Prove that $SL(2,\mathbb{R})$ acts transitively on the upper half plane 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. 
 A: 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 $z$.
A: With some geometry this can be intuitively obvious, and we can say some other things.
$\mathrm{SL}(2,\mathbb{R})$ has an Iwasawa decomposition $KAN$ where $K$ is the subgroup of rotations, $A$ is the subgroup of positive diagonal matrices, and $N$ is the subgroup of unitriangular matrices. That is, every $2\times 2$ real matrix with determinant $1$ may be uniquely expressed as
$$\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \phantom{-}\cos\theta\end{pmatrix} \begin{pmatrix}\lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix} \begin{pmatrix} 1 & h \\ 0 & 1\end{pmatrix} $$
for some angle $\theta$, scalar $\lambda>0$ and $h\in\mathbb{R}$. Note that $KAN$ does not refer to a direct product as groups, but it does mean $\mathrm{SL}(2,\mathbb{R})\simeq K\times A\times N\simeq S^1\times \mathbb{R}^2$ as manifolds.
The subgroup $N$ acts on $\mathbb{H}$ by translating points horizontally. The subgroup $A$ acts by homotheties (scaling every ray from the origin by a factor of $\lambda^2$). To interpret the action of $K=\mathrm{Stab}(i)$, one can interpret $\mathbb{H}$ as the upper-half plane model of the hyperbolic plane, in which case these are hyperbolic rotations around $i$.
While it is traditional to write $KAN$, to prove the decomposition more easily it is probably easier to note that $AN$ (the positive-diagonal upper triangulars) is itself a subgroup, and that $G=PQ$ is equivalent to $G=QP$ for subgroups $P,Q\le G$ (just apply inversion to all elements). This way we can prove that $AN=NA$ and then prove $\mathrm{SL}(2,\mathbb{R})=(NA)K$ instead.
Observe every element in $NA$ is uniquely expressible as $na$ with $n\in N$, $a\in A$. We can also see that $NA$ acts transitively on $\mathbb{H}$: given any two points $z,w\in\mathbb{H}$, we can go from $z$ to $w$ by first applying a dilation $a\in A$ so that $az$ and $w$ are at the same height (have the same imaginary part), then we apply a horizontal translation $n\in N$ so that $naz=w$. In fact, the choice of $a$ and $n$ is unique; $NA$ acts on $\mathbb{H}$ regularly (i.e. $\mathbb{H}$ is a principal homogeneous space, a.k.a. a torsor). Note already proves $\mathrm{SL}(2,\mathbb{R})$ acts transitively.
Finally, try the following exercise: if $H,K\le G$ are subgroups and $G$ acts on $X$ such that $H$ acts regularly and $K$ is a point-stabilizer then $G=HK$. Apply with $G=\mathrm{SL}(2,\mathbb{R})$ and $H=NA$ to conclude the Iwasawa decomposition.
A: Suppose $\;z\in\Bbb C\,,\;w\in\Bbb H\;$, and write $\;w=n+im\;,\;\;n,m\in\Bbb R\;,\;m>0\;$:
$$\frac{az+b}{cz+d}=w\iff (a-cw)z=dw-b\iff z=-\frac{dw-b}{cw-a}$$
and
$$-\frac{dw-b}{cw-a}\cdot\frac{c\overline w-a}{c\overline w-a}=-\frac{cd|w|^2-adw-bc\overline w+ab}{|cw-a|^2}=$$
$$=-\frac{ab+cd|w|^2+(ad+bc)n}{|cw-a|^2}+\frac{\overbrace{(ad-bc)}^{=1}m}{|cw-a|^2}i$$
so also $\;z\in\Bbb H\;$ and we're done
