Proof that groups of prime order are cyclic

Proof

1. Let $G$ be a group of order $p$, where $p$ is a prime number.
2. Since $p \ge 2$, $G$ contains an element $g$ that is not the identity $e$.
3. The order of $g$ divides $p$. Thus the order of $g$ is $1$ or $p$.
4. Only the identity has order $1$, and $g \ne e$. Thus $g$ has order $p$.
5. Thus the cyclic subgroup $\langle g \rangle$ has $p$ elements, so $\langle g \rangle = G$.
6. Thus $G$ is cyclic, with generator $g$.

Idea

A prime has only two divisors. An element that is not the identity cannot have order one, so its order is the full order of the group.