How "abelian" can a non-abelian group be? Something I have been wondering: in general, is there a bound for how many elements in a finite non-abelian group $G$ can commute with every other element? Equivalently, is there is a bound for the order of the center relative to the order of the group?
And, if $G$ is non-abelian and infinite, does this question make any sense? (ie does it make sense to consider the ratio $\frac{|Z(G)|}{|G|}$?)
 A: Note that if $A,B$ are two groups then $Z(A \times B) = Z(A) \times Z(B)$. This implies that the size of the center of a non-abelian group can be arbitrarily large (take $A$ an abelian group of a desired size and $B$ a non-abelian group with trivial center, for example any non-abelian simple group), and still the bound $|Z(G)| \leq |G|/4$ (deriving from the fact that if $G/Z(G)$ is cyclic then $G$ is abelian) cannot be improved (consider e.g. the group of quaternions, $Q_8$). So, both $|G|$ and $|Z(G)|$ can be arbitrarily large (even infinite), and hence you don't have much control on that. I guess you could ask about specific numbers instead. For example, your question could be the following:
(1) what are the numbers $n$ for which there exists a group $G$ with $Z(G)=\{1\}$ and $|G|=n$ ?
I believe this is a very non-trivial question. For instance, it is known that a number $n$ is "cyclic", i.e. such that every group of order $n$ is cyclic, if and only if $(n,\varphi(n))=1$. And there is a similar result concerning nilpotent groups (which always have non-trivial center), see here. My guess would be that the answer to (1) is: the non-nilpotent numbers. But I don't know.
A: You can consider the index of $Z(G)$ in $G$, which may be finite even if both $G$ and $Z(G)$ are infinite. Other than that, if $G$ is not abelian, that index is at least $2$, i.e. the center contains at most half the group elements.
A: Although what Hagen von Eitzen says is true, there is a much stronger restriction. The index of the center of a nonabelian group can never be a prime. So if the center is of index 2, then that group is actually abelian and so the center is actually the whole group.
Let's see why this is true.

Claim: The center $Z(G)$ of a group $G$ can never have index $p$ in $G$ for any prime $p$.

proof: As we so often do, suppose not; i.e. suppose we have a group $G$ with center $Z(G)$ of index $p$. Then $Z(G)$ is normal, so we can consider the quotient group $G/Z(G)$, which is a group of size $p$. Thus it is cyclic, say generated by an element $g$, and abelian.
In particular, this means that every element of $G$ can be written as $g^iz$ for some element $z$ in the center. So take two elements $a = g^iz_1$ and $b = g^jz_2$ in $G$, and consider their product.
$$ab = g_iz_1g^jz_2 = g^ig^jz_1z_2 = g^jg^iz_2z_1=g^jz_2g^iz_1 =ba$$
Where I used that the $g^i$ commute, and elements in the center commute with everything. Thus this shows that the group is abelian, contrary to the fact that the center was not the whole group. Contradiction. $\diamondsuit$
So the center cannot contain more than half the elements of the group, and in fact can't contain exactly half the elements of the group. But there is no reason why the index of the center couldn't be $6$ infinitely often, visibly by taking semidirect products of the symmetric group on 3 symbols with larger and larger abelian groups.
