I am trying to prove that the action of $SL_2(\mathbb{R})$ on $\mathbb{H}$ via $$ \left( \begin{array}{ c c } a & b \\ c & d \end{array} \right)z\rightarrow \frac{az+b}{cz+d} $$ is transitive and the Stabilizer of $i$ is $SO_2(\mathbb{R})$. For stabilizer of $i$ I took an arbitrary matrix and through some algebra I got that $a^2+b^2=1$ and $c^2+d^2=1$. It is clear that $b$ or $c$ has to be negative but I cannot find a reason to eliminate the case when $a,d<0$. How can we eliminate the case as I want to show that $a=cos\theta$, $b=-sin\theta$, $c=sin\theta$, $d=cos\theta$.
Next I am stuck at proving the part that the action is transitive. I am trying the similar strategy as above but the calculations just get messier
For the stabilizer calculation, you need as well to use that, by virtue of the matrix being in $SL(2, \mathbb{R})$, you have $$ad - bc = 1.$$ It may also be helpful to note that because, e.g., $a^2 + b^2 = 1$, there is some angle $\theta$ such that $$a = \cos \theta, b = -\sin \theta.$$
To show that the action of a group $G$ on a space $X$ is transitive, it's enough to pick an element $x_0 \in X$ and show that for all $x \in X$ there is some $g \in G$ such that $x = g \cdot x_0$; then, one can map any element $x$ to any other element $y$ by mapping the $x$ to $x_0$ and then $x_0$ to $y$. Often, this turns out to be cleaner than showing directly that one can map an arbitrary element to another arbitrary element, especially if one can choose an element $x_0$ on which the group acts particularly nicely.