Let $(G,\cdot)$ be a group and let $N$ be normal subgroup of the group $G$, which is also an Abelian group. Prove that for every subgroup $H$ of the group $G$, $N\cap H$ is normal subgroup to the group $NH$.
Attempt: I managed to prove that $N\cap H$ is a subgroup of $NH$, but now I have to prove that for every element $x$ of $NH$ and every element $y$ of $N\cap H$ there is an $xyx^{-1}$ element of $N\cap H$. So basically we need to prove that $xyx^{-1}$ is in $N$ and in $H$. That it is in $N$ is obvious, since $N$ is the normal subgroup of $G$, but I don't know how to prove that it is in $H$.
Write $x=nh$ with $n\in N, h\in H$. Then:
$xyx^{-1}=nhyh^{-1}n^{-1}$
Note that $hyh^{-1}\in N$, because $N$ is normal in $G$. Since $N$ is abelian it follows that $n(hyh^{-1})n^{-1}=hyh^{-1}$. And this is an element of $H$, as a product of elements of $H$.