Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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 at 15:46
@HJRW I wrote a similar but more complicated example at the same time. –  Benjamin Steinberg Aug 1 at 15:55
Whoops, I should have written 'take $Q$ to be an infinite finitely generated group with no finite quotients...' –  HJRW Aug 1 at 16:11
every group has a finite quotient –  YCor Aug 1 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 at 2:14

1 Answer 1

up vote 5 down vote accepted

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.)

share|improve this answer
Great! I need this to give an elementary proof of the Karras-Solitar theorem and this seems to be the only point where I need to invoke some theorem. Is there an argument which does not use Hall's theorem? –  Pablo Aug 1 at 15:04
Which Karrass-Solitar Thm? They have many. –  Benjamin Steinberg Aug 1 at 15:42
@Pablo, Hall's theorem is the basic result in this area (and easy to prove). You won't be able to avoid it without essentially reproving it---and furthermore, you should certainly aim to understand its proof if you want to do anything new. I outlined a proof that's usually attributed to Stallings, though it's essentially the same as Hall's original proof, here: ldtopology.wordpress.com/2008/12/01/… . –  HJRW Aug 1 at 15:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.