Groups of prime order are cyclic

Proposition (Prime order implies cyclic). Let $G$ be a finite group with $|G|=p$ where $p$ is prime. Then $G$ is cyclic; more precisely, for every $g\in G$ with $g\neq e$, one has $G=\langle g\rangle$.

Context. This is one of the first applications of Lagrange's theorem : subgroup orders must divide the group order.

Proof sketch. Pick $g\neq e$. The cyclic subgroup $\langle g\rangle$ is a subgroup of $G$, hence its order divides $|G|=p$. Since $g\neq e$, $\langle g\rangle$ is nontrivial, so $|\langle g\rangle|\neq 1$, forcing $|\langle g\rangle|=p$. Therefore $\langle g\rangle=G$.