Let $G$ be a group. If $H\leq G$ is a subgroup and $N\vartriangleleft G$, then $HN$ is a subgroup of $G$. Let $H$ be a subgroup of $G$ and $N$ a normal subgroup of $G$. Prove that $HN=\{hn\:|\:h \in H, n \in N\}$ is a subgroup.
Here what I have so far. It is not really much I understand what I need to show but I am stuck in one little thing which I can not by pass.
Proof:
I claim that $HN$ is not empty.
Since $G$ is a group, $1_{g} \in G$. Since both $H$ and $N$ are subgroups of $G$, $1_{g} \in H$ and $1_{g} \in N$. Therefore $1_{g} \in HN$.
I claim that $HN$ is closed.
Let $h_{1}$$n_1$ and $h_{2}$$n_2$ $\in HN$. $(h_{1}n_1)(h_{2}n_2)=(h_{1}n_1h_{2}n_2)=h_{1}(n_1h_{2})n_2$.
Now this is where my difficulty is I cannot swap $n_1$$h_{2}$ to get $h_{2}$$n_1$ so what do I do. Help!!!

It is true that $h^{-1}nh=n'\in N\;\implies nh=hn'$ , because $\;N\lhd G\;$ , then
$$h_1n_1h_2n_2=h_1(n_1h_2)n_2=h_1(h_2n')n_2=(h_1h_2)(n'n_2)\in HN$$
