Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $S_{\infty}'$?

share|improve this question
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
add comment

2 Answers

up vote 11 down vote accepted

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.

share|improve this answer
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
add comment

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.