Action of special orthogonal group on Upper half complex plane. When group $G=SL_{2}(\mathbb{R})$ acts on upper half plane I am able to find the stabiliser subgroup of $i$ is nothing but special orthogonal group $\{ \begin{bmatrix} 
\cos x & \sin x \\
-\sin x & \cos x \\
\end{bmatrix}  \mid x \in \mathbb{R} \} $ .
The statement I am not getting is " any element of the stability subgroup of $i$ acts on $\mathbb{H}$ as rotation at $i$ of angle $2x$". I have tried to get general idea by taking example of some specific elements but not getting how rotation works. Any help or hint would be fine. Thanks in advance.
 A: OK. Let's look at the map
$$
f(z) = \frac{az + b}{cz + d},
$$
where we'll plug in specific values for $a,b,c,d$ soon. The derivative of this is
\begin{align}
f'(z) &= \frac{ad - bc}{(cz+d)^2};
\end{align}
evaluating this at $z = i$ (the point fixed by your stabilizer subgroup) gives
\begin{align}
f'(i) &= \frac{ad - bc}{(ci+d)^2};
\end{align}
Now let's substitute in the values $\sin x$ and $\cos x$ where appropriate to get
\begin{align}
f'(i) 
&= \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{((-\sin x)i+ \cos x)^2}\\
&= \frac{1}{(\cos x - i \sin x)^2} \\
&= \frac{1}{\cos 2x - i \sin 2x}\\
&= \cos 2x + i \sin 2x.
\end{align}
This derivative is the best local linear approximation to the transformation at $z = i$. In other words, if $v$ is a vector based at $i$, then $f'(i)\cdot v$ is where $v$ would point "after applying $f$". And because multiplying by $\cos t + i \sin t$ amounts to rotation by $t$ in the complex plane, we see that in our case, the transformation described ends up rotating (locally) by $2x$.
