(related to Theorem: Classification of Cyclic Groups)

- By hypothesis, $(G,\ast)$ is a cyclic group with a generator $g\in G$ with $G=\langle g\rangle.$
- Define the function $f:(\mathbb Z, + )\to (G,\ast),$ by $f(k):=a^k$ for all $a\in (G,\ast)$ and all elements $k$ of the additive group of integers $k\in (\mathbb Z,+).$
- $f$ is a group homomorphism, since $f(k+j)=a^{k+j}=a^k\ast a^j=f(k)\ast f(j).$
- In particular, the image $\operatorname{im}(f)$ equals the group $G$, since $\operatorname{im}(f)=\{a^k\mid k\in\mathbb Z\}=\langle g\rangle=G.$
- On the other hand, the kernel $\ker(f)=\{a\in G\mid f(a)=e_G\}$ is a subgroup of $(Z,+)$.
- Let $G$ have an infinite order $|G|=\infty.$
- Then $\ker(f)=\{0\}.$
- It follows from the isomorphism theorem for groups that $(G,\ast)$ is isomorphic to $\mathbb Z/\{0\}.$
- But $\mathbb Z/\{0\}=\mathbb Z.$
- Therefore, $(G,\ast)$ is isomorphic to $(\mathbb Z, + ).$

- Let $G$ have an finite order $|G|=n$ with $n > 0.$
- Note that all additive subgroups of integers have the form $(\mathbb Z_n, + )$ for a natural number $n\in\mathbb N.$
- In particular, $\ker(f)$ has this form.
- It follows from the isomorphism theorem for groups that $(G,\ast)$ is isomorphic to $(\mathbb Z/\ker(f), + ) = (\mathbb Z_n, + ).$∎