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

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.

• A Tarski Monster is a counterexample. Oct 11 '19 at 19:01
• 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? Oct 11 '19 at 19:22
• There are trivial examples without going into Tarski monsters. Check out infinite dihedral groups. This may better fit MathSE.
– YCor
Oct 11 '19 at 20:27
• (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.
– YCor
Oct 11 '19 at 22:59