Prove that the centralizer subgroup is normal in the normalizer subgroup

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 $nz(n^{-1}sn)z^{-1}n^{-1}=n(n^{-1}sn)n^{-1}=s$.

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.