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 $G$?

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), 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.
 A: 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.
