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!
Best Answer
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.