Timeline for General bound for the number of subgroups of a finite group
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 3, 2019 at 10:11 | answer | added | Adam P. Goucher | timeline score: 3 | |
| May 3, 2019 at 9:15 | history | edited | YCor |
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| Feb 10, 2015 at 13:51 | answer | added | Kasper Andersen | timeline score: 16 | |
| S Jun 8, 2013 at 16:28 | vote | accept | CommunityBot | ||
| Jun 4, 2013 at 10:52 | answer | added | Nick Gill | timeline score: 11 | |
| Jun 4, 2013 at 7:38 | comment | added | Name | My comment above is wrong because according to the link @Gerry-Myerson sent, the symmetric group $S_3$ has $6$ subgroups and the cyclic group $\mathbb{Z}/{6\mathbb{Z}}$ has $4$ subgroups. | |
| Jun 4, 2013 at 6:50 | answer | added | Geoff Robinson | timeline score: 25 | |
| Jun 4, 2013 at 5:09 | comment | added | user13040 | It seems that elementary abelian $p$-groups dominate. Since these groups have $\Theta (p^{\frac{(\log n)^2}{4}}) = \Theta(n^{c \log n})$, it is plausible that some sub-exponential bound of this sort holds for all finite groups, but I am not aware of any relevant work. | |
| Jun 4, 2013 at 5:00 | vote | accept | CommunityBot | ||
| S Jun 8, 2013 at 16:28 | |||||
| Jun 3, 2013 at 23:53 | comment | added | Gerry Myerson | Maximal number of subgroups of a group with $n$ elements is tabulated at oeis.org/A018216 | |
| Jun 3, 2013 at 23:47 | answer | added | P Vanchinathan | timeline score: 5 | |
| Jun 3, 2013 at 22:13 | comment | added | Name | It seems that among all groups of order $n$, the abelian group whose $p$-sylow subgroups are $p$-elementary has the largest number of subgroups and it is not difficult task to count them. | |
| Jun 3, 2013 at 21:54 | comment | added | Gerhard Paseman | For n bigger than 4, you can even take a quarter of that. Gerhard "Except For Exceptions, Of Course" Paseman, 2013.06.03 | |
| Jun 3, 2013 at 21:21 | comment | added | Stefan Kohl♦ | Yes. -- You can take $f(n) := 2^n$, which is the total number of subsets of $G$. | |
| Jun 3, 2013 at 20:52 | history | asked | user13040 | CC BY-SA 3.0 |