Quotient Group of Infinite Cyclic Group by Subgroup

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

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

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:
 * $\displaystyle \Z, \frac \Z \Z, \frac \Z {2 \Z}, \frac \Z {3 \Z}, \frac \Z {4 \Z}, \ldots$

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.