The question:
Let $\Sigma$ be an alphabet and $L\subseteq \Sigma^*$ be a non-empty language over that alphabet. Prove that if $L=L^2$ then $e \in L$ (where $e$ represents the empty string).
My Solution Sketch:
I first assume $L = L^2$. That is, every element in $L$ is also in $L^2$ and vice versa.
Because $e$ concatenated with itself ($ee$) is just $e$, then the only way for $L = L^2$ to hold is if $e \in L$. Otherwise, not every element in $L$ would be in $L^2$ and vice versa.
My concern:
It's only a sketch of a proof but I feel like it's very handwavy and not very rigorous. It is my first foray into automata so if anyone could provide some tips on how to formalize this argument and add more rigor to it, it would be very appreciated!
Unfortunately, the answer is wrong. Indeed, if $L = \emptyset$, then $L = L^2$, but $L$ does not contain the empty word. The correct statement should assume that $L$ is nonempty.