Whether $L=\{(a^m,a^n)\}^*$ is regular or not? I am condidering the automatic structure for Baumslag-Solitar semigroups. And I have a question. For any $m,n \in Z$, whether the set $L=\{(a^m,a^n)\}^*$ is regular or not. Here a set is regular means it can be recognized by a finite automaton.
Since the operations:union, intersection, complement, concatenation and Kleene star for the regular sets are closed (see here), I have tried to represent $L$ to be the result of some sets under the operations mentioned above. If it is successfully represented, $L$ will be regular. But I failed. So I want to ask for some clues for this question.
Thanks for your assistance. 

As Boris has pointed out in the comments, so far your question doesn't make complete sense, because you haven't specified a finite alphabet $\Sigma$ such that $L\subseteq \Sigma^*$.
What I think you probably want is to consider $L$ as a language over $\Sigma = (A\cap\{\epsilon\})\times(A\cap\{\epsilon\})$, where we are simplifying strings over $\Sigma$ and writing them as pairs of strings over $A$, for example $(a,a)(a,\epsilon) = (a^2,a)$.
If this is what you mean, then for a given $m,n\in \mathbb{N}$, $L = \{(a^m,a^n)\}^*$ is indeed a regular language over $\Sigma$, which is obvious, since $(a^m,a^n)$ is just (the shorthand form of) a particular word over $\Sigma$, and so $(a^m,a^n)^*$ is a regular expression.
