Show if $N$ is normal subgroup of $G$ and $H$ is a subgroup of $G$, then $N \cap H$ is normal subgroup of $H$. Show if $N$ is normal subgroup of $G$ and $H$ is a subgroup of $G$, then  $N \cap H$ is normal subgroup of $H$.
attempt:  Then recall $N \cap H$ is normal if and only if $h(N \cap H) h^{-1} \subset N \cap H$.
Then suppose $j \in (N \cap H)$ and $h \in H$. , so $hjh^{-1} = (h^{-1})^{-1} j(h^{-1}) = k \in (N \cap H),$ for some $k$, then we can solve for $j$ so we get $j = h^{-1}kh \in h^{-1}(N\cap H)h. $ 
Hence $N \cap H $ is cointained in $h^{-1}(N \cap H)j$ so $N \cap H = h^{-1}(N \cap H)h$. So $N\cap H$ is normal subgroup of $H$.
Can someone please verify this?
My professor said I need to choose an element from one side and show the element is in the other side too. So containment in both sides. My professor said I can't assume $h(N \cap H) h^{-1}  = hNh^{-1} \cap hHh^{-1} = N \cap H.$
Can someone please help me if this is wrong. Thank you very much!

To show $N ∩ H $ is normal in $H$ we must show that the conjugate of $N ∩ H$ by an element
of $h$ is $N ∩ H$. So let $h ∈ H$. Then $h(N ∩ H)h
^{−1} = hNh^{−1} ∩ hHh^{−1}$ but
$hNh^{−1} = N$ since $N$ is normal in $G$, and $hHh^{−1} = H$ since $H$ is a subgroup of $G$,
so $g(N ∩ H)g^
{−1} = N ∩ H$, as desired.
