For any non-empty set $X$ let $\text{Sym}(X)$ denote the group of bijections $f:X\to X$ with composition.
Is there an infinite set $X$ and a surjective group homomorphism $\pi: \text{Sym}(X)\to \mathbb{Z}$?
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asked Apr 21, 2018 at 8:08
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12$\begingroup$ No. Vitali proved in 1915 that every element of $\mathrm{Sym}(X)$, for $X$ infinite, is a product of squares, and therefore $\mathrm{Hom}(\mathrm{Sym}(X),\mathbf{Z}/2\mathbf{Z})=\{0\}$ $\endgroup$– YCorCommented Apr 21, 2018 at 8:23
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2$\begingroup$ (I should say that precisely Vitali stated and proved that $\mathrm{Hom}(\mathrm{Sym}(X),\mathbf{Z}/2\mathbf{Z})=0$, and that his method consists in proving that every element is a product of, say 4, squares.) $\endgroup$– YCorCommented Apr 21, 2018 at 9:14
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6$\begingroup$ Just a side remark: the definition of $\mathrm{Sym}(X)$ works just fine for empty $X$! Including that unnecessary “non-empty” in so many definitions is nothing but a superstitious habit. $\endgroup$– Peter LeFanu LumsdaineCommented Apr 21, 2018 at 9:24
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12$\begingroup$ The answer is: no. You do not need to know anything fancy, only that every element in $Sym(X)$ is conjugate to its inverse -- which is obvious from looking at the cycle decomposition. What is less trivial is Vitali's theorem, but that is not needed to exclude existence homomorphisms to $Z$. $\endgroup$– Andreas ThomCommented Apr 21, 2018 at 9:49
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2$\begingroup$ Nice, it's indeed even simpler than Vitali's (which is not fancy however, but takes a few more lines). Vitali's paper (which is 2-3 pages) was up to my knowledge, the first where the the group of all permutations of an infinite set was defined and considered as a group. The question of looking at homomorphisms to $Z/2Z$ was natural since the initial question was whether one can extend the signature homomorphism. $\endgroup$– YCorCommented Apr 21, 2018 at 10:21
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1 Answer
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The answer here is negative. In fact, any non-trivial quotient group of the symmetric group $\mathrm{Sym}(X)$ contains a copy of $\mathrm{Sym}(X)$. Indeed, by the Baer-Schreier-Ulam Theorem, any normal subgroup $N\ne \mathrm{Sym}(X)$ is contained in the subgroup $\mathrm{Sym}_<(X)$ of permutations having support of cardinality $<\kappa:=|X|$. Let $q:\mathrm{Sym}(X)\to \mathrm{Sym}(X)/N$ be the quotient homomorphism.
Since $X$ is infinite, we can choose a family of pairwise distinct $\{x_{p}\}_{p\in \kappa\times\kappa}$ in $X$.
For every permutation $\pi\in \mathrm{Sym}(\kappa)$ of $\kappa$ define the permutation $\bar\pi\in \mathrm{Sym}(X)$ letting $\bar\pi(x_{\alpha,\beta})=x_{(\pi(\alpha),\beta)}$ for $(\alpha,\beta)\in \kappa\times \kappa$ and $\bar\pi(x)=x$ for any $x\in X\setminus\{x_{p}:p\in \kappa^2\}$. It is clear that $e:\mathrm{Sym}(\kappa)\to \mathrm{Sym}(X)$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(\mathrm{Sym}(\kappa))$ in $\mathrm{Sym}(X)$ is disjoint with the subgroup $\mathrm{Sym}_{<}(X)\supset N$ and hence the composition $q\circ e:\mathrm{Sym}(\kappa)\to \mathrm{Sym}(X)/N$ is injective.
answered Apr 21, 2018 at 8:32
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$\begingroup$ It would be nice if you could use a consistent notation for the symetric group over a set. $\endgroup$ Commented Apr 21, 2018 at 10:42
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$\begingroup$ @LucGuyot Done! Thank you for your comment. $\endgroup$ Commented Apr 21, 2018 at 11:12