Is there a faithful transitive action of $G = \mathrm{PSL}_2(\mathbb{Z})$ on $\mathbb{Z}$ such that orbits under each $g \in G$ are finite?

  • $\begingroup$ What does "orbits under each $g$" mean? Do you mean that for all $n \in \mathbb{Z}$ and for all $g$ the set $\{g^k \cdot n \mid k \in \mathbb{Z}\}$ is finite? Is there some reason why you require $G$ to act on $\mathbb{Z}$? Or would you be happy with an action on any countable set? $\endgroup$ – Sam Nead Jan 18 '15 at 17:39
  • $\begingroup$ @SamNead I would be happy with an action on any countable set and you got the local finiteness condition correctly. $\endgroup$ – Pablo Jan 18 '15 at 17:41
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    $\begingroup$ The existence of such an action of $G = \text{PSL}_2\mathbb{Z}$ on a countable set (call in $C$) implies the existence of such an action for the rank two free group $G = F_2$, because $\text{PSL}_2 \mathbb{Z}$ has a normal $F_2$ subgroup of index 6. Any finite index normal subgroup $N$ would do. That is, by restricting the action of $\text{PSL}_2\mathbb{Z}$ to the subgroup $N$, and then restricting the action of $N$ to any single orbit $C_0$ of the finitely many orbits of $N$ on $C$, one obtains the desired action of $N$ on $C_0$. $\endgroup$ – Lee Mosher Jan 18 '15 at 23:41
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    $\begingroup$ The action of $N$ on $C_0$ is faithful because, if there were a nontrivial kernel $K$, then $K$ would also be the kernel of the conjugate actions of $N$ on the other orbits of $N$ on $C$, and so $K$ would be in the kernel of the whole action of $G$ on $C$. $\endgroup$ – Lee Mosher Jan 18 '15 at 23:44
  • $\begingroup$ @LeeMosher Do you have an idea of what happens if we take $G = \mathrm{Out}(F_n)$ instead? Even for $n = 3$? $\endgroup$ – Pablo Jan 19 '15 at 8:18

Say that a finitely generated group $G$ has property Z if it acts transitively on an infinite countable set $X$ such that the orbits of each element are finite.

  1. If a finite index subgroup $H$ of $G$ has property Z, then $G$ has property Z. Namely, given $H$ acting on $X$, take the action of $G$ on the "induced permutation representation"

$$Y = \coprod_{G/H} gX$$

This is transitive, because the action of $G$ acts transitively on the factors, and then $H \simeq g H g^{-1}$ acts transitively on $gX$ by assumption. Because the action is transitive, to check the orbits of all $g \in G$ are finite, it suffices to check it for a single element $x \in X$. Also, to check that the orbit of any $g \in G$ is finite, it suffices to show that the orbit of some finite power of $g$ is finite. Yet, because $H$ has finite index, a finite power of $g$ lands in $H$, which then has a finite orbit on $x \in X$ by assumption.

  1. If a quotient $G/H$ has property Z, then $G$ has property $Z$. Explicitly, if $G/H$ acts on $X$, then there is a corresponding action of $G$ on $X$ which factors through $G/H$.

I claim that $G =\mathrm{PSL}_2(\mathbf{Z})$ has property $Z$. The group $G$ is finitely generated and virtually free, so, by (1), we may reduce to the case when $G$ is a finitely generated free group with at least any given number of generators. By (2), we may then reduce to the case of any finitely generated group we like. On the other hand, we may now take $G$ to be any finitely generated infinite (hence countable) group such that the order of any element $g \in G$ is finite, and to take $X = G$ with the tautological left action. Many such groups exist (Burnside groups, Grigorchuk group, Tarski Monsters...)

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    $\begingroup$ Note that I wanted the action to be faithful. $\endgroup$ – Pablo Jan 18 '15 at 20:36

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