Finitely many commutators implies finite derived subgroup

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

 * 1) uses::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).
 * 2) uses::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: