Is the centralizer of a Subgroup abelian? Let $G$ be a group, and let $H$ a subgroup of $G$.
Let: $$c(H)=\{x\in G : xh=hx, \forall h\in H \}$$
I have already proved that this is a subgroup of $G$, but I'm not sure if it's abelian (I've been looking for a counterexample of this but without success so far).
Another question, how can I express the Center of G $Z(G)=\{z\in G : zx=xz, \forall x\in G \}$ in terms of $c(H)$? Is the center abelian?
Thank you for your help, I'm new to groups and I'm kind of lost.
 A: A strategy when trying to understand such a notion is to consider extreme cases. This is not always helpful, but it cannot hurt and sometimes it is useful. 
You have a notion defined for every subgroup. There are two subgroups present for any group $G$, the full group $G$ and the trivial subgroup $\{e\}$.


*

*What is $c(\{e\})$? It is the set of all elements $x \in G$ such that $xe=ex$. But that's of course true for all $x\in G$! So $c(\{e\}) = G$. Thus if  $G$ is not abelian this is not abelian. (For a more concise version of this part see the post mentioned by Zev.) 

*What is $c(G)$? It is defined as $\{x\in G : xh=hx, \forall h\in G \}$. But, this looks familiar, doesn't it? it is exactly the definition of $Z(G)$ just using different names for the variables.  
Thus considering these extreme cases, the two questions almost answered themselves. 
Finally on whether $Z(G)$ is abelian.  We need to check if $ab =ba $ for all $a,b \in Z(G)$. But, when $a \in Z(G)$ then by definition it commutes with every element of $G$ thus in particular with $b$. Hence the center is abelian.  
