Let $G$ be a group and assume that the derived subgroup (ie, commutator subgroup) of $G$ is finite. Does the center of $G$ have finite index in $G$?

Some background: In Centralizer of a finite normal subgroup has finite index I showed that if the derived subgroup is finite and the quotient has a set of generators whose representatives in $G$ commute pairwise, then the center of $G$ has finite index (since in this case, we get $G = G'H$ where $H$ is the subgroup generated by those representatives).

My feeling is that this is not the case, but I have not been able to think of a counterexample (even though looking at the proof for the above case does suggest how such a counterexample should look, I can't seem to construct it).

  • $\begingroup$ @DonAntonio: By transfer, you mean what is known as Verlagerung in german? In that case, I have seen it, but only seen it used to deal with finite groups. $\endgroup$ Jan 7, 2013 at 15:08
  • $\begingroup$ @DonAntonio: I suspect that you are thinking of the converse question: does $|G:Z(G)|$ finite imply $G'$ finite? The answer to that is yes, and proofs use the transfer. The answer to the question posed is no, although it is true for finitely generated groups. $\endgroup$
    – Derek Holt
    Jan 7, 2013 at 15:12
  • $\begingroup$ Derek, you're completely right: my bad. Thanks $\endgroup$
    – DonAntonio
    Jan 7, 2013 at 16:02

1 Answer 1


A counterexample is an infinitely generated extraspecial $p$-group for a prime $p$. For example:

$G = \langle x_i,y_i,z\ (i \in {\mathbb Z}) \mid$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [z,x_i]=[z,y_i]=1, x_i^p=y_i^p=z^p=1, [x_i,y_i]=z, [x_i,y_j]=1\ (i \ne j) \rangle$

In general, $|G'|$ finite implies that $G/Z(G)$ has bounded exponent, and hence it is finite if $G$ is finitely generated. To prove that, note that, since $|G:C_G(G')|$ is finite, we can assume $G' < Z(G)$, and so the commutator map is a bilinear map from $G \times G$ to the finite group $G'$, and hence $g^{|G'|} \in Z(G)$ for all $g \in G$.

  • 1
    $\begingroup$ But there are finitely generated, infinite groups of bounded exponent. The statement that the center of $G$ is of finite index in $G$ is true nevertheless. However it seems a different argument is required. $\endgroup$ Mar 24, 2015 at 21:54
  • $\begingroup$ If I am not mistaken one also needs to add the relations $[x_i, x_j] = 1 = [y_i, y_j]$ for all $i$ and $j$, right? $\endgroup$ Mar 25, 2015 at 20:38

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