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 $N H \ne \varnothing$.


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

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