Timeline for Generating random finite groups
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 5 at 6:28 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| S Jul 17, 2015 at 0:12 | history | bounty ended | CommunityBot | ||
| S Jul 17, 2015 at 0:12 | history | notice removed | CommunityBot | ||
| Jul 13, 2015 at 13:04 | answer | added | Achim Krause | timeline score: 4 | |
| Jul 13, 2015 at 12:32 | answer | added | Russ Woodroofe | timeline score: 6 | |
| Jul 9, 2015 at 1:29 | comment | added | Gjergji Zaimi | This is related in spirit :) mathoverflow.net/questions/36735/… | |
| S Jul 8, 2015 at 22:55 | history | bounty started | Joseph O'Rourke | ||
| S Jul 8, 2015 at 22:55 | history | notice added | Joseph O'Rourke | Draw attention | |
| May 2, 2014 at 22:04 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Open problems
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| May 2, 2014 at 14:32 | comment | added | Geoff Robinson | As other have said, getting a uniform distribution on groups of order $n$ looks unlikely. Most groups of order $n$ are nilpotent, and most nilpotent ones have class at most $2$- these statements can be made reasonably precise. | |
| May 2, 2014 at 13:47 | comment | added | Stefan Kohl♦ | When I was a student, I thought about the problem of generating a random group in search of an improvement of the Elliptic Curves Method (ECM) for factoring integers. The idea was that since most finite groups are 2-groups, in particular one would almost always hit a group with smooth order, and hence under reasonable circumstances obtain a factorization almost instantly -- in contrast to ECM, which uses abelian groups only. However, thinking a bit more about the issue, I realized the problems with feasibility ... . | |
| May 2, 2014 at 13:16 | comment | added | Derek Holt | This sounds very pessimistic, but I would be very surprised if there were any reasonably accurate way of doing that except in cases where it is possible to produce a complete list of all groups of order $n$. It would be challenging to come up with any plausible ideas for this for groups of order a power of $2$, and even if you could do so, the results would be dependent on unproved conjectures that "almost all" such groups are $2$-step nilpotent with elementary abelian layers. | |
| May 2, 2014 at 11:58 | comment | added | Michael Zieve | You may want to weight groups by their automorphism groups. That is, if $a(G)$ denotes $1/\#\text{Aut}(G)$, and $b(n)$ is the sum of $a(G)$ over all groups $G$ of order $n$, then have an order-$n$ group $G$ occur with probability $a(G)/b(n)$. This is the philosophy behind the Cohen-Lenstra heuristics. For instance, see Qiaochu Yuan's answer to mathoverflow.146861. | |
| May 2, 2014 at 11:47 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
More precision in language.
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| May 2, 2014 at 11:41 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |