Q:

Proving if $H<G$ and $N\triangleleft$ G then $H\cap N \triangleleft H$

I want to prove that if $H<G$ and $N\triangleleft$ G then $H\cap N \triangleleft H$.
Here is what I have done:
$N\triangleleft G \iff aN=Na \quad \forall a\in G$ 
$hN=Nh \quad \forall h\in H$ since $H<G$
Then it follows that $h(H\cap N) = (H\cap N)h$ since $H\cap N \subset N$.
Therefore by definition $H\cap N \triangleleft H$.
This is correct right? If not what why and what do I need to do instead?

A:

It isn't entirely clear why we should have $h(H\cap N)=(H\cap N)h$ for each $h\in H$ using only your reasoning. It's certainly true, but you haven't really justified it (unless you have some prior results that you're referencing). You could use an "element-chasing" argument to show double-inclusion, but I actually recommend an alternate approach, using the following result (which is fairly simple to prove).

Let $G$ be a group, $N\subseteq G$. Then $N\trianglelefteq G$ if and only if $gng^{-1}\in N$ for all $g\in G$ and $n\in N.$

Using this approach, take any $h\in H$ and any $n\in H\cap N$. Since $h,n\in H,$ then $hnh^{-1}\in H$. Since $h\in H\subseteq G$ and $n\in N\trianglelefteq G,$ then $hnh^{-1}\in N.$ Hence, $hnh^{-1}\in H\cap N$ for all $h\in H$ and all $n\in H\cap N,$ whence $H\cap N\trianglelefteq H.$

