Subset Product of Normal Subgroups is Normal

Theorem
Let $\struct {G, \circ}$ be a group.

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

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

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

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

Let $g \in G$.

From Subgroup is Normal iff Contains Conjugate Elements:
 * $g n g^{-1}\in N, g n' g^{-1}\in N'$

So:
 * $\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n n' g^{-1} \in N N'$

So $N N'$ is normal.