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!
 A: It's way easier (algebraically) than you're making it out to be, but it's tricky to get into the right mindset.
Given that $N \lhd G$ with $H$ some subgroup of $G$, we'd like to show that $N \cap H \lhd H$.
In other words, we need to show that conjugating anything in $N \cap H$ by something in $H$ lands us back in $N \cap H$.
So, let $j \in N \cap H$, and let $h \in H$. You just need to show that $h^{-1}jh$ is in $H$ and $N$ also (hence in $N \cap H$). Think about it in these terms before reading on. Seriously. 
Why should conjugating $j$ by $h$ land us back in $H$? 

Well, because $j \in H$, and $H$ is a subgroup.

Why should conjugating $j$ by $h$ land us back in $N$? 

 Well, because $j \in N$ and $N \lhd G$, with $h \in G$; conjugating anything in $N$ by anything in $G$ (of which $H$ is a subgroup) lands us back in $N$.

So, we started with $j \in N \cap H$, and we ended up with $h^{-1}jh \in N \cap H$ so we're done. Easy on algebra, but it definitely takes a certain viewpoint. When I first learned algebra, I would have tried to use way more equations and junk, for what it's worth.
A: The thing we need to show is that given any $h\in H$, and $x\in N\cap H$, $y:=hxh^{-1}\in N \cap H$.
Clearly $y$ is in $H$ since $h,x,h^{-1}$ are all in $H$.
$y$ is also in $N$ since $N$ is a normal subgroup and hence conjugating $x$ by anything will still remain in $N$.
Thus, $y\in N\cap H$.
A: 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.
