# 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

1. 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. 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 $H \le G$, 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.