Hello,


Let $G$ be an infinite finitely generated discrete group. I call an infinite set $S$ irregular iff for every $g\in G$, $g\neq 1$, we have that $S\cap gS$ is finite. For example $\{z^3|z\in\mathbb{Z}\}$ is irregular in $\mathbb{Z}$. Now my easy to state question: Does every free ultrafilter on $G$ contain at least one irregular subset?


The following is true and might probably be useful: $G$ acts freely on the space of all ultrafilters (the Stone-Cech compactification of $G$ as a discrete space).


free ultrafilters: <http://en.wikipedia.org/wiki/Ultrafilter> 



Free ultrafilters on groups and irregularity
No, a free ultrafilter on the additive group of integers need not contain an irregular set. The Galvin-Glazer proof of Hindman's theorem (which is nowadays the standard proof of that theorem) begins by showing the existence of idempotent ultrafilters $U$ on $\mathbb N$. I won't bother to define "idempotent" here, since what I need is not the definition but the following consequence of it. If $X$ is any set in $U$, then there is an infinite set $\{x\_0<x\_1<\dots\}$ of positive integers such that any finite sum of distinct $x\_i$'s is in $X$. In particular, the intersection of $X$ and $x\_0+X$ contains $x\_0+x\_i$ for all $i>0$, and therefore $X$ is not irregular.


