December 14th
Today I learned that, for $K$ the fraction field of $\mathcal O_K$ Dedekind, $\op{SL}_2(\mathcal O_K)$ acts transitively on $K\cup\{\infty\}$ as fractional linear transformations if and only if $K$ has class number 1, from Keith Conrad . Amazingly, the main idea is that all ideals in Dedekind domains can be written with two generators.
The main reduction is to note that having $\op{SL}_2(\mathcal O_K)$ act transitively is equivalent to every element of $K$ being of the form $\alpha/\beta$ for $(\alpha,\beta)=(1)$ and $\alpha,\beta\in\mathcal O_K.$ In other words, we can write elements of $K$ in "reduced form.'' Equivalently, every element is in the orbit of $\infty$ if and only if we can write elements of $K$ in reduced form.
In one direction, suppose we can pick up an arbitrary element of $\alpha\in K$ and write it as $a/b$ with $(a,b)=(1).$ To show all elements are in the orbit of $\infty,$ note that $\infty$ is certainly in its own orbit, and for other elements $\alpha=a/b,$ we note $(a,b)=(1)$ implies we can write\[ad-bc=1\]for some $c,d\in\mathcal O_K.$ But then\[\begin{bmatrix} a & c \\ b & d\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix}\]shows what we need. In particular, the matrix is in $\op{SL}_2(\mathcal O_K).$
In the other direction, if all elements are in the orbit of $\infty,$ then for any $\alpha\in K,$ there exists a matrix in $\op{SL}_2(\mathcal O_K)$ such that\[\begin{bmatrix} a & c \\ b & d\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix},\]where $\alpha=a/b.$ Certainly $a,b\in\mathcal O_K$ by construction, and $ad-bc=1$ (by construction) again implies that $(a,b)=1.$ So we're done with the reduction.
To finish, we have to show that being able to write elements of $K$ in reduced form is equivalent to having class number $1.$ Well, if we have class number $1,$ then $\mathcal O_K$ is a PID and therefore a UFD. So when we write elements $\alpha\in K$ as $a/b$ for any $a,b\in\mathcal O_K,$ we can prime factor $a$ and $b$ to make the numerator and denominator coprime. Alternatively, letting $(c)=(a)+(b)$ because $\mathcal O_K$ is a PID, the fraction we're looking for is $(a/c)/(b/c).$
The other direction is harder. Again suppose that every fraction can be written in reduced form, and we'll show that $\mathcal O_K$ is a principal ideal domain. Pick up some ideal, and it's at least generated by $(a,b)$ for some $a,b\in\mathcal O_K.$ If $b=0,$ then $(a,b)=(a)$ and principal; the same holds for $a=0.$ So assume $ab\ne0.$ We attempt to extract a greatest common denominator of $a$ and $b$ using reducing fractions. Suppose that\[\frac ab=\frac cd\]where $(c,d)=(1).$ Experience with $\ZZ$ tells us that $b/d$ should be our greatest common denominator, so let's extract it. The above implies that $ad=bc,$ so $bc\in(d).$ In fact, $(c,d)=(1)$ lets us write $cx+dy=1,$ implying that\[b=bcx+bdy\in(d)\]as well, so we may finally let $g=b/d\in\mathcal O_K.$ Additionally, we see $a/c=b/d=g$ as well. To finish, we compute\[(a,b)=(g)(a/g,b/g)=(g)(c,d)=(g)(1)=(g),\]so $(a,b)$ was principal after all.