Timeline for For the symmetric group on an infinite set, is there a generating set of strictly smaller cardinality? [closed]
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2011 at 21:31 | comment | added | Igor Rivin | Actually, the question is not great, but it is very close to some quite interesting questions (google "the Bergman property") | |
| Nov 5, 2011 at 14:50 | history | closed |
Bill Johnson user6976 Gjergji Zaimi Alain Valette Martin Brandenburg |
too localized | |
| Nov 5, 2011 at 10:10 | comment | added | Felix Denis | Just realised this myself. Sorry for wasting your internet. | |
| Nov 5, 2011 at 10:04 | vote | accept | Felix Denis | ||
| Nov 5, 2011 at 4:59 | comment | added | user6976 | This is not a research level question. Voted to close. | |
| Nov 5, 2011 at 4:06 | comment | added | Bill Johnson | Why is a completely trivial question like this getting action? | |
| Nov 5, 2011 at 1:59 | answer | added | Todd Trimble | timeline score: 4 | |
| Nov 5, 2011 at 1:30 | answer | added | Boris Novikov | timeline score: 1 | |
| Nov 5, 2011 at 1:25 | comment | added | Steve D | And of course Tom's comment shows the general case, too (assuming AoC): a generating set of size $\omega$ gets you a group only as big as $\omega\cdot |\mathbb{N}|$. | |
| Nov 5, 2011 at 1:13 | comment | added | Tom Goodwillie | No: If a group has a countable generating set then it is countable. But there are uncountably many permutations of a countably infinite set. | |
| Nov 5, 2011 at 0:54 | history | asked | Felix Denis | CC BY-SA 3.0 |