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For a finite subgroup of an infinite group $G$, does its normal closure have infinite index in $G$?

Or if not, is it true when we replace $G$ by some subgroup? That is:

Let $H$ be a finite subgroup of an infinite group $G$. Is there always a subgroup $F \subseteq G$, such that the normal closure $\overline{H\cap F}^F$ of $H$ in $F$ has infinite index in $F$?

Edit: As Derek points out, the Tarski monster is a counterexample. But the known Tarski monsters seem not to be finitely presented, see this answer: https://mathoverflow.net/a/26073/61732. So one could ask the above question for $G$ finitely presented. According to the last comment on the answer in the link, it is an open question if there are finitely presented Tarski monsters.

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    $\begingroup$ A Tarski Monster is a counterexample. $\endgroup$
    – Derek Holt
    Commented Oct 11, 2019 at 19:01
  • $\begingroup$ Ok, this is really good to know, thank you! According to this answer: mathoverflow.net/a/26073/61732, the Tarski monster is finitely generated, but not finitely presented. So what if we assume $G$ to be finitely presented? $\endgroup$
    – Lukas Braun
    Commented Oct 11, 2019 at 19:22
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    $\begingroup$ There are trivial examples without going into Tarski monsters. Check out infinite dihedral groups. This may better fit MathSE. $\endgroup$
    – YCor
    Commented Oct 11, 2019 at 20:27
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    $\begingroup$ (My previous comment concerns the 1st question.) Derek solves both with Tarski monsters for finitely generated groups. But for the second question most groups yield a positive answer, for instance (a) residually finite groups (take $H$ of finite index with $H\cap F=1$), (b) non-torsion groups (take $H$ infinite cyclic). Since no infinite finitely presented torsion group is known, there is no counterexample known to the second question; also it is not known that there is no finitely presented quasi-finite group, and hence a positive answer to the second question is not known either. $\endgroup$
    – YCor
    Commented Oct 11, 2019 at 22:59

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