Question

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 usefulin 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 $S_\infty / \cup_n A_n$.

Answer 1

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 $z$".

Answer 2

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.

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

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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.

Answer 3

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.