Quotient Group of Cyclic Group

Theorem

Let $G$ be a cyclic group which is group generated by $g$.

Let $H$ be a subgroup of $G$.

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

Proof 1

Let $G$ be a cyclic group generated by $g$.

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 $G / H$.

$\blacksquare$

Proof 2

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 Cyclic Group, $g H$ generates $G / H$.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 47 \gamma$