# L* regular -> L regular?

If language L* (Kleene Star) is regular, does it imply that L is also regular?

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No. Pick a nonregular language L over an alphabet A that contains every letter of A as one-letter words. (for example, A = {0,1}, L = {w such that $|w|_0 - |w|_1 = \pm 1$}).

Then L* = A* so L* is regular, but L is not.

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Generalizing, you can even pick $L$ non-recursive. – Yuval Filmus Feb 27 '11 at 20:49

A nice exercise is to show that, for any $L \subseteq \{0\}^*$, $L^{*}$ is regular.

(Picking $SQR = \{0^{n^2}\ |\ n \in \mathbb{N}\}$ provides a counterexample to your statement. That $SQR$ is not regular, can be proved using the Pumping Lemma.)

Spoiler:

This follows from the Frobenius problem for two coins (which was discoverd by J.J Sylvester).

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