When is the centralizer equal to the normalizer? I have a very basic understanding so correct me if I'm wrong. If the group is abelian, then ever element in that group commutes with a set as well as every element in that set. So would that be the only requirement? 
 A: One specific case;
Let $a$ be an element with order $2$ in $G$ and let $H=<a>$ then $C_G(H)=N_G(H)$.
Result come from the fact that since it has only one nontrivial element, any element normalize $H$ must centralize.
Another spesific case;
There are some sylow$-p$ sumbroup P s.t. $N_G(P)=P$ if $P$ is abelian then $P\leq C_G(P)\leq N_G(P)\implies N_G(P)=C_G(P)$
A: One can classify this property in certain groups. For example, in torsion-free hyperbolic groups (I think if you add torsion it will boil down to finite groups, as "infinite cyclic" is replaced by "virtually cyclic" in the following proof).
Theorem: Let $H$ be a torsion-free hyperbolic group. Then a subgroup $K$ is such that $C_H(K)=N_H(K)$ if and only if $K$ is infinite cyclic.
For example, $H$ could be a free group or the fundamental group of a compact, non-orientable surface of genus at least two.
Proof: To see that this holds, begin by opening the book Metric spaces of non-positive curvature by Bridson and Haefliger. This book tells us that in such a group if $C_H(K)$ is non-trivial then $K$ is necessarily infinite cyclic. Therefore, $C_H(K)$ is also infinite cyclic, and indeed is the unique maximal infinite cyclic subgroup containing $K$. Another result we need to know about torsion-free hyperbolic groups to prove this is that maximal infinite cyclic groups are malnormal, that is, if $M$ is maximal infinite cyclic then the following holds.
$$M^g\cap M=1\Rightarrow g\notin M$$
Therefore, $N_H(K)$ is the maximal infinite cyclic subgroup containing $K$. Hence, $N_H(K)=C_H(K)$.
