1
$\begingroup$

Let $G$ be a group generated by $S=\{x_1,\cdots,x_n\}$ and let its lower central series be defined as $\Gamma_1=G$, $\Gamma_m=[\Gamma_{m-1},G]$ for $m\geq 2$.

By definition $\Gamma_m$ is generated by all iterated commutators of the form $[[g_1,g_2],\cdots,g_m]$, where $g_1,\cdots,g_m\in G$.

Is it true that $\Gamma_m$ is generated by all iterated commutators of the form $[[x_{i_1},x_{i_2}],\cdots,x_{i_m}]$, where $x_{i_1},\cdots,x_{i_m}\in S$?

$\endgroup$
  • $\begingroup$ It's true if $G$ is nilpotent. $\endgroup$ – Derek Holt Dec 6 '14 at 16:24
  • $\begingroup$ Thanks @DerekHolt! Where can I find a reference of this property? $\endgroup$ – Zuriel Dec 6 '14 at 16:32
  • $\begingroup$ @DerekHolt, Actually I am interested in pure braid groups $P_n$. I think that they are residually nilpotent but not nilpotent. In this case, is the conclusion still true for pure braid groups? $\endgroup$ – Zuriel Dec 6 '14 at 16:35
  • $\begingroup$ It's not true for residually nilpotent groups in general, because free groups are residually nilpotent. For the pure braid group, I think that $[P_n,P_n]$ is probably not finitely generated. $\endgroup$ – Derek Holt Dec 6 '14 at 19:10
  • 1
    $\begingroup$ Sorry, what I said was wrong. It is true that $\Gamma_m$ is generated modulo $\Gamma_{m+1}$ by iterated commutators of length $m$. That is true for any group, and you can prove it by induction on $m$. So, for a nilpotent group, $\Gamma_m$ is generated by iterated commutators of length greater than or equal to $m$. $\endgroup$ – Derek Holt Nov 13 '18 at 14:05
2
$\begingroup$

This is not true, even for $\Gamma_2$. Let $G$ be the free group on $S$; it is known that $[G,G]$ is not finitely generated (provided that $\vert S\vert\ge 2$), hence it cannot be generated by all commutators of the form $[x_{i_1}, x_{i_2}]$ since there are only finitely many such.

$\endgroup$
  • $\begingroup$ Thank you for your reply! $\endgroup$ – Zuriel Nov 13 '18 at 14:38

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.