# If $S_\mathbb N$ is partitioned into finitely many pieces, must one piece contain a “skew copy” of every countable group?

Suppose $$G$$ and $$H$$ are groups. $$C \subseteq H$$ is called a skew copy of $$G$$ in $$H$$ if $$C = hK$$ for some $$h \in H$$ and some subgroup $$K$$ of $$H$$ with $$K \cong G$$.

Question 1: Suppose the infinite symmetric group $$S_\mathbb N$$ is partitioned into finitely many pieces. Must one of these pieces contain a skew copy of every countable group?

Given two groups $$G$$ and $$H$$, let us write $$H \rightarrow G$$ to mean that whenever $$H$$ is partitioned into finitely many pieces, one of the pieces contains a skew copy of $$G$$.

Question 2: If the answer to Question 1 is negative, is it nonetheless true that there is some (possibly very large) group $$H$$ such that $$H \rightarrow G$$ for every countable group $$G$$?

$$(1)$$ The answer to Question 1 is positive if and only if $$S_\mathbb N \rightarrow G$$ for every countable group $$G$$.

$$(2)$$ $$S_\mathbb N \rightarrow G$$ for every finite group $$G$$. In fact, something a little stronger is true: for every finite group $$G$$ and every $$r \in \mathbb N$$, there is a finite group $$H$$ such that if $$H$$ is partitioned into $$r$$ pieces, then one of them contains a skew copy of $$G$$.

$$(3)$$ If $$S_\mathbb N$$ is partitioned into finitely many Borel pieces (or, a little more liberally, pieces with the Property of Baire), then one of the pieces contains a skew copy of $$S_\mathbb N$$, and therefore contains a skew copy of every countable group. I do not know whether the Property of Baire can be replaced with Lebesgue measurability.

The answer to both questions is no.

For every group $$G$$, by induction on $$|G| = \kappa$$, we'll construct a partition $$G = A \sqcup B$$ such that for every $$g, h \in G$$, if $$g$$ has infinite order, then $$h\langle g\rangle$$ meets both $$A, B$$ on an infinite set -- Call such a partition of $$G$$ good. If $$\kappa = \aleph_0$$, this is clear so assume $$|G| = \kappa \geq \aleph_1$$ and fix a continuously increasing sequence $$\langle H_{\alpha}: \alpha < \kappa \rangle$$ of subgroups of $$G$$ such that $$|H_{\alpha}| = |\alpha + \omega|$$ and $$G = \bigcup \{H_{\alpha}: \alpha < \kappa\}$$. For each $$\alpha < \kappa$$, let $$H_{\alpha} = A'_{\alpha} \sqcup B'_{\alpha}$$ be a good partition of $$H_{\alpha}$$. For $$\alpha \leq \kappa$$, define $$A_{\alpha}, B_{\alpha}$$ as follows. $$A_0 = A'_0$$ and $$B_0 = B'_0$$, $$A_{\alpha + 1} = A_{\alpha} \cup (A'_{\alpha + 1} \setminus H_{\alpha})$$ and $$B_{\alpha + 1} = B_{\alpha} \cup (B'_{\alpha + 1} \setminus H_{\alpha})$$ and if $$\gamma \leq \kappa$$ is limit, then $$A_{\gamma} = \bigcup \{A_{\alpha}: \alpha < \gamma\}$$ and $$B_{\gamma} = \bigcup \{B_{\alpha}: \alpha < \gamma\}$$. Let us check, by induction on $$\alpha \leq \kappa$$, that $$A = A_{\alpha}$$, $$B = B_{\alpha}$$ form a good partition of $$H_{\alpha}$$. If $$\alpha = 0$$ or a limit, this is clear. So assume $$A_{\alpha}, B_{\alpha}$$ form a good partition of $$H_{\alpha}$$ and we'll show that $$A_{\alpha+1}, B_{\alpha+1}$$ form a good partition of $$H_{\alpha + 1}$$. Fix $$g, h \in H_{\alpha + 1}$$ such that $$g$$ is torsion free. We have the following cases.

Case 1: $$g \in H_{\alpha}$$. If $$h \in H_{\alpha}$$, this is clear. If $$h \notin H_{\alpha}$$, then note that $$h \langle g\rangle \subseteq H_{\alpha + 1} \setminus H_{\alpha}$$ and $$A'_{\alpha + 1}, B'_{\alpha + 1}$$ form a good partition of $$H_{\alpha + 1}$$.

Case 2: Not Case 1. If $$|h\langle g\rangle \cap H_{\alpha}| \leq 1$$, then this is clear as $$A'_{\alpha + 1}, B'_{\alpha + 1}$$ form a good partition of $$H_{\alpha + 1}$$. If not, then for some $$n \neq m$$, $$hg^n, hg^m \in H_{\alpha}$$ and so $$g^{n - m} \in H_{\alpha}$$. Now apply Case 1.