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.
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 $2x$.