# Subgroup of finite index with only one finite conjugacy class

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?

• 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 FC-center and denoted $FC(G)$. $G$ is called icc if $FC(G)=1$. You ask whether $FC(G)$ finite implies $G$ virtually icc. – YCor Sep 8 '16 at 7:03

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 finite-index subgroup which has only one finite conjugacy class, since the center lies in the kernel of any homomorphism to a finite group.
• 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 $H$. – YCor Sep 8 '16 at 6:59
• 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. – Ian Agol Sep 8 '16 at 16:25