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, we have that $N H = H N \le G$.

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 $.