Timeline for Does the hypergraph structure of the set of subgroups of a finite group characterize isomorphism type?
Current License: CC BY-SA 4.0
10 events
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| May 28, 2019 at 11:53 | history | edited | Max Horn | CC BY-SA 4.0 |
added 783 characters in body
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| Jul 10, 2016 at 19:40 | comment | added | Ángel Valencia | I found a better answer!: math.stackexchange.com/questions/14588/… | |
| Aug 19, 2010 at 8:39 | vote | accept | Chris Beck | ||
| Aug 18, 2010 at 10:13 | comment | added | Max Horn | I just modified ElementsInNiceOrder to randomly permute the subgroups of order 5 (equivalently, the conjugacy class $g_1^G$), and run the program a couple hundred times. That empirically confirms my claim. To prove it in general, you have to show that any permutation of the order 5 groups extends to an automorphism of the subgroup lattice. Every subgroup of order 5 is in exactly 2 groups of order 55. There are 22 subgroups of these in total, split into 2 conjugacy classes of size 11. Pick any pair, their intersection defines a unique group of order 5 -- so we get all $11^2=121$ of them! | |
| Aug 18, 2010 at 9:57 | comment | added | Max Horn | The construction for the groups is general, so you can construct infinitely many. All these are (super)solvable. So, what about simple groups? My hunch: They won't work (there are few pairs of simple groups of equal order); using the existing knowledge of the maximal subgroups of simple groups one should be able to resolve this. My example seems to be pretty "rare", the "equal conjugacy classes" property is strong. The order chosen by GAP is deterministically "random". It works so well because all the subgroups of order 5 intersect each other trivially, and also the subgroup of order $11^2$. | |
| Aug 18, 2010 at 8:54 | comment | added | Chris Beck | Wow, I can't believe that ordering works! It is completely arbitrary dependent on how GAP chooses to order the elements, right? Out of curiosity, do you have reason to believe that a random bijection that respects the orders of elements should work, for most large enough groups, if not all groups, or something along these lines? I intend to accept the answer, I'd just like to work out the pencil and paper proof myself first :) | |
| Aug 17, 2010 at 19:23 | history | edited | Max Horn | CC BY-SA 2.5 |
Tried to improve language a bit, correct spelling
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| Aug 17, 2010 at 18:00 | comment | added | Cam McLeman | Very nice answer! | |
| Aug 17, 2010 at 14:47 | history | edited | Max Horn | CC BY-SA 2.5 |
Simplified computation of Gsubs and Hsubs
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| Aug 17, 2010 at 13:10 | history | answered | Max Horn | CC BY-SA 2.5 |