Let $\text{E}(2)$ be the group of isometries of the plane $\mathbb R^2$. Then $\text{E}(2)=\text{O}(2)\times\mathbb R^2$ as a topological space and is the semi-direct product as groups. Let $G$ be a discrete subgroup of $\text{E}(2)$ such that $\mathbb E^2/G$ is compact. Let $p$ be any point of the plane which is not left fixed by any non-identity element of $G$. Then $$A=\{x\in\mathbb R^2\mid \|x-p\|\leq \|x-g(p)\|\text{ for all }g\in G\}$$ is a convex polygon. I believe this but don't see how to prove it. This must be a basic result somewhere but I can't seem to find it. Thank you for any insight (or full answer).
If $x,y\in A$ to show convexity I have to show $tx+(1-t)y\in A$, $t\in[0,1]$. So I need to show for $g\in G$ that $$\|tx+(1-t)y-p\|\leq\|tx+(1-t)y-g(p)\|$$
For any points $p$ and $q$, $$ \{x\in\mathbb R^2 \mid \|x-p\|\le\|x-q\| \} $$ is a halfplane (or the whole plane if $p=q$), so your set is an intersection of halfplanes, hence convex.