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?
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

$$\begin{bmatrix} b &amp; a \\ 0 &amp; 1 \end{bmatrix} $$

evaluated on the point $i$ in $\mathbb{H}$. This matrix may not be in $SL_2 \mathbb{R} $, but you can easily fix that.
ANSWER: 

If $A \in SL(2,\mathbb R)$, then

\begin{align*}
 A\cdot i&amp;=\frac{ai+b}{ci+d}\\
 &amp;=\frac{ai+b}{ci+d}\cdot \frac{(-ci+d)}{(-ci+d)}\\
 &amp;=\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 $y&gt;0$. Now $a^2=y \implies a=\sqrt y$. Therefore, $ab=x \implies b=\frac{x}{\sqrt y}$.

So, $\begin{pmatrix} \sqrt y &amp; \frac{x}{\sqrt y} \\ 0 &amp; \frac{1}{\sqrt y} \end{pmatrix} \in SL(2,\mathbb R)$ and 
$$\begin{pmatrix} \sqrt y &amp; \frac{x}{\sqrt y} \\ 0 &amp; \frac{1}{\sqrt y} \end{pmatrix} \cdot i =z$$
implies transitive.