Décomposition d'Iwasawa Let $$SL_2(\mathbb R)=\{A\in \mathcal M_{2\times 2}(\mathbb R)\mid \det A=1\}.$$ We denote $$N=\left\{n(x)=\begin{pmatrix}1&x\\0&1\end{pmatrix}\mid x\in \mathbb R\right\}\quad, \quad A=\left\{a(y)=\begin{pmatrix}\sqrt{y}&0\\0&\frac{1}{\sqrt{y}}\end{pmatrix}\mid y>0\right\}$$
and $$K=SO_2(\mathbb R)=\left\{\begin{pmatrix}a&-b\\b&a\end{pmatrix}\mid a^2+b^2=1\right\}.$$
Finally we denote $$\mathbb H=\{x+iy\mid y>0\}.$$
We want to show that all $g\in SL_2(\mathbb R)$ can be written as $g=n\cdot a\cdot k$ where $n\in N$, $a\in A$ and $k\in K$. We will use the action $SL_2(\mathbb R)$ on $\mathbb H$ defined by $\gamma . z=\frac{az+b}{cz+d}$.
1) Show that $K$ is the stabilizer of $i$ in $SL_2(\mathbb R)$. Deduce that $$SL_2(\mathbb R)/K\cong \mathbb H.$$
2) Show that $N$ and $A$ are subgroup of $SL_2(\mathbb R)$ and that $$N\cong (\mathbb R,+)\quad \text{and}\quad A\cong (\mathbb R_{>0},\times ).$$
3) Show that $N.A:=\{n.a\mid n\in N,a\in A\}$ is a subgroup of $SL_2(\mathbb R)$ and that $N.A\cap K=\{id\}$.
4) Let $n\in N$ and $a\in A$. What are the transformation $$z\longmapsto n.z\quad \text{and}\quad  z\longmapsto a.z\ ?$$
5) If $z=x+iy$, show that there is a unique $n\in N$ and a unique $a\in A$ s.t. $n.a.i=z$. What are the orbits of $N$ and $A$ in $\mathbb H$ ?
6) Show that the orbits of $K$ in $\mathbb H$ are circle that contain $i$.
7) Let $g\in SL_2(\mathbb R)$ and $z=g.i$. Show that there is $k\in K$ s.t. $g=n.a.k$ where $n\in N$ and $a\in A$. 
8) Show that this representation is unique.
My work
1), 2) and 3) are done.
4) I would say a translation for the first one and a homothetic for the second one. Do you agree ?
5) I have absolutely no idea for $z=n.a.i$. I tried to use what I did previously, but with no success. Of the orbits, I would say for $N$ that for $N$, the orbit of $z\in \mathbb H$, it's all the line parallel to the vector $(O,(x,y))$ (the notation is not good, but I mean the vector of origine $O$ and extremity $(x,y)$.) And for the homothetic, the orbits of $z$ is a triangle ?
6) Done.
7),8) I don't know.
Thanks for helping. 
 A: So for part 7 we won't do anything fancy. 
We that $SL_2(\mathbb{R})$ are those matrices 
$$ \begin{bmatrix} q & w \\ e& r\end{bmatrix}$$
Whose determinant is 1, meaning 
$$ qr - ew = 1$$
Now consider the product $n . a . k$
$$\begin{bmatrix} 1 & x \\ 0 & 1\end{bmatrix} \begin{bmatrix} \sqrt{y} & 0 \\ 0& \frac{1}{\sqrt{y}}\end{bmatrix} \begin{bmatrix} a & -b \\ b& a\end{bmatrix} $$ 
Where $a^2 + b^2 =1, y > 0, x \in \mathbb{R}$
We create then a product which yields 
$$\begin{bmatrix} 1 & x \\ 0 & 1\end{bmatrix} \begin{bmatrix} a \sqrt y & -b \sqrt{y} \\ \frac{b}{\sqrt{y}} & \frac{a}{\sqrt{y}}\end{bmatrix} =$$ 
$$\begin{bmatrix} a\sqrt{y} + \frac{bx}{\sqrt{y}} & \frac{a}{\sqrt{y}} - bx \sqrt{y} \\ \frac{b}{\sqrt{y}} & \frac{a}{\sqrt{y}}\end{bmatrix} $$ 
Its trivial from the product that this has determinant 1 (so that's a good sign). Now observe the following,
suppose you are given $q,w,e$ for a matrix M then there is exactly one $r$ that you can take on, so that 
$$ M = \begin{bmatrix} q & w \\ e& r\end{bmatrix} \in SL_2(\mathbb{R})$$
That $r$ being $\frac{1+ew}{q}$. So if a matrix $M$ is matched on for its first 3 components by another $B$ and we know that both $M,B \in SL_2(\mathbb{R})$ then it must be the case that $M = B$. 
So $SL_2(\mathbb{R}) \in \lbrace B | B = n.a.k, n \in N, a \in A, k \in K \rbrace$
just component wise equate the terms to yield a system of polynomials whose solutions are the desired entries (warning, I didn't show that there always are real solutions, this should be checked, and might actually be the real bulk of the proof).
Now the tricky part is uniqueness. Observe that 
$$\begin{bmatrix} a\sqrt{y} + \frac{bx}{\sqrt{y}} & \frac{a}{\sqrt{y}} - bx \sqrt{y} \\ \frac{b}{\sqrt{y}} & \frac{a}{\sqrt{y}}\end{bmatrix} $$ 
has exactly 3 free components
and that 
$$ \begin{bmatrix} q & w \\ e& r\end{bmatrix}$$
has 3 free parameters (we showed earlier the 4th is fixed once 3 are fixed). So that means in any equation of the form 
$$ \begin{bmatrix} q & w \\ e& r\end{bmatrix} = \begin{bmatrix} a\sqrt{y} + \frac{bx}{\sqrt{y}} & \frac{a}{\sqrt{y}} - bx \sqrt{y} \\ \frac{b}{\sqrt{y}} & \frac{a}{\sqrt{y}}\end{bmatrix} $$
There are a finite number of possible combinations of $y,x,a,b$ that can work (since it is just a system of polynomials). So we know there won't be infinitely many representations, but now the challenge remains to show that there is only ONE solution to this system of polynomials that satisfies our constraints. Not sure yet how to do this but hopefully that is a hint. 
