Timeline for How many rearrangements must fail to alter the value of a sum before you conclude that none do?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 17, 2017 at 18:08 | history | edited | Michael Hardy | CC BY-SA 3.0 |
added 2 characters in body
|
| Aug 14, 2015 at 13:38 | history | edited | Michael Hardy | CC BY-SA 3.0 |
added 12 characters in body
|
| Aug 14, 2015 at 9:46 | comment | added | Jeremy Rickard | Ah, sorry! I'd misread the quantifiers in the question (in a way that makes no sense whatsoever). | |
| Aug 14, 2015 at 9:03 | comment | added | Aaron Meyerowitz | It is not clear to me that every set of permutations which generates the full symmetric group works. We are not allowed to compose. Consider the full symmetric group on the even integers along with all the involutions consisting of a finite or infinite number of pairs (2k-1 2k). I'm not sure that they generates the full symmetric group, but they certainly don't change the sum of the series 1,0,-1,0,1/2,0,-1/2,0, | |
| Aug 14, 2015 at 8:31 | comment | added | Jeremy Rickard | The set of all involutions generates the full symmetric group, so that's another reason that the class $C$ of involutions works. I'm not sure whether you still have a generating set if you add condition $(*)$ and/or fixed-point freeness. | |
| Aug 14, 2015 at 8:07 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |