Proof. Let $G$ be a cyclic group and $g$ be a generator of $G$ Define $\varphi \colon {\mathbb Z} \to G$ by $\varphi (c)=g^c$ Since $\varphi(a+b)=g^{a+b}=g^ag^b=\varphi(a)\varphi(b)$ $\varphi$ is a group homomorphism. If $h \in G$ then there exists $x \in {\mathbb Z}$ such that $h=g^x$ Since $\varphi (x)=g^x=h$ $\varphi$ is surjective.

Note that $\ker \varphi =\{c \in {\mathbb Z}\, : \, \varphi(c)=e_G\}=\{c \in {\mathbb Z}\, : \, g^c=e_G\}$

If $G$ is infinite, then $\ker \varphi = \{0\}$ and $\varphi$ is injective. Hence, $\varphi$ is a group isomorphism, and $G \cong {\mathbb Z}$

If $G$ is finite, then let $|G|=n$ Thus, $|g|=|\langle g \rangle |=|G|=n$ If $g^c=e_G$ then $n$ divides $c$ Therefore, $\ker \varphi =n{\mathbb Z}$ By the first isomorphism theorem, $G \cong \mathbb{Z}/n\mathbb{Z}=\mathbb{Z}_n$

Let $H$ and $K$ be cyclic groups of the same order. If $H$ and $K$ are infinite, then, by the above argument, $H \cong {\mathbb Z}$ and $K \cong {\mathbb Z}$ If $H$ and $K$ are finite of order $n$ then, by the above argument, $H \cong {\mathbb Z}_n$ and $K \cong {\mathbb Z}_n$ In any case, it follows that $H \cong K$