# fundamental groups of complements to countable subsets of the plane

This question is a follow-up of this MSE post and a comment by Henno Brandsma:

Question 1. Let $$S$$ be the set of isomorphism classes of fundamental groups $$\pi_1(E^2 - C)$$, where $$C$$ ranges over all countably infinite subsets of the Euclidean plane $$E^2$$. What is the cardinality of $$S$$?

All what I can say is that $$S$$ contains at least two non-isomorphic groups:

• One is the free group $$F_\omega$$ of countably infinite rank, the fundamental group of the complement to a closed discrete countably infinite subset of $$E^2$$, does not matter which one, say, $$C={\mathbb Z}\subset {\mathbb R}\subset {\mathbb R}^2$$.

• The other is $$G_{{\mathbb Q}^2}=\pi_1(E^2-C)$$, where $$C$$ is a dense countable subset of $$E^2$$, again, does not matter which one, for instance, $$C={\mathbb Q}^2$$. This group is not free since it contains, for instance, the fundamental group of the Hawaiian Earrings. (Actually, it contains $$\pi_1$$ of every nowhere dense planar Peano continuum.)

The natural expectation is that $$S$$ is uncountable (more precisely, has the cardinality of continuum: It is clear that the cardinality of $$S$$ cannot be higher than that).

Edit. Following Yves' suggestion:

Question 2. Let $${\mathbb H}$$ denote the Hawaiian Earrings. Is the fundamental group $$\pi_1({\mathbb H})$$ essentially freely indecomposable? (Here a group $$G$$ is essentially freely indecomposable if in every free product decomposition $$G\cong G_1\star G_2$$, one of the free factors $$G_1, G_2$$ is free of finite rank.) One can also ask for the weaker property of $$G=\pi_1({\mathbb H})$$, namely, that $$G$$ does not admit free product decompositions $$G\cong G_1\star G_2$$ with two uncountable factors.

The only relevant result I could find in the literature is the theorem (due to Higman) which (according to "The combinatorial structure of the Hawaiian earring group" by Cannon and Conner) implies that that every freely indecomposable free factor of $$G=\pi_1({\mathbb H})$$ is either trivial or infinite cyclic. Maybe Higman's methods prove more, but his paper ("Unrestricted Free Products, and Varieties of Topological Groups", Journal of LMS, 1952) is behind the paywall.

If Q2 (even in the weaker form) has positive answer, then in Q1 at least one can say that $$S$$ is infinite.

• Note that is $C$ is closed then the $\pi_1$ is free and countable (better denoted $F_\omega$ than $F_\infty$, since uncountable free groups such as $F_{\mathfrak{c}}$ come into the game, at least as subgroup of some of these groups). Also if $C=\{1/n:n\ge 1\}$, then $\mathbf{C}-C$ is maybe homotopy equivalent to the Hawaiian earring space. – YCor Feb 8 at 15:38
• If $D$ is a nonempty compact totally disconnected subset of the sphere $S$, one can produce a closed subset $W_D$ containing $D$, such that $D$ precisely consists of the set of accumulation points in $W_D$. Of course $W_D$ is not unique, but it might be unique up to homeomorphism of $S$ (I'm also only interested in $D$ up to homeomorphism, or [nontrivially] equivalently homeomorphism of $S$).Define $C_D=W_D\smallsetminus D$. Then one could wonder whether the isomorphism class of $\pi_1(\mathbf{R}^2-C_D)$ determines $D$ up to homeomorphism (there are continuum such $D$ up to homeomorphism). – YCor Feb 8 at 15:48
• @YCor Yes, in the first example $E^2 -C$ is h.e. to the H.E. The difficulty with the question, as I currently see it, is that I see very few invariants to distinguish isomorphism types of groups appearing as $\pi_1(E^2- C)$. For instance, I cannot even tell is $G_{{\mathbb Q}^2}$ splits nontrivially as a free product (I am quite sure, it does not) or is not isomorphic to the $\pi_1$ of the H.E (I am sure, it is not). – Moishe Kohan Feb 8 at 21:52
• One can also wonder whether the HE-group can be split as free product of two uncountable groups. If the answer is no, then it's not isomorphic to the $\pi_1$ of the bouquet of two HE along a non-singular point. (This question is maybe more accessible, because this group is well-studied and quite well-understood.) – YCor Feb 9 at 0:40