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.