Subset Product with Normal Subgroup as Generator

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

Let $$H \le G, N \triangleleft G$$.

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

Proof

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

So we need to prove that $$N H \le G$$, and then by the definition of a Group Generator, $$\left \langle {N, H} \right \rangle$$ will be the smallest subgroup containing $$N H$$ and the result will follow.


 * 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.


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