# Signatures on the infinite symmetric group

I have a question about the symmetric group. Taking signatures of permutations defines a surjective homomorphism $S_n \rightarrow \mathbb{Z}/2$. This is compatible with the natural inclusions $S_n \hookrightarrow S_{n+1}$, so we get a surjection $S_{\infty} \rightarrow \mathbb{z}/2$. Here $S_{\infty}$ is the direct limit of the $S_n$. In other words, $S_{\infty}$ is the group of finitely supported permutations of a countable set.

This brings me to my question. let $S_{\infty}'$ be the set of all permutations of a countable set. We have an inclusion $S_{\infty} \hookrightarrow S_{\infty}'$. Does the signature map $S_{\infty} \rightarrow \mathbb{Z}/2$ extend to $S_{\infty}'$?

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See also mathoverflow.net/questions/12291/sign-of-infinite-permutations. this question actually starts with this observation here and tries to replace Z/2 by another group. – Martin Brandenburg Feb 5 '11 at 9:00

The answer to the question is "no". In fact, $S'_{\infty}$ is a perfect group, so there are no maps from it to an abelian group. Even more is true -- every element of $S'_{\infty}$ can be expressed as a commutator! This is much stronger than simply saying that $[S'_{\infty},S'_{\infty}] = S'_{\infty}$.

For these results, see Theorem 6 of the following paper.

MR0040298 (12,671e) Ore, Oystein Some remarks on commutators. Proc. Amer. Math. Soc. 2, (1951). 307–314.

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Thank you very much! – Ben S Feb 4 '11 at 22:36
For EVEN MORE, see the paper of M. Droste and I. Rivin (on arxiv.org, though it has now appeared online in Bulletin of the AMS). – Igor Rivin Feb 5 '11 at 3:35
Sorry, that's bulletin of the LMS above. – Igor Rivin Feb 5 '11 at 16:10

Here is a direct way to see the answer is 'no'.

Let our countable set be the set of all integers $\mathbb{Z}$.

What is the sign of $(1,2)(3,4)(5,6)(7,8)\dots$? (Note that we're fixing all nonpositive integers here.)

Whatever it is, you can multiply by the transposition $(1,2)$ to get a permutation with the opposite sign, then you can conjugate, which doesn't affect the sign, by $(\dots,-3,-2,-1,0,1,2,3,\dots)^{-2}$ giving back the element you started with. Contradiction.

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Another way to look at it: the above argument expresses $(1,2)$ as a commutator. – ndkrempel Feb 5 '11 at 3:38