TITLE: Prove that $SL(2,\mathbb{R})$ acts transitively on the upper half plane
QUESTION [9 upvotes]: I want to prove $\mathrm{SL}(2,\mathbb{R})$ acts transitively on the upper half plane $\mathbb{U}:=\{z\in \Bbb C\ |\ \mathrm{Im}(z)>0\}$ by 
$$z\longmapsto \dfrac{az+b}{cz+d}.$$
 Is it enough to say that since $\begin{pmatrix} a & b\\c & d \end{pmatrix}\in \text{SL}(2,\mathbb{R})$ is always invertible, for $x,y\in \mathbb{U}$, we can always find a matrix $A\in \text{SL}(2,\mathbb{R})$ such that $Ax =y$?
I feel like there should be more to it.

REPLY [19 votes]: Let $z=x+iy$ be a given point in the upper half plane. Then the $SL(2,\mathbb{R})$ matrix
$$
 \begin{pmatrix}
\sqrt{y}&x/\sqrt{y} \\
0&1/\sqrt{y}
\end{pmatrix}
$$
maps $i$ to $z$. Since $z$ was arbitrary, the orbit of $i$ is the whole upper half plane. Furthermore any point $w$ can be mapped to $i$
by using the inverse of this matrix, so any $w$ can be mapped to any $z$.