Difference between centralizer and center groups? This is probably stupid question, but I can't see the difference between the two subgroups:
$$C_G(A)=\{g\in G| gag^{-1}=a,\forall a\in A\}$$
$$Z(G)=\{g\in G| ga=ag,\forall a\in G\}$$
Is the difference that the centralizer takes a subset $A\in G$ and the center always uses the entire group $G$?
 A: The normalizer of a subset of a group is the part of the group that commutes with the subset (but not necessarily holding the element of the subset fixed): $g$ in the normalizer of $H \subset G$ means $gH = Hg$, or more specifically, $gh_1 = h_2 g$ for some $h_1, h_2 \in H$.
The centralizer of a subset of a group is the part of the group that commutes with the subset elementwise (i.e., holding the element of the subset fixed): $g$ in the centralizer of $H \subset G$ means $gh = hg$ for all $h \in H$.  Clearly, commuting elementwise implies commuting with the subset, so the normalizer of a subset contains the centralizer of that subset.
The center of a group is the part of the group that commutes with everything in the group.  Commuting with everything implies commuting with elements of some subset, so the centralizer of a subset contains the center of the group.
Putting these together: $Z(G) \subset C_G(H) \subset N_G(H)$.
Regarding "$\leq$"...  The centralizer and normalizer are defined on subsets.  It may turn out that a subset is also a subgroup, but this is not required.  (It will essentially always be the case that the subsets are subgroups once you are past the material introducing these definitions.)  This is why I have used "$\subset$" in the above.
A: Converting the comments into an actual answer:
Yes, we have the special case $C_G(G)=Z(G)$, so the notion of centralizer can be thought of as a generalization of centers, since the centralizer $C_G(A)$ works for every subset $A$ of $G$ (note that it doesn't have to be a subgroup).
As for the normalizer, $Z(G)\subseteq C_G(A)\subseteq N_G(A)$ for every subgroup $A\leq G$, and $N_G(A)$ is the maximal subgroup of $G$ in which $A$ is normal.  The normalizer can also be defined for an arbitrary subset, and the above still holds other than that we must now say that $\langle A \rangle$, the subgroup generated by $A$, is normal in $N_G(A)$, and the normalizer is maximal with respect to this property.
