Comparing/contrasting hyperbolic and Euclidean geometry - or, on how ${\rm PSO}_2(\Bbb R)$ sits inside ${\rm PSL}_2(\Bbb R)$ I am studying hyperbolic geometry, in particular comparing and contrasting it with familiar Euclidean geometry. Let $\Bbb E$ be the Euclidean plane, and $G={\rm Iso}^+(\Bbb E)$ be the group of orientation-preserving isometries of the plane. Then there is a subgroup $\Lambda\le G$ such that: $\Bbb E$ is a $\Lambda$-torsor, orbits of one-parameter subgroups are geodesics, and $G=\Lambda\rtimes{\rm Stab}(p)$ is an internal semidirect product for any choice of point $p\in\Bbb E$. Indeed $G={\rm Aff}(\Bbb R^2)$ and $\Lambda=\Bbb R^2$ the subgroup of translations works. Note that the stabilizer, of the point $p=0$ wlog, is ${\rm Stab}(0)={\rm SO}_2(\Bbb R)$ (rotations).
I am wondering to what extent any of this is true for $G={\rm Iso}^+(\Bbb H)$, where $\Bbb H$ is the hyperbolic plane. We know that $G={\rm PSL}_2(\Bbb R)$ acts via linear fractional transformations on the upper half plane model, and we can compute $S={\rm Stab}(i)={\rm PSO}_2(\Bbb R)$ relatively easily. So, does there exist a $\Lambda\le G$ for which (i) $\Bbb H$ is a $\Lambda$-torsor, and/or (ii) one-parameter subgroups of $\Lambda$ have geodesic orbits, and/or (iii) we can write $G=\Lambda S$ and $\Lambda\cap S=1$, and if so if (iv) $\Lambda\triangleleft G$ is normal? (Note that (iv) is analogous to the internal semidirect product condition, and (iii) is a weakening of it without the normality condition.)
I believe (ii) is a failure: wlog (I think) assume a one-parameter subgroup $\Lambda$ has $i\Bbb R^+$ (the positive imaginary axis) as an orbit. Then it must stabilize it, so after some computation we see it consists of matrices that are each of the form $[\begin{smallmatrix}\lambda&0\\0&\lambda^{-1}\end{smallmatrix}]$ or $[\begin{smallmatrix}0&-\lambda\\ \lambda^{-1}&0\end{smallmatrix}]$. I think some geometric reasoning in terms of connected components forces them to be of the first form, and those basically scale as $z\mapsto\lambda^2z$, whose trajectories are Euclidean rays pointing outward from the origin, which are not hyperbolic geodesics.
I am not sure how to approach the other parts though. This is not part of the discussion or exercises of any notes or text or other material, I'm just trying to better visualize the symmetry of $\Bbb H$.
 A: I'd already shown (ii) was impossible and Travis in the comments pointed out why (iv) is impossible, as well as gave the $\Lambda$ which satisfies (i) and (iii).
(i) The subgroup $\Lambda$ represented by upper triangular matrices works. It is comprised of horizontal translations, homotheties (Euclidean dilations centered at the origin, which are actually isometrices of hyperobolic space) and their compositions, which clearly acts sharply transitively on $\Bbb H$.
(ii) A one-parameter subgroup of homotheties applied to an element of hyperbolic space in the upper half-plane model traces out a Euclidean ray. The hyperbolic geodesics are circular arcs, so if we pick a nonhorizontal ray this contradicts the second hypothesized condition.
(iii) Elements of ${\rm PSO}_2(\Bbb R)$ look like $(\begin{smallmatrix}a & -b \\ b & a\end{smallmatrix})$ for real $a,b$ with $a^2+b^2=1$. If such a matrix is upper triangular, then $b=0$ and $a=\pm1$ so the matrix is $\pm$ the identity, both of which represent the trivial isometry. Therefore $\Lambda\cap S=1$. Note that if $G$ acts transitively on a set $X$, and $H$ a subgroup also acts transitively, then $G=H{\rm Stab}(x)$ for any $x\in X$, so the other part of condition three is automatic.
(iv) $G={\rm PSL}_2(\Bbb R)$ is simple so $\Lambda$ is not normal.
