A quick topology question.
If $X$ is some subset of the plane $\mathbb R^2$ (equipped with the subspace topology) and we consider the fundamental group $G = \pi_1(X,x)$ for some $x \in X$, is the following reasonable to say?
All non-trivial elements of $G$ have infinite order.
There is no subgroup of $G$ isomorphic to $\mathbb Z^2$.
If not, what else does one need to assume?
Take care, and thanks for reading!
Your first question has been answered in the affirmative in Is the fundamental group of every subset of $\mathbb{R}^2$ torsion-free?. However, this is a highly non-trival result.
The second assertion is true for compact $X$. See
Fischer, Hanspeter, and Andreas Zastrow. "The fundamental groups of subsets of closed surfaces inject into their first shape groups." Algebraic & Geometric Topology 5.4 (2005): 1655-1676
https://www.emis.de/journals/UW/agt/ftp/main/2005/agt-5l67.pdf
See Corollary 7 in this paper. I guess it is true also for non-compact $X$, but I do not know a proof.
