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I was reading a paper and find this statement in the abstract, "If $H$ has finite index in $F_m$, then $H$ has non-trivial intersection with each of the non-trivial subgroups of $F_m$" where $F_m$ is a free group of rank m. The author claims that it's a obvious statement but I don't see how. All I know is, as $[F_m:H]<\infty$, $H$ is finitely generated and free(being subgroup of a free group $F_m$).

Thanks for any help!

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  • $\begingroup$ It's true and somewhat immediate in arbitrary torsion-free groups. More generally in a group the intersection of an infinite subgroup with a finite index subgroup is infinite. $\endgroup$ – YCor Feb 9 '20 at 13:47
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Let $M$ be the core of $H$; then $[F_m:M]=k\lt \infty$. Let $K$ be any nontrivial subgroup, and let $x\in K$, $x\neq e$. Then $x^k\in K$ is nontrivial, but has trivial image in $F_m/M$, since the order of $F_m/M$ is $k$. Thus, $x^k\in K\cap M\subseteq K\cap H$. Hence $K\cap H$ is nontrivial, as claimed.

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  • $\begingroup$ Thanks for a quick solution. I was wandering where did you use that fact that group $F_m$ is free? $\endgroup$ – Infinity Feb 9 '20 at 3:12
  • $\begingroup$ does the result still holds even when the group is not free?? @Arturo Magidin $\endgroup$ – Infinity Feb 9 '20 at 3:15
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    $\begingroup$ @Infinity: Well, I used the fact that $F_m$ is torsionfree; if you don’t know that the group is torsion free, then $x^k$ could be trivial. But other than that, I don’t think you need to assume the groups are free. $\endgroup$ – Arturo Magidin Feb 9 '20 at 3:26
  • $\begingroup$ alright, that makes lot more sense now! thanks! $\endgroup$ – Infinity Feb 9 '20 at 3:43
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    $\begingroup$ @Infinity: No. If $m\gt 1$, take the subgroup generated by one free generator, and then the subgroup generated by a different free generator. $\endgroup$ – Arturo Magidin Feb 9 '20 at 4:31

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