If in group G there are finitely many finite conjugacy classes, is it possible to find some subgroup of G of finite index which has only one finite conjugacy class? ( the class {e}). Maybe there is some class of groups which possesses this property?

$\begingroup$ Classical terminology: the union of finite conjugacy classes of a group $G$ forms a subgroup (it's the set of elements whose centralizer has finite index, it is called its FCcenter and denoted $FC(G)$. $G$ is called icc if $FC(G)=1$. You ask whether $FC(G)$ finite implies $G$ virtually icc. $\endgroup$– YCorSep 8, 2016 at 7:03
1 Answer
No, there is a theorem of Deligne that the universal central extension of $Sp(2g,\mathbb{Z})$ is not residually finite. If one takes a finite central extension (which is also not residually finite), then the elements in the center have finite conjugacy class, and all other elements have infinite conjugacy class, but there is no finiteindex subgroup which has only one finite conjugacy class, since the center lies in the kernel of any homomorphism to a finite group.
It might be reasonable to consider your question for the class of residually finite groups.

$\begingroup$ In the residually finite case, this is trivially yes (take a normal subgroup of finite index $H$ with trivial intersection with $FC(G)$; then infinite conjugacy classes of $G$ contained in $H$ split into finitely many infinite conjugacy classes of $H$. $\endgroup$– YCorSep 8, 2016 at 6:59

$\begingroup$ thank you! Yes, for residually finite groups it is trivial. Are there some known examples of classes of groups for which finite FC(G) implies virtually ICC? $\endgroup$ Sep 8, 2016 at 11:17

$\begingroup$ Okay, I understand, I didn't know about $FC(G)$. If $FC(G)$ is finite, then the action of $G$ on $FC(G)$ by conjugation gives a homomorphism to a finite symmetric group whose kernel has $FC(G)$ central. So all counterexamples will come from the Deligne type: a group whose center is finite and not separable. $\endgroup$– Ian AgolSep 8, 2016 at 16:25