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Let $G$ be a group of polynomial growth, let $S$ be a finite generating set for $G$ and let $B_n$ be the set of elements of $G$ given by words of length $\leq n$ in the generating set $S$. Is there an upper bound for the numbers $\displaystyle\frac{\#B_{2n}}{\#B_n}$, with $n\in\mathbb{N}$?

Context: I am looking for sufficients conditions for a group to have a certain kind of Folner sequence, and if this is true then polynomial growth groups would have such Folner sequences. In case anyone is curious, the Folner sequences I am looking for are left-Folner sequences $(F_N)_{N\in\mathbb{N}}$ such that $\exists\varepsilon>0$ $\forall N\in\mathbb{N}$ there exist infinitely many values of $L\in\mathbb{N}$ such that there is a set $A\subseteq F_L$ which is a disjoint union of right translates of $F_N$ and satisfies $\frac{\#A}{\#F_L}>\varepsilon$. I call such a Folner sequence "self-covering".

Definition of polynomial growth I am using: A finitely generated group has polynomial growth if there exists a generating set $S$ and some big constant $N\in\mathbb{N}$ such that for all $n\in\mathbb{N}$, the ball $B_n$ defined above has $\leq N\cdot n^N$ elements.

Saúl RM
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    Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Shaun Jan 21 '24 at 22:36
  • Start by defining polynomial growth for groups. – Moishe Kohan Jan 21 '24 at 23:19
  • Do you know about lower volume bounds for groups of polynomial growth? Do you know Gromov's theorem? – Moishe Kohan Jan 21 '24 at 23:53
  • I don't know about lower volume bounds. It seems reasonable that one could use Gromov's theorem to prove one, but I asked in case it was a well-known fact (I am not an expert in these topics) – Saúl RM Jan 21 '24 at 23:57
  • Gromov's theorem implies a lower growth bound of the form $an^N$. Can you finish the proof now? – Moishe Kohan Jan 22 '24 at 00:06
  • Yes, using that it's obvious. I think I see why that's true, I should write a detailed answer one of these days – Saúl RM Jan 22 '24 at 00:07

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According to Gromov's polynomial growth theorem, each finitely generated group of polynomial growth is virtually nilpotent. According to the Bass–Guivarc’h Theorem (see references in the same link) the for virtually nilpotent groups the growth function is $\sim n^d$ for some $d$ that can be computed in terms of the lower central series of the finite index nilpotent subgroup, meaning that $$ a n^d \le |B_n|\le b n^d $$ for all $n$ and some fixed positive constants $a, b$. From this, it is immediate that $$ \sup_n \frac{|B_{2n}|}{|B_n|} <\infty. $$

Moishe Kohan
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