Subset Product of Normal Subgroups is Normal

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $N$ and $N'$ be normal subgroups of $G$.

Then $N N'$ is also a normal subgroup of $G$.

Proof
From Subgroup Product with Normal Subgroup is Subgroup‎, we already have that $N N'$ is a subgroup of $G$.

Let $n \in N, n' \in N'$.

Let $g \in G$.

Then from Normal Subgroup Equivalent Definitions, $g n g^{-1}\in N, g n' g^{-1}\in N'$.

So $\left({g n g^{-1}}\right) \left({g n' g^{-1}}\right) = g n n' g^{-1} \in N N'$.

So $N N'$ is normal.