Show that $H$ is a normal subgroup of $G$. 
Let $N$ be a normal subgroup of $G$. Let $H$ be the set of all elements $h$ of $G$ such
that $hn = nh$ for all $n \in N$. Show that $H$ is a normal subgroup of $G$.

My attempt: I've been trying to do this exercise and so far I've been able to show that $H$ is a normal subgroup of $N$ (the proof is similar to showing that the center is a subgroup), but that $H$ is a normal subgroup of $N$ does not always imply that $H$ be a normal subgroup of $G$. How could I show in this case that $H$ is indeed a normal subgroup of $G$?
 A: Let $h\in H, n\in N$ for every $g\in G$, $ghg^{-1}n(ghg^{-1})^{-1}$
$=ghg^{-1}ngh^{-1}g^{-1}$, since $N$ is normal, $g^{-1}ng\in N$, we deduce that
$hg^{-1}ngh^{-1}=g^{-1}ng$, this implies that
$ghg^{-1}n(ghg^{-1})^{-1}=gg^{-1}ngg^{-1}=n.$
A: We have to prove that if $h \in H$ and $g \in G$, then $g^{-1}hg \in H$. Note that $x \in H$ iff $xnx^{-1} = n$ for every $n \in N$ (i.e., iff conjugation by $x$ fixes $N$). For any $g \in G$, $h \in H$ and $n \in N$, we have:
$$
\begin{align*}
(g^{-1}hg)n(g^{-1}hg)^{-1}
   &= (g^{-1}hg)n(g^{-1}h^{-1}g) \\
   &= g^{-1}(h(gng^{-1})h^{-1})g \\
   &= g^{-1}(gng^{-1})g \tag*{because $gng^{-1} \in N$ and $h \in H$} \\
   &= (g^{-1}g)n(g^{-1}g) \\
   &= n
\end{align*}
$$
So with $x = g^{-1}hg$, we have $xnx^{-1} = n$ for every $n \in N$, so that $x \in H$, which is what we had to prove.
A: \begin{alignat}{1}
H &= \{g\in G\mid gn=ng, \forall n\in N\} \\
&= \bigcap_{n\in N}\{g\in G\mid gn=ng\} \\
&= \bigcap_{n\in N}\{g\in G\mid gng^{-1}=n\} \\
&= \bigcap_{n\in N}G_n \\
\end{alignat}
where $G_n$ is the stabilizer of $n\in N$ under the $G$-action on $N$ by conjugation. Therefore, $H$ is the kernel of such an action, and hence it is a normal subgroup of $G$.
A: You need to start by showing $H\le G$.
I will use the two-step subgroup test.
Since for any $n\in N$, we have $en=n=ne$, so $e\in H$. Hence $H\neq\varnothing$.
By definition, since $H$ consists only of elements of $G$, we have $H\subseteq G$.
Let $h,k\in H$. Then $nh=hn$ and $nk=kn$, so $$k^{-1}n^{-1}=(nk)^{-1}=(kn)^{-1}=n^{-1}k^{-1},$$ but, multiplying by $n$ on the left and right, we have $nk^{-1}=k^{-1}n$, so $k^{-1}\in H$. Also,
$$\begin{align}
n(hk)&=(nh)k\\
&=(hn)k\\
&=h(nk)\\
&=h(kn)\\
&=(hk)n,
\end{align}$$
so $hk\in H$.
Hence $H\le G$.
For normality, for $n\in N$, $g\in G$, $h\in H$, we have
$$\begin{align}(ghg^{-1})n(ghg^{-1})^{-1}&=(ghg^{-1})n(gh^{-1}g^{-1})\\
&=g(h(g^{-1}ng)h^{-1})g^{-1}\\
&=g(g^{-1}ng)g^{-1}\\
&=(gg^{-1})n(gg^{-1})\\
&=n
\end{align}$$
by definition of $H$ and $g^{-1}ng\in N\unlhd G$. But then $(ghg^{-1})n=n(ghg^{-1})$. Hence $ghg^{-1}\in H$.
Hence $H\unlhd G$.
