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Let $G$ be a group, $Z(G)$ the center of $G$, and $[G, G]$ the commutator subgroup of $G$. How do I show that if $[G:Z(G)]<\infty$, then $[G.G]$ is finite?

How to prove this? I cannot think of a good approach.

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It is good form to include the question in the body of the post, even if you have asked it in the title. I have edited the post accordingly. –  Aaron Feb 26 '12 at 1:16
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2 Answers

up vote 4 down vote accepted

A full proof is here. The one in the question comes from Passman's The algebraic structure of group rings, and the answer contains an improvement.

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Let $T$ be a transversal for $Z(G)$ in $G$, noting that $|T|$ is finite. Then $[G,G]$ is generated by the commutators $[t_i,t_j]$, where $t_i,t_j\in T$. Thus $[G,G]$ is finitely generated.

The transfer map from $G$ to $Z(G)$ is given by $g\mapsto g^n$, with $n=[G:Z(G)]$. Since $Z(G)$ is abelian, $[G,G]$ is in the kernel, and hence for every $x\in [G,G]$, $x^n=1$.

Now $Z(G)\cap [G,G]$ has finite index in $[G,G]$; it is abelian, finitely generated, and has finite exponent. It is thus a finite group. Since it has finite index in $[G,G]$, $[G,G]$ is finite as well.

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Why is $Z(G) \cap [G,G]$ finitely generated? –  Ted Feb 26 '12 at 5:27
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It is a finite index subgroup of a finitely generated group. –  user641 Feb 26 '12 at 5:32
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