# Ends of a group and Spherical growth function

I am trying to get some intuition about a spherical growth function of a group, using the notation from A. Mann $$s_G(m)=\sum_{n=0}^m a_G(n)$$, where $$s_G(m)$$ is the cumulative growth function and $$a_G(n)$$ the strict growth function.

For example for $$G=\mathbb{Z}$$ with one generator, $$a_G(n)=2$$ for all $$n$$. And in general, I can not assume that $$a_G(n)$$ is not eventually constant, for example for free groups, $$a_G(n)$$ increases as long as $$n$$ does, but as I mentioned, in $$\mathbb{Z}$$ it doesn't. And my intuition tells me that if $$G$$ has one end or infinite ends, $$a_G(n)$$ has enough "space" to grow together with $$n$$. This is something like:

CLAIM: If $$G$$ is a group with one end or infinite ends, then $$a_G(n)$$ is not eventually constant. Does this make sense? I don't have any ideas trying to prove this statement. Or there are groups that have different behaviors? Any intuition about this would be appreciated.

Edit: Maybe I need another hypothesis similar to: If $$\Gamma$$ is the Cayley Graph of $$G$$ with a finite group of generators, and it has the property that $$\Gamma - B_n(u_0)$$ is a connected graph where $$B_n(u_0)$$ is the ball with center $$u_0$$ and $$n\in \mathbb{N}$$ for any vertex $$u_0$$ and $$n\in \mathbb{N}$$??

• It sounds like the growth function depends on the choice of generating set, right? I wonder if even $\mathbb{Z}$ might have a periodic, rather than constant, growth function if you choose a clever generating set. Commented Sep 3, 2021 at 2:04

## 1 Answer

Suppose $$a_G(n)$$ is eventually constant, and that constant is not zero. Then the cumulative growth function has linear bounds above and below.

By Gromov's polynomial growth theorem, $$G$$ has a finite index nilpotent subgroup.

The Bass-Guivarch theorem gives the exact degree of polynomial growth of a finitely generated nilpotent group, and in particular a nilpotent group that has a cumulative growth function with linear bounds above and below has a finite index subgroup isomorphic to $$\mathbb Z$$.

The group $$G$$ therefore has a finite index subgroup isomorphic to $$\mathbb Z$$, and so $$G$$ has two ends.

• Thanks! Do you know any book or article where I can learn more about this Bass-Guivarch theorem? I am not so familiarized with central series and I would like to understand this better Commented May 25, 2021 at 12:25
• In that link I gave, next to Bass' name there is a citation to his paper. Commented May 25, 2021 at 13:19