Finitely many commutators implies finite derived subgroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term group with finite derived subgroup
View a complete list of pages giving proofs of equivalence of definitions

Statement

Suppose $G$ is a group such that the set of commutators in $G$ is a finite set. Then, the derived subgroup of $G$ (i.e., the subgroup of $G$ generated by the set of commutators) is a finite group, and hence, $G$ is a group with finite derived subgroup.

Caveats

Note that this statement is not saying that if the derived subgroup is finitely generated, then it is finite. That statement is in fact false -- the derived subgroup of the infinite dihedral group is an infinite cyclic group. The statement is really about a very specific choice of generating set for the derived subgroup, namely, the set of all commutators.

Facts used

Finitely generated and FC implies FZ: The relevant part is that any finitely generated group that is also a FC-group (every conjugacy class is finite) is a FZ-group (the center has finite index).
FZ implies finite derived subgroup (this result is also called the Schur-Baer theorem).

Proof

Given: A group $G$ , with only finitely many elements that are commutators, say $g_{1}=[h_{1},k_{1}],g_{2}=[h_{2},k_{2}],\dots g_{n}=[h_{n},k_{n}]$ (note that the $g_{i}$ s are unique, but the $h_{i},k_{i}$ are not uniquely determined).

To prove: The derived subgroup $G'$ of $G$ is finite.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Let $H=\langle h_{1},k_{1},\dots ,h_{n},k_{n}\rangle$ . Then, $H$ is a finitely generated subgroup of $G$ whose set of commutators is the same as the set of commutators of $G$ , and it is a finite set.		The whole given statement		$H$ is finitely generated by definition. Since $H\leq G$ , every commutator of elements of $H$ is a commutator of elements of $G$ . The other direction is obvious from the given: we've already written every commutator of elements of $G$ as a commutator of elements of $H$ .
2	$H$ is a FC-group, i.e., the size of the conjugacy class of every element $x\in H$ is finite.				For any $y\in H$ , we have $yxy^{-1}=yxy^{-1}x^{-1}x=[y,x]x$ . Thus, every conjugate of $x$ is a commutator of elements of $H$ times the element $x$ . Step (1) tells us that the set of commutators of elements of $H$ is finite, so the set of conjugates of $x$ is the image of a finite set and hence finite.
3	$H$ is a FZ-group, i.e., the center of $H$ has finite index in $H$ .	Fact (1)		Steps (1), (2)	By Step (1), $H$ is finitely generated. By Step (2), $H$ is a FC-group. Thus, by Fact (1), $H$ is a FZ-group.
4	The derived subgroup $H'$ is finite.	Fact (2)		Step (3)	Step-fact combination direct.
5	$G'$ is finite.			Steps (1), (4)	Step (1) says that $G'=H'$ . Combining with Step (4) gives the result.

References

Math Stackexchange question about the subject. The proof presented on this page is an adaptation of that proof.