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Let $G$ be an infinite, discrete, countable group. Can $G$ have a translation-invariant ultrafilter? An ultrafilter $\mathcal{F} \subset 2^G$ is translation-invariant if $A \in \mathcal{F}$ implies $g A \in \mathcal{F}$ for all $g \in G$.

The existence of a translation-invariant ultrafilter can be easily seen to be impossible if (i) $G$ is non-amenable, or (ii) $G$ has finite index subgroups, or more generally (iii) $G$ admits a subset that has finitely many disjoint translates whose union is the entire group.

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2 Answers 2

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Every nontrivial group $G$ satisfies your condition (iii).

To see why, note that for any subgroup $H$ of $G$, the left action of $H$ on $G$ is free. So if $X$ is a set of representatives of the orbits of said action, then $G=\sqcup_{h\in H}hX$.

If $H$ is a nontrivial subgroup of finite order, then $G=\sqcup_{h\in H}hX$ satisfies condition (iii).

If $G$ has no elements of finite order, let $H$ be the subgroup generated by some $h\in G\setminus\{e\}$ and let $H_2$ be the subgroup generated by $h^2$. Then $H=H_2\sqcup hH_2$, so $G=\sqcup_{h\in H}hX=(H_2X)\sqcup h(H_2X)$ satisfies condition (iii) again.

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    $\begingroup$ Thanks! Would you mind adding a few more details? Why are these two sets disjoint, and why their union is $G$? $\endgroup$
    – Vladimir
    Commented Jun 10 at 2:26
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    $\begingroup$ I added a bit more detail. Another way to view $X$ is as a set of representatives of the orbits of the left action of $H$ on $G$ (that way it becomes clear that $G=\sqcup_{h\in H}hX$, checking each orbit separately) $\endgroup$
    – Saúl RM
    Commented Jun 10 at 2:49
  • $\begingroup$ Whoops yes. I will edit $\endgroup$
    – Saúl RM
    Commented Jun 11 at 1:52
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I had considered this in the (currently in revision) paper "Near actions", in the setting of group actions:

A $G$-action on a set $X$ preserves an ultrafilter iff every finitely generated subgroup of $G$ fixes a point.

For $X=G$, action by translation, this happens iff $G=\{1\}$.

Copy of the proof from my draft: "Let $f$ be an injective self-map of a set $X$ with no fixed point. Then there exists, by an easy argument [$*$], a partition $X=X_1\sqcup X_2\sqcup X_3$ such that $f(X_i)\cap X_i=\emptyset$.

A straightforward consequence is that $f$ preserves no ultrafilter. Hence, if $f$ is an arbitrary permutation, the set of ultrafilters fixed by $f$ are those ultrafilters on the fixed-point-set of $f$. Hence, if $G$ is generated by a finite subset $S$ and $X$ is a $G$-set, the set of ultrafilters fixed by $G$ equals the set of ultrafilters on the fixed-point-set of $S$, i.e., of $G$. This proves the forward implication in the first assertion."


If $X=G\neq\{1\}$, the whole can be simplified: just choose $g\in G\smallsetminus\{1\}$. The 1st paragraph can be simplified, since $g$ acts on $G$ through cycles all of the same length. And the second paragraph also simplifies since then $g$ itself preserves no mean on $G$.


Proof of [$*$] when $f$ is a permutation (enough here). Arguing orbitwise, we can suppose that $f$ is a single cycle (by assumption, of length $\ge 2$). If the cycle has even or infinite length, enumerate it; let $X_1$ resp. $X_2$ be the set of points with odd resp. even index, and $X_3$ be empty. Then the partition works. If the cycle has odd length, say $(x_0,\dots,x_n)$, $n\ge 2$ even, choose $X_1=\{x_1,x_3,\dots,x_{n-1}\}$, $X_2=\{x_2,x_4,\dots,x_n\}$ and $X_3=\{x_0\}$.

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  • $\begingroup$ [*] is (a slightly special case of) the well-known three-set lemma. There were some posts about it recently. $\endgroup$ Commented Jun 11 at 5:41
  • $\begingroup$ @EmilJeřábek yes I'm aware it's well-known (I remember a graph-theoretic generalization of it by Erdös). I don't have full references in mind. $\endgroup$
    – YCor
    Commented Jun 11 at 6:55

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