3
$\begingroup$

Let $F_n$ be a free profinite group of finite rank $n$ and let $V_k$ denote the intersection of all open subgroups of $F_n$ of rank at most $k$ ($k \in \mathbb{N}$).

My questions are:

Can I explicitly compute the index of $V_k$ in $F_n$ for all $k \in \mathbb{N}$?

Or if not, maybe for some special $k$, e.g. $k = p$, $p$ prime?

I came to that question when I looked up the proof that the automorphism group of a finitely generated free profinite group is profinite (see e.g. Proposition 4.4.3 in Profinite Groups by Ribes-Zalesskii).

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Could you clarify which notion of rank you are using? $\endgroup$
    – Ian Agol
    Jan 21, 2014 at 16:09
  • 1
    $\begingroup$ By "rank" I mean the smallest (topologically) generating set $\endgroup$
    – user45818
    Jan 21, 2014 at 17:29
  • 2
    $\begingroup$ I think there is work of Bou Rabee et al on this. $\endgroup$
    – Benjamin Steinberg
    Jan 21, 2014 at 21:51

1 Answer 1

Reset to default
3
$\begingroup$

Ok, I think I understand the question now. To clarify, the free profinite group of rank $n$ may be regarded the profinite completion $\hat{K}_n$ of $K_n=\mathbb{Z}^{\ast n}$ the free group of rank $n$. An open subgroup $H< F_n$ must be finite index since $F_n$ is compact, and the cosets of $H$ cover $F_n$. Thus, $H$ induces a finite-index subgroup $J< K_n$. Then $J$ is a free group of rank $$rank(J)=1+[K_n:J](n-1)=1+[F_n:H](n-1).$$ Since $J$ is finite-index in $K_n$, $H=\hat{J}$, the profinite completion of $J$, and therefore $H$ is a free group of rank $rank(H)=k=1+[F_n:H](n-1)$. Thus, your question is equivalent to asking for the index of the intersection of all subgroups of $K_n$ of index $\leq \frac{k-1}{n-1}$. I'm not sure this has been worked out, except possibly for $k=2n-1$ corresponding to index $2$ subgroups, where of course the answer is index $2^n$. One may obtain a crude upper bound by counting homomorphisms of $K_n$ to $S_j$ (where $j=\frac{k-1}{n-1}$), which is at most $j!^n$. So one has a bound on the index of the intersection of stabilizers of all of these homomorphisms of $j^{j!^n}$ :).

Improve this answer
$\endgroup$
6
  • $\begingroup$ Can we do a little better by taking into account that the intersection is necessarily normal? $\endgroup$
    – HJRW
    Jan 21, 2014 at 21:40
  • $\begingroup$ it´s normal, and even characteristic $\endgroup$
    – user45818
    Jan 22, 2014 at 10:14
  • 1
    $\begingroup$ @user45818: you don't have to check this answer - a more complete one might appear. In any case, I suspect this might be hard in general, except possibly for some other small cases such as index 3 or 4. You might have a look at the book on subgroup growth by Lubotzky & Segal. link.springer.com/book/10.1007%2F978-3-0348-8965-0 $\endgroup$
    – Ian Agol
    Jan 22, 2014 at 18:09
  • 3
    $\begingroup$ Theorem 6.1 in our paper, www-personal.umich.edu/~khalidb/intgrowthfree.pdf, (joint with Martin Kassabov, Ian Biringer, and Francesco Matucci) seems relevant here. In particular, we compute the growth of the function defined to be the index of the intersection of all subgroups of index n in a fixed nonabelian free group (the answer is (e^(n^n)). $\endgroup$
    – Khalid Bou-Rabee
    Feb 3, 2014 at 18:09
  • 1
    $\begingroup$ @KhalidBou-Rabee: Cool, thanks for the link. I did a quick web search, but didn't notice it. I suppose even obtaining an exact inductive formula is probably prohibitive. $\endgroup$
    – Ian Agol
    Feb 3, 2014 at 20:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .