FC not implies BFC

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., FC-group) need not satisfy the second group property (i.e., BFC-group)
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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.