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.



Does the Hawaiian Earring Group embed into the permutation group of $\mathbb N$?
$\DeclareMathOperator\S{\mathfrak{S}}\DeclareMathOperator\N{\mathbf{N}}$Yes, because:


1. as a subgroup of a projective limit of a sequence of finitely generated free groups $F\_n$, it embeds into the product $\prod\_n F\_n$.
2. each countable group embeds into $\S(\N)$ (just consider the left action)
3. 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 $\S(\N)^{\N}$.




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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 $\S(\N)/\S\_{\mathrm{fin}}(\N)$.


