proof that all cyclic groups of the same order are isomorphic to each other

The following is a proof that all cyclic groups of the same order are isomorphic to each other.

Proof.

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

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

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

If $G$ is finite, then let $|G|=n$. Thus, $|g|=|⟨g⟩|=|G|=n$. If $g^c=e_G$, then $n$ divides $c$. Therefore, $\ker \phi =n\mathbb{Z}$. 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 , $H\cong \mathbb{Z}$ and $K\cong \mathbb{Z}$. If $H$ and $K$ are finite of order $n$, then, by the above , $H\cong {\mathbb{Z}}_n$ and $K\cong {\mathbb{Z}}_n$. In any case, it follows that $H\cong K$.∎