Timeline for Which subgroup order of the symmetric group is the most frequent?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2015 at 0:31 | vote | accept | Daniel Soltész | ||
| Jul 11, 2015 at 1:25 | answer | added | Igor Rivin | timeline score: 7 | |
| Jul 11, 2015 at 0:25 | comment | added | Daniel Soltész | @EricWofsey Well I can't prove it I just read it at OEIS. Here is the proof, (jstor.org/stable/2946623?seq=8#page_scan_tab_contents) I think the lower bound suggests that we look at the cases where $k$ is a power of two. (Maybe this effect is the reason Gerhard Paseman had two in his mind.) | |
| Jul 11, 2015 at 0:22 | comment | added | Alex R. | @EricWofsey: It seems like a difficult question actually, see the top answer here: math.stackexchange.com/questions/76176/… | |
| Jul 10, 2015 at 23:44 | comment | added | Eric Wofsey | How do you prove there are $c^{n^2}$ subgroups? It seems likely to me that by inspecting such a proof it would not be too hard to find the $k$ that maximizes $a_k$. | |
| Jul 10, 2015 at 23:25 | comment | added | Daniel Soltész | @GerhardPaseman Well I don't have any OEIS sequence for a_k, but I have one for the total number of subgroups (oeis.org/A005432). But k can not be small by a counting argument. The number of $2$ element subgroups is trivially at most $n!$, but the total number of subgroups has order of magnitude $c^{n^2}$, so the most frequent order by the pidgeon-hole must be at least $c^{n^2}/n!$ which is larger. | |
| Jul 10, 2015 at 23:03 | comment | added | Gerhard Paseman | I would guess k=2. Do you have any data or OEIS sequences? Gerhard "Yet Another Use For Mode" Paseman, 2015.07.10 | |
| Jul 10, 2015 at 23:03 | history | edited | Daniel Soltész | CC BY-SA 3.0 |
added 110 characters in body
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| Jul 10, 2015 at 23:00 | comment | added | Daniel Soltész | In terms of number of subgroups. I will edit the question to make this obvious. | |
| Jul 10, 2015 at 22:59 | comment | added | Stefan Kohl♦ | "Most frequent" in terms of number of conjugacy classes of subgroups, or really in terms of number of subgroups? | |
| Jul 10, 2015 at 22:56 | history | asked | Daniel Soltész | CC BY-SA 3.0 |