Timeline for Surjective group homomorphism from $\text{Sym}(X)$ onto $\mathbb{Z}$
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2018 at 5:15 | comment | added | YCor | @LSpice G. Vitali. Sostituzioni sopra una infinità numerabile di elementi. Bollettino Mathesis 7: 29–31, 1915. | |
| Apr 21, 2018 at 23:14 | comment | added | LSpice | @YCor, what is the Vitali paper? | |
| Apr 21, 2018 at 22:32 | history | reopened |
Derek Holt Benjamin Steinberg Daniel Loughran Johannes Hahn David Handelman |
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| Apr 21, 2018 at 21:53 | comment | added | Andreas Thom | @Qfwfq: This is a question which can be answered by any competent google search and thus is not of research level. We see a lot of questions asked for the sake of asking (I believe) and this should be discouraged in my opinion. | |
| Apr 21, 2018 at 16:50 | comment | added | YCor | More explicitly, it's a sub-duplicate of mathoverflow.net/questions/281289/… | |
| Apr 21, 2018 at 16:12 | comment | added | YCor | It also has many sort-of duplicates: one already mentioned math.stackexchange.com/questions/846209; it is already answered in mathoverflow.net/questions/12291 (although the question is distinct), etc (search "Schreier-Ulam") | |
| Apr 21, 2018 at 15:41 | review | Reopen votes | |||
| Apr 21, 2018 at 22:32 | |||||
| Apr 21, 2018 at 15:36 | comment | added | Qfwfq | Okay to close the question as "no longer relevant". But... is it really so offtopic? | |
| Apr 21, 2018 at 15:10 | history | closed |
YCor Emil Jeřábek Peter LeFanu Lumsdaine Andreas Thom R. van Dobben de Bruyn |
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| Apr 21, 2018 at 12:58 | vote | accept | Dominic van der Zypen | ||
| Apr 21, 2018 at 10:27 | comment | added | YCor | Andreas' argument also shows that it has no nontrivial homomorphism into any torsion-free abelian group (while Vitali's fact does not discard homomorphisms into $\mathbf{Q}$). Actually it shows that the abelianization has exponent $\le 2$. So the combination of the two arguments shows that the group is perfect, which of course follows from Onofri's later (and more difficult) classification of normal subgroups. | |
| Apr 21, 2018 at 10:21 | comment | added | YCor | 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. | |
| Apr 21, 2018 at 9:49 | comment | added | Andreas Thom | 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 $Z$. | |
| Apr 21, 2018 at 9:24 | comment | added | Peter LeFanu Lumsdaine | 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. | |
| Apr 21, 2018 at 9:14 | comment | added | YCor | (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.) | |
| Apr 21, 2018 at 8:59 | review | Close votes | |||
| Apr 21, 2018 at 15:11 | |||||
| Apr 21, 2018 at 8:32 | answer | added | Taras Banakh | timeline score: 12 | |
| Apr 21, 2018 at 8:31 | comment | added | Derek Holt | Or see this MSE question for example. | |
| Apr 21, 2018 at 8:23 | comment | added | YCor | No. Vitali proved in 1915 that every element of $\mathrm{Sym}(X)$, for $X$ infinite, is a product of squares, and therefore $\mathrm{Hom}(\mathrm{Sym}(X),\mathbf{Z}/2\mathbf{Z})=\{0\}$ | |
| Apr 21, 2018 at 8:08 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |