 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. Bella

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,
$G=⟨a⟩=\left\{{a}^{i}\mid i\in \mathbb{Z}\right\}$
By definition of $\frac{G}{H}$, we have
$\frac{G}{H}=\left\{xH\mid x\in G\right\}$
Use the hypothesis that G is cyclic, hence it gives
$\frac{G}{H}=\left\{xH\mid x\in G\right\}=\left\{{a}^{i}H\mid {a}^{i}\in G\right\}=\left\{aH{\right)}^{i}\mid {a}^{i}\in G\right\}=⟨aH⟩$
Hence, it is proved that $\frac{G}{H}$ is cyclic.

Do you have a similar question?