Subset Product with Normal Subgroup is Subgroup

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

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