# Group with a translation invariant ultrafilter

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.

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.

• Thanks! Would you mind adding a few more details? Why are these two sets disjoint, and why their union is $G$? Commented Jun 10 at 2:26
• 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) Commented Jun 10 at 2:49
• Whoops yes. I will edit Commented Jun 11 at 1:52

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 $$X_3=\{x_0\}$$.

• [*] is (a slightly special case of) the well-known three-set lemma. There were some posts about it recently. Commented Jun 11 at 5:41
• @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.
– YCor
Commented Jun 11 at 6:55