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


 * Now we need to prove that $$N \triangleleft N H$$.