If language L* (Kleene Star) is regular, does it imply that L is also regular?
2026-04-25 22:21:41.1777155701
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L* regular -> L regular?
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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).
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.