Cyclic groups of the same order are isomorphic

Statement

Let $G$ and $H$ be cyclic groups of the same order, finite or infinite. Then $G$ and $H$ are isomorphic: there is a bijection $\varphi \colon G \to H$ with $\varphi(ab) = \varphi(a) \varphi(b)$ for all $a, b \in G$.

Remarks

If $g$ generates $G$ and $h$ generates $H$, an isomorphism is $\varphi(g^k) = h^k$ for $k \in \mathbb{Z}$. This map is well defined because $g^a = g^b$ exactly when $h^a = h^b$, by the order of a cyclic subgroup. Thus each cyclic group of order $n$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$, and each infinite cyclic group is isomorphic to $\mathbb{Z}$.

Explanations

L1 human

Up to the names of the elements, there is only one cyclic group of each size. To match two of them, send the generator of one to the generator of the other, and send each power to the same power. For example, the rotations of a hexagon and the integers modulo $6$ are the same group in two costumes.