Using closure properties, prove that $L=\{a^kb^ra^m|k,r,m\ge0 \text{ and } m=k+r\}$ is not regular

Question

I'm trying to prove that $L=\{a^kb^ra^m|k,r,m\ge0 \text{ and } m=k+r\}$ is not regular and, although it's trivial to prove it via pumping lemma, I'm having troubles trying to find a way to prove it via closure properties (for example a union of $L$ and another regular language which may give a non-regular language).

It simply doesn't occur to me any language to use alongside $L$ (I guess I lack some imagination...)

Answer 1

Consider the intersection of your language $L$ with the regular language $L' = \{b^r a^m \mid r,m \ge 0 \}$. Then $L \cap L'$ contains all the words of the form $a^k b^r a^m$ where $k=0$ and $m=k+r=r$. That is, $L \cap L' = \{ b^r a^r \mid r \ge 0 \}$, which is a well-known non-regular language.