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.

Problem.
Let $H$ and $K$ be normal subgroups of a group $G$. Suppose that $H < K$ and the quotient group $G/H$ is abelian. Then prove that $G/K$ is also an abelian group.

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Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.
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(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]

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Let $K, N$ be normal subgroups of a group $G$. Suppose that the quotient groups $G/K$ and $G/N$ are both abelian groups. Then show that the group
\[G/(K \cap N)\]
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Hint.
We use the following fact to prove the problem.
Lemma: For a […]

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Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.
Let $N$ be a subgroup of $G$.
Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.
Definitions.
Recall that for any $a, b \in G$, the […]

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Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.
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We give a […]

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Show that any subgroup of index $2$ in a group is a normal subgroup.
Hint.
Left (right) cosets partition the group into disjoint sets.
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 […]

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Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

If Quotient $G/H$ is Abelian Group and $H < K \triangleleft G$, then $G/K$ is Abelian
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Then prove that $G/K$ is also an abelian group.
Solution.
We will give two proofs.
Hint (The third isomorphism theorem)
Recall the third […]

[…] that [G/K cong (G/H)/(G/K).] Since the group $G/H$ is abelian by assumption, and in general a quotient group of an abelian group is abelian, it follows $(G/H)/(G/K)$ is an abelian […]

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[…] that [G/K cong (G/H)/(G/K).] Since the group $G/H$ is abelian by assumption, and in general a quotient group of an abelian group is abelian, it follows $(G/H)/(G/K)$ is an abelian […]