$D(G)$ is finite if $G$ has finitely many commutators

I read somewhere that if a group $G$ has only finitely many commutators then the commutator subgroup $D(G)$ is itself finite.

Do you know a proof of this result?

Solution 1:

I am sorry but I do not have a complete answer at the moment.

I will show that if $G$ is finitely generated and there are only finitely many commutators, then $D(G)$ is finite.

We will prove that the center of $G$, $Z(G)$ has finite index in $G$, then a theorem by Schur will allow us to conclude.

If $G$ is finitely generated, take $g_1,\ldots,g_n\in G$ to be a generating set. We have $$Z(G)=\bigcap_{i=1}^n Z(g_i),$$ where $Z(g)$ is the centralizer of $g$ in $G$.

It suffices to prove that every $Z(g_i)$ has finite index in $G$; this is the same as asking to verify that every element $g_i$ has finitely many conjugates.

Take $h\in G$. We have $hgh^{-1}=g[g^{-1},h]$. Since there are only finitely many commutators, every $g\in G$ has finitely many conjugates.

Hence $Z(g_i)$ is of finite index in $G$, for every $i=1\ldots n$.

Now, we have the following theorem

Theorem (Schur) If $[G:Z(G)]<\infty$, then $D(G)$ is finite.

I learned about this theorem in the italian book Gruppi by Antonio Machì. There is now a translated edition where it is Theorem 2.38 on page 84.