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?

  • $\begingroup$ Let me figure this out: You want to show that $D(G)$ is finite if $G$ has finitely many commutators, right? But is $D(G)$ not defined as the subgroup of commutators? Am I misreading something? $\endgroup$
    – awllower
    Jan 24, 2013 at 14:22
  • 3
    $\begingroup$ @awllower: $D(G)$ is the subgroup generated by the commutators. The commutators themselves do not form a subgroup in general, see here for a discussion. $\endgroup$
    – Martin
    Jan 24, 2013 at 14:27
  • $\begingroup$ @Martin I see. Thanks for pointing out this. $\endgroup$
    – awllower
    Jan 24, 2013 at 14:43
  • $\begingroup$ @awllower $\endgroup$
    – Alexander Gruber
    Jan 24, 2013 at 17:33

1 Answer 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.

  • 9
    $\begingroup$ But if $G$ has only finitely many commutators $[g_i,h_i]$, then $D(G) = D(H)$ with $H = \langle g_i,h_i \rangle$ finitely generated, so the result is true for arbitrary $G$. $\endgroup$
    – Derek Holt
    Jan 24, 2013 at 16:30
  • $\begingroup$ You are right! Thanks $\endgroup$ Jan 24, 2013 at 17:27
  • 2
    $\begingroup$ I added a precise reference. Here are screen shots of the relevant passage: part 1 and part 2. $\endgroup$
    – Martin
    Apr 29, 2013 at 19:25
  • $\begingroup$ The same proof can be found in Problems in Group Theory by J. Dixon (problems 5.21 to 5.24). $\endgroup$
    – Seirios
    May 6, 2013 at 16:18

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