The following is a proof that every group of prime order is cyclic.

Let $p$ be a prime and $G$ be a group such that $|G|=p$ Then $G$ contains more than one element. Let $g \in G$ such that $g \ne e_G$ Then $\langle g \rangle$ contains more than one element. Since $\langle g \rangle \le G$ by Lagrange's theorem, $|\langle g \rangle |$ divides $p$ Since $|\langle g \rangle |>1$ and $|\langle g \rangle |$ divides a prime, $|\langle g \rangle |=p=|G|$ Hence, $\langle g \rangle =G$ It follows that $G$ is cyclic.