Timeline for On the sum of the subgroup orders of a finite group
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2020 at 0:14 | comment | added | R. van Dobben de Bruyn | Another way to get the same bound: every element is contained in the subgroup it generates and in the whole group (and these are different if $G$ is not cyclic). (By double counting, $\sigma(G)/|G|$ is the average number of groups containing a given element. I haven't been able to put this method to any further use.) | |
| Jun 29, 2020 at 14:54 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
removed superfluous text
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| Jun 29, 2020 at 11:08 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
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| Jun 29, 2020 at 10:56 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
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| Jun 29, 2020 at 10:43 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
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| Jun 29, 2020 at 10:40 | comment | added | Sebastien Palcoux | Good! Your redaction can be shorten by writting: $G$ non-cyclic iff $G$ equals the union of its maximal subgroups (which share the trivial element). Then for $G$ non-cyclic, the sum of the order of its maximal subgroups must be greater than $|G|$, so that $\sigma(G) > 2|G|$ . | |
| Jun 29, 2020 at 10:16 | history | answered | Geoff Robinson | CC BY-SA 4.0 |