# FC not implies finite derived subgroup

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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., group with finite derived subgroup)
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## Statement

It is possible for a group to be a FC-group but not to be a group with finite derived subgroup.

## 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.
• $G$ has infinite derived subgroup: If we view things in terms of internal direct products, the derived subgroup is the internal direct product of the derived subgroups of each copy. Viewed externally, it is isomorphic to the restricted external direct product of a countably infinite number of copies of $K'$, which by assumption is nontrivial. Hence, it is infinite.