# Free ultrafilters on groups and irregularity

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

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In the definition of irregular, I guess you want $g$ to be non-identity? –  Joel David Hamkins Nov 17 '12 at 11:31
To save people searching through the wikipedia article: free = non-principal. –  HJRW Nov 17 '12 at 12:03
@Joel: Yeah, I'll correct that, thx. –  Werner Thumann Nov 17 '12 at 14:05

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.
Any idempotent ultrafilter on $\mathbb N$ generates (by closing it under supersets) an idempotent ultrafilter on $\mathbb Z$. More generally, the canonical embedding $\beta\mathbb N\to\beta\mathbb Z$ is a homomorphism of semigroups. –  Andreas Blass Nov 19 '12 at 1:37