Timeline for What's the maximum probability of associativity for triples in a nonassociative loop?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2018 at 7:35 | answer | added | Robert Furber | timeline score: 5 | |
| Oct 2, 2018 at 4:16 | history | edited | John Baez | CC BY-SA 4.0 |
fixed typo
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| Oct 2, 2018 at 4:15 | comment | added | John Baez | I'll fix the typo. | |
| Oct 1, 2018 at 2:37 | comment | added | Robert Furber | @benblumsmith You are right. The probabilities $\frac{1}{2}$ and $\frac{1}{4}$ should be the other way round, as they are in John's linked blog post. | |
| Sep 27, 2018 at 19:01 | comment | added | benblumsmith | Confused by the claim that 1/2 the elements in a group can lie in its center. The max is 1/4, maybe this is what you meant? (This is what is attained by $Q_8$ and $D_4$.) In general, the center cannot have prime index since the quotient by the center cannot be cyclic (or $G$ would be generated over the center by a single element and then it would be abelian, contradiction), and the smallest composite number is 4. | |
| Sep 26, 2018 at 1:33 | comment | added | John Baez | After Gjegji gave his nice answer I considered adding the Moufang condition; it's possible general loops are too "floppy" to give an interesting answer. Mounfang loops of small order have been listed, so someone could compute their probabilities of associativity. | |
| Sep 25, 2018 at 1:31 | comment | added | Robert Furber | The other thing to mention is that the octonion 16-loop is not just any kind of loop, but a Moufang loop (and the Ježek-Kepka loops are not Moufang). Moufang loops have a Lagrange theorem, so the argument about the maximal size of the nucleus (the associativity analogue of the centre) might work. I have no idea about the second part of the argument. | |
| Sep 25, 2018 at 1:27 | comment | added | Robert Furber | lies on the line and four that don't, so the probability of associativity given distinct elements different from 1 is $\frac{1}{5}$. All together, the answer is $\frac{169}{512} + \frac{343}{512} \cdot \left(\frac{19}{49} + \frac{30}{49} \cdot \frac{1}{5}\right) = \frac{43}{64}$. | |
| Sep 25, 2018 at 1:25 | comment | added | Robert Furber | For what it's worth, the probability is $\frac{43}{64}$. Proof: We can restrict to dealing with positive signs in a triple (a,b,c) because the signs don't affect whether it associates or not. The probability of having at least one 1 is $1 - \left(\frac{7}{8}\right)^3 = \frac{169}{512}$. The probability, given that there's no 1, that at least two elements of the triple are the same is $1 - 1 \cdot \frac{6}{7} \cdot \frac{5}{7} = \frac{19}{49}$. If all three elements are different, then they associate iff they lie on a line in the Fano plane. Given the first two points, there is one point that.. | |
| Sep 22, 2018 at 23:59 | answer | added | Gjergji Zaimi | timeline score: 28 | |
| Sep 22, 2018 at 23:32 | history | asked | John Baez | CC BY-SA 4.0 |