Product of a subgroup and a normal subgroup is a subgroup

Statement

Let $G$ be a group, let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$. Then the product $HN$ is a subgroup of $G$, and $HN = NH$.

Remarks

If $H$ is also normal, then $HN$ is normal in $G$. The proof uses $hN = Nh$ for each $h \in H$, so that $HN = \bigcup_{h \in H} hN = \bigcup_{h \in H} Nh = NH$. This product appears in the second isomorphism theorem.

Explanations

L1 human

The product of two subgroups can fail to be a subgroup, but not when one of them is normal. A normal subgroup lets you move its elements past other elements, at the cost of changing them inside the subgroup. So products of the form $hn$ stay of that form when you multiply them.