Question regarding the normality of a certain subgroup of a group Let $G$ be a group and $N$ a normal subgroup of $G$. Let $H=\{g\in G\mid gn=ng\space \forall n\in N\}$. Prove that $H$ is a normal subgroup of $G$. 
I've tried seeing if we can write $H$ as the intersection or product of normal subgroups, but haven't succeeded so far. I'd be grateful if someone could provide a hint in this direction. 
 A: I assume that you already proved that $H$ is a subgroup of $G$.
For the normality, you can go back to the definition of "normal subgroup" : let $h \in H, g \in G$. You want to prove that $ghg^{-1} \in H$, that is :
$$ ghg^{-1}n = nghg^{-1} \quad \forall n \in N $$
using that $N$ is a normal subgroup of $G$.
Hint :

 Try to make "$g^{-1}ng$" appear in the equation above, in order to use the normality of $N$...

Hint 2 :

 $ghg^{-1}n = (ghg^{-1}n)gg^{-1} = gh(g^{-1}ng)g^{-1} = g(g^{-1}ng) h g^{-1} = nghg^{-1}$ because $(g^{-1}ng) \in N$ (so it commutes with $h \in H$).

A: *

*If $a\in H$ and $n\in N$ then $a^{-1}n=\left(n^{-1}a\right)^{-1}=\left(an^{-1}\right)^{-1}=na^{-1}$
implying that $a^{-1}\in H$.

*If $a,b\in H$ and $n\in N$ then $abn=anb=nab$ implying that $ab\in H$.

*If $a\in H$, $g\in G$ and $n\in N$ then $gag^{-1}n=gag^{-1}ngg^{-1}=gg^{-1}ngag^{-1}=ngag^{-1}$
implying that $gag^{-1}\in H$.
In the second equality of the last step the normality of $N$ is used: $n\in N\implies g^{-1}ng\in N$.
A: Another way to look at it, which is a common concept in Group Theory: for any subgroup $H$ of $G$, we define the centralizer, and normalizer respectively as
$$C_G(H)=\{g \in G: gh=hg \text{ for all } h \in H\}$$ and 
$$N_G(H)=\{g \in G: gH=Hg\}.$$
It is an easy exercise to see that $C_G(H) \unlhd N_G(H)$. For a normal subgroup $N$, obviously $N_G(N)=G$, whence your answer to your post. It is also worthwhile knowing that $N_G(H)/C_G(H)$ embeds homomorphically in Aut$(H)$ (this can be seen by considering the conjugation action of $N_G(H)$ on $H$ - the kernel of this action is precisely $C_G(H)$.
