The centre of a group $G$ can be described as the set of all elements $g\in G$ whose conjugacy class consists just of $g$ itself. The FC-centre of a group $G$ is the union of all finite conjugacy classes; it is a normal (and even characteristic) subgroup of $G$. Is having a non-trivial FC-centre equivalent to having either a non-trivial centre or a non-trivial finite normal subgroup? [Obviously, each of the latter conditions implies the former.]



Centre, FC-centre and finite normal subgroups
The answer was already given in the negative by a trivial counterexample, but there is a way to get a result in the spirit of the expected result.


Indeed, it is known that in an FC-group $G$, the set $G\_\mathrm{Tor}$ of torsion elements is a subgroup (then obviously characteristic) and the quotient is a torsion-free abelian group. 


It follows that if $G$ is a group, with nontrivial FC-center $\mathrm{FC}(G)$,



> 
> * either $\mathrm{FC}(G)\_\mathrm{T}$ is non-trivial, which means that $G$ has a nontrivial finite normal subgroup;
> * or $\mathrm{FC}(G)\_\mathrm{T}$ is trivial, in which case $\mathrm{FC}(G)$ is abelian and torsion-free. Choosing the subgroup generated by a nontrivial conjugacy class in $\mathrm{FC}(G)$, this even implies that $G$ has a normal subgroup $N\neq 1$ that is free abelian of finite rank, and the conjugation action of $G$ on $N$ factors through a finite quotient of $G$.
> 
> 
> 


(And of course, conversely, the existence of such a normal subgroup implies having nontrivial FC-center.)


