Let $F$ be a free finitely generated group, $H \leq F$ of infinite index. Let $c : F \rightarrow \hat{F}$ be the embedding in the profinite completion. Denote by $\tilde{F}, \tilde{H}$ the closure of $c(F), c(H)$ respectively. Is it possible that $[\tilde{F} : \tilde{H}] < \infty$?


Does it change anything if $H$ is finitely generated?



Can a closure make the index finite?
If $H$ is fg then by Marshall Hall's theorem it is closed and so $\overline{H}\cap F=H$. But intersecting a finite index closed subgroup (=open subgroup) with $F$ gives a finite index subgroup. So this is impossible if $H$ is fg. 


**Added**. If $H$ is infinitely generated then the closure could be finite index. Choose $n$ so that the free Burnside group of exponent $n$ is infinite. Let $H$ be the verbal subgroup of $F$ associated to the word $x^n$. Then H has infinite index but its closure has finite index by Zelmanov's solution if the restricted Burnside problem. (The closure is the kernel of the map to the free group in the variety generated by finite groups of exponent $n$ which is a finite group by Zelmanov.)


