nagasenaz

2021-01-25

Suppose G is a group and H is a normal subgroup of G. Prove or disprove ass appropirate. If G is cyclic, then $\frac{G}{H}$ is cyclic.

Definition: A subgroup H of a group is said to be a normal subgroup of G it for all$a\in G$ , aH = Ha

Definition: Suppose G is group, and H a normal subgruop og G. THe froup consisting of the set$\frac{G}{H}$ with operation defined by (aH)(bH)-(ab)H is called the quotient of G by H.

Definition: A subgroup H of a group is said to be a normal subgroup of G it for all

Definition: Suppose G is group, and H a normal subgruop og G. THe froup consisting of the set

Bella

Skilled2021-01-26Added 81 answers

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

To prove:$\frac{G}{H}$ is cyclic.

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

Let us suppose,

$G=\u27e8a\u27e9=\{{a}^{i}\mid i\in \mathbb{Z}\}$

By definition of$\frac{G}{H}$ , we have

$\frac{G}{H}=\{xH\mid x\in G\}$

Use the hypothesis that G is cyclic, hence it gives

$\frac{G}{H}=\{xH\mid x\in G\}=\{{a}^{i}H\mid {a}^{i}\in G\}=\{aH{)}^{i}\mid {a}^{i}\in G\}=\u27e8aH\u27e9$

Hence, it is proved that$\frac{G}{H}$ is cyclic.

To prove:

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

Let us suppose,

By definition of

Use the hypothesis that G is cyclic, hence it gives

Hence, it is proved that