Quotient Group of Cyclic Group

Theorem

Let $G$ be a cyclic group which is 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 $G = \gen g$.

Then by Homomorphism of Powers for Integers:

$\forall n \in \Z: \map {q_H} {g^n} = \paren {\map {q_H} g}^n = \paren {g H}^n$

As $G = \set {g^n: n \in \Z}$, we conclude that:

$G / H = q_H \sqbrk G = \set {\paren {g H}^n: n \in \Z}$

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

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \gamma$