$Sl(2, \Bbb R)$ is transitive on the upper half-plane. Here I have a question on the conclusion.
I have proved that for any $w \in \Bbb H$ there exists $z \in \Bbb H$ s.t $A⋅z=w$ for some $A \in Sl(2, \Bbb R)$. Should I have my conclusion as transitive? How do I get it? I can also prove that it is an action.
For the definition of transitive: if $X$ is non-empty and if for each pair $x, y \in X$ there exists a $g \in G$ such that $g⋅x = y.$ This is also equivalent to that $Orbit(x)=X$ for every $x \in X$.
I am unable to conclude that. Few years ago I think I figured it out but today suddenly I am stuck here for an hour.
By the way, I proved with an alternate way to creating $A_x \in Sl(2, \Bbb R)$ s.t $A_x⋅ i=x$. But I want to conclude from the first argument as well. Please help.
Edit: I also found this link when my previous question was closed Prove that $SL(2,\mathbb{R})$ acts transitively on the upper half plane. Here also, the accepted answer concluded at the same portion where I ended my attempt and unable to move from there. Maybe it is a small leap from there. So any help would be appreciated. I will also post my doubt at the comment there but probably I won't be able to write everything. My earlier question was closed and it was written that "Your post has been associated with a similar question. If this question doesn’t resolve your question, ask a new one." So I become confused I edited my previous question https://math.stackexchange.com/questions/4178280/a-question-on-the-conclusion-of-the-proof-of-sl2-bbb-r-is-transitive-on-t. I will delete that and ask a new(this) one as well.