Consider $G$ - finitely generated nilpotent group. Is it true that all members of the lower central series $\gamma_i G=[\gamma _{i-1} G, G],\ \gamma_0 = G$ are finitely generated as well?

  • 1
    $\begingroup$ Tensor products in Robinson's textbook for a Course in the Theory of Groups is very nice. Section 5.2.5 on page 126 of the 1st edition. $\endgroup$ – Jack Schmidt May 29 '14 at 18:05
  • $\begingroup$ All subgroup of $G$ are finitely generated. $\endgroup$ – Derek Holt May 29 '14 at 19:10
  • $\begingroup$ Derek Holt, why? $\endgroup$ – user1928091 May 29 '14 at 19:18

Yes. You check it by induction on the nilpotency length. Starting the induction is clear. Suppose $G$ $n$-step nilpotent with $n\ge 2$. Using the inductive hypothesis, $G/\gamma_nG$ is an iterated extension of finitely generated abelian groups and hence is finitely presented. Hence $\gamma_nG$ is finitely generated as a normal subgroup, and therefore, being central, $\gamma_nG$ is finitely generated. Since $\gamma_i(G)$ is extension of $\gamma_n(G)$ by $\gamma_i(G/\gamma_nG)$ which are both finitely generated, we conclude that $\gamma_i(G)$ is finitely generated for all $i$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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