What's the maximum probability of associativity for triples in a nonassociative loop?

Question

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:

• The 5/8 theorem.

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.

Answer 1

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.

Answer 2

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.