How to understand the Möbius transform as a group action? The group $SL(2,R)$ acts on the upper half-plane by the formula 
$$ \left(\begin{array}{cc} a & b \\ c & d \end{array}  \right) z =  \frac{az + b}{cz + d} .$$
It is indeed straightforward to check that it is an action. But is there any smart way to see it without calculation? 
It is kind of miracle for me.
 A: The Cayley transform
$$
z\mapsto\frac{z-i}{z+i}
$$
sends the complex upper half plane ${\cal H}$ conformally onto the unitary disc $D$. Under this transform, the Möbius transformations corresponding to elements in the maximal compact group $K={\rm SO}_2$ become rotations of $D$. In order to understand the remaining tranformations recall the decomposition ${\rm SL}_2=NAK$ where $N$ are the unitary upper triangular matrices and $A$ are the diagonal matrices. 
When the "translation" is completed it becomes quite clear that we have an action, although it may be argued that the direct verification is actually simpler.
A: You can view $\widehat{\Bbb{C}}=\Bbb{C} \cup \{\infty\}$ as $\Bbb{CP}^1$, from which you get a natural action of $SL_2(\Bbb{C})$ on $\Bbb{\widehat{C}}$. You can check that this action coincides with Mobius transformations, which explains the "miracle". Now if you restrict attention to the upper half plane $H$, it is easy to see that at least $SL_2(\Bbb{R})$ preserves the boundary (that is, the real line). Its image turns out to be exactly the Mobius transformations preserving $H$. 
