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 $K \cong G$.



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> **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?
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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 $G$. 



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> **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 $G$?
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What I know so far about this question is: 


$(1)$ 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](https://en.wikipedia.org/wiki/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.



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


