# Can a closure make the index finite?

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?

• If H is not finitely generated it is possible. For instance, take $Q$ to be a finitely generated group with no finite quotients (there are many such), and let $H$ be the kernel of some surjection $F\to Q$. – HJRW Aug 1 '14 at 15:46
• @HJRW I wrote a similar but more complicated example at the same time. – Benjamin Steinberg Aug 1 '14 at 15:55
• Whoops, I should have written 'take $Q$ to be an infinite finitely generated group with no finite quotients...' – HJRW Aug 1 '14 at 16:11
• every group has a finite quotient – YCor Aug 1 '14 at 17:41
• As Yves helpfully points out, I should in fact have written 'take $Q$ to be an infinite finitely generated group with no non-trivial finite quotients...' – HJRW Aug 2 '14 at 2:14

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.)