Consider the group $\mathrm{SL}_{2}(\mathbb{R})$ acting on the set $T=\{z\in \mathbb{C}|\text{Im}\hspace{0.1cm}(z)>0\}$. The action is defined by
$$ \begin{pmatrix} a & b\\ c & d \end{pmatrix}\cdot z:= \frac{az+ b}{cz+d} $$
Lets show that is a group action.
In fact $$\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \cdot z := \frac{1\cdot z + 0}{0\cdot z+1}=z$$ thus holds the first action condition.
Furthermore, $$\left(\begin{pmatrix} a_{1} & b_{1}\\ c_{1} & d_{1} \end{pmatrix} \begin{pmatrix} a_{2} & b_{2}\\ c_{2} & d_{2} \end{pmatrix}\right)\cdot z=\begin{pmatrix} (a_{1}a_{2}+b_{1}c_{2}) & (a_{1}b_{2}+b_{1}d_{2})\\ (c_{1}a_{2}+d_{1}c_{2}) & (c_{1}b_{2}+d_{1}d_{2}) \end{pmatrix}\cdot z=\dfrac{(a_{1}a_{2}+b_{1}c_{2})z+ (a_{1}b_{2}+b_{1}d_{2})}{(c_{1}a_{2}+d_{1}c_{2})z+(c_{1}b_{2}+d_{1}d_{2})} $$ On the other hand, $$\begin{pmatrix} a_{1} & b_{1}\\ c_{1} & d_{1} \end{pmatrix}\left( \begin{pmatrix} a_{2} & b_{2}\\ c_{2} & d_{2} \end{pmatrix}\cdot z\right)=\begin{pmatrix} a_{1} & b_{1}\\ c_{1} & d_{1} \end{pmatrix}\cdot \frac{a_{2}z+ b_{2}}{c_{2}z+d_{2}}=\frac{a_{1}\frac{a_{2}z+ b_{2}}{c_{2}z+d_{2}}+ b_{1}}{c_{1}\frac{a_{2}z+ b_{2}}{c_{2}z+d_{2}}+d_{1}}=\frac{\frac{a_{1}(a_{2}z+ b_{2})+b_{1}(c_{2}z+d_{2})}{c_{2}z+d_{2}}}{\frac{c_{1}(a_{2}z+ b_{2})+d_{1}(c_{2}z+d_{2})}{c_{2}z+d_{2}}}$$ $$=\frac{a_{1}(a_{2}z+ b_{2})+b_{1}(c_{2}z+d_{2})}{c_{1}(a_{2}z+ b_{2})+d_{1}(c_{2}z+d_{2})}= \dfrac{(a_{1}a_{2}+b_{1}c_{2})z+ (a_{1}b_{2}+b_{1}d_{2})}{(c_{1}a_{2}+d_{1}c_{2})z+(c_{1}b_{2}+d_{1}d_{2})}$$ Thus $$\left(\begin{pmatrix} a_{1} & b_{1}\\ c_{1} & d_{1} \end{pmatrix} \begin{pmatrix} a_{2} & b_{2}\\ c_{2} & d_{2} \end{pmatrix}\right)\cdot z= \begin{pmatrix} a_{1} & b_{1}\\ c_{1} & d_{1} \end{pmatrix}\left( \begin{pmatrix} a_{2} & b_{2}\\ c_{2} & d_{2} \end{pmatrix}\cdot z\right)$$ This show the second group action condition.
Therefore that is a group action of $\mathrm{SL}_{2}(\mathbb{R})$ on the set $T=\{z\in \mathbb{C}|\text{Im}\hspace{0.1cm}(z)>0\}$.
My question is: Do you know about the orbits? Thank you.
It is transitive. So there is only one orbit. Take $z=x+iy$ then we have $$\begin{pmatrix} \sqrt{y} & x/\sqrt{y}\\ 0 & 1/\sqrt{y} \end{pmatrix}i = z.$$
By the way: The stabilizer of $i$ is $$\operatorname{SO}_2(\mathbb R) = \left\{\begin{pmatrix} a & b\\ -b & a \end{pmatrix} : a^2+b^2=1 \right\},$$ so we have $$ T \cong \operatorname{SL}_2(\mathbb R)/\operatorname{SO}_2(\mathbb R).$$