# L* is in P but L is not?

I am trying to show that $L^* \in P$ does not necessarily mean that $L \in P$. My idea was to find a language $L \notin P$ , apply $*$ operator and show that $L^* \in P$ by showing that $L^*$ can be decided with a polynomially bounded Turing machine, which will be a contradiction to $L^* \in P \Rightarrow L \in P$. However, I am having trouble in finding such $L$. Is there a trivial $L$ that I'm not thinking of? Or should change my method? Any help will be appreciated.

• The method is ok. A hint: what about unary languages? What happens if $\{1\}$ is included in an unary language? – Vor May 17 '17 at 12:00

Yes, your method is fine and, yes, there's a fairly trivial way to complete your proof attempt: it's easy to arrange that $L^*=\Sigma^*$, which is in P.