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Hello? I have a simple question about combinatorial group theory.

For a group $G$, I saw $[G_k, G_m] \subset G_{k+m}$ and these two subgroups need not be equal. Then is there any known condition that they can be equal? How about the case that $G$ is (f.g.) free?

Actually, my first question was that: for a f.g. free group $F$, $F_m \subset [F_k,F_k]$ for some big $m$?

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  • $\begingroup$ Can you say what $G_k$ etc. means? $\endgroup$
    – Steve D
    Commented Oct 26, 2011 at 10:21
  • $\begingroup$ Presumably the lower central series $\endgroup$
    – Derek Holt
    Commented Oct 26, 2011 at 11:50
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    $\begingroup$ As $D_\infty$ is metabelian but not nilpotent, it follows that $F_m$ is not contained in $[F_k,F_k]$ for any $k$. $\endgroup$
    – HJRW
    Commented Oct 26, 2011 at 11:51
  • $\begingroup$ Sorry for you to confuse. Derek Holt is right. And what is $D_\infty$? $\endgroup$
    – qkqh
    Commented Oct 26, 2011 at 13:19
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    $\begingroup$ @qkqh: I suppose $D_{\infty}$ is the infinite dihedral group $\mathbb{Z}\rtimes\mathbb{Z}/2$. $\endgroup$
    – Qfwfq
    Commented Oct 26, 2011 at 13:34

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Presumably $G=G_1 \ge G_2 \ge G_3 \ge \cdots$ is the lower central series of $G$?

So, for a free group $F$ of rank $r$, $F_2$ = $[F,F]$, and $F/[F_2,F_2]$ is the free $r$-generator metabelian group.

$F_m < [F_2,F_2]$ would imply that $F/[F_2,F_2]$ is nilpotent of class at most $m$, but since there are 2-generator metabelian groups that are not nilpotent, this is not possible for $r \ge 2$.

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