Does the Hawaiian Earring Group embed into the permutation group of $\mathbb N$? Recall that the Hawaiian earring group, $\mathbb G$, is the fundamental group of the Hawaiian Earing using the point at the origin. It can be understood more combinatorially as a subgroup of the inverse limit $\varprojlim F_n$ where $F_n$ is the free group, say with  $n$ generators, call them $a^n_0,...,a^n_{n-1}$ and $p_{n+1, n}:F_{n+1} \to F_n$ is the map induced by sending $a^{n+1}_i$ to $a^n_i$ for $i < n$ and $p(a^{n+1}_n) = a^n_0$ (say). The permutation group of the natural numbers, $S(\mathbb N)$, is the group of permutations of $\mathbb N$.
My question is simply, does $\mathbb G$ embed into $S(\mathbb N)$?
Note that $S(\mathbb N)$ does not embed into $\mathbb G$ since the later is torsion free whereas $S(\mathbb N)$ has elements of finite order. However $S(\mathbb N)$ embeds each $F_n$ hence the question.
 A: $\DeclareMathOperator\S{\mathfrak{S}}\DeclareMathOperator\N{\mathbf{N}}$Yes, because:

*

*as a subgroup of a projective limit of a sequence of finitely generated free groups $F_n$, it embeds into the product $\prod_n F_n$.


*each countable group embeds into $\S(\N)$ (just consider the left action)


*If $(G_n)$ is a sequence of groups, each embedding into $\S(\N)$, then $\prod_n G_n$ also embeds into $\S(\N)$: just use the fact that the component-wise of the horizontal partition of $\S(\N^2)\simeq\S(\N)$ is isomorphic to $\S(\N)^{\N}$.

Side notes:
A necessary condition for a group to embed into $\S(\N)$ is to have cardinal $\le c$. For abelian groups it's necessary and sufficient. However, there are groups of cardinal $\le c$ that don't embed into $\S(\N)$. Example are the restricted direct product of any uncountable family of non-abelian groups, say, $\S_3^{(\aleph_1)}$.
Under ZFC+CH it's an open question whether every group of cardinal $\le c$ embeds into $\S(\N)/\S_{\mathrm{fin}}(\N)$.
