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 group on $d$ generators.

Denote by $F_{k}$ the $k$-th term in $F$'s derived series. Put $G = F/F_k$. What is the normal subgroup growth of $G$? Explicitly, for each natural number $n$, what is the number of normal subgroups of index not greater than $n$ in $G$? An asymptotic result will also be interesting of course.

For example, the case $k=2$ is covered in the book about subgroup growth.

More generally, Let $C$ be a formation of finite groups ($C$ is closed for quotients and subdirect products). Let $G$ be the quotient of $F$ by the intersection of all normal subgroups $N \unlhd F$ for which $F/N$ is a $C$-group. Assume that $G$ is not trivial ($F$ is not residually $C$). The question is the same as before: What is the normal subgroup growth of $G$?

share|improve this question
    
You first question is: "what is the normal subgroup growth of free $k$-step solvable groups?" In your second question, I guess that $G$ is not the intersection, but the quotient by the given intersection, is it? –  YCor Apr 16 at 16:25
    
Yes, of course. I will fix that. –  Pablo Apr 16 at 17:01
    
If you are only interested in showing that it is sub-exponential, I think this follows from Corollary 2.8 in Lubotzky-Segal's book? Corollary 2.8 states that the normal subgroup growth of a free group is of strict type n^(log(n)). –  Khalid Bou-Rabee Apr 16 at 19:00

1 Answer 1

up vote 5 down vote accepted

As is stated in Lubotzky-Segal's book on this topic, very little is known about normal subgroup growth for soluble groups. However, we can say that a rank d free $k$-step solvable group, $k \geq 2$, has normal subgroup growth that is very close to $n^{\log(n)}$. Indeed, this group maps onto any metabelian group generated by <= d generators, and it is known that metabelian groups have growth bounded below by $n^{b(log(n))^{1-2/\kappa(G)}}$ (here $\kappa(G)$ is the Krull dimension, see Theorem 9.2 in Lubotzky-Segal). This gives a lower bound. For the upper bound, use Corollary 2.8 as in my comment.

I believe we can sharpen the above argument a bit to demonstrate that normal subgroup growth can distinguish $F/F_3$ from $F/F_2$. Indeed, if $F$ is a free group of rank greater than 2, then $F/F_3$ maps onto $(\mathbb{Z} \wr \mathbb{Z}^{(k)}) \rtimes C_k$ (where the action of $C_k$ in the semidirect product is the natural one) for any natural number $k$. The group $\mathbb{Z} \wr \mathbb{Z}^{(k)}$ has Krull dimension $k+1$, and so we can, from this, demonstrate that the normal subgroup growth of $F/F_3$ is never bounded below (asymptotically) by $n^{a (\log(n))^{1-1/t}}$ for any natural number $t$. However, Theorem 9.2 in Lubotzky-Segal shows that this is true for $F/F_2$.

share|improve this answer
    
The bound from above is clear to me. I do not understand what "basically"means in your answer because you are able to produce only an exponent for the logarithm which is strictly less than 1. Therefore, your method doesn't show the lower bound as you have claimed. There is a gap. –  Pablo Apr 16 at 23:30
1  
The number of generators is fixed so may be not every metabelian group is an image as it can have more generators. So it's not completely clear that you can get an arbitrary value for $k(G)$. Furthermore, it is still possible that the growth rate will be strictly smaller than $n^{log(n)}$. For example, it could be $n^{log(n)/log(log(n))}$. –  Pablo Apr 17 at 7:35
    
Thanks Pablo. I clarified my answer. I am sorry for the confusion. –  Khalid Bou-Rabee Apr 17 at 12:50

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.