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Let $F$ be a nonabelian free finitely generated group, and $F = G_0 \rhd G_1 \rhd G_2 \dots$ a strictly descending subnormal chain of subgroups ($G_n \lhd G_{n-1}$ for each $n \in \mathbb{N}$) each closed in $F$ (with respect to the profinite topology). Set $$H = \bigcap_{n \in \mathbb{N}} G_n$$

Is it possible for $H$ to be nontrivial and finitely generated?

Since it seems possible, I further assume that $[F : G_n] < \infty$ for $n \in \mathbb{N}$ making the assumption on being closed in the profinite topology superfluous.

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    $\begingroup$ You can obtain $H=\{1\}$. Are you looking for $H$ infinite? $\endgroup$
    – YCor
    Commented Sep 15, 2014 at 17:01
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    $\begingroup$ Any subgroup closed in the pro-$p$ topology, for some prime $p$, will be an intersection open pro-$p$ subgroups, each of which is a subnormal subgroup of $p$-power index. For example any free factor (or a retract) is closed in pro-$p$ topology. $\endgroup$
    – Ashot Minasyan
    Commented Sep 15, 2014 at 19:54
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    $\begingroup$ Of course the profinite topology is finer than the pro-$p$ topology, so any closed set in the latter is also closed in the former. Not every f.g. subgroup is an intersection of subnormal subgroups in a free group, though. For example any cyclic subgroup of $S_3$, of order $2$, is not subnormal. Taking its preimage in the free group of rank $2$ one gets a subgroup of index $3$ which is maximal but not (sub)normal. $\endgroup$
    – Ashot Minasyan
    Commented Sep 15, 2014 at 20:03
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    $\begingroup$ @Pablo, Ashot has already provided counterexamples to your conjecture. $\endgroup$
    – HJRW
    Commented Sep 16, 2014 at 9:08
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    $\begingroup$ Right, missed that. @Ashot: I would gladly accept your comment as an answer. $\endgroup$
    – Pablo
    Commented Sep 16, 2014 at 9:42

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