Let $G$ be a non-trivial cyclic group with an unique generator.
Show that $G$ has exactly two elements.
Let $g$ be a generator of $G$. My first idea to approach the problem was to show that if $|G|=n > 2$ then there is an integer $k$ between 1 and $n$ such that $\langle g^k \rangle = G$ or, equivalently, $g \in \langle g^k \rangle$.
My second idea was to analyse the problem in 2 separate cases. For the first one, if $n$ is prime, let $k$ be a positive integer then, by Lagrange's Theorem, $|g^k| \mid n$ and, therefore, $\langle g^k \rangle = n$. Hence, we must conclude that, if $n > 2$ the generator cannot be unique.
For the second case, where $n$ is composite, I was hoping to get to a contradiction, but could not advance any further. Is there a way to develop either of these ideas or is it necessary a new approach?