Timeline for General bound for the number of subgroups of a finite group
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 21 at 13:52 | comment | added | Geoff Robinson | @MikhailBorovoi : Yes thanks, I suppose that is true- if you pick ( the integer part of) log2(n) elements of G allowing replacement, and allow some of the chosen elements to be the identity, you will get at least one generating set for every possible subgroup of G (with much repetition). | |
| Jul 19 at 13:08 | comment | added | Mikhail Borovoi | Thank you, @GeoffRobinson. Your argument gives a better crude upper bound: $n^{\log_2(n)}$. | |
| Jul 19 at 9:31 | comment | added | Geoff Robinson | My answer gave a crude upper bound, certainly not optimal, but the example of elementary Abelian 2-groups show that even that crude bound was not too far from the truth. | |
| Jul 16 at 19:15 | comment | added | Geoff Robinson | @MikhailBorovoi : See Kasper Andersen's answer below for the "right" optimal bound proved by Borovik, Pyber and Shalev. | |
| Jul 16 at 18:34 | comment | added | Mikhail Borovoi | Or even $n^{\log_2(n)}$, as Borovik, Pyber and Shalev mention without proof? | |
| Jul 16 at 18:19 | comment | added | Mikhail Borovoi | Dear Geoff, do I understand correctly that we have an exact (rather than approximate) upper bound $\log_{2}(n)n^{\log_{2}(n)}$ for the total number of subgroups of a group of order $n$? | |
| Jun 8, 2013 at 18:27 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
inserted elementary Abelian, as intended; deleted 2 characters in body
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| Jun 8, 2013 at 18:12 | comment | added | user13040 | Regarding the last sentence, did you intend "of an elementary abelian group" instead? | |
| Jun 8, 2013 at 16:28 | vote | accept | CommunityBot | ||
| Jun 5, 2013 at 22:30 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Expanded
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| Jun 4, 2013 at 6:50 | history | answered | Geoff Robinson | CC BY-SA 3.0 |