Timeline for Number of non-Abelian groups of order $2^n$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 15 at 6:01 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question has been bumped anyway)
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| S Nov 9, 2013 at 7:20 | history | suggested | Abhimanyu Pallavi Sudhir | CC BY-SA 3.0 |
reasrch.att2.com/~njas/sequences
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| Nov 9, 2013 at 6:53 | review | Suggested edits | |||
| S Nov 9, 2013 at 7:20 | |||||
| Jul 4, 2010 at 0:21 | vote | accept | HYYY | ||
| Jul 3, 2010 at 2:32 | answer | added | john mangual | timeline score: 5 | |
| Jul 3, 2010 at 1:30 | answer | added | Matthew Kahle | timeline score: 16 | |
| Jul 3, 2010 at 1:03 | answer | added | The Mathemagician | timeline score: -1 | |
| Jul 3, 2010 at 0:23 | comment | added | HYYY | @Qiaochu Yuan: Ah, okay. That would work. And if I can apply the fundamental theorem correctly, the number of Abelian $p$-groups of order $p^n$ is the partition function of $n$. Thank you for the idea. | |
| Jul 3, 2010 at 0:21 | answer | added | Daniel Mehkeri | timeline score: 4 | |
| Jul 2, 2010 at 23:48 | comment | added | Qiaochu Yuan | I don't think you'll be able to do any better than subtracting the number of abelian groups from the number of groups, since the latter is complicated and the former is simple. As a hint for what the former looks like: en.wikipedia.org/wiki/… | |
| Jul 2, 2010 at 23:41 | history | asked | HYYY | CC BY-SA 2.5 |