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If I have a subgroup $S$ of a free group $\mathcal{F}_m$, what can I say about the behaviour of the descending sequence of subgroups $\left< S, \Gamma_c(\mathcal{F}_m) \right>$ (where $\Gamma_c(\mathcal{F}_m)$ is the $c$-th term of the lower central series)?

In particular, does this sequence always converge to $S$ or not?

(Below is what I've tried so far)

If after some point $S$ contains $\Gamma_c$, then this is obviously true, but that's only the case if the corresponding f.g. group $\mathcal{F}_m/S$ is nilpotent, so there's still a lot to go. If after some point $\Gamma_c \cap S =\{1\}$ then it's also true, but that would mean $S$ is nilpotent which is impossible since $S$ is free as a subgroup of a free group.

I've tried to use the diamond isomorphism theorem to use that $S \Gamma_c/S \cong \Gamma_c/S \cap \Gamma_c$, but it doesn't seem clear to me that my sequence converging to $S$ has much of a connection with $S \Gamma_c/S$ converging to the trivial group - indeed it's not entirely clear what "converging" means in the context of these quotients.

Thanks a lot!

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  • $\begingroup$ (By the way: an answer only in the case where S is finitely generated is fine for my purposes) $\endgroup$ May 5, 2020 at 11:54
  • $\begingroup$ In general you can't say much. For example, if $m \ge 2$ and $S \lhd \mathcal{F}_m$ is such that $\mathcal{F}_m/S \cong A_5$, then $\langle S,\Gamma_c \rangle = \mathcal{F}_m$ for all $c$. $\endgroup$ May 5, 2020 at 20:06
  • $\begingroup$ Terms of the lower central series define a basis of neighborhoods of $1$, which gives rise to a pro-nilpotent topology on $\mathcal{F}_m$. Your sequence of subgroups converges to $S$ iff $S$ is closed in this topology. $\endgroup$ May 5, 2020 at 20:11
  • $\begingroup$ Thanks a lot for your very complete answer! $\endgroup$ May 6, 2020 at 13:42

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