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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Hint.
Left (right) cosets partition the group into disjoint sets.
Consider both left and right cosets.
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Suppose that $H < K$ and the quotient group $G/H$ is abelian.
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Hint (The third isomorphism theorem)
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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 […]

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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 […]