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 $x_{i_1},\cdots,x_{i_m}\in S$?

  • $\begingroup$ 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
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    $\begingroup$ 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

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.

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

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