Prove that the centralizer subgroup is normal in the normalizer subgroup To my dear friends with gratitude.
I want to get help proving centralizer of a nonempty subset of a group is a normal subgroup in the normalizer of that set in the mentioned group.symbolically:
$C_G (S)\trianglelefteq N_G (S)$
 A: Of course you can try to prove it directly, by using the definitions of normalizer and centralizer nonetheless there's a different approach to the problem. 
Here's a little hint (to prove with the different approach).
Generally a good technique in proving that some subgroup is normal is to show that it's the kernel of some homomorphism, proving that centralizer of a subgroup is normal in the normalizer of the same subgroup can be done in this way.
The trick is to find the right homomorphism, but here's a second hint:
every action of group $G$ on a set $X$ is the same as an homomorphism of type $G \to \text{Aut}(X)$, i.e. a permutation representation of the group.
The normalizer of a subgroup have a natural action on the same subgroup.
Hope this helps.
A: Let  $z \in  C_G(S)$ and let $n \in N_G(S)$. We want to show that $nzn^{-1} \in C_G(S)$.
If $s \in S$, then $(nzn^{-1})s(nzn^{-1})^{-1}=nz(n^{-1}sn)z^{-1}n^{-1}$.
Now $n^{-1}sn \in S$  and so $nz(n^{-1}sn)z^{-1}n^{-1}=n(n^{-1}sn)n^{-1}=s$.
