In a finite nonabelian group, the probability that two randomly chosen elements commute cannot exceed 5/8. One easy proof also makes it easy to find the smallest groups that attain this bound, namely the two nonabelian groups of order 8:

This makes me wonder: how large can the probability be that three randomly chosen elements $a,b,c$ of a finite nonassociative loop associate, i.e. obey $(ab)c = a(bc)$?

You can prove the 5/8 theorem for groups by separately settling two questions:

  • What is the largest possible fraction of elements of a noncommutative finite group that lie in the center? (Answer: 1/4)

  • Given a noncentral element of a finite group, what's the largest possible fraction of elements that commute with it? (Answer: 1/2)

The nonabelian groups of order 8 achieve both these upper bounds. We could try a similar strategy for my question, attempting to settle these:

  • What is the largest possible fraction of elements $a$ in a finite loop such that $(ab)c = a(bc)$ for all elements $b,c$?

  • If an element $a$ of a finite loop does not have $(ab)c = a(bc)$ for all elements $b,c$, what is the largest possible fraction of elements $b$ such that $(ab)c = a(bc)$ for all $c$?

  • If a pair $a,b$ does not have $(ab)c = a(bc)$ for all elements $c$, what is the largest possible fraction of elements $c$ such that $(ab)c = a(bc)$?

Unfortunately I don't know how to settle these.

Since the quaternion 8-group $$Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$$ attains the 5/8 bound for commutativity of pairs in a nonabelian group, one might hope that the octonion 16-loop $$O_{16} = \{\pm 1, \pm e_1, \dots, \pm e_7\}$$ attains the maximum probability of associativity for triples in a nonassociative loop. Does it?

I'm afraid I haven't even worked out the probability that a triple in $O_{16}$ associates, though it would be easy to do.

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    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.. – Robert Furber Sep 25 at 1:25
  • 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 $\frac{169}{512} + \frac{343}{512} \cdot \left(\frac{19}{49} + \frac{30}{49} \cdot \frac{1}{5}\right) = \frac{43}{64}$. – Robert Furber Sep 25 at 1:27
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    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. – Robert Furber Sep 25 at 1:31
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    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. – John Baez Sep 26 at 1:33
  • 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. – benblumsmith Sep 27 at 19:01

I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19 (Example 2.1):

Suppose $Q(+)$ is an abelian group of even order $n\geq 6$. Let $a,b\in Q-\{0\}$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $x\notin \{b,a+b\}$ or $y\notin \{b,a+b\}$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.

Then $Q(\cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.

Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.

This is a bit too long for a comment, so it's an answer. The Loops package for Gap, by Gabor Nagy and Petr Vojtechovsky contains implementations of all the nonassociative Moufang loops of order $\leq 64$ and order equal to $81$ or $243$. So I wrote a gap script to calculate the association probabilities of triples of elements. I made it as simple-minded as possible to reduce the possibility of bugs, and because I'd never written anything in Gap before.

None of the association probabilities exceeded $\frac{43}{64}$, the association probability for the octonion loop, so the conjecture is correct for these particular Moufang loops (the script took about half an hour on my laptop).

Since the package also has some Bol loops, I checked them, and the left Bol loop of order 8 with the following Cayley table has association probability $\frac{13}{16} = \frac{52}{64} > \frac{43}{64}$:

  1 2 3 4 5 6 7 8
  ---------------
1|1 2 3 4 5 6 7 8
2|2 1 4 3 7 8 5 6
3|3 4 1 2 6 5 8 7
4|4 3 2 1 8 7 6 5
5|5 6 7 8 1 2 3 4
6|6 8 5 7 3 1 4 2
7|7 5 8 6 2 4 1 3
8|8 7 6 5 4 3 2 1

and many of the left Bol loops of order 16 also have association probabilities exceeding $\frac{43}{64}$. Therefore, if the conjecture is correct for Moufang loops, the proof must use an argument that fails for left Bol loops.

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