I did my last exam task and wonder if it's ok, please check me. ;)
Let $G$ be a group of isometries of a plane $\mathbb{R}^2$ generated by $f(x,y)=(x+1,y)$ and $g(x,y)=(1-x,y+1)$. Prove that $G$ acts on $R^2$ in completely discrete way. What kind of space is $\mathbb{R}^2/G$?
By completely discrete I mean that for every $x\in X$ there is a neighbourhood $x\in U_x\subset X$ such that for every $e\neq g\in G$ there is $U_x\cap g(U_x)=\emptyset$.
Ok, so... it's gonna be short. It suffices to make a drawing or simply see that $\mathbb{R}^2/G\approx S^1\times S^1$, so a torus, am I right?
So for every x, if we take $U_x=B(x,\epsilon)$ with $\epsilon=0.0001$ it would cover the above definition. If think that every $\epsilon<0.5$ would be ok. We slide by 1 down/up or left/right in practice (we can use $g$ to slide up/down and then correct with $f$ in left/right direction). Every other isometry would be an integer multiplicity of those 'slides'. So I think that's enough to say that G acts on a plane in a discrete way.
It's because $f$ makes every point $p$ to be in one orbit with a point $p+\mathbb{Z}$, so we can even forget about moving in X axis with $g$, which does simmilar thing, but with Y axis. At first I thought that it would be $S^1\times\mathbb{R}$, and I solve like that on the exam.. but that's what I claim now. Is that correct?
Thanks!
First, visualize the fundamental domain of the action, $Q=\{(x,y) \in \mathbb{R}^2 : 0\leq x\leq 1, 0 \leq y \leq 1\}$.
$f$ shifts $Q$ horizontally, $g$ flips it and shifts it vertically.
Note that $fgf = g \implies $ each $h\in G$ can be written as $h=f^a g^b$ for suitable $a,b\in \mathbb{Z}$.
If $1\neq h\in G$, and $p=(x,y)\in \mathbb{R}^2$, then $$hp=f^a g^b p = \begin{cases} (1-x+a,y+b) & \text{ if $b$ is odd; } \\ (x+a,y+b) & \text{ if $b$ is even; } \end{cases} $$
This implies that $(p-hp)^2 \geq 1 $ for each $h$ with $(a,b) \neq (0,0)$. The same expression shows that $Q$ is the fundamental domain of the action.
Note now that the vertical sides of $Q$ are identified by $f$ (orientation-preserving), while the horizontal sides by $g$ (orientation-reversing). Then, the quotient space $\mathbb{R}^2/G$ is homeomorphic to the Klein bottle.
Remark: Being $\mathbb{R}^2$ simply connected, $G$ is isomorphic to the fundamental group of the quotient space $\mathbb{R}^2/G$. Therefore, it cannot be that $\mathbb{R}^2/G$ is a Torus, since $G$ is non-abelian while the fundamental group of the Torus is abelian.