Corollary 10.11

Let $G$ be of prime order $p$, and let $a$ be an element of $G$ different from the identity. Then the cyclic subgroup $\langle a \rangle$ of $G$ generated by $a$ has at least two elements, $a$ and $e$. But by Theorem 10.10, the order $m \geq 2$ of $\langle a \rangle$ must divide the prime $p$. Thus we must have $m=p$ and $\langle a \rangle = G$, so $G$ is cyclic.