# finite index subgroups in free group non-trivial intersection with each of the non-trivial subgroups of the free group.

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!

• 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. – YCor Feb 9 '20 at 13:47

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.
• Thanks for a quick solution. I was wandering where did you use that fact that group $F_m$ is free? – Infinity Feb 9 '20 at 3:12
• @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. – Arturo Magidin Feb 9 '20 at 3:26
• @Infinity: No. If $m\gt 1$, take the subgroup generated by one free generator, and then the subgroup generated by a different free generator. – Arturo Magidin Feb 9 '20 at 4:31