International Journal of Mathematics and Mathematical Sciences
Volume 2016 (2016), Article ID 2759090, 7 pages
http://dx.doi.org/10.1155/2016/2759090

Research Article

Natural Partial Orders on Transformation Semigroups with Fixed Sets

Yanisa Chaiya, Preeyanuch Honyam, and Jintana Sanwong

Department of Mathematics, Chiang Mai University, Chiang Mai 50200, Thailand

Received 20 April 2016; Revised 1 July 2016; Accepted 5 July 2016

Academic Editor: Pentti Haukkanen

Copyright © 2016 Yanisa Chaiya et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

Let be a nonempty set. For a fixed subset of , let be the set of all self-maps on which fix all elements in . Then is a regular monoid under the composition of maps. In this paper, we characterize the natural partial order on and this result extends the result due to Kowol and Mitsch. Further, we find elements which are compatible and describe minimal and maximal elements.

1. Introduction

For any semigroup , the natural partial order on , the set of all idempotents on , is defined by

In 1980, Hartwig [1] and Nambooripad [2] proved that if is a regular semigroup, then the relation

is a partial order on which extends the usual ordering of the set .

Later in 1986, the natural partial order on a regular semigroup was further extended to any semigroup by Mitsch [3] as follows:

Let be a set and denote the semigroup of binary relations on the set under the composition of relations. A partial transformation semigroup is the collection of functions from a subset of into with composition which is denoted by . Let be the set of all transformations from into itself and it is called the full transformation semigroup on . Then and are subsemigroups of . It is well known that and are regular semigroups.

In 1986, Kowol and Mitsch [4] characterized the natural partial order on in terms of images and kernels. They also proved that an element is maximal with respect to the natural order if and only if is surjective or injective; is minimal if and only if is a constant map. Moreover, they described lower and upper bounds for two transformations and gave necessary and sufficient conditions for their existence.

Later in 2006, Namnak and Preechasilp [5] studied two natural partial orders on and characterized when two elements of are related under these orders. They also described the minimality, maximality, left compatibility, and right compatibility of elements with respect to each order.

Let be a subset of . Recently, Fernandes and Sanwong [6] defined

where denotes the image of . Moreover, they defined to be the set of all injective transformations in . Hence and are subsemigroups of .

In [7], Sangkhanan and Sanwong described natural partial order on and in terms of domains, images, and kernels. They also compared with the subset order and characterized the meet and join of these two orders. Furthermore, they found elements of and which are compatible and determined the minimal and maximal elements.

Let be a fixed subset of and

In 2013, Honyam and Sanwong [8] proved that is a regular semigroup and they also determined its Green’s relations and ideals. Moreover, they proved that is never isomorphic to for any set when , and every semigroup is isomorphic to a subsemigroup of for some appropriate sets and with . Note that this also follows trivially from the fact that embeds in for any set with . Recently, the authors in [9] proved that there are only three types of maximal subsemigroups of and these maximal subsemigroups coincide with the maximal regular subsemigroups when is a finite set with . They also gave necessary and sufficient conditions for to be factorizable, unit-regular, and directly finite.

In this paper, we characterize the natural partial order on and find elements which are compatible under this order in Section 3. In Section 4, we describe the minimal elements, the maximal elements, and the covering elements. Moreover, we find the number of upper covers of minimal elements and the number of lower covers of maximal elements.

2. Preliminaries and Notations

In [8], the authors proved that is a regular subsemigroup of . Note that contains , the identity map on . If , then ; and if or , then consists of one element, . So, throughout this paper we will consider the case and .

For any , the symbol denotes the partition of induced by the map , namely,

For , , and , we say that refines if for each there exists such that .

Throughout this paper, unless otherwise stated, let .

For each , we have for all . So . If , then we write

and take as understood that the subscripts and belong to the index sets and , respectively, such that , , and . Thus for all , for all and . Here can be an empty set.

An idempotent in a semigroup is said to be minimal if has the property and implies .

In [8] the authors showed that

is the set of all minimal idempotents in and it is an ideal of . We note that is simply the set and is an idempotent in if and only if for all .

3. Natural Partial Order on

Kowol and Mitsch [4] gave a characterization of the natural partial order on . Later in 1994, Higgins [10] showed that if is a regular subsemigroup of a semigroup , then the natural partial order on is the restriction to of the natural partial order on . Here we describe the natural partial order on which is a regular subsemigroup of without making use of Higgins’ result and when we take , we recapture the result above by Kowol and Mitsch.

We note that if and for some , then refines .

Since is regular, we use to study the natural partial order on this semigroup.

Theorem 1. Let . Then if and only if the following statements hold:

(1);(2) refines ;(3)if , then .

Proof. Suppose that . Then, by , we have

for some . Thus . Since , we get that refines . Now, let . Then for some and thus . Hence and then since is an idempotent.
Conversely, assume that conditions (1)–(3) hold. By condition , we can write

where , , and . Since and refines , we obtain for all . If , then define and thus . If , then, for each , let So . By condition , ; that is, and hence . Define

We get and .
If , then . If , then, for each , we choose .
Case  1. Consider . Then . We define by

To prove that , let . If for some or for some , then it is clear that . Now, if for some , then and thus since refines . So . Hence . It remains to show that is an idempotent. Let . Then for some . Thus since .
Case  2. Consider We choose and define by

By the same prove as given in Case  1, we get and for all . If , then . So is an idempotent. Therefore, by .

Remark 2. If , then , and we have the characterization of on which first appeared in [4, Proposition  2.3].

As a direct consequence of Theorem 1, we get the following corollary.

Corollary 3. Let with . If , then .

Let be a semigroup. An element is said to be left (right) compatible with respect to the partial order if    whenever

The following results describe all the left compatible and right compatible elements in when . We also write instead of and for .

Theorem 4. Assume that and let . Then is left compatible if and only if is a minimal idempotent or is surjective.

Proof. Suppose that is left compatible. Assume by contrary that is not a minimal idempotent and is not surjective. So there are and . Define

Then with and thus since is left compatible. However, since but , a contradiction.
Conversely, let . If is a minimal idempotent, then . Now, assume that is surjective. So . Let . So for some . Since , we have that refines and hence for some . Since , we get for some and for some because is surjective. Hence ; that is, . Further, , thus refines . Let . So and then . By Theorem 1, we have which implies that is left compatible.

Theorem 5. The following statements hold.

(1)If , then is right compatible if and only if is a minimal idempotent or is injective.(2)If , then is right compatible if and only if is injective.

Proof. Assume that and is right compatible. Suppose in the contrary that is not a minimal idempotent and is not injective. So we can write

where and . Since is not injective, two cases arise.
Case  1. Consider . Choose and for some . Let and define by

we get . Moreover, we have , hence there is such that . However, for all since This means that does not refine . By Theorem 1, we get , a contradiction.
Case  2. Consider for some . Choose such that . Let . Define by

we get . Since , there is such that . However, for all since . So does not refine . By Theorem 1, we get , a contradiction.
Conversely, let be such that . If is a minimal idempotent, then and ; that is, is right compatible. Now, assume that is injective. Since , we get . Let . So for some and hence for some . So for some . Since is injective, for some and ; that is, . Thus which implies that refines . Let . So for some . Since is injective, and then . Thus since and that . Therefore, , and we conclude that is right compatible.
Suppose that is right compatible and is not injective. Write

where and . Since is not injective, two cases arise.
Case  1. for some . Choose and . Let and define

Then and hence . We can see that , but for all since . This means that does not refine , a contradiction.
Case  2. for some . This is virtually identical to Case  2 of above.

4. Minimal and Maximal Elements

Let be a semigroup together with the partial order . is said to be directed downward if every pair of elements has a lower bound. In other words, for any and in , there exists in with and . A directed upward semigroup is defined dually.

If , then and it has neither minimum nor maximum elements under the natural order (see [4]). So, in Lemmas 6 and 7 we assume that .

Lemma 6. Assume that . Then the following statements are equivalent.

(1) has a minimum element.(2) is directed downward.(3).

Proof. (1)⇒(2) This is clear.
(2)⇒(3) Assume that is directed downward. Let and . Consider

We have and there is such that and . By Theorem 1, refines and refines . Then there is such that and . Thus and hence . Since are arbitrary elements in , we obtain that .
(3)⇒(1) Assume that . It is easy to see that is the minimum element in .

Lemma 7. Assume that . Then the following statements are equivalent.

(1) has a maximum element.(2) is directed upward.(3).

Proof. (1)⇒(2) This is clear.
(2)⇒(3) Assume that is directed upward. Let and . Define

Then there is such that and . Since and are bijective, is also bijective and thus . So Since are arbitrary elements in , we get .
(3)⇒(1) Assume that . It is easy to see that is the maximum element in .

We now describe minimal and maximal elements in when . If , then has a minimum element by Lemma 6 and it is minimal. In the same way, if , then has a maximum element by Lemma 7 and it is maximal.

Theorem 8. Assume that and let . Then is minimal if and only if is a minimal idempotent.

Proof. Assume that is minimal but is not a minimal idempotent. So we can write

where . Choose and . Let , and define by

Hence , which contradicts the minimality of .
Conversely, assume that is a minimal idempotent and . Since , we get and hence . By Corollary 3, we obtain .

Theorem 9. Assume that and let . Then is maximal if and only if is injective or is surjective.

Proof. Let be maximal. Assume that is not injective and surjective. So there are such that with and . Write

Case  1. for some We may assume that . Let and define

Then and which contradicts the maximality of .
Case  2. for some Then we let and define

Then and which contradicts the maximality of .
Conversely, assume that is injective or is surjective and for some . Then and . Consider the case where is injective, by letting . Then for some and ; that is, for some . So and by Theorem 1. Since is injective, we get and thus , whence . Hence, in this case, and by Corollary 3 we obtain . In the case is surjective, we get ; that is, . Again by Corollary 3, we have that . Therefore, is maximal.

Figure 1 shows the diagram of when and . The notation for means that , , , and .

Figure 1

An element is called an upper cover for if and there is no such that ; lower covers are defined dually.

Lemma 10. Assume that and let . Then the following statements hold.

(1)If is not minimal in , then there is some lower cover of in .(2)If is not maximal in , then there is some upper cover of in .

Proof. Let be not minimal. By Theorem 8, is not a minimal idempotent. So we can write

where . Define as in the proof of Theorem 8, we get . Suppose that there is such that . Then by Theorem 1, and thus . Since which implies or , thus or by Corollary 3. Therefore, is a lower cover of .
The proof is similar to , using or from the proof of Theorem 9 as appropriate.

Now, we aim to find the number of upper covers of minimal elements and the number of lower covers of maximal elements when is a finite set. The following lemma is needed in finding such numbers.

Lemma 11. Assume that and let with . Then is an upper cover of if and only if .

Proof. Write

Since , we can write

where , , and is contained in either for some or for some . We get .
Assume that is an upper cover of . If , then which implies that , a contradiction. For the case , we choose and hence for some or for some . Assume that (the other case being similar). Let and Define

Since , we get , a contradiction. Therefore, .
The converse is proved in similar fashion to Lemma 10 (1).

Let be a finite set with elements and a nonempty proper subset of with elements. If , then has unique minimal element, say . By Lemma 11, each of upper covers of is of the form , where and . Since there are ways to choose and choice of , in this case there are in total upper covers of .

If , then has unique maximal element, the identity map. Let , , and . Then each of lower covers of is of the form , where . Since can be chosen from , there are in total lower covers of .

Theorem 12. Assume that and let . Then the following statements hold.

(1)If is minimal, then there are

 upper covers of .(2)If is maximal, then there are lower covers of .

Proof. Since is a finite set with elements, and .
Let be minimal in and an upper cover of . Then by Lemma 11; that is, for some . Since must refine , we can write

where and for some . We claim that for all . Assume by contrary that there is such that . Let . So for some , but ; that is , a contradiction. So we can write

where . Since there are ways to choose and choices of , in this case can have forms, but can be chosen from , so that there are in total upper covers of .
Assume that is maximal. Then is a bijection and we can write

where and . Let be a lower cover of . Then ; that is, for some . Let and . So , then since . Hence for all and for some or for some . Thus

where , or

where . For the first form and the second form, the numbers of ways of placing is and , respectively. So the total number of ways of placing is . But varies in the index set ; hence there are in total lower covers of .

Competing Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research was supported by Chiang Mai University.

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