Show that the action of $PSL_2(\mathbb{R})$ on $\mathbb{H}$ is transitive , i.e for all $z,w \in \mathbb{H}$ there exists some $g \in PSL_2(\mathbb{R}) $ such that $gz=w$ .
It suffices to find a $g \in PSL_2(\mathbb{R})$ with $$gz=i \ \ (1)$$ for any z $ \in \mathbb{H}$ . For $w \in \mathbb{H}$ there is also a $ h \in PSL_2(\mathbb{R}) $ such that $$hw=i $$. Thus we have $$(h^{-1}g) z=w $$
(1) is equivalent to $$ g^{-1}i=z \iff \frac{ai+b}{ci+d}=z \iff \frac{ac+bd}{c^2+d^2}+\frac{i}{c^2+d^2}=x+iy $$
Choosing $c=0 $ leads to $$ \frac{1}{d^2}=y $$
Thus we find $d=\pm \frac{1}{\sqrt{y}}$ and $a=\pm \sqrt{y}$ and $b=\pm \frac{x}{\sqrt{y}}$ .
I think if you find this g then you can calculate $h^{-1}g$ .
Thanks for help .