Regular Language and Unknown Operation

Question

I read the $L$ is a Regular language. Define $L'$ as following:

$$L'=\{a_2a_1a_4a_3\ldots a_{2n}a_{2n-1}\mid a_1a_2\ldots a_{2n} \in L\}.$$

why $L'$ is Regular?

any hint or idea would be highly appriciated.

Answer 1

Let $A$ be the alphabet and let $B = \{[a_1a_2] \mid a_1, a_2 \in A \} $. Let $\alpha, \beta: B^* \to A^*$ be the monoid homomorphisms defined by $\alpha([a_1a_2]) = a_1a_2$ and $\beta([a_1a_2]) = a_2a_1$, for each letter $[a_1a_2]$ of $B$. Now $L' = \beta(\alpha^{-1}(L))$. Since regular languages are closed under homomorphisms and inverses of homomorphisms, $L'$ is regular.

Note. The same proof shows that if $L$ is context-free, then $L'$ is context-free.