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 $S_\infty / \cup_n A_n$.

  • $\begingroup$ Could you clarify the question a bit? In particular, what do you mean by a homomorphism that "restricts to the signum"? $\endgroup$ – Thorny Jan 19 '10 at 9:51
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    $\begingroup$ "signum" sounds to me like the kind of thing a native German speaker might call "sign" (of a permutation). The question seems pretty clear to me. S_infty contains the subgroup G of permutations that only move finitely many elements. Such permutations have a sign. That gives us a map s:G-->Z/2. The OP claims there's no group hom S_infty-->Z/2 that extends s, and wants to know whether there's an "explicit" group hom s:S_infty-->(some group containing Z/2) that restricts to s. And he doesn't want an argument of the form "by Zorn's Lemma blah blah done", he wants an explicit construction. $\endgroup$ – Kevin Buzzard Jan 19 '10 at 11:02
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    $\begingroup$ And "reminds on the signum" probably means that he wants the extension to S_infty to have a definition which is reminiscent of the definition of "sign" on G. $\endgroup$ – Kevin Buzzard Jan 19 '10 at 11:03
  • $\begingroup$ perhaps I should take another english course :) $\endgroup$ – Martin Brandenburg Jan 19 '10 at 17:44
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    $\begingroup$ "Signum function" is perfectly good English, but is used, as far as I know, to refer to the function $\mathrm{sgn}:\mathbb R\to\{-1,0,1\}$ that we all know from Calculus. $\endgroup$ – Mariano Suárez-Álvarez Jan 19 '10 at 18:13

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.

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    $\begingroup$ @Pete: ouch. Doesn't this mean that my answer is the best that Martin will get?? The kernel of his sign map will be a non-trivial proper normal subgroup, and it can't contain a transposition, so it must be my A (your a_infty). $\endgroup$ – Kevin Buzzard Jan 19 '10 at 13:04
  • $\begingroup$ ok I should have written that I knew this result (I used it in my claim that Z/2 does not work), but you have made it clear that my problem is really to get a good description of $S_\infty / \mathfrak{a}_\infty$ - is there a good one? $\endgroup$ – Martin Brandenburg Jan 19 '10 at 17:45
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    $\begingroup$ The "Schreier-Ulam" theorem was initially proved by Onofri (1929) and then rediscovered by Schreier-Ulam (1933), the first reference being widely forgotten. Details: math.stackexchange.com/a/2645097/35400 $\endgroup$ – YCor Feb 10 '18 at 22:08
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    $\begingroup$ ... and the weaker fact that there is no nontrivial homomorphism from $S_\infty$ to $\mathbf{Z}/2\mathbf{Z}$ was initially proved by Vitali in 1915. Up to my knowledge, this is the first explicit appearance of the infinite symmetric group. $\endgroup$ – YCor Dec 20 '18 at 23:59

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 $z$".

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    $\begingroup$ If Pete's statement of Schreier-Ulam is correct (and I have no reason to believe it's not) then S_\infty/A has no non-trivial quotients which are non-trivial on z. So it looks like this "answer" is the best you're going to get, modulo finding a nicer way of defining the quotient space. $\endgroup$ – Kevin Buzzard Jan 19 '10 at 13:05
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    $\begingroup$ "...(and I have no reason to believe it's not)" Thanks for the ringing endorsement, KB. I have no reason to believe that you're not wearing a "Rehab Is For Quitters" T-shirt and parachute pants...no reason at all. $\endgroup$ – Pete L. Clark Jan 20 '10 at 4:12
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    $\begingroup$ FWIW today I'm wearing fashionablygeek.com/wp-content/uploads/2007/12/usheep-shirt.jpg in olive green. $\endgroup$ – Kevin Buzzard Jan 20 '10 at 10:28

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.


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.)

  • $\begingroup$ There has been a lot of work in set theory on the classification of Borel equivalence relations on Polish spaces (= separable, completely metrizable spaces). On the basis of my admittedly incomplete knowledge of this work, I expect that there is no Borel function from $S$ into a Polish space that is constant on exactly the cosets of $F$ (or of $A$). That is, one cannot "nicely" embed $S/F$ (or $S/A$) into a "nice" space. $\endgroup$ – Andreas Blass Feb 26 '19 at 2:05
  • $\begingroup$ @AndreasBlass yes indeed, but with this too restricted interpretation of "nice". I mentioned the embedding of $S/F$ into $\mathrm{Homeo}(\beta X-X)$ as possibly nice, and the latter is not a Polish group, even if $X$ is countable. Furthermore it's too restrictive to consider Borel functions: the induced homomorphism $S\to\mathrm{Homeo}(\beta X-X)$ (vanishing on $F$) is not Borel when $S$ it endowed with its standard topology (induced by action on $X$). $\endgroup$ – YCor Feb 26 '19 at 8:29
  • $\begingroup$ In addition, using automatic continuity for $S_\infty$, one has, for every infinite set $X$ that there is no nontrivial (abstract) group homomorphism of $S_X/A_X$ to any Polish group. Here $S_X=S$ and $A_X=A$ (I just emphasize the dependency on $X$.) $\endgroup$ – YCor Feb 26 '19 at 11:15

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