Timeline for Order-increasing bijection from arbitrary groups to cyclic groups
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| May 13 at 16:59 | answer | added | Mohsen | timeline score: 5 | |
| May 7, 2019 at 11:40 | comment | added | Nick Gill | This paper -- arxiv.org/pdf/1812.04167.pdf -- purports to give a positive answer to this question. I say "purports" because I have not read it; equally I have no reason to doubt its validity. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Oct 1, 2014 at 17:06 | history | suggested | Hagen von Eitzen | CC BY-SA 3.0 |
The set-braces were not rendered
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| Oct 1, 2014 at 16:59 | review | Suggested edits | |||
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| Sep 17, 2012 at 15:34 | answer | added | Frieder Ladisch | timeline score: 12 | |
| Aug 30, 2012 at 17:02 | comment | added | Frieder Ladisch | The statement also holds for groups of prime power order (proof: easy induction) and thus for nilpotent groups, so @Max no need to check groups of order 512 with GAP ;-) | |
| Aug 29, 2012 at 19:16 | comment | added | Max Horn | For what it's worth, I just verified with GAP that the statement holds for all groups of order at most 511. | |
| Aug 28, 2012 at 20:29 | comment | added | Marty Isaacs | I had not known about Lindsey's theorem that among groups of order n, the cyclic group has the largest possible average (equivalently sum) of element orders. This appeared (as mentioned by Dickman) as a solution to a Monthly problem in Dec. 1991. Independently, and very much later, I found a different proof of this result. (See my paper with Amiri and Amiri in Comm. in Alg. 37 (2009).) Of course, if the Lemma that is the subject of this posting is true, it provides a third proof. It would also prove the corresponding result about the geometric mean. (I have an unpublished proof of that.) | |
| Aug 25, 2012 at 8:54 | comment | added | Benjamin Dickman | For any finite group $G$, let $f(G)$ denote the average order of an element of $G$. If the fact you've stated is true, then for any group $G$ with $n$ elements, we'd have $f(G) \leq f(\mathbb{Z}/n\mathbb{Z})$. This alone should have been enough of a tip-off that the short solution linked to in AMM 10775 was incomplete: the above conclusion about $f$ does indeed hold, but it's AMM 6636(ii), and took a fair bit of work to prove. Source: 6636, F. Schmidt, Amer. Math. Monthly, Vol. 98, No. 10 (Dec., 1991), pp. 970-972, jstor.org/stable/2324168. | |
| Aug 24, 2012 at 12:19 | history | edited | Tom De Medts | CC BY-SA 3.0 |
edited body
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| Aug 24, 2012 at 6:19 | history | edited | Tom De Medts | CC BY-SA 3.0 |
corrected mistake |G_6|=6 instead of 12
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| Aug 23, 2012 at 22:56 | comment | added | Frieder Ladisch | In the "example", it should say $|G_6|= 6$, while '6 elements of order 12' is correct. The point of this "example" is that when proceeding according to the "proof" that appeared in Amer. Math. Monthly, $G_6$ would be exhausted by $S_1$, $S_2$, $S_3$ and $S_4$, so no place left for $S_6$. | |
| Aug 23, 2012 at 22:09 | comment | added | Marty Isaacs | I think it was Ladisch who mentioned the Hall marriage theorem. In fact, using Hall's theorem, the problem is equivalent to the following: Given a set S of divisors of n = |G|, show that the total number of elements of G having order dividing some member of S is at least as large as for a cyclic group of order n. For example, if S = {a}, then for a cyclic group with order divisible by a the count is a, and by Frobenius' theorem, the count for G is a multiple of a. The inequality thus holds if |S| = 1. I have now found a proof in the case |S| = 2, but I don't (yet) see a general argument. | |
| Aug 23, 2012 at 17:05 | comment | added | HJRW | Shouldn't that be '6 elements of order 6'? | |
| Aug 23, 2012 at 15:35 | history | edited | Tom De Medts | CC BY-SA 3.0 |
added 2302 characters in body
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| Aug 11, 2012 at 22:54 | comment | added | Gjergji Zaimi | I deleted my "answer" and will think about this some more when I get some time. I hope this doesn't turn out to be another false belief :) | |
| Aug 7, 2012 at 9:03 | history | asked | Tom De Medts | CC BY-SA 3.0 |