Timeline for Why can't a nonabelian group be 75% abelian?
Current License: CC BY-SA 3.0
11 events
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| Sep 10, 2018 at 15:06 | comment | added | LSpice | @ArturoMagidin's reference: Givens - The probability that two semigroup elements commute can be almost anything. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jul 20, 2015 at 5:04 | comment | added | Arturo Magidin | On the other hand, note that the probability that two semigroup elements commute can be any rational number, as shown by Givens and by Ponomarenko and Selinski (Givens, B. The probability that two semigroup elements commute can be almost anything, College Math J. 39 (5), 399-400, 2008; and also the paper by Michelle Soule). | |
| Jul 9, 2015 at 23:23 | comment | added | Geoff Robinson | @j.p.: Notice that the two facts are interrelated. If we had $[G:Z(G)] < 4$, then every element $x \in G \backslash Z(G)$ would commute with at least $2|Z(G)|$ elements, so with more than $\frac{|G|}{2}$ elements, a contradiction. | |
| Jul 9, 2015 at 22:25 | vote | accept | Joseph O'Rourke | ||
| Jul 9, 2015 at 19:10 | answer | added | Geoff Robinson | timeline score: 66 | |
| Jul 9, 2015 at 17:18 | comment | added | Johannes Hahn | You should post this as a proper answer. | |
| Jul 9, 2015 at 16:53 | comment | added | Joseph O'Rourke | Thank you, GeoffRobinson and @j.p.! Together your comments provide a lucid logic, just what I sought. | |
| Jul 9, 2015 at 16:12 | comment | added | j.p. | If you combine Geoff's comment with the fact that the index of the center cannot be smaller than $4$ ($G/Z(G)$ cannot be cyclic), then you get exactly the bound $\frac{5}{8}$: The central elements commute with all other elements, the non-central elements with at most half the elements. At most a quarter of the elements are central, so the probability to commute is bounded by $\frac{1}{4}\cdot 1 + \frac{3}{4}\cdot\frac{1}{2} = \frac{5}{8}$. So the intuition is a single element cannot commute with too many other elements without being central and the center itself cannot be too big. | |
| Jul 9, 2015 at 15:42 | comment | added | Geoff Robinson | There are lots of papers in the literature on this topic, including one by Bob Guralnick and myself. I suppose that one answer to your general question (at least for finite groups), an element $x$ which commutes with more than half the elements of the group $G$ has to be central, using Lagrange's Theorem. So once the probability that a single element $x$ commutes with other group elements gets above 1/2, it has to be 1. | |
| Jul 9, 2015 at 15:36 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |