Finitely many commutators implies finite derived subgroup

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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


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.


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).


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_is are unique, but the h_i,k_i are not uniquely determined).

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


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.