The group $\mathrm{PSL}\_2(\mathbf{Q})$ of fractional linear transformations $x \mapsto (ax+b)/(cx+d)$ such that $a,b,c,d \in \mathbf{Q}$ and $ad-bc = 1$ acts on $\mathbf{Q} \cup \lbrace \infty\rbrace$. The weakest version of my question is the following:



> 
> Is there a short proof that no element of $\mathrm{PSL}\_2(\mathbf{Q})$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$?
> 
> 
> 


It seems obvious to me that no element of $\mathrm{PSL}\_2(\mathbf{Q})$ can have finitely many orbits on $\mathbf{Q} \cup \lbrace\infty\rbrace$, but I have not been able to prove this.



> 
> Is this stronger claim true, and if so, does it have a short proof?
> 
> 
> 


**Some remarks and motivation:** (1) The subgroup of $\mathrm{PSL}\_2(\mathbf{Q})$ generated by the maps $x \mapsto x+1$ and $x \mapsto x/(x+1)$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$; these elements are used to generate the Calkin–Wilf tree of rational numbers.


(2) By conjugating by elements of $\mathrm{PGL}\_2(\mathbf{Q})$ it suffices to look at maps of the form $g : x \mapsto 1/(b-x)$. If $|b| > 2$ then $g$ is conjugate in $\mathrm{PGL}\_2(\mathbf{R})$ to a map of the form $x \mapsto \lambda x$. It follows that the iterates of $g$, starting at the initial value $x\_0$ are of the form


$$ \frac{1}{b-\frac{1}{b-\ldots\frac{1}{b-x\_0}}} $$


and converge to the continued fraction $1/(b-1/(b- \ldots ))$. The iterates of $g^{-1}$ behave similarly. So in this case each orbit of $\langle g \rangle \le \mathrm{PSL}\_2(\mathbf{Q})$ has precisely two limit points in $\mathbf{R}$, namely the two roots of $x^2-bx+1$. Clearly this implies that $\langle g \rangle$ has infinitely many orbits on $\mathbf{Q} \cup \lbrace \infty\rbrace$. 
However if $|b| < 2$ then I have not been able to make this approach work.


(3) If $\mathbf{Q}$ is replaced with the finite field $\mathbf{F}\_p$ then
is not too hard to show that no element of $\mathrm{PSL}\_2(\mathbf{F}\_p)$ acts transitively on $\mathbf{F}\_p \cup \lbrace \infty \rbrace$: see Lemma 8.2 in [this paper](http://arxiv.org/abs/1205.4960) with John Britnell. This leads to a proof of the result for $\mathbf{Q}$ by reduction mod $p$.



Orbits of the projective special linear group on $\mathbf{Q} \cup \{\infty\}$ 
If your matrix is scalar it acts trivially on the projective line, so assume it's not scalar; hence it's conjugate in $\text{GL}\_2(\mathbf{Q})$ to a companion matrix $A=\pmatrix{0 & -1 \cr 1 & z}$. 


If $z$ is integer with $|z|\ge 3$ then the dynamics of $A$ on the real projective line $\mathbf{P}^1(\mathbf{R})$ has a repelling point $a\_-$ and a attractive point $a\_+$; in particular, if $K$ is an infinite compact subset of $\mathbf{P}^1(\mathbf{R})\smallsetminus (a\_-,a\_+)$, then its intersection with each $A$-orbit $\{A^nx:n\in\mathbf{Z}\}$ is finite and hence $K$ intersects infinitely many $A$-orbits. Since $K$ can be chosen to be contained in $\mathbf{P}^1(\mathbf{Q})$, it follows that $A$ admits infinitely many orbits on $\mathbf{P}^1(\mathbf{Q})$. If $|z|=2$ a similar argument works (with $a\_+=a\_-$). If $|z|<2$ then $A$ has finite order.


Assume now $z$ is not an integer. Let $p$ be a prime divisor of the denominator of $z$. Let $|z|\_p$ be the norm of $z$ in $\mathbf{Q}\_p$. An easy calculation shows that the function $x\mapsto -1/(x+z)$ maps $\mathbf{Z}\_p$ into itself and is $(1/|z|\_p^2)$-contracting on 
$\mathbf{Z}\_p$,
so it admits a unique fixed point which I denote by $a\_+$. Thus $A$ admits a north-south dynamics on $\mathbf{P}^1({\mathbf{Q}\_p})$, with repelling and attracting points $a\_-$ and $a\_+$. So the same argument as in the real case, using density of $\mathbf{P}^1({\mathbf{Q}})$ in $\mathbf{P}^1({\mathbf{Q}\_p})$ shows 
that $A$ has infinitely many orbits. I assumed (as you do) that $A$ has determinant 1 but the argument carries over with minor modification for arbitrary $A$.


