Quotient Group of Infinite Cyclic Group by Subgroup

Theorem
Let $$C_n$$ be the cyclic group of order $n$.

Then $$C_n \cong \frac {\left({\mathbb{Z}, +}\right)} {n \left({\mathbb{Z}, +}\right)} = \frac {\mathbb{Z}} {n \mathbb{Z}}$$.

Proof
Let $$C_n = \left \langle {a: a^n = e_{C_n}} \right \rangle$$, that is, let $$a$$ be a generator of $$C_n$$.

Let us define $$\phi: \left({\mathbb{Z}, +}\right) \to C_n$$ such that $$\forall k \in \mathbb{Z}: \phi \left({k}\right) = a^k$$.

Then from the First Isomorphism Theorem: $$\mathrm{Im} \left({\phi}\right) = C_n = \left({\mathbb{Z}, +}\right) / \mathrm{ker} \left({\phi}\right)$$

We now need to show that $$\mathrm{ker} \left({\phi}\right) = n \mathbb{Z}$$.

We have $$\mathrm{ker} \left({\phi}\right) = \left\{{k \in \mathbb{Z}: a^k = e_{C_n}}\right\}$$.

Let $$x \in \mathrm{ker} \left({\phi}\right)$$. Then $$a^x = e_{C_n}$$ and thus $$n \backslash x$$.

The result follows.