Timeline for How to generate all finite groups of order n? [closed]
Current License: CC BY-SA 2.5
13 events
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| Nov 24, 2010 at 18:36 | comment | added | HJRW | (cont'd) A better question might be something like: 'The XYZ algorithm outputs the multiplication tables of all abelian groups of order n in time f(n). What's the best known complexity of an algorithm that works for all groups?' | |
| Nov 24, 2010 at 18:35 | comment | added | HJRW | habitmelon - thanks for clarifying what you mean by 'generate'. But I still don't think the question is clear enough. There are some very naive algorithms for generating multiplication tables - just list all possible tables, and check each one to see whether it satisfies the axioms of a group, for instance - but this answer isn't very useful. One way to clarify things further would be to explain what you already know about 'how to generate all Abelian groups of order n'... | |
| Nov 24, 2010 at 18:21 | history | edited | tlehman | CC BY-SA 2.5 |
added 78 characters in body
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| Nov 23, 2010 at 17:39 | vote | accept | tlehman | ||
| Nov 23, 2010 at 3:31 | history | closed |
Martin Brandenburg user6976 Simon Thomas HJRW Victor Protsak |
not a real question | |
| Nov 22, 2010 at 22:58 | comment | added | Jim Humphreys | I share Henry's concern about the meaning of this question, as my comment above indicates. Unless it is formulated more exactly it should probably be closed. | |
| Nov 22, 2010 at 18:39 | comment | added | HJRW | Unless the OP explains what s/he means by 'generate', this is not a real question. Voting to close. | |
| Nov 22, 2010 at 14:25 | comment | added | Jim Humphreys |
Given the way in which $p$-groups proliferate as the power of $p$ goes up, there will always be practical constraints on what can be effectively computed for a given $n$. It may be more interesting to ask what can be learned by listing all groups of a given order. For example, early projects focused instead on finding all simple groups of order less than a million; this had a clear theoretical motivation.
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| Nov 22, 2010 at 2:46 | comment | added | S. Carnahan♦ | Here is an algorithm: Take all $n$-tuples of distinct elements in the symmetric group on $n$ letters (with one of them being identity), and check to see if they are closed under multiplication. For those that are, check all set-theoretic bijections between them to see if they are homomorphisms. This will give you a list of isomorphism class representatives, but it might take a while. | |
| Nov 21, 2010 at 21:07 | answer | added | Derek Holt | timeline score: 18 | |
| Nov 21, 2010 at 20:21 | comment | added | Andy Putman | What do you mean by "generate"? Other than "it's built out of simple groups of appropriate orders, but the extensions are a mess" or "write down every multiplication table and check that it gives a group", I don't think that there is going to be an easy answer. | |
| Nov 21, 2010 at 20:20 | comment | added | Kevin Buzzard | There are electronic tables of such things. In general it's quite a hard problem. The last time I looked it wasn't even known how many groups of order $n$ there were for several $n$ less than 10000 (I might be out of date here, but I would feel pretty confident saying that the list of groups of order at most a million is almost certainly still not known). | |
| Nov 21, 2010 at 20:13 | history | asked | tlehman | CC BY-SA 2.5 |