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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).

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@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. –  Tobias Kildetoft Jan 7 '13 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. –  Derek Holt Jan 7 '13 at 15:12
Derek, you're completely right: my bad. Thanks –  DonAntonio Jan 7 '13 at 16:02

1 Answer 1

up vote 4 down vote accepted

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 $g \in G$.

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