Is the Language $L=\{(ab)^{3n}\:|\:n\in\mathbb N\}$ regular?

Question

Is the language regular?

My application of the pumping lemma suggests: splitting it in $xyz$: $$ x = \emptyset \mid y= (ab)^{j} \mid z=(ab)^{3n-j} $$

Pumping up $y$: $$ xyyz = (ab)^{3n+j} \mid (ab)^{3n+j}\notin L $$

Is that correct? Using "The Pumping Game" by Prof. Dr. Edmund Weitz, I couldn't find a solution for this specific language. So I'm not 100% convinced of my solution.

Answer 1

This language is regular:

$L=(ababab)^*$ Where $\{ababab\}$ is trivially regular, and closure of regularity under Kleene star gives us that $L$ is regular.

Alternatively one can construct a regular grammar that generates the language $L$.