2
$\begingroup$

This question is a development of my previous question.

Let $G$ be a finitely generated group acting transitively on an infinite set $X$ so that for every $g\in G$ and $x\in X$ the $g$-orbit $\{g^nx:n\in\mathbb Z\}$ of $x$ is finite.

Let $S=S^{-1}$ be a finite set of generators of $G$ and $\Gamma=(X,E)$ be the "Cayley" graph of the action. In this graph a doubleton $\{x,y\}\subset X$ is an edge of the graph $\Gamma$ iff $y\in Sx$.

Problem. Has the graph $\Gamma$ one end (which means that for any finite subset $F\subset X$ the graph $\Gamma\setminus F$ has only one unbounded connected component)?

$\endgroup$
1
  • 2
    $\begingroup$ Note: a finitely generated group $G$ for which every infinite connected Schreier graph $G/H$ is 1-ended is said to have Property FW. $\endgroup$ – YCor Apr 10 '19 at 7:16
6
$\begingroup$

No, the Grigorchuk group, which is torsion, admits a 2-ended connected Schreier graph. See for instance http://www.math.tamu.edu/~yvorobet/Research/Schreier.pdf


Edit: Here's a more elementary example, just assuming the bare existence of an infinite finitely generated torsion group, and also produces examples with bounded torsion, and examples with an infinite space of ends.

Let $\Gamma,C$ be any finitely generated groups (in a first reading, one can assume that $\Gamma$ is 1-ended and that $|C|=2$). We consider the Schreier graph $X$ of the wreath product $G=C\wr\Gamma=\Gamma\ltimes C^{(\Gamma)}$ with respect to the subgroup $C^{(\Gamma\smallsetminus\{1\})}$. We can view $X$ as the product $\Gamma\times C$, where $\Gamma$ and $C$ act as $$g\cdot (g',c')=(gg',c');\quad c\cdot (g',c')=(g',c') \;(g\neq 1)\quad c\cdot (1,c')=(1,cc').$$

One fixes finite generating subsets $S_\Gamma,S_C$, so that $S=S_\Gamma\cup S_C$ generates the wreath product $G$.

If $|C|=2$, the Schreier graph $X$ can therefore be described as the disjoint union of two copies of the Cayley graph of $G$, joined by a single edge joining their basepoints. If $G$ is 1-ended, this is 2-ended.

In general ($C$ arbitrary) $X$ is obtained from the Cayley graph of $C$ by "planting" Cayley graphs of $G$ at each point.

In particular, say when $C$ has finitely many ends, the number of ends of $X$ is equal to $|C|$ times the number of ends of $G$. This can be chosen to be infinite, say when both $C$ and $\Gamma$ are both infinite Burnside groups, in which case $G$ even has bounded torsion.

If $\Gamma,C$ are both finitely-ended, the space of ends of $X$ is countable anyway: if $C$ is infinite with 1 or 2 ends, then this is just a 1-point or 2-point compactification of an infinite countable set.

So, when $G$ is torsion, the resulting space of ends we thus obtain are all finite set, and the 1-point compactification of an infinite countable set.

In general, this can also produce $X$ to have a space of ends that is uncountable with isolated points (e.g., when $\Gamma=\mathbf{Z}$ and $C$ is $\infty$-ended).

$\endgroup$
4
  • $\begingroup$ I'm not sure that I know right now an example with infinitely many ends (countably or uncountably), although I tend to guess both exist. $\endgroup$ – YCor Apr 9 '19 at 9:46
  • $\begingroup$ Thank you for the quick answer. By the way, Vorobets is by origin from my native city Lviv and I even have 3 joint papers with him, see e.g. (arxiv.org/abs/0902.1556). $\endgroup$ – Taras Banakh Apr 9 '19 at 9:53
  • $\begingroup$ For my purposes I need just 1 end actions (in this case I can connect some distant points by paths that miss a selected finited set). $\endgroup$ – Taras Banakh Apr 9 '19 at 9:55
  • $\begingroup$ The above answers your questions anyway, I'm just mentioning that I don't know about some weakening. Concerning, I don't claim proper attribution so I lazily posted a link with a picture. Many people have studied the Schreier graph of branched groups on the boundary of the rooted tree. $\endgroup$ – YCor Apr 9 '19 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.