# Transitive action of $SL_2(\mathbb{R})$ on $\mathbb{H}$

I'm studying Modular Forms and I'm not understanding why the action of $SL_2(\mathbb{R})$ on $\mathbb{H}$ defined by $\begin{pmatrix} a & b \\ c & d \end{pmatrix}z=\frac{az+b}{cd+d}$ is transitive.

The author of the notes I'm reading just says that for $z=x+iy \in \mathbb{H}$ we have $z=\begin{pmatrix} y^{\frac{1}{2}} & xy^{\frac{-1}{2}} \\ 0 & y^{\frac{-1}{2}} \end{pmatrix}i$, so it's clear that the action is transitive.

Am I missing something obvious here? How does transitivity follows from that observation?

It's a group. The author explains why for every $z$ there exists a group element mapping $i$ to $z$. Composition of one of these maps with the inverse of another shows that for every $z$ and $w$ there is a group element mapping $w$ to $z$.
• Say $\phi_z(i)=z$. Then $\phi_z\circ\phi_w^{-1}(w)=z$. – David C. Ullrich Aug 14 '15 at 17:23
In fact ${\rm Iso}^+(\Bbb R)$ (the maps $z\mapsto ax+b$ with $a,b\in\Bbb R$ and $a>0$) alone acts transitively on $\Bbb H$. Given any two complex numbers $z,w\in \Bbb H$, one can move $z$ to $w$ by first scaling it by $a$ until $az$ has the correct imaginary part, and then translate it horizontally by $b$ until $az+b$ is at $w$.