TITLE: Find a "natural" group that contains the quotient of the infinite symmetric group by the alternating subgroup
QUESTION [18 upvotes]: Let $S_\infty$ the group of permutations of $\mathbb{N}$. It can be shown that there is no homomorphism $S_\infty \to \mathbf{Z}/2$ extending the sign on the finite symmetric groups. Is it possible to write down a homomorphism (an unexplicit one won't be useful in my application) of $S_\infty$ into another (infinite) group, which restricts to the sign? Perhaps we should also require that the homomorphism somehow also reminds of the sign in the infinite case. Thus perhaps we should formalize something like $(-1)^M$, where $M$ is an infinite set (as you might guess, this is related with my question about Infinite Tensor Products).
EDIT: As was pointed out by Pete, the question is equivalent to: Find a nice, "natural" group which contains $S_\infty / \cup_n A_n$.

REPLY [9 votes]: This group is quite fascinating and I've thought quite a lot about this question. Although I'm unable to answer it in a sensible way (an example of non-sensible way would be to embed this group in the group of permutations on itself through left translation), let me say a few things.
Let $S$ be the whole symmetric group (on an infinite set $X$), $F$ its finitary subgroup (finitely supported permutations), and $A$ its subgroup of index 2 of even permutations. The question is about $S/A$, which lies in a central extension
$$ 1\to F/A(\simeq \mathbf{Z}/2\mathbf{Z}) \to S/A\to S/F\to 1;$$ 
let $\omega^X\in H^2(S/F,\mathbf{Z}/2\mathbf{Z})$ be the cohomology class of this extension. It follows from Vitali's result $\mathrm{Hom}(S,\mathbf{Z}/2\mathbf{Z})=0$ that $\omega\neq 0$. Sergiescu observed, using acyclicity of $S$ (la Harpe- McDuff) that $H^2(S/F,\mathbf{Z}/2\mathbf{Z})$ is reduced to $\{0,\omega^X\}$, and indeed that $H_2(S/F)\simeq\mathbf{Z}/2\mathbf{Z}$.
Given a group $G$ and a homomorphism $f:G\to S/F$ (I call this a balanced near action), one thus gets, by pullback, a cohomology class $f^*\omega^X\in H^2(G,\mathbf{Z}/2\mathbf{Z})$, whose nonvanishing is on obstruction for $f$ to lift to a homomorphism $G\to S$, i.e., a genuine action on $X$. 
The first nontrivial explicit computations of such classes were done by Kapoudjian in the context of Higman-Thompson and Neretin's groups and their natural near actions on trees. For this reason, I call $f^*\omega^X$ Kapoudjian class of the near action, and I address it here, §8.6 (self-advertisement warning, so I cw this answer).
Concerning the initial question, it has a kind of easier analogue which deserve some comment, namely finding a "natural" embedding of $S/F$ (rather than $S/A$): the simplest answer seems to be to embed it into the group of self-homeomorphisms of the Stone-Cech boundary of $X$. For $X$ countable, there is a significant literature around how large is the image within the whole self-homeomorphism group (Rudin-Shelah problem). In any case such ideas do not seem to provide embeddings of $S/A$. I insist anyway, because any embedding of $S/A$ kind of induces an embedding of $S/F$ (not literally speaking, but after slightly modifying the target group), so first a good understanding of how to embed $S/F$ would be useful (there are not so many understood ways), and second a good understanding of the central extension would be useful too, and this is well encoded in the Kapoudjian class.

Added: I made Vitali's 1915 rare paper (in Italian) available here. Reference info: G. Vitali. Sostituzioni sopra una infinità numerabile di elementi. Bollettino Mathesis 7: 29-31, 1915. (Any suggestion for a more standard repository is welcome.)