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 $\mathbb Z/n\mathbb Z$.