Signatures on the infinite symmetric group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:16:10Z http://mathoverflow.net/feeds/question/54371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54371/signatures-on-the-infinite-symmetric-group Signatures on the infinite symmetric group Ben S 2011-02-04T22:25:18Z 2011-02-04T23:45:22Z <p>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.</p> <p>This brings me to my question. let <code>$S_{\infty}'$</code> be the set of all permutations of a countable set. We have an inclusion <code>$S_{\infty} \hookrightarrow S_{\infty}'$</code>. Does the signature map $S_{\infty} \rightarrow \mathbb{Z}/2$ extend to <code>$S_{\infty}'$</code>?</p> http://mathoverflow.net/questions/54371/signatures-on-the-infinite-symmetric-group/54372#54372 Answer by Andy Putman for Signatures on the infinite symmetric group Andy Putman 2011-02-04T22:29:20Z 2011-02-04T22:29:20Z <p>The answer to the question is "no". In fact, <code>$S'_{\infty}$</code> is a perfect group, so there are no maps from it to an abelian group. Even more is true -- every element of <code>$S'_{\infty}$</code> can be expressed as a commutator! This is much stronger than simply saying that <code>$[S'_{\infty},S'_{\infty}] = S'_{\infty}$</code>.</p> <p>For these results, see Theorem 6 of the following paper.</p> <p>MR0040298 (12,671e) Ore, Oystein Some remarks on commutators. Proc. Amer. Math. Soc. 2, (1951). 307–314. </p> http://mathoverflow.net/questions/54371/signatures-on-the-infinite-symmetric-group/54374#54374 Answer by ndkrempel for Signatures on the infinite symmetric group ndkrempel 2011-02-04T23:22:44Z 2011-02-04T23:45:22Z <p>Here is a direct way to see the answer is 'no'.</p> <p>Let our countable set be the set of all integers $\mathbb{Z}$.</p> <p>What is the sign of $(1,2)(3,4)(5,6)(7,8)\dots$? (Note that we're fixing all nonpositive integers here.)</p> <p>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.</p>