Quotient Group of Infinite Cyclic Group by Subgroup

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

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

where:
 * $$\Z$$ is the Additive Group of Integers;
 * $$n \Z$$ is the Additive Group of Integer Multiples;
 * $$\Z / n \Z$$ is the quotient group of $$\Z$$ by $$n \Z$$.

Thus, every cyclic group is isomorphic to one of:
 * $$\Z, \frac {\Z} {\Z}, \frac {\Z} {2 \Z}, \frac {\Z} {3 \Z}, \frac {\Z} {4 \Z}, \ldots$$

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

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

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

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

Let $$x \in \ker \left({\phi}\right)$$.

Then $$a^x = e_{C_n}$$ and thus $$n \backslash x$$.

The result follows.