I want to find a group $G$ and an action of $G$ on $\mathbb{R}^2$ such that $\mathbb{R}^2 / G \approx M \setminus \partial M$, where $M$ is the Mobious strip, and $\partial M$ is its boundary, a circle.
Some thoughts: when $\mathbb{Z} \times \mathbb{Z}$ acts on $\mathbb{R}^2$, with translation, we get a torus. Also, if we act with $\mathbb{Z}_{2}$ on the torus $S^{1} \times S^{1}$ by $m \times (z_{1}, z_{2}) = (z_{2}, z_{1})$ if $m=1$ and $(z_{1}, z_{2})$ if $m=0$, then we get a Mobius strip. So I thought maybe acting on $\mathbb{R}^2$ with $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}_{2}$ would be a good idea, by making each group ($\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Z}_{2}$) act accordingly. However, this action does not produce the desired results. Is there a better one that does?
If you allow me to consider the open Mobius strip as the nontrivial fiber bundle of the circle $S^1$ with fiber $\mathbb{R}$, having one twist, then I can give one such group $G$. Let $G:=\mathbb{Z}$. The action of $\mathbb{Z}$ on $\mathbb{R}^2$ is given by $$n\cdot (x,y):=(x+n,(-1)^{n}y)$$ for all $x,y\in\mathbb{R}$ and for every $n\in\mathbb{Z}$.