Question

If $\mathit{N}$ is a subgroup of an abelian group $\mathit{G}$, prove that $\mathit{G}\mathbf{/}\mathit{N}$ is abelian.

Short answer

The group $G/N$ is abelian.

Step-by-step solution

Normal subgroup and Quotient group 

Let $N$ be the normal subgroupof role="math" localid="1654500823236" $G$. Then

1. $G/N$ is a group under the operation defined by $\left(Na\right)\left(Nc\right)=Nac$.

2. If $G$ is finite, then the order of role="math" localid="1654500875043" $G/N$ is role="math" localid="1654500878145" $\left|G\right|/\left|N\right|$.

3. If $G$ is an abelian group, then so is role="math" localid="1654500881538" $G/N$.

The group $G/N$ is called the quotient group or factor group of $G$ by $N$.

G/N is abelian

Let $N$ be a subgroup of an abelian group $G$.

Since $G$is abelian, all the subgroups of G are normal. Let $gN,hN\in G/N$; then

$$\begin{gathered} \left(gN\right)\left(hN\right)=ghN \\ \begin{array}{c}=hgN\\ =\left(hN\right)\left(gN\right)\end{array} \end{gathered}$$

Therefore, $G/N$ is abelian.