Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
    
@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
add comment

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 $g \in G$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.