Subset Product with Normal Subgroup as Generator

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

Let:
 * $H$ be a subgroup of $G$
 * $N$ be a normal subgroup of $G$.

Then:
 * $N \triangleleft \left \langle {N, H} \right \rangle = N H = H N \le G$.

where:
 * $\le$ denotes subgroup
 * $\triangleleft$ denotes normal subgroup
 * $\left \langle {N, H} \right \rangle$ denotes a group generator
 * $N H$ denotes subset product.

Proof
From Subset Product is Subset of Generator:
 * $N H \subseteq \left \langle {N, H} \right \rangle$

From Subgroup Product with Normal Subgroup is Subgroup:
 * $N H = H N \le G$

Then by the definition of a group generator, $\left \langle {N, H} \right \rangle$ is the smallest subgroup containing $N H$ and so:
 * $\left \langle {N, H} \right \rangle = N H = H N \le G$

From Normal Subgroup of Subgroup Product we have that:
 * $N \triangleleft N H$