Question

# Suppose G is a group and H is a normal subgroup of G. Prove or disprove ass appropirate. If G is cyclic, then G/H is cyclic. Definition: A subgroup H

Abstract algebra
Suppose G is a group and H is a normal subgroup of G. Prove or disprove ass appropirate. If G is cyclic, then $$\displaystyle\frac{{G}}{{H}}$$ is cyclic.
Definition: A subgroup H of a group is said to be a normal subgroup of G it for all $$\displaystyle{a}\in{G}$$, aH = Ha
Definition: Suppose G is group, and H a normal subgruop og G. THe froup consisting of the set $$\displaystyle\frac{{G}}{{H}}$$ with operation defined by (aH)(bH)-(ab)H is called the quotient of G by H.

2021-01-26
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.