Suppose $N\unlhd G$ and let $H\subseteq G$ such that $hn=nh$ for all $n \in N, h \in H$. Show $H\unlhd G$. 
Suppose $N\unlhd G$ and let $H\subseteq G$ such that $hn=nh$ for all $n \in N, h \in H$. Show $H\unlhd G$.

It was not hard to show $H$ is a subgroup of $G$. For identity, $en=n=ne$, where $e$ is the identity in $G$. For closure, $(h_1h_2)n=h_1nh_2=n(h_1h_2)$ by the fact that all $h$ communte with all $n$. For inverse, we already have $hn=nh$. Multiply $h^{-1}$ to the left and right we will get $nh^{-1}=h^{-1}n$.
The question is, I have been spending a long time trying to show $H$ is indeed a normal subgroup of $G$.
My thoughts were $g^{-1}ng \in N$ as $N$ is a normal subgroup of $G$.
I have tried let $g^{-1}ng = n_1 \in N$. This forms connections between $N$ and $G$.
Also, $hn=nh, hn_1=n_1h$. This forms connections between $N$ and $H$.
So I decided to rewrite those relations and substitute into the term $g^{-1}hg$, and tried to show $g^{-1}hg \in H$.
Thing is, no matter how did I calculate it, it just ended as getting an identity equation. Any help would be much appreciated.
 A: Others have noted that the result is true if we assume that $H$ is the set of all elements that centralize $N$; that is, that
$$H = C_G(N) = \{g\in G\mid gn=ng\text{ for all }n\in N\}.$$
This seems implied by the arguments given in the original post "proving" that $H$ is a subgroup.
However, as written, the statement literally just asks to show that any subgroup of $C_G(N)$ is normal in $G$. This gives a counterexample to the statement as written.
Let $G=C_2\times S_3$, $N=C_2\times\{e\}$, and $H=\{e\}\times\langle (12)\rangle$. Then $H$ centralizes $N$, but $H$ is not normal in $G$, as it is not normal in $\{e\}\times S_3$.
A: Note that $H=C_G(N)$ is known to be the centralizer of the subgroup $N$. Since $N$ is a normal subgroup of $G$, we have $gNg^{-1}=N$ for any $g\in G$. In other words, the function $\varphi_g\colon N\to N$ such that $\varphi_g(n)=gng^{-1}$ defines an automorphism, for every $g\in G$. So, the function $\Psi\colon G\to{\rm Aut}(N)$ such that $\Psi(g)=\varphi_g$ is well defined, and it is easy to prove that $\Psi$ is a homomorphism. Knowing that the kernel of any homomorphism is a normal subgroup, one can show that, in fact, $\text{ker}\Psi=C_G(N).$
