# If $L_1L_2$ is regular language, is $L_2L_1$ regular as well?

If $$L_1L_2$$ is regular language, is $$L_2L_1$$ regular as well?

I think the answer is no, but I'm not sure how to contradict.

• Perhaps there's a simple example where $L1$ is a subset of the words of the form $a^n$ and $L2$ is a subset of words of the form $a^nb$. This way the sequences of $a$s get merged in the first case but are separated in the second. Dec 18, 2022 at 21:53

Let $$L_1 = a^*$$ and $$L_2 = \{a^{n^2}b \mid n \geqslant 0\}$$. Then $$L_1L_2 = a^*b$$ is regular, but $$L_2L_1= \{a^{n^2}ba^m \mid n,m \geqslant 0\}$$ is not regular since its intersection with the regular language $$a^*b$$ is equal to $$L_2$$, which is not regular.
Let $$L_1 = \{a^ib^j | i \ge j \ge 0\}$$ and $$L_2 = \{b^k c | k \ge 0\}$$.
Now $$L_1L_2$$ is $$\{a^i b^{j+k} c | i \ge j, j \ge 0, k \ge 0\}$$, in which there is no relationship between $$i$$ and $$j+k$$, so that's equivalent to $$\{a^ib^jc | i, j \ge 0\}$$, which is the regular language $$a^*b^*c$$.
But $$L_2L_1$$ is $$\{b^k c a^ib^j | i \ge j \ge 0, k \ge 0\}$$ which is not regular (as can easily be seen by intersecting it with $$ca^*b^*$$).