Every Group of Prime Order Cyclic

Proof

Let $G$ be a finite group of prime order. Then, for any element $g\in G$, $⟨g⟩\le G$ and hence by Lagrange's theorem

Given that $G$ is prime, the only possible values for $|⟨g⟩|$ are $1$ or $|G|$. If $|⟨g⟩|=1$, then $g=\mathrm{id}$. In this case we select a different element $g^{\prime }$, which must exist since $G$ is of prime order and so has at least two elements, and the identity is unique. This element therefore must satisfy $|⟨g^{\prime }⟩|=|G|$. As a subgroup of equivalent order, $⟨g^{\prime }⟩=G$.

A direct consequence of this is the fact that every group of prime order is abelian.