Quotient Group of Cyclic Group

Theorem
Every quotient group of a cyclic group is cyclic.

Let $H$ be a subgroup of a cyclic group $G$ group generated by $g$.

Then $g H$ generates $G / H$.

Proof
Let $G$ be a cyclic group generated by $g$ and let $H \le G$.

We need to show that every element of $G / H$ is of the form $\left({g H}\right)^k$ for some $k \in \Z$.


 * Suppose $x H \in G / H$.

Then, since $G$ is generated by $g$, $x = g^k$ for some $k \in \Z$.

But $\left({g H}\right)^k = \left({g^k}\right) H = x H$.

So $g H$ generates $G / H$.


 * Alternatively, we take this approach:

Let $H$ be a subgroup of the cyclic group $G = \left \langle {g} \right \rangle$.

Then by Homomorphism of Powers for Integers:
 * $\forall n \in \Z: q_H \left({g^n}\right) = \left({q_H \left({g}\right)}\right)^n = \left({g H}\right)^n$

As $G = \left\{{g^n: n \in \Z}\right\}$, we conclude that:
 * $G / H = q_H \left({G}\right) = \left\{{\left({g H}\right)^n: n \in \Z}\right\}$

Thus, by Epimorphism from Integers to a Cyclic Group, $g H$ generates $G / H$.