Finite cyclic group is isomorphic to ℤ/nℤ

Proposition (Finite cyclic groups). Let $G$ be a group . Suppose $G$ is cyclic of finite order $n$, i.e. $G=\langle g\rangle$ and $|G|=n$. Then $G$ is isomorphic to the additive group $\mathbb Z/n\mathbb Z$. Concretely, the map

is a well-defined isomorphism.

Context. This identifies finite cyclic groups up to unique isomorphism by their order. Many computations about cyclic groups can therefore be reduced to modular arithmetic in $\mathbb Z/n\mathbb Z$.

Proof sketch. Well-definedness uses $g^{k+n}=g^k$ since $g^n=e$. The map is a homomorphism because $\varphi(\overline{k+\ell})=g^{k+\ell}=g^kg^\ell$. It is surjective because $G=\langle g\rangle$, and injective because the kernel is exactly $\overline{0}$ when $g$ has order $n$.