Deduce that every element $g$ of $SL_2(\mathbb R)$ may be expressed in the form $g=hk$ where $h\in H$ and $k$ is a rotation. 
Let $G$ be $SL_2(\mathbb R)$ the group of real $2\times 2$ matrices of
  determinant $1$, acting on $\mathbb C\cup \{\infty\}$ by Möbius transformations. Compute the orbit and and stabiliser of each
  of $0,i,-i$.
Compute the orbit of $i$ under the subgroup
  $$H=\begin{Bmatrix}\begin{pmatrix}a & b\\0 & d \end{pmatrix}:a,b,d\in
 \mathbb R, ad=1\end{Bmatrix}.$$ Deduce that every element $g$ of $G$
  may be expressed in the form $g=hk$ where $h\in H$ and for some
  $\theta \in \mathbb R$, $$k=\begin{pmatrix}\cos \theta & -\sin
 \theta\\\sin \theta & \cos\theta \end{pmatrix}$$

I get the orbits in
$G$
and
$H$
as
$\{z\in \mathbb C:\Im(z)>0\}, \{z\in \mathbb C:\Im(z)=0\}, \{z\in \mathbb C:\Im(z)<0\}$.
I can kind of see where the result will come from but I can't quite get it out.
Any help is appreciated, thank you
 A: Try and do the exercise in its abstract form. Suppose


*

*$G$ acts on a set $X$

*$H$ is a subgroup of $G$

*$H$ acts transitively on $X$

*$K$ is the stabilizer of a point of $X$


Use these hypotheses to prove $G=HK$, i.e. every $g\in G$ is expressible as $hk$.
Hint: Suppose $K$ is the stabilizer of $x$. Since $H$ acts transitively, every $g$ sends $x$ to a place that some $h\in H$ sends it to too, in which case you want to show $g$ and $h$ differ by an element of $K$.
In fact, we can go further: if we restrict to $H_0$, the subgroup of your $H$ with positive diagonal entries, then $H_0$ acts regularly on your $X$, in which case $H_0\cap K=1$, which means every $g\in G$ is uniquely expressible as $hk$ with $h\in H_0,k\in K$.
Note that $G=H_0K$ is equivalent to $G=KH_0$ (apply reciprocation to both sides) and the latter can be proved by showing for every $g\in G$ there is a $k\in K$ for which $k^{-1}g\in H_0$, which can be done by seeing that left-multiplying by rotation matrices simply rotates each column, and it will be possible to rotate the first column vector to be in the positive $x$-axis.
Moreoever, this $H_0$ of positive upper triangulars may be further decomposes as $AN$, where $A$ is comprised of diagonal matrices with positive entries and $N$ is comprised of unitriangulars. This gives the decomposition $G=KAN$, for which we know that the map $K\times A\times N\to G$ given by $(k,a,n)\mapsto kan$ is a diffeomorphism. This is called the Iwasawa decomposition of $\mathrm{SL}_2(\mathbb{R})$.
