Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$.
Then prove that the quotient group $G/N$ is also an abelian group.

Proof.

Each element of $G/N$ is a coset $aN$ for some $a\in G$.
Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in G$.

Then we have
\begin{align*}
(aN)(bN)&=(ab)N \\
&=(ba)N && \text{since $G$ is abelian}\\
&=(bN)(aN).
\end{align*}
Here the first and the third equality is the definition of the group operation of $G/N$.

Remark

Since $N$ is a normal subgroup of $G$, the set of left cosets $G/H$ becomes a group with group operation
\[(aN)(bN)=(ab)N\] for any $a, b\in G$.

Related Question.

As an application, try the following problem.

The proof of this problem is given in the post ↴
If quotient $G/H$ is abelian group and $H < K \triangleleft G$, then $G/K$ is abelian.

The post Quotient Group of Abelian Group is Abelian first appeared on Problems in Mathematics.

Show that any subgroup of index $2$ in a group is a normal subgroup.

Hint.

1. Left (right) cosets partition the group into disjoint sets.
2. Consider both left and right cosets.

Proof.

Let $H$ be a subgroup of index $2$ in a group $G$.
Let $e \in G$ be the identity element of $G$.

To prove that $H$ is a normal subgroup, we want to show that for any $g\in G$, $gH=Hg$.
If $g \in H$, then this is true. So we assume that $g \not \in H$.

Note that left cosets partition $G$ into two disjoint sets since the index is $2$.
Since $g \not \in H$, these are $eH$ and $gH$. (If $gH=H$, then $g \in H$.)

Similarly right cosets partition $G$ into two disjoint sets.
These disjoint right cosets are $He$ and $Hg$.

Because of these partitions, we have as sets
\[gH=G – eH=G-H=G-He=Hg.\] Therefore $H$ is a normal subgroup in $G$.

The post Any Subgroup of Index 2 in a Finite Group is Normal first appeared on Problems in Mathematics.