The symmetric group of $\mathbb{Z}$ is isomorphic to which group?

Question

The question: The symmetric group of $\mathbb{Z}$ (denoted as $\text{Sym}(\mathbb{Z})$), also known as the group of all bijections from $\mathbb{Z}$ to $\mathbb{Z}$, is isomorphic to which "usual" group?

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By "usual group", I mean a group that was already researched and analyzed through at least some means. In addition, I already knew that $|\text{Sym}(\mathbb{Z})| = |\mathbb{R}|$ and (the following is a wrong claim) $$\text{Sym}(\mathbb{Z}) = \langle G \; | \; x^2 = e \; \; \forall x \in G\rangle$$, whereas $e$ is the group's identity and $G$ is the set of all swapping permutations having the form $(0 \; \; n) \; (n \in \mathbb{Z} \setminus \{0\})$.

Thanks in advance.

Answer 1

$\mathrm{Sym}(\mathbb{Z})$ is isomorphic to $\mathrm{Sym}(\mathbb{N}).$ Since $\mathbb{N}$ is an initial ordinal, usually denoted $\omega$, this is probably as good as we can really expect to do. I don't think it has a truly useful presentation by generators and relations; transpositions won't do it, of course.

Answer 2

The subgroup of finite permutations has some sort of useful representation theory developed by the Russian school. See for example http://arxiv.org/abs/math/0312270 .

If you want all permutations there is nothing to say. The group is not expressible in terms of anything simpler than itself. The full symmetric group on an infinite set does not have many features in common with $S_n$ for finite $n$.