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> Let $G$ be group and $ N \triangleleft G$ , $H < G$. Prove $NH < G$.
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I am beginner in this so I am not sure if I proved it right.

Let $n\_{1}h\_{1}, n\_{2}h\_{2} \in NH$. We want to show that $ n\_{1}h\_{1}( n\_{2}h\_{2} )^{-1} \in NH$.

$ n\_{1}h\_{1}( n\_{2}h\_{2} )^{-1} = n\_{1}h\_{1}( h\_{2}^{-1}n\_{2}^{-1} ) = n\_{1}(h\_{1} h\_{2}^{-1})n\_{2}^{-1} $. 

Because $H < G$, $h\_{1} h\_{2}^{-1} \in H$ and because ( I am not sure about this part) $ N \triangleleft G$ (then $\forall x \in G$, $ xN = Nx$ ) so $(h\_{1} h\_{2}^{-1})n\_{2}^{-1} = n\_{3}(h\_{1} h\_{2}^{-1})$ for some $n\_{3} \in N$ and then 

$n\_{1}(h\_{1} h\_{2}^{-1})n\_{2}^{-1} = n\_{1}n\_{3}(h\_{1} h\_{2}^{-1}) \in NH$.