A group of prime order is cyclic
= A group of prime order is cyclic
Let $G$ have <group order> equal to a <prime number> $p$. Choose any $g\ne e$. By <Lagrange's theorem>, the <order of a group element> $g$ divides $p$, and it is greater than one. It is therefore $p$, so the <cyclic subgroup> generated by $g$ is all of $G$. This proves that every nonidentity element generates $G$, and $G$ is isomorphic to the additive group of integers modulo $p$.