As far as I have heard, centralizers play an important role in the theory of groups. My question arises from curiosity and the desire to understand how much control centralizers have over the group. Here we go:
If $G$ is group such that centralizer of every element (except the identity) of $G$ is finite, then is $G$ necessarily a finite group?
If the answer is "No", then such an infinite group will be heavily non-abelian (intuitively). Well, for one thing, $Z(G)=1$. Also since every element commutes with its own powers, it looks like every element would have to have a finite order. So my question seems to be stronger version of this.
I appreciate any pointers :)
As Tobias mentioned, Tarski monsters are infinite $p$-groups $G$ such that every proper subgroup has order $p$. If $1\neq x,y \in G$, $\langle x \rangle \neq \langle y \rangle$, and $C_G(x)$ is infinite, then $C_G(x) = G$ contains $y$, so $\langle x, y \rangle$ has order $p^2$ a contradiction. Hence $C_G(x) = \langle x \rangle$ for every non-identity element of $G$.
Centralizers do control a lot of the structure of the group, but they do a much better job when there are elementary abelian subgroups of sufficient rank. In that case, you get a nice partially overlapping system of centralizers, and you can see more of the group. When there are no non-cyclic abelian $p$-subgroups, then different methods come into play that usually rely on some sort of counting since a finite group with tiny centralizers has to be put together very carefully. Infinite groups get to cheat in a way, since they can stay a $p$-group without having a non-identity center.
As an example of finite: A finite group in which each centralizer of an non-identity element has prime order is necessarily a non-abelian group of order $pq$ for primes $p<q$ with $p$ dividing $q-1$. There is a unique such group up to isomorphism for each such pair of primes, and all such groups have the centralizer property.