How to show the action of $SL_2(\mathbb R)$ on the complex upper half plane is transitive? I've seen other answers here and here but I still am not understanding how to show that this action is transitive. I want to show that for all $z \in \mathbb H$ there exists a matrix $g \in SL_2(\mathbb R)$ such that $gi=z$.
If I start with $\frac{ai+b}{ci+d}=z=x+iy$, then I get
$$\frac{ai+b}{ci+d}=z \implies ai+b=z(ci+d) \implies i=\frac{dz-b}{a-cz}$$ 
I don't know what to do with this or what this tells me.
If I do $\frac{az+b}{cz+d}=i$ I don't get anywhere either.
How can I go about this?
 A: Here's a hint. You can write 
$$ z= a+bi = \frac{b\cdot i + a}{0\cdot i +1} ,$$
which is the transformation corresponding to the matrix
$$\begin{bmatrix} b & a \\ 0 & 1 \end{bmatrix} $$
evaluated on the point $i$ in $\mathbb{H}$. This matrix may not be in $SL_2 \mathbb{R} $, but you can easily fix that. 
A: First show that for any $x\in\mathbb R$ there exists a matrix that gives the transformation $z\mapsto z+x$. Then do the same for $z\mapsto\lambda z$ (for $\lambda>0$). Composing those gets you from $i$ to anywhere in the upper halfplane (first take a dilation to get the imaginary part right, then translate to fix up the real part).
A: This is quite straightforward using your equations; you have just forgotten to write $z$ in terms of its real and imaginary parts.  Using $z=x+iy$ with your equation $ai+b=z(ci+d)$, you get $$ai+b=(xc+dy)i+(xd-yc).$$  So if you fix any $c,d\in \mathbb{R}$, you can make this equation true by choosing $a=xc+dy$ and $b=xd-yc$.  The only additional restriction is that $ad-bc=1$.  To satisfy this, just plug in the formulas for $a$ and $b$ in terms of $c$ and $d$ and the condition becomes $$(xc+dy)d-(xd-yc)c=1,$$ or $$yd^2+yc^2=1.$$  Since $y>0$ (your point is in the upper half plane), it is possible to choose $c,d\in\mathbb{R}$ which make this equation true.
A: ANSWER: 
If $A \in SL(2,\mathbb R)$, then
\begin{align*}
 A\cdot i&=\frac{ai+b}{ci+d}\\
 &=\frac{ai+b}{ci+d}\cdot \frac{(-ci+d)}{(-ci+d)}\\
 &=\frac{ac+bd+i(ad-bc)}{c^2+d^2}.
\end{align*}
We're in $SL(2, \mathbb R$), so $ad-bc=1$, and
$$\frac{ac+bd+i(ad-bc)}{c^2+d^2}=\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}.$$
If $c=0$ then $a \ne 0$, $d\ne 0$ and $ad-bc=1$ becomes $ad=1$. So, $d=1/a$. Also,
$$\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}=\frac{bd}{d^2}+i\frac{1}{d^2}=\frac{b}{d}+i\frac{1}{d^2}=ab+a^2i.$$
Since $z \in \mathbb H$, then $z=x+iy$ where $y>0$. Now $a^2=y \implies a=\sqrt y$. Therefore, $ab=x \implies b=\frac{x}{\sqrt y}$.
So, $\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \in SL(2,\mathbb R)$ and 
$$\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \cdot i =z$$
implies transitive.
