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

Hi all,

This is a question about the groups $H_{n,s}$ introduced by Völklein in his book "Groups as Galois groups", §7.1, and defined as follows: let $N$ be the intersection of all normal subgroups of index $\le n$ in $F_s$, the free group on $s$ generators, and define $H_{n,s} = F_s / N$.

Have these groups been studied? Do they have a name? Is it possible to compute their orders, at least in some cases?

For example for $s=1$, then $H_{n,1}$ is the cyclic group of order $lcm(1, 2, \ldots, n)$.

In Völklein's book, these are introduced primarily to avoid talking about profinite groups (the inverse limits of the $H_{n,s}$, with fixed $s$, is the free profinite group of $s$ generators).

Any information you may have on these will be great appreciated.

Thanks!

Pierre

share|improve this question
add comment

2 Answers

I am preparing a paper with Ian Biringer, Martin Kassabov, and Francesco Matucci, where we study the growth of the index of the intersection of all normal subgroups of index at most $n$ in a given group. We call this the study of intersection growth of the group. In your notation, for the free group of rank $s$, $F_s$, and every natural number $n$, the intersection growth function, $i_{F_s}(n)$, is defined to be the order of $H_{n,s}$. As general motivation for studying this growth, for a general group $\Gamma$, we show that the growth of $i_\Gamma(n)$ determines the dimension of the profinite completion of $\Gamma$.

This paper (which we may split into two) has some examples worked out: we have precise calculations for this growth for nilpotent groups and certain arithmetic groups. In the case of a rank $s$ free group, we found the lower bound $e^{n^{s-2/3}}$ (which we compute by finding the precise growth when one only intersects maximal subgroups).

share|improve this answer
1  
Nice! I'm eager to read that paper. –  Pierre Apr 19 '13 at 9:50
    
The paper should appear on the arxiv this week (and is available on my website). –  Khalid Bou-Rabee Sep 30 '13 at 21:52
add comment

Yes, you can compute their orders in a few easy cases. For example, if $p$ is prime, then $H_{p,s}$ is elementary abelian of order $p^s$, and $H_{p^2,s}$ is homocyclic abelian of order $p^{2s}$. I expect it would not be too hard to describe the structure when $n=p^3$, or when $n=pq$ for distinct primes $p,q$.

For small $n$ you could compute $H_{n,s}$ directly. For example, when $n=6$, it has order 972.

But it would be very difficult to compute the order more generally. Some interesting quotients of $H_{n,s}$ have been studied. For example, if we take $s=2$, $n=60$, and let $K$ be the intersection of the kernels of homomorphisms of $F_2$ onto $A_5$, then $F_2/K$ is a direct product of 19 copies of $A_5$.

share|improve this answer
2  
I am afraid that misread the definition of $H_{n,s}$ in your question. I took the definition to be $F_s/N$ where $N$ is the intersection of all normal subgroups of index exactly $n$ (rather than at most $n$, which is what you wrote). So Khalid Bou-Rabee's answer is more useful than mine for the question that you asked! –  Derek Holt Apr 18 '13 at 17:40
add comment

Your Answer

 
discard

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.