Let G be a group, and H be a normal subgroup of G. Assume that G is cyclic.

To prove: \(\displaystyle\frac{{G}}{{H}}\) is cyclic.

Proof: As H is normal and G is cyclic, then G is Abelian.

Let us suppose,

\(\displaystyle{G}={\left\langle{a}\right\rangle}={\left\lbrace{a}^{{i}}{\mid}{i}\in\mathbb{Z}\right\rbrace}\)

By definition of \(\displaystyle\frac{{G}}{{H}}\), we have

\(\displaystyle\frac{{G}}{{H}}={\left\lbrace{x}{H}{\mid}{x}\in{G}\right\rbrace}\)

Use the hypothesis that G is cyclic, hence it gives

\(\displaystyle\frac{{G}}{{H}}={\left\lbrace{x}{H}{\mid}{x}\in{G}\right\rbrace}={\left\lbrace{a}^{{i}}{H}{\mid}{a}^{{i}}\in{G}\right\rbrace}={\lbrace}{a}{H}{)}^{{i}}{\mid}{a}^{{i}}\in{G}\rbrace={\left\langle{a}{H}\right\rangle}\)

Hence, it is proved that \(\displaystyle\frac{{G}}{{H}}\) is cyclic.

To prove: \(\displaystyle\frac{{G}}{{H}}\) is cyclic.

Proof: As H is normal and G is cyclic, then G is Abelian.

Let us suppose,

\(\displaystyle{G}={\left\langle{a}\right\rangle}={\left\lbrace{a}^{{i}}{\mid}{i}\in\mathbb{Z}\right\rbrace}\)

By definition of \(\displaystyle\frac{{G}}{{H}}\), we have

\(\displaystyle\frac{{G}}{{H}}={\left\lbrace{x}{H}{\mid}{x}\in{G}\right\rbrace}\)

Use the hypothesis that G is cyclic, hence it gives

\(\displaystyle\frac{{G}}{{H}}={\left\lbrace{x}{H}{\mid}{x}\in{G}\right\rbrace}={\left\lbrace{a}^{{i}}{H}{\mid}{a}^{{i}}\in{G}\right\rbrace}={\lbrace}{a}{H}{)}^{{i}}{\mid}{a}^{{i}}\in{G}\rbrace={\left\langle{a}{H}\right\rangle}\)

Hence, it is proved that \(\displaystyle\frac{{G}}{{H}}\) is cyclic.