# Normal subgroups of finite index in free groups

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

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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).

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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

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 $A_5$.

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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