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?

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).
