${\{a\}^{\ast}}^2$ is not ${\{a\}^{\ast}}$

Question

I'd like to understand why the language $$(1)\ \ \ ({\{a\}^{\ast}})^2$$ is not the same as $$(2)\ \ \ {\{a,b\}^{\ast}}$$ I even don't know what ${\ }^2$ means here. And why the first language is something like $$\mathbb{( N,+)\times (N,+) }$$ EDIT I have omitted $b$ in the formula (2)!

Answer 1

Since you have the tag monoid, my guess is that in your question, the two occurrences of the word language should be changed to monoid.

Then your question makes much more sense: the notation $(\{a\}^*)^2$ stands for the monoid $\{a\}^* \times \{a\}^*$, which is indeed isomorphic to the commutative monoid $(\Bbb{N}, +) \times (\Bbb{N}, +)$, but not to the non-commutative monoid $\{a,b\}^*$.

Just for the record, if you really mean languages, then I would interpret $(\{a\}^*)^2$ as the concatenation product $a^*a^*$, which is equal to $a^*$.