Let $N$ be a normal subgroup of $G$, i.e., $g^{-1}Ng = N$ for all $g\in G$.
How to show that for all $g,\, h\in G$, we have
$\{xy: x\in Ng \text{ and } y\in Nh\} = Ngh.$
My effort is that I know
$$Ngh = \{ngh : n \in N\}.$$
Now $$xy = (n_1g)(n_2h) = n_1gh h^{-1}n_2h,$$ since $h^{-1}n_2h\in N,$ we let $h^{-1}n_2h = n_3$, so we get
$$xy = (n_1g)(n_2h) = n_1gh h^{-1}n_2h = n_1ghn_3,$$ then I still have no idea how to show that $n_1ghn_3$ is in $Ngh$