Table of Contents Author Guidelines Submit a Manuscript
International Journal of Mathematics and Mathematical Sciences
VolumeΒ 2011, Article IDΒ 615014, 18 pages
http://dx.doi.org/10.1155/2011/615014
Research Article

The Order of Hypersubstitutions of Type (2,1)

Department of Mathematics, Khon Kaen University, Khon Kaen 40002, Thailand

Received 30 December 2010; Accepted 27 March 2011

Academic Editor: H.Β Srivastava

Copyright Β© 2011 Tawhat Changphas and Wonlop Hemvong. 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

Hypersubstitutions are mappings which map operation symbols to terms of the corresponding arities. They were introduced as a way of making precise the concept of a hyperidentity and generalizations to -hyperidentities. A variety in which every identity is satisfied as a hyperidentity is called solid. If every identity is an -hyperidentity for a subset of the set of all hypersubstitutions, the variety is called -solid. There is a Galois connection between monoids of hypersubstitutions and sublattices of the lattice of all varieties of algebras of a given type. Therefore, it is interesting and useful to know how semigroup or monoid properties of monoids of hypersubstitutions transfer under this Galois connection to properties of the corresponding lattices of -solid varieties. In this paper, we study the order of each hypersubstitution of type (2,1), that is, the order of the cyclic subsemigroup of the monoid of all hypersubstitutions of type (2,1) generated by that hypersubstitution.

1. Preliminaries

Let denote the set of all positive integers. Let be a type. Let be a countably infinite alphabet of variables such that the sequence of the operation symbols is disjoint with , and let be an -element alphabet where . Here, is -ary for a natural number . An -ary () term of type is inductively defined as follows: (i)every variable is an -ary term,(ii)if are -ary terms and is an -ary operation symbol, then is an -ary term.

Let be the smallest set containing and being closed under finite application of (ii). The set of all terms of type over the alphabet is defined as the disjoint union .

Any mapping is called a hypersubstitution of type if is an -ary term of type for every . Any hypersubstitution of type can be uniquely extended to a map on as follows: (i) if ,(ii) if .

A binary operation is defined on the set of all hypersubstitutions of type by for all -ary operation symbols . This binary associative operation makes into a monoid, with the identity hypersubstitution which maps every to as an identity element. For a submonoid of an identity of a variety of type () is called an -hyperidentity of if for every hypersubstitution , the equation holds in . A variety is called -solid if every identity of is an -hyperidentity of . If is a submonoid of , then the collection of all -solid varieties of type is a complete sublattice of the lattice of all varieties of type [1].

Let and let be the cyclic subsemigroup of generated by . The order of is defined as the order of the semigroup . If is finite, then the order of the hypersubstition is finite, otherwise the order of is infinite. The hypersubstitution is idempotent if and only if the order of is 1.

The order of a hypersubstitution of type (2) is 1, 2, or infinite [2]. The order of a hypersubstitution of type (3) is 1, 2, 3, or infinite [3]. The order of a hypersubstitution of type (2,2) is 1, 2, 3, 4, or infinite [4]. We are interested in type (2,1). The main result is as follows

Main Theorem. Any hypersubstitution of type has order either infinite or less than or equal to 3.

Throughout this paper, let and be the binary operation symbols and the unary operation symbols of type , respectively. For a binary term and an unary term of type , the hypersubstitution which maps the operation symbol to the term and the operation symbol to the term will be denoted by .

For a binary term , we introduce the following notations:β€”the first variable (from the left) occurring in ,β€”the last variable occurring in ,β€”the set of all variable occurring in ,β€”the total number of all operation symbols occurring in ,β€”the set of all operation symbols occurring in ,β€”the first operation symbol (from the left) occurring in .

For , let denote the left path from the root to the leaf which is labelled by the leftmost variable in and denote the right path from the root to the leaf which is labelled by the rightmost variable in . The operation symbols occurring in and will be denoted by and , respectively. If such that or , we define and

For and of the hypersubstitution , we break our analysis into the following six cases, which cover all possibilities for and : (I) and ,(II) and ,(III) and ,(IV) and ,(V) and ,(VI) and .

In case (I), we have term , and . For we have twenty-four cases, for which the orders can be verified by simple calculations. The case and gives order 3, six other cases give order 2, and the remaining 17 cases give order 1.

2. Case II: and

In this section, we consider the order of a hypersubstitution where and . We consider three subcases of Case II: , , , , , .

The following formula for the operation symbol count of the compound term for some was proved in [5]: where = the number of occurrences of variable in the term  .

Using the facts (see [5]) that for all if is regular, that is, , for all , and the formula above, we obtain the following theorem.

Theorem 2.1. Let , . If , , , , then the order of is infinite.

Proof. Since is regular, by induction we obtain . Then, the order of is infinite.

The following lemmas are easy to prove.

Lemma 2.2. Let , be such that , . If , then is not a variable for all .

Lemma 2.3. Let , . If for some , then for all .

Lemma 2.4. Let , be such that , , and , for some , . Then, the following hold: (i)if and , then the order of is infinite;(ii)if and , then the order of is infinite.

Proof. (i) Assume and . Then, . Let . By Lemma 2.2, . Since , by Lemma 2.3, . Now, This shows that for all . Hence, the order of is infinite.
(ii) Assume and , then . Let . By Lemma 2.2, . Since , by Lemma 2.3   for all . Then, Therefore, for all . Hence we have the claim.

Throughout the rest of this paper, we assume that when is not a variable term it has the form or , for some terms , and that when is not a variable term it has the form or , for some terms , .

In Case (II-2), we have , . We consider the following four subcases: , ; , ; , ; , .

In Case (2.1), we can separate into four subcases: ; , ; , , ; , , .

Theorem 2.5. Let , be such that , , , and , for some , . Then, the following hold: (i)if and satisfy (2.1.1) or (2.1.4), then has infinite order;(ii)if and satisfy (2.1.2) or (2.1.3), then the order of is less than or equal to 3.

Proof. (i) Assume and satisfy (2.1.1). By Lemma 2.2, we have for all . Since , by Lemma 2.3   for all . Therefore, This shows that for all . Hence, has infinite order.
Assume and satisfy (2.1.4). Since , for all . Since , we have for all . Hence, the order of is infinite.
(ii) Assume and satisfy (2.1.2). Since , and , we obtain . This gives Since , and , we have . So, Hence, . This shows that the order of is less than or equal to 2.
Assume and satisfy (2.1.3). Because of (2.1.2), we have , which implies . Since , we have for some with and . It follows that . Since , we obtain Thus, . Hence, the order of is less than or equal to 3.

In Case (2.2), we have and . Since and , and . Then, we obtain the following.

Theorem 2.6. Let , be such that , , , and for some , . Then, has infinite order.

Proof. Assume that , , , , . Then, , . It can be proved, as in the proof of Theorem 2.5(i), that the order of is infinite.

In Case (2.3), we have and . Since , . Then, we obtain the following.

Theorem 2.7. Let , be such that , , , and for some , . Then, has infinite order.

Proof. By , . Since , and Lemma 2.4(ii), we have has infinite order.

In Case (2.4), we have and . Since and , and or . We consider the following three subcases: ; , ; , .

Theorem 2.8. Let , be such that , , , , and , for some , . Then, the following hold: (i)if and satisfy (2.4.1) or (2.4.3), then has infinite order;(ii)if and satisfy (2.4.2), then the order of is less than or equal to 2.

Proof. (i) Assume that and satisfy (2.4.1). By Lemma 2.2, we have for all . Since , by Lemma 2.3   for all . Therefore, This shows that for all . Hence, has infinite order.
Assume that and satisfy (2.4.3). This case can be proved as in (2.4.1).
(ii) Assume that and satisfy (2.4.2). Then, and . We have and . Hence, the order of is less than or equal to 3.

In Case (II-3), we have , . We consider the following four subcases: , ; , ; , ; , .

In Case (3.1), we can separate into four subcases: ; , ; , , ; ,, .

Theorem 2.9. Let , be such that , , , , and , for some , . Then, the following hold: (i)if and satisfy (3.1.1) or (3.1.4), then has infinite order;(ii)if and satisfy (3.1.2) or (2.1.3), then the order of is less than or equal to 2.

Proof. (i) Assume that and satisfy (3.1.1). By Lemma 2.2, we have for all   in . Since , by Lemma 2.3   for all . Therefore, This shows that for all . Hence, has infinite order.
Assume that and satisfy (3.1.4). Then, for all . Since and , we have This implies that for all . Hence, the order of is infinite.
(ii) Assume that and satisfy (3.1.2). Then, we get , or , . In the first, the order of is equal to 2. For the latter, is idempotent.
Assume that and satisfy (3.1.3). Then, , , which can be proved the same way as the first case of (ii).

In Case (3.2), we have and . Since and , and . Then, we obtain the following.

Theorem 2.10. Let , be such that , , , and for some , . Then has infinite order.

Proof. Since and , we have Hence, the order of is infinite.

In Case (3.3), we have and . Since , . Then, we obtain the following.

Theorem 2.11. Let , be such that , , , and for some , . Then has infinite order.

Proof. Since    and Lemma 2.4(ii), we have has infinite order.

In Case (3.4), we have and . Since and , and . We have the following result.

Theorem 2.12. Let , be such that , , , , and , for some , . Then, has infinite order.

Proof. Since , . We have, by Lemma 2.2, for all  in . Since , by Lemma 2.3   for all . We obtain Hence, has infinite order.

3. Case III: and

We consider three subcases: , ; , ; , .

In Case (III-1), we separate into the following subcases: (1);(2);(3);(4).

We have the following results. Subcases (2) and (4) give infinite order, by Lemma 2.4(ii). The next proposition deal with subcases (1) and (3).

Proposition 3.1. Let for some be such that and . Then, the following hold: (i)if , then the order of is equal to 1 or is infinite;(ii)if , then the order of is less than or equal to 3 or infinite.

Proof. (i) Assume that . Then, . We consider the following.
(iβ€²) If , then . We get that is idempotent.
(iβ€²β€²) If , suppose that . Let . Then .
Now, A similar argument works for . This shows that the order of is infinite.
(ii) If , then and . This gives . Then, the order of is less than or equal to 3. If , then it can be proved as in (iβ€²β€²) that the order is infinite.

In Case (III-2), we have , . We consider the following subcases: , ; ;, ;, ; , , then ;, , then ;, ;;.

Proposition 3.2. Let , , , and . Then, the following hold: (i)if satisfies or , then the order of is less than or equal to 2;(ii)if satisfies , then the order of is infinite.

Proof. (i) If and satisfy (3.2.1.1), then . Assume that and satisfy (3.2.1.3). Since and , we have . Clearly, . Then, . Hence, the order of is less than or equal to 2.
(ii) Assume that and satisfy (3.2.1.2). Since , and , it follows that for some . For , we have Hence, the order of is infinite.

Proposition 3.3. Let or , , , and . Then, the order of is infinite.

Proof. Assume that . Since , , . For any natural number , makes have infinite order.
Assume that . Since , , . For , a similar proof to those above gives . Hence, the order of is infinite.

Proposition 3.4. Let , , and . Then the following hold: (i)if satisfies , then the order of is equal to 2;(ii)if satisfies , then the order of is infinite.

Proof. (i) If and satisfy (3.2.4.1), then and . Hence, the order of is equal to 2.
(ii) Assume that and satisfy (3.2.4.2). Since , , it follows that . For . As in the proof of Proposition 3.3, we have . Hence, the order of is infinite.

In Case (III-3), we have , . We consider the following subcases: , ;;; , ..;;.

Proposition 3.5. Let , , , and . Then, the following hold: (i)if satisfies , then the order of is equal to 2;(ii)if satisfies , then the order of is infinite.

Proof. (i) If and satisfy , then and . Hence the order of is equal to 2.
(ii) This can be proved in the same way as Proposition 3.2 (ii).

Proposition 3.6. Let or , , , and . Then, the order of is infinite.

Proof. Assume that . Since , , . The same argument works as in Proposition 3.3. Hence, the order of is infinite. The case that can be proved similarly.

Proposition 3.7. Let , , , and . Then, the following hold: (i)if satisfies , then the order of is equal to 2;(ii)if satisfies , then the order of is infinite.

Proof. (i) If and satisfy (3.3.4.1), then and . Hence, the order of is equal to 2.
(ii) Assume and satisfy (3.3.4.2), Since , , . By Lemma 2.2, we have . Then, This shows that, for all . Consequently, for all . Then, order of is infinite.

4. Case IV: and

We consider three subcases: , ; , ;, .

In Case (IV-1), we separate into four cases: , ; , ; , ; , .

Theorem 4.1. Let , be such that , , , . Then, the following hold: (i)if satisfies or or , then the order of is infinite;(ii)if satisfies , then the order of is equal to 1 or is infinite.

Proof. (i) If and satisfy , then we let for some . Since , we may assume that . By Lemma 2.2, we have . For any natural number , makes have infinite order. Cases and can be proved similarly.
(ii) Assume that and satisfy . If , then and . Hence, is idempotent. The case that can be proved as in (i).

In Case (IV-2), we have , . If for some term , then means that is not variable.

Proposition 4.2. Let or , , for some and . Then, the order of is infinite.

Proof. Assume , for some and . Since . By Lemma 2.2 we have . It can be proved, as in the proof of Theorem 4.1(i), that the order of is infinite.

In Case (IV-3), we have , . If , then . We consider the following cases: , ; , .

Proposition 4.3. Let , , and , . If and satisfy , then the order of is infinite.

Proof. The case (4.3.1) can be verified as in Proposition 4.2.

Proposition 4.4. Let , , and , . If and satisfy , then the order of is equal to 1 or is infinite.

Proof. Assume that and satisfy . If , , and , then and . Hence, is idempotent. Assume that . This case can be proved as in Proposition 4.3.

5. Case V: and

In this case, we have or , . We consider three subcases: , ; , ; , .

In Case (V-1), we separate into the following subcases: , , ; , , ; , , ;