Let $G$ be a nilpotent group generated by $d$ elements.

Is there a function $r(d)$ such that every (necessarily finitely generated) subgroup $H$ of $G$ can be generated by at most $r(d)$ elements?

Thanks in advance!

  • $\begingroup$ You can simplify your question by removing "and $r(d)$ is the least such integer". Finding the minimal $r(d)$ (if finite) might be rather more difficult. $\endgroup$ – Yuval Filmus May 24 '11 at 1:41
  • $\begingroup$ Even «is there a bound on the number of elements needed to generate a finitely generated subgroup of a nilpotent group generated by $d$ elements?», without introducing unneeded notation :) $\endgroup$ – Mariano Suárez-Álvarez May 24 '11 at 1:52
  • $\begingroup$ A f.g. nilpotent group is in fact supersolvable, so all subgroups are finitely generated. But I think the answer is "no" if $r$ is only a function of $d$. You might want to consider letting $r$ be a function of the nilpotency class as well (and then I believe the answer is yes). $\endgroup$ – user641 May 24 '11 at 1:55
  • $\begingroup$ @Mariano: Very elegant! Please, feel free to edit the question. @Steve: I just wanted to emphasize on the number of generators. I also believe the answer is yes! $\endgroup$ – Florian Pei May 24 '11 at 3:03

If your function depends only on the number of generators of $G$, then the answer is "no", except in the trivial case of $d=1$.

Let $G$ be the relatively free nilpotent group of rank $2$ and class $c$. By the Hall Basis Theorem, the basic commutators of weight $m$, $1\leq m\leq c$, are a basis for $G_m/G_{m+1}$. By a Theorem of Witt, the number of basic commutators of weight $m$ in 2 generators is $$\frac{1}{m}\sum_{d|m}\mu(d)2^{m/d}$$ where $\mu(k)$ is the Moebius function ($\mu(1)=1$, $\mu(m)=0$ if $m$ is divisible by the square of a prime, and $\mu(m)=(-1)^k$ if $m$ is the product of $k$ distinct primes).

(More generally, the number of basic commutators in $r$ generators of weight $n$ is given by $$M_r(n) = \frac{1}{n}\sum_{d|n}\mu(d)r^{n/d},$$ a formula I will use below.)

In particular, letting $m=c=p$ be a prime, the number of basic commutators of weight $p$ is $$\frac{1}{p}\left(\mu(1)2^p + \mu(p)2\right) = \frac{2^p-2}{p}.$$ As $p\to\infty$, this quantity goes to $\infty$, so for any $N$ we can find a subgroup of a $2$-generated nilpotent group and a subgroup of it that cannot be generated by $N$ or fewer generators. Replace the relatively free group of rank $2$ with the relatively free group of rank $d$ to show that for every prime $p$ there is a $d$-generated nilpotent group with a free abelian group of rank $(d^p-d)/p$.

If you want to stick to finite nilpotent groups, you can still do it, either by considering $p$ groups with $p\gt c$ (the small class case, where the groups are regular and things work out); or if you want to fix $p$, then taking generators of appropriately high order guarantees enough nontrivial linearly independent generators in the $c$th term of the lower central series; the basic commutators are no longer necessarily linearly independent in this situation, but there aren't enough relations to swamp out the exponential growth in the relatively free case.

If you replace your function $f$ with a function of two arguments, one being the number of generators of $G$ and the other being the class $c$ of $G$, then you are in better luck. Let $H$ be a subgroup of $G$, and consider the central series for $H$ given by $$1\lt H\cap G_c \lt H\cap G_{c-1} \lt\cdots \lt H\cap G_2\lt H.$$ For $k$, $1\leq k\leq c$, we have $$\frac{H\cap G_k}{H\cap G_{k+1}} = \frac{H\cap G_k}{(H\cap G_k)\cap G_{k+1}} \cong \frac{(H\cap G_k)G_{k+1}}{G_{k+1}}.$$ The last group is a subgroup of $G_k/G_{k+1}$; since $G_k/G_{k+1}$ is abelian generated by at most $$M_r(k) = \frac{1}{k}\sum_{t|k}\mu(t)r^{k/t}$$ elements (the basic commutators of weight $k$ in $r$ generators), then $(H\cap G_k)/(H\cap G_{k+1})$ is generated by at most $M_d(k)$ elements. Putting together generating sets for each of these slices of $H$ gives that $H$ can be generated by at most $$\sum_{k=1}^c M_d(k)$$ elements.

In fact, this estimate is almost certainly overkill, but it shows that there is an upper bound for the number of generators that subgroups of $G$ may require. By finding the the corresponding number for the relatively free group of rank $d$ and class $c$ over all its subgroups and taking the maximum, you obtain the value of $r(d,c)$.

Note also that there is no bound for the nilpotency of a nilpotent group in terms of the number of generators (except in the trivial case where the number of generators is $1$), even if we require the group to be a finite $p$-group and specify the prime. If you restrict both the $p$ and the exponent of the group, then of course the Restricted Burnside Problem comes to your rescue to give you a bound for the nilpotency, so that you can also find a function of the form $r(d,p^m)$ where $d$ is the number of generators, $p$ is a prime, $m$ is a positive integer, and the group is required to be a $d$-generated finite $p$-group of exponent $p^m$.

  • $\begingroup$ That's the answer I was looking for. Thank you so much! $\endgroup$ – Florian Pei May 24 '11 at 4:23

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