For which $n$ can the plane with $n$ points removed be equipped with a Lie group structure?

Question

Let $X=\mathbb{R}^2-\{p_1, p_2,..., p_n\}$, then if $n=0$ then clearly $X=\mathbb{R}^2$ is a Lie group under addition. If $n=1$, then $X=\mathbb{R}^2-\{p_1\}$ is isomorphic to $\mathbb{C}\setminus\{0\}$ which has a Lie group structure under multiplication. But I don't know how to proceed in the case in which $n>1$. I have thought of using that the tangent bundle of a Lie group is trivial but I don't know how to do this. I would appreciate any help or suggestions. Thank you.

Answer 1

Mnifldz has the right idea. The fundamental group of $\mathbb{R}^2$ with $n$ points removed is a free group on $n$ generators (I guess there is a MSE question explaining that if you look for it). In particular these groups are non-abelian for $n > 1$

There is however a really nice and surprising result involving the (non)-abelianness of fundamental groups of Lie groups. Namely that The fundamental group of a topological group is abelian.

So this gives your answer: $\mathbb{R}^2 - \{p_1, \ldots, p_n\}$ can only be homeomorphic to (the underlying topological space of) a topological group if $n \leq 1$.