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 $S\_{\infty}'$?



Signatures on the infinite symmetric group
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. 


