TITLE: How "abelian" can a non-abelian group be?
QUESTION [22 upvotes]: 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 $\frac{|Z(G)|}{|G|}$?)

REPLY [4 votes]: 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.