# If the derived subgroup is finite, does the center have finite index?

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).

• @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. Jan 7, 2013 at 15:08
• @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. Jan 7, 2013 at 15:12
• Derek, you're completely right: my bad. Thanks Jan 7, 2013 at 16:02

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 $g \in G$.
• 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. Mar 24, 2015 at 21:54
• 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? Mar 25, 2015 at 20:38