$\mathbb{R}^2$ as a quotient of a group

Question

I am looking for a locally compact group $G$ with a closed subgroup $H$ such that $G/H$ is homeomorphic with $\mathbb{R}^2$ but $G$ does decompose into a semidirect and/or direct product which possesses $\mathbb{R}^2$ as a component.

Answer 1

Presumably since the hypotheses on $G$ and $H$ are topological, one wants to require that $G / H$ be homeomorphic to $\mathbb{R}^2$, and not simply in bijection with it.

Example 1 (Mobius transformations of the upper half-plane) The Lie (and hence locally compact) group $SL(2, \mathbb{R})$ acts transitively on the upper half-plane $\mathbb{H} := \{z : \Im z > 0\}$ of the complex plane (which is, needless to say, homeomorphic to $\mathbb{R}^2$) by Mobius transformations: $$\begin{pmatrix} a & b \\ c & d\end{pmatrix} \cdot z := \frac{a z + b}{c z + d};$$ since $-\mathbb{I}_2$ acts by the identity, this group action descends to $PSL(2, \mathbb{R})$, which like $SL(2, \mathbb{R})$ is simple and hence does not factor as a semidirect product. Computing directly shows that the stabilizer in $PSL(2, \mathbb{R})$ of $i$ is isomorphic to $SO(2)$, which in particular is closed.

Example 2 (Affine real transformations) An example of a different (and trivial) flavor is the following: Consider the Lie group $\mathbb{R} \rtimes \mathbb{R}_+$ (again, clearly homeomorphic to $\mathbb{R}^2$) of orientation-preserving affine transformations of the line $\mathbb{R}$, that is, the group of maps $\mathbb{R} \to \mathbb{R}$, $t \mapsto a t + b$, $a > 0$, $b \in \mathbb{R}$. Like all groups, it acts transitvely and freely on itself by left-multiplication, and so the stabilizer of this action contains only the identity, $t \mapsto t$. Though this topological group is homeomorphic to $\mathbb{R}^2$, it is nonabelian and hence not isomorphic to $(\mathbb{R}^2, +)$.