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To clarify, as requested by Community
I am looking for a proof that the total number of idempotent elements over all quasigroups of order n equals the number of quasigroups of that order.
Could be provided by:

  1. an alternative to the sketch below,or
  2. proving the assumption in the sketch or
  3. pointing relevant papers etc dealing with the matter.

When scanning all quasigroups of order 5 and counting the occurrences of quasigroups with 0, 1, 2, 3, 4 or 5 idempotent elements I obtained the following result:

The number of quasigroups with 0, 1, 2, 3, 4 or 5 idempotent elements were respectively 50112, 70640, 31680, 8160, 640 and 48. The number of idempotent elements involved is then respectively 0, 70640, 63360, 24480, 2560 and 240 summing to a total of 161280 which is also the total number of quasigroups and is consistent with the published number of order 5 quasigroups and Latin squares.

Being quite surprised by this equality I checked for order 3 and order 4 and found equality of the corresponding totals. I also extracted (using GAP) large random samples of quasigroups of higher orders and obtained results which seemed consistent with this result.

Is this a known result? That is: the total number of idempotent elements over all quasigroups of order n equals the total number of quasigroups of that order

I would be surprised if such a simple result is not well know but have been unable to find any reference to it

If it is a known result is it proven or a conjecture?

Following is a sketch of an incomplete proof:
Consider the cell (i,j) in the set of N multiplication tables containing all the quasigroups of order n.
There are n symbols related to a quasigroup of order n.
Let us assume the number of times any one symbol occurs in position (i,j) is equal to the number of times any other symbol occurs. (this seems a reasonable assumption but needs to be proved for a complete proof)
Thus over the set of quasigroups a given symbol will occur N/n times in position (i,j) in the set of all multiplication tables of order n.
For a position (n,n) on the diagonal the symbol n will therefore occur N/n times which gives the number of idempotent elements which occur in position. Given there are n diagonal positions there are therefore N idempotent elements over the complete set of quasigroups.

Can someone provide a proof the assumption above which seems intuitively obvious i.e. that, over the set of all multiplication tables for quasigroups of a specified order, the number of times any one symbol occurs in position (i,j) is equal to the number of times any other symbol occurs in that position.

Incidentally, when I extracted a quasigroup representative for each isomorphism class for order 5 quasigroups and counted idempotent elements in the extracted set the above equality did not hold.

John Palmer
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 04 '21 at 11:10
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    An alternative way of stating the result is "The average, over all quasigroups of order $n$, of the number of idempotents of a quasigroup is exactly $1$ for all $n>0$". – Somos Sep 05 '21 at 02:35
  • Clarification, as requested has been added at the beginning of the entry. – John Palmer Sep 05 '21 at 06:08
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    Thank you Somos for your alternative statement . It is an interesting way of restating the problem. Does this way of looking at the problem lead to a different approach to proof? – John Palmer Sep 05 '21 at 06:15
  • I think that the original statement is a bit confusing with one number equaling another number and it is good that you gave a detailed account when $n=5$. My alternative statement in terms of an average makes it obvious that there must exist quasigroups with no idempotents when $n>1$. I have no idea of an average based proof but one might exist. – Somos Sep 07 '21 at 02:25

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