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:
 * $$\forall n \in \mathbb{Z}: q_H \left({g^n}\right) = \left({q_H \left({g}\right)}\right)^n = \left({g H}\right)^n$$

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 $$G / H$$.