FC not implies BFC

Statement
It is possible to have a FC-group $$G$$ (i.e., every conjugacy class in $$G$$ is finite) that is not a BFC-group (i.e., there is no finite upper bound on the sizes of conjugacy classes).

Proof
Suppose $$K$$ is a finite group that is not abelian, so the derived subgroup of $$K$$ is nontrivial. Let $$G$$ be the restricted external direct product of a countably infinite number of copies of $$K$$. We can verify that:


 * $$G$$ is a FC-group: any element of $$G$$ has all but finitely many coordinates equal to the identity element of $$K$$, and therefore all its conjugates all have those coordinates as the identity element, so there can be only finitely many conjugates.
 * There is no finite upper bound on the sizes of conjugacy classes in $$G$$: Let $$x \in K$$ be a non-central element, with a conjugacy class of size $$s > 1$$. Then, an element with $$n$$ coordinates that are copies of $$x$$, and the remaining coordinates the identity element, has $$s^n$$ conjugates. As $$n$$ can be arbitrarily large, this allows for arbitrarily large conjugacy class sizes.