A semigroup is nilpotent of degree 3 if every product of 3 elements gives the same result. In 2012, Andreas Distler and James D. Mitchell wrote that:

It is part of the folklore of semigroup theory that almost all finite semigroups are nilpotent of degree 3.

Here's a way to make this precise:

Conjecture. If $S(n)$ is the number of isomorphism classes of semigroups with $n$ elements, and $S_3(n)$ is the number of isomorphism classes of semigroups with $n$ elements that are nilpotent of degree 3, then

$$ \lim_{n \to \infty} \frac{S_3(n)}{S(n)} = 1. $$

Has anyone proved this, or a similar result?

Semigroup theorists seem to enjoy counting semigroups up to equivalence, meaning up to isomorphism or anti-isomorphism, so maybe someone has proved a similar result but counting semigroups up to equivalence rather than up to isomomorphism.

In 2012, Andreas Distler, Chris Jefferson, Tom Kelsey, and Lars Kottho counted the semigroups with 10 elements and found

$$ 12,418,001,077,381,302,684 $$

of them. Of these, all but

$$ 718,981,858,383,872 $$

were nilpotent of degree 3. So, about 99.994% were nilpotent of degree 3.

By the way, this calculation still took a lot of work.

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    $\begingroup$ I believe if you count labeled multiplication tables rather than isomorphism then the result is known. See ams.org/journals/proc/1976-055-01/S0002-9939-1976-0414380-0/… for labeled semigroups. $\endgroup$ – Benjamin Steinberg Apr 14 at 20:33
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    $\begingroup$ Since 3-nilpotent is anti-isomorphism-invariant and passing from isomorphism classes to equivalence classes is $\le 2$-to-1, counting up to isomorphism or equivalence doesn't matter. $\endgroup$ – YCor Apr 14 at 20:37
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    $\begingroup$ Sorry. I deleted too many old comments. I'll rewrite. Nobody has ever proved this folklore conjecture. It is like the conjecture that almost all finite groups of order less than n are 2-groups. Everybody believes it but no ideas how to prove it. $\endgroup$ – Benjamin Steinberg Apr 14 at 20:40
  • $\begingroup$ For isomorphism classes, it is not even known how many there are semigroups of order $n$, $n>20$. So 99% for all $n$ seems unrealistic to prove. Something like $>50\%$ seems possible but it is not obvious. But, yes, I do not know any semigroup theorist who does not believe 99.9% (in fact $\to 1$ as $n\to \infty$ and the convergence is more than exponentially fast). $\endgroup$ – user6976 Apr 14 at 20:41
  • $\begingroup$ Of course, there are also finite semilattices of height 3. Their number is large. I wonder if anybody has computed the number. They are somewhat similar to 3-nilp semigroups. $\endgroup$ – user6976 Apr 14 at 21:51

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