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.


2 Answers 2


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.

  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ 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

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 $X_3=\{x_0\}$.

  • $\begingroup$ [*] 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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