Timeline for Injective proof about sizes of conjugacy classes in S_n
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 3, 2009 at 18:12 | vote | accept | Jonah Ostroff | ||
| Nov 3, 2009 at 18:12 | history | bounty ended | Jonah Ostroff | ||
| Nov 3, 2009 at 16:48 | answer | added | Michael Lugo | timeline score: 0 | |
| Nov 3, 2009 at 16:22 | answer | added | user631 | timeline score: 5 | |
| Nov 2, 2009 at 15:59 | comment | added | Jonah Ostroff | (Append the word "fixed" after "gets" in the last sentence above.) Something that seems to work here is fixing whichever element follows the 1 in Sonia's cycle ordering below. In the second-biggest conjugacy classes, everything but 1 meets this condition exactly (n-2)! times, which is what we'd hope if there are (n-1)! permutations total in that class. Where to go from there, though, eludes me. Perhaps someone else can figure it out. (Now with bounty!) | |
| Nov 2, 2009 at 15:56 | comment | added | Jonah Ostroff | I'm still pretty interested in finding an answer to this, though I haven't really made any progress myself. Here's something you might think about, though. The second-largest conjugacy classes (there are a few of these for small n) are the ones of size (n-1)! (by contrast, there are n!/(n-1) (n-1)-cycles). So to make an obviously non-surjective mapping from these second-biggest sets, we might try thinking about which element gets fixed in the image. It would be nice if exactly one element never gets, so that the functions are clearly not surjective.... | |
| Nov 2, 2009 at 15:49 | history | bounty started | Jonah Ostroff | ||
| Oct 28, 2009 at 8:09 | answer | added | Sonia Balagopalan | timeline score: 2 | |
| Oct 27, 2009 at 20:34 | history | asked | Jonah Ostroff | CC BY-SA 2.5 |