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Let $F$ be a nonabelian finitely generated free profinite group, and let $H \leq F$ be a finitely generated closed subgroup. Must there be some open subgroup $H \leq U \leq F$, and a closed normal subgroup $N \lhd U$ such that $H \cap N = \{1\}$ and $HN = U$?

Note that the analogous statements are true if $F$ is free pro-$p$ group for any prime number $p$, if $F$ is a free group, if $F$ is a surface group and probably in other cases (which?).

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  • $\begingroup$ In answer to your '(which?)', as usual the right hypothesis on the subgroup is not 'finitely generated' but 'quasiconvex'. Haglund and Wise showed that any quasiconvex subgroup of any virtually special group is a virtual retract. $\endgroup$
    – HJRW
    Commented Jan 28, 2015 at 16:37
  • $\begingroup$ Do you have a reference for the free pro-p case? $\endgroup$
    – HJRW
    Commented Jan 28, 2015 at 16:38
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    $\begingroup$ @HJRW For the pro-$p$ case this is a Frattini argument given in Ribes-Zalsskii (2-nd edition) Theorem 9.1.19 which shows that $H$ is a free factor of some open $H \leq U \leq F$ so it is a retract since we can map $H$ to itself and its free complement in $U$ to $1$ producing a retraction. $\endgroup$
    – Pablo
    Commented Jan 28, 2015 at 17:15
  • $\begingroup$ One obstruction might come from finitely generated (abstract) subgroups of $F$ which are not themselves free. Parafree groups give examples of these. $\endgroup$
    – HJRW
    Commented Jan 28, 2015 at 19:35

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