Is there a name for the quotient of the symmetric group by the finitary symmetric group? The finitary symmetric group on a set $S$ is the group of permutations that only move a finite set points. That is:
$$FSym(S) = \{\phi:\{s : s \in S, \phi(s) \neq s\}\text{ is finite}\}$$
This is a subgroup of the symmetric group on $S$. Moreover, it is a normal subgroup. To see this, let $\phi \in FSym(S)$ and $\psi \in Sym(S)$ be permutations. Then
$$\psi\phi\psi^{-1}(x) = x \iff \phi(\psi^{-1}(x)) = \psi^{-1}(x)$$
$\phi \in FSym(S)$ implies that there are only finitely many $\psi^{-1}(x)$ that do not satisfy the above condition, which implies that there are only finitely many $x$ that do not satisfy it as well.
Does the group $Sym(S) / FSym(S)$ have a name?

The above construction can also be slightly generalized. Let $I$ be an ideal on $S$. Then we define $$Sym_I(S) = \{\phi : \{s : s \in S, \phi(s) \neq s\} \in I\}$$
We will show that $Sym_I(S)$ is a subgroup of $Sym(S)$ in more detail than we did above.


*

*$Sym_I(S)$ contains the identity permutation since $\emptyset \in I$.

*For any $\phi \in Sym_I(S)$, $\phi^{-1}$ moves the same points as $\phi$. Therefore, $\phi^{-1} \in Sym_I(S)$.

*For any $\phi, \psi \in Sym_I(S)$, we have that $$\phi \psi (x) \neq x \implies \psi(x) \neq x \lor \phi(x) \neq x$$ so the points moved by $\phi \psi$ is a subset of the union of the points moved by $\psi$ and the points moved by $\phi$. This set of points will be in the ideal.


If $I \subseteq J$, then $Sym_I(S)$ is a subgroup of $Sym_I(J)$. If it is a normal subgroup, then the quotient group $Sym_J(S) / Sym_I(S)$ will exist.
If $I$ is the ideal of finite sets and $J$ is the ideal of all subsets, then
$$Sym_I(S) = FSym(S)$$
$$Sym_J(S) = Sym(S)$$
 A: There's a group which deserves a name more than this quotient, which is called the near symmetric group $\mathfrak{S}^\star(S)$, namely the group of "bijection between cofinite subsets" modulo cofinite coincidence. It can be viewed as the group of "germs at infinity" of permutations of $S$.
There is a canonical "index homomorphism" from $\mathfrak{S}^\star(S)$ onto $\mathbf{Z}$, which maps an element with representative a bijection $S\smallsetminus F_1\to S\smallsetminus F_2$ ($F_1,F_2$ finite), to $|F_2|-|F_1|$.
The kernel $\mathfrak{S}^\star_0(S)$ of the index homomorphism can be identified to the quotient of the symmetric group $\mathfrak{S}(S)$ by its normal subgroup of finitary permutations.
It can be shown that the canonical homomorphism $\mathfrak{S}^\star(S)\to\mathrm{Aut}(\mathfrak{S}^\star_0(S))$ is an isomorphism. It also equals the derived subgroup of $\mathfrak{S}^\star(S)$.
I call $\mathfrak{S}^\star_0(S)$ (the group you're asking about) the index 0 near symmetric group, or the balanced near symmetric group.
