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International Journal of Mathematics and Mathematical Sciences
Volume 2008, Article ID 794013, 11 pages
http://dx.doi.org/10.1155/2008/794013
Research Article

Regularity and Green's Relations on a Semigroup of Transformations with Restricted Range

1Department of Mathematics, Chiang Mai University, Chiangmai 50200, Thailand
2Department of Mathematics, Chiangmai Rajabhat University, Chiangmai 50300, Thailand

Received 14 May 2008; Accepted 18 September 2008

Academic Editor: Robert Redfield

Copyright © 2008 Jintana Sanwong and Worachead Sommanee. 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 the full transformation semigroup on the set and let . Then is a sub-semigroup of determined by a nonempty subset of . In this paper, we give a necessary and sufficient condition for to be regular. In the case that is not regular, the largest regular sub-semigroup is obtained and this sub-semigroup is shown to determine the Green's relations on . Also, a class of maximal inverse sub-semigroups of is obtained.

1. Introduction

Let be a nonempty subset of and let denote the semigroup of transformations from into itself. We consider the sub-semigroup of defined by when denotes the range of . In fact, if , then contains exactly one element (namely, the constant map with range ) and if then .

In 1975, Symons [1] described all the automorphisms of and found that the most difficult case occurs when . He also determined when is isomorphic to and, surprisingly, the answer depends on the cardinals and , not on for . Here, we study other algebraic properties of this semigroup. Recall that an element of a semigroup is called regular if for some in . A semigroup is regular if every element of is regular. It is already known that is a regular semigroup (see [[2], page 33]). But is not regular in general. So, in Section 2, we prove that is regular if and only if or . We also prove that if is not regular, the set is the largest regular sub-semigroup of . In Section 3, we characterize the Green's relations on and find that its and relations are surprising, but they reduce to those on when . And in Section 4, we give a class of maximal inverse sub-semigroups of , of the form and is injective on . When , the set and is injective on is a class of maximal inverse sub-semigroups of given in [4].

Note that throughout the paper, we write functions on the right; in particular, this means that for a composition , is applied first.

2. Regularity of

To give a necessary and sufficient condition for the semigroup to be regular, we first note the following.

(1)If , say then contains exactly one element (namely, the constant map with range ), so is regular.(2)If , then which is a regular semigroup.(3)If , then or , and is regular by (1) and (2). Now, we need some notation. We adopt the convention introduced in ([[3], page 241]), namely, if , then we write and take as understood that the subscript belongs to some (unmentioned) index set , the abbreviation denotes , and that = and .

Theorem 2.1. is a regular semigroup if and only if or .

Proof. Assume that and . Let be such that and choose . Let be any element in where and .
We define , and it is clear that .
Since , so for all and So, we conclude that for all , this implies that is a nonregular element in . Therefore, is not a regular semigroup. The converse is clear by the previous note.

Now, we consider the set It is easy to see that . Since , there exists and we see that the constant map with range satisfies the condition in , therefore, and so . And for each and , we have , and thus . This proves the following.

Lemma 2.2. is a right ideal of .

In general, is not a left ideal of as shown in the following example.

Example 2.3. Let denote the set of positive integers, let denote the set of all positive even integers, and define Then and , but Thus is not a left ideal of .

Theorem 2.4. is the largest regular sub-semigroup of .

Proof. From Lemma 2.2, we see that is a sub-semigroup of . Let and write where and . For each , choose , so and , for all such that . Choose and let . Define where . Then and . Since , we have . Hence is a regular sub-semigroup of . Now, let be any regular element in . Then for some , so , and thus . Therefore, is the largest regular sub-semigroup of as required.

Note that if then for each we have which implies that    consists of only one element and so is a regular semigroup, and if then which is also a regular semigroup.

3. Green's Relations on

Let be a semigroup. Then we define to be a semigroup of adding an identity to if does not already have an identity element in it and if contains an identity. The following definitions are due to J. A. Green. For any , we define or equivalently; if and only if for some .

Dually, we define or equivalently; if and only if for some .

And we define or equivalently; if and only if for some .

Finally, we define and .

In [2, 3], Clifford and Preston characterized Green's relations on the full transformation semigroup where is an arbitrary set. They proved that Here, we do the same for the semigroup and we obtain results which generalize the same results on

Lemma 3.1. Let . If , then if and only if for some .

Proof. Let be an element of . It is clear that if for some , then . Now, we assume that and write where . For each (since ), we get for some which implies and thus . Choose , so and . Since is the disjoint union of the , we can define Then and .

From now on, the notations denote the set of all elements of which are -related (-related, -related, -related) to where .

Theorem 3.2. For , the following statements hold.
(1)If , then .(2)If , then .

Proof. Let be any element in and let . Then which implies that and for some .
(1) Assume that . If , then and . If , then and both belong to . Thus , and hence . From and , we get by Lemma 3.1. Similarly, from and we get . Therefore, . Now, if and then it is clear by Lemma 3.1 that .
(2) Assume that . If , then . Thus which is a contradiction, so or and .

We note that for any , is an equivalence on and . The relation is usually called the kernel of .

Theorem 3.3. Let . Then if and only if for some . Hence if and only if .

Proof. It is clear that if for some , then . Now, suppose that . If , then for some , so we define by Then is well defined (since ) and . For each , let , so by the definition of . Thus as required, and the remaining assertion is clear.

Lemma 3.4. Let . If then either both and are in , or neither is in .

Proof. Assume that and suppose that is false. So one of or is not in , we suppose that . Thus , so there is such that for all . Thus for all . If , then , so for some which contradicts . Therefore, .

Using Theorem 3.3 and Lemma 3.4, we have the following corollary.

Corollary 3.5. For , the following statements hold.
(1)If , then .(2)If , then .

As a direct consequence of Theorems 3.2 and 3.3, we have the following.

Theorem 3.6. For , the following statements hold.
(1)If , then .(2)If , then .

In [2, 3], volume 1, Clifford and Preston proved that two elements of are -related if and only if they have the same rank (i.e., the ranges of the two elements have the same cardinality). But for we have the following.

Theorem 3.7. For , the following statements hold.
(1)If , then .(2)If , then .

Proof. Let be any element in and let . Then and for some .
(1) If , then since we must have and . From , we get and for some . Since is a right ideal of , so . And . Conversely, assume that and . Then there is a bijection . We let , then and . Since implies , so , hence . Since and , so by Theorem 3.2. Now, since and is injective on we get , so . Therefore, and are -related and .
(2) If , then (since ) and thus which implies that . So by Lemma 3.4 we must have . The other containment is clear since .

In order to characterize the -relation on , the following lemma is needed.

Lemma 3.8. Let . If for some and , then .

Proof. If for some and . Then , which implies that and so . If , then and so . If , then . Thus as required.

Theorem 3.9. Let . Then

Proof. First, assume that . Then and for some If , then and which imply and thus . If or , then we conclude that and for some and . For example, if and , then and imply . By using Lemma 3.8, we get that , so it follows that
Conversely, if , then which implies that since . If , then by applying Lemma 2.7 in [2, 3], to and , we get there are such that , ; and , . From and , we get and , so and for some . And from , we write , so For each , choose and define by Then and . Similarly, from we can prove that for some .
Therefore, and which implies that as required.

Recall that on any semigroup and on ; but in , this is not always true, as shown in the following example.

Example 3.10. Let denote the set of positive integers and let denote the set of all positive even integers. Then we define Hence and , so . Since , we have and are not -related on .
As a consequence of Theorems 3.7 and 3.9, we see that on the sub-semigroup of .

Corollary 3.11. If , then on if and only if on .

Proof. In general, we have . Let and on . Then or If , then . Thus, both cases imply and on by Theorem 3.7.

If we replace with in the above corollary, we then get on . Next, we will consider the case when is a finite subset of .

Theorem 3.12. If is a finite subset of , then on .

Proof. Let be a finite subset of and let be such that and are -related. Then or We note that if , then . For if then is a finite set (since is finite) which implies that and thus . Now, if and then but , so which contradicts Lemma 3.4. Therefore, either both and are in , or neither is in . If , then by Corollary 3.11. If , then which implies that and thus by Theorem 3.7. Therefore, and the other containment is clear.

4. Maximal Inverse Sub-Semigroups on

We first recall that a semigroup is said to be an inverse semigroup if it is regular and any two idempotents commute. In this section, we give one class of maximal inverse sub-semigroups on . If , then there is only one element in , the constant map. Hence, in this case, there is no maximal inverse sub-semigroup on . Therefore, from now on, we assume that .

In 1976, Nichols [4] gave a class of maximal inverse sub-semigroups of . Later in 1978, Reilly [5] generalized Nichols' result. Here, with some mild modifications of the proof given in [4], we get one class of maximal inverse sub-semigroups of which generalizes Nichols' result.

Let be a set and a nonempty subset of . For each , define We see that , since the constant map . To describe maximal inverse sub-semigroups on , we first prove the following.

Theorem 4.1. Let and . Then if and only if and is injective on .

Proof. If , then and , thus if and only if , and is injective on .
Now, we prove for the case . Assume that . So and is injective on . We show that . Let , so there exists such that . Thus since . Hence for some . So, and . By the definition of , we must have . Therefore, and is injective on . Conversely, assume that the conditions hold. Since and , we get and thus . So and therefore . Since is injective on and , it follows that is injective on , and so .

Recall that for each , is an idempotent in if and only if for all . Since is a sub-semigroup of , we conclude that is an idempotent in if and only if for all . And by Theorem 4.1, if is an idempotent in then for all .

Lemma 4.2. Let be a regular sub-semigroup of such that and suppose that . Then
(i) on for some idempotent .(ii)If , then is not -related on to any element in .

Proof. (i) We write where , and define by where . Then , is an identity map on and . So and is injective on . Since , so and . From , we get , so . Thus and it is also an idempotent. From the fact that is a regular sub-semigroup of , it follows from Hall's theorem that where means there exist in such that . Since , we must have by Theorem 3.2 that on .
(ii) Assume that and suppose that on for some . Thus by Theorem 3.3, and hence . Now, let be such that . Then implies (since ), so and since is injective on , we get which implies that is injective on , which contradicts . Therefore, is not -related to any element in .

Theorem 4.3. is a maximal inverse sub-semigroup of .

Proof. First, we prove that is a sub-semigroup of .
Let be elements in . Then , and are injective on and , respectively. Since is a right ideal of , it follows that . Clearly , and is injective on . Therefore, .
Next, we show that is a regular sub-semigroup of . For each , and for all (see Theorem 4.1). Let and write for all , thus where , and define by where and . Since , we get , and so . And for each , for all since Thus , and so . Since and is injective on , it follows that . And, it is clear that .
Now, we prove that any two idempotents in commute, which is enough to show that is an inverse semigroup. Assume that are idempotents in . Then for all and for all . Let be any element in .
Case 1. . Then . So, if , we get and . But if , then , and . Thus in this case .
Case 2. . Then . So by the same proof as given in Case 1, we get .
Therefore, is an inverse sub-semigroup of .

To prove the maximality, we suppose that is properly contained in an inverse sub-semigroup of where and let . Let be the constant map with range , so is an idempotent in , and thus is an idempotent in . Since and every two idempotents in commute, it follows that and . Since is regular, for some and on such that is an idempotent in . Let , then by Lemma 4.2 we must have and for some idempotent . Since every idempotent in a semigroup is a right identity for , we have which is a contradiction since but . Therefore, as required.

As an application of Theorem 4.3, we get the following corollary which first appeared in [4].

Corollary 4.4. and is injective on is a maximal inverse sub-semigroup of .

Proof. By taking in Theorem 4.3, we get and and is injective on which is a maximal inverse sub-semigroup of .

Recall that the number of combinations of distinct things taken at a time written is given by That is, is the number of ways that objects can be chosen from distinct objects.

In the next result, we use the above information to find the number of elements in when is a finite subset of .

Theorem 4.5. Suppose that is an arbitrary set and is a nonempty subset of such that . Then for each , .

Proof. Let and . Then by Theorem 4.1 we see that where , and . If , then can have only one form, the constant map . If has elements, where , then can have choices and for each choice of can have choices, thus there are ways to choose and . Since the restriction of to is a permutation, for each choice of and , the map has possible forms. Hence in this case can have forms.
Therefore, as required.

We observe that the number of elements in depends only on the elements of , and when we replace with in Theorem 4.5, we have the following corollary which first appeared in [4].

Corollary 4.6. If is a finite set with , then the number of elements in and is injective on equals .

Acknowledgments

Most of the work in this paper formed part of an MS thesis written by W. Sommanee under the supervision of J. Sanwong. W. Sommanee greatly appreciates the help of his supervisor in this work; and both authors are very grateful to the referee for his/her careful reading of the original manuscript and helpful comments.

References

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