), respectively. Then,
1)
if and only
;
2)
if and only
.
Proof. 1) Let
,
and
be as expressed in (10) and (11) such that
. Since
, then by Corollary 2.1,
and
. This means that there is an isometry from
to
for all
. Notice that
,
and since there is an isometry from
to
and from
to
, then
. Inductively, we see that
, for all
. Hence,
.
Conversely, suppose
. Now let
. Clearly,
,
and
.
2) The result follows directly from Lemma 2.2 and Lemma 2.4.
□
Consequently, we have the following Corollaries.
Corollary 2.6. On the monoid
,
.
As a consequence of the above Corollary, we deduce the following characterization of Green’s relations on the semigroup
.
Theorem 2.7. Let
. Then,
is
-trivial and therefore, the semigroup
is non-regular.
Next, we deduce the characterization of the regular element in
below:
Corollary 2.8. Let
be as expressed in Equation (10). Then,
is regular if and only if
is an idempotent.
Proof. The result follows from the fact that
is an
-trivial semigroup. □
Now, as a consequence of Corollary 2.6, we readily have the following result.
Theorem 2.9. Every
(
) is a maximal subgroup of
and is isomorphic to the trivial group
.
The next thing is to investigate when the whole semigroup
is regular. First, we note the following.
Remark 2.10. 1) Obviously if
, then
, which is a regular semigroup.
2) Moreover, if
, then
and
. Evidently, the semigroup
is regular, since every element in
is a regular idempotent element.
3) However, if
, then
is in one of the following forms
or
. Thus,
and
are non-regular monoids due to the fact that the elements
and
are not regular. Furthermore, the monoids
,
and
are regular.
Further, we investigate when the whole semigroup
is regular for
, in the Theorem below.
Theorem 2.11. For
, the monoid
is regular if and only if
.
Proof.
is regular if and only if every element in
is regular if and only if every element in
is an idempotent (by Corollary 2.8) if and only if the kernel class of every element in
has a transversal
which is equal to
if and only if
or
or
or
.
Thus, we have the following remark:
Remark 2.12. For
, the monoid
is not regular if and only if
or
.
Starred Green’s Relations
There are five starred Green’s equivalences defined on a semigroup
, namely
,
,
,
and
. In a semigroup
and for
,
if and only if
in some over semigroup of
say
. The relation
is defined dually. We shall use the notation
to mean
in
and similarly,
to mean
in
. The relations
and
have the following characterizations as in [24]:
(12)
and
(13)
The join of the relations
and
is
while their intersection is
. For basic definitions of starred Green’s relations, we refer the reader to [24] [25]. If a semigroup
is not regular, then there is a need to characterize its starred Green’s relations in order to investigate the class to which it belongs. Now, in this section, we shall consider
which does not belong to the set
.
We now record the following result from [19].
Lemma 2.13. ([19], Ex. 2.6 (16)) Let
. Then
1)
if and only if
;
2)
if and only if
;
3)
if and only if
;
4)
.
Now, we give the characterization of the relations
and
on the monoid
in the theorems below.
Theorem 2.14. Let
be expressed as in (10) and (11), respectively. Then,
if and only if
.
Proof. Suppose
. Now define
by
Then, obviously
and one can easily verify that
if and only if
(by (12)). Obviously,
Thus,
. Similarly, one can show that
. Therefore,
.
Conversely, if
, then by Lemma 2.13,
. Thus, by definition it follows that
. □
Theorem 2.15. On the monoid
,
.
Proof. The result follows from definition and Theorem 2.5-2). □
Theorem 2.16. Let
. Then,
if and only if
.
Proof. Suppose
. This means that there exists
such that
and
. Then, by Theorem 2.14,
and by Theorem 2.15,
. Thus,
, as required.
Conversely, if
. Then, since
is reflexive, it follows that
. □
Now, we have the following remark:
Remark 2.17. On the semigroup
,
1)
;
2)
.
A semigroup
is said to be a left abundant (resp., right abundant) if both the
-class (resp.,
-class) contains an idempotent and
is said to be abundant if both
-class and
-class contains an idempotent [24]. A semigroup
is said to be left quasi-adequate (resp., right quasi-adequate) if it is left abundant (resp., right abundant) and its set of idempotent elements forms a subsemigroup, and it is quasi-adequate if it is both left and right quasi-adequate [24].
We now present the following results.
Theorem 2.18. The monoid
is left abundant.
Proof. Let
and
. Denote
to be the
-class of
. Now either
or
or there exist
such that
.
Case 1. If
, then
, so that
where
. Thus, define
as
Obviously,
and clearly
.
Case 2. If
, then
so that
is of the form
where
. So, define
Clearly,
.
Case 3. If
and
. Then, define
The element
is clearly in
. Hence, in all the three cases
and by Theorem 2.14,
, as such
, as required. □
Theorem 2.19. For
, the monoid
is not right abundant.
Proof. If
, then obviously by Lemma 2.4,
. □
The next lemma shows that the collection of idempotents in
is a subsemigroup of
.
Lemma 2.20.
is a semilattice.
Proof. The proof is the same as the proof of (11], Theorem 7). □
Consequently, we have proved the following theorem.
Theorem 2.21. The monoid
is left adequate.
3. The Combinatorial and Rank Properties of
In this section, we determine the cardinality and rank of the monoid
. These properties heavily depend on the size of the subset
, as we shall see below.
We now present the following proposition, which counts the number of elements each of height
in
.
Proposition 3.1. Let
such that
, and let
be as defined in (6). Then, for
Proof. To count the number of elements of height
in
, first is to partition the set
(which obviously has
elements since
) into
parts (since the rank of each of the elements is
). This is equivalent to selecting
elements out of
elements. The result now follows. □
Theorem 3.2. Let
such that
. Then
Proof. Using Proposition 3.1, we see that
□
Rank of
The semigroup
is
-trivial (from Corollary 2.6 and Theorem 2.7) and therefore, in line with [26], it admits a minimum generating set. Now, let
be defined as in (5). The next result shows that the collection of all the elements in
is the minimum generating set of
.
Lemma 3.3. Let
. Then,
if and only if
and
.
Proof. Let
and suppose
. Thus,
, as such
. Moreover,
implies
is one of the transversals of
and
. Thus, by Lemma 2.4 we have
.
The converse is obvious. □
From Proposition 3.1, we record the following remark:
Remark 3.4. For each
,
.
We now present the following theorem:
Theorem 3.5. Let
be as defined in (7). Then, the rank of
is given by:
Proof. Notice that
is the minimum generating set, as stated by Lemma 3.3, and it’s order is given by Remark 3.4. □
The following lemma plays a crucial role in determining the rank of
.
Lemma 3.6. If
then
for
.
Proof. Let
be as expressed in (10). Now since
, it means that there exists
for some
such that
is a fixed point of
, i.e.,
. Now either
or
or
. We consider the following cases:
Case 1: If
. Then,
. We may without loss of generality suppose
, where
. If
, then
is of the form
In this case, define
and
Now if
. Then,
has the form
and so
and
are defined as
and
respectively.
Case 2. If
. Then,
. We may without loss of generality suppose
, where
. If
then
is of the form
Now, define
and
If
and
(in this case
), then
has the form
and so
and
are defined as
and
respectively.
Also, if
and
, then
is of the form:
Define
and
as
and
respectively.
Case 3: If
, then
must be singleton
. Thus,
is of the form
Now, define
and
as
and
Now, clearly in each case,
and also,
. The proof is now complete. □
Consequently, we obtain the rank of the two-sided ideal
as stated in the following result:
Theorem 3.7. Let
be as defined in (4), and let
be as defined in (7). Then, the
.
Proof. The result follows from Lemma 3.6, Theorem 3.5 and Lemma 3.3. □
Next, we deduce the following result.
Corollary 3.8. Let
such that
, and
be as expressed in equation (3). Then,
.
Proof. The result follows from Theorem 3.7 and the fact that
□
We conclude the paper with the following isomorphism result:
Theorem 3.9. Let
and
be nonempty subsets of
. Then,
if and only if
=
.
Proof. The result follows easily from Lemma 1.7.
Acknowledgements
The authors would like to thank Prof. Abdullahi Umar for technical and financial support.