Find a "natural" group that contains the quotient of the infinite symmetric group by the alternating subgroup 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 $S_\infty / \cup_n A_n$.
 A: 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.)
A: This is not an answer per se [Edit: OK, maybe it is! I was a little fuzzy on exactly what was being asked for when I wrote this, and in the past Martin has expressed unhappiness with responses which he feels have not answered his questions.] but it should be useful for those who are thinking about the problem (c.f. Kevin Buzzard's answer) to know the following classic result.
Theorem (Schreier-Ulam): The only nontrivial proper normal subgroups of $S_{\infty}$ are $\mathfrak{s}_{\infty} = \bigcup_{n \geq 1} S_n$ and $\mathfrak{a}_{\infty} = \bigcup_{n \geq 1} A_n$, i.e. the "little symmetric group" of all permutations which move only finitely many elements and its index two alternating subgroup.  

Reference: J. Schreier and S. Ulam, 
Über die Permutationsgruppe der natürlichen Zahlenfolge. Stud. Math. 4, 134-141 (1933).

Addendum: Certainly this theorem implies that any homomorphism from $S_{\infty}$ into a group $G$ which restricts to the sign homomorphism on $\mathfrak{s}_{\infty}$ must have kernel precisely equal to $\mathfrak{a}_{\infty}$.  Whether this answers the question depends, I suppose, on how much you care about what the induced monomorphism $S_{\infty}/\mathfrak{a}_{\infty} \hookrightarrow G$ looks like.
A: Let $A$ denote the subgroup of $S_\infty$ consisting of permutations that only move finitely many elements, and have even signature. Then $A$ is a normal subgroup of $S_\infty$, and the quotient $S_\infty/A$ is a candidate group. It contains a central element $z$ of order 2, namely the image of $G/A$, where $G$ is all the permutations which only move finitely many elements. The quotient map $S_\infty\to S_\infty/A$ has all the properties you want---except that it doesn't look anything like the signature/sign map. Will this abstract but not-using-the-axiom-of-choice construction work for you or do you need a much more concrete target group?
If $S_\infty/A$ is no good for you, then my answer arguably reduces your question to "write down a nice quotient of $S_\infty/A$ which is non-trivial on $z$".
A: If one considers the distinguishing feature of the sign homomorphism $S_n \to \mathbb{Z}/2$ to be that it is the canonical map from $S_n$ to its abelianization, then there is nothing analogous for $S_\infty$ in the sense that the abelianization of $S_\infty$ is trivial.  The abelianization of a group $G$ is also the group homology $H_1(G, \mathbb{Z})$, and in fact for $G = S_\infty$, all the homology groups $H_i(S_\infty, \mathbb{Z})$ vanish for $i > 0$; $S_\infty$ is an acyclic group.  See Acyclic groups of automorphisms whose first page contains a statement of this result and similar ones.
