Subset Product with Normal Subgroup is Subgroup

Theorem
Let $G$ be a group whose identity is $e$.

Let:
 * $(1): \quad H$ be a subgroup of $G$
 * $(2): \quad N$ be a normal subgroup of $G$.

Let $H N$ denote subset product.

Then $H N$ and $N H$ are both subgroups of $G$.

Proof
It is clear that $e \in N H$, so $N H \ne \O$.

Suppose $n_1, n_2 \in N$ and $h_1, h_2 \in H$.

Then:

Since $N$ is normal in $G$:
 * $\exists n \in N: n = h_1 n_2 h_1^{-1}$

Thus:

Also:

so from the Two-Step Subgroup Test, $N H$ is a subgroup of $G$.

The fact that $N H = H N$ follows from Subset Product of Subgroups.