Count of Commutative Quasigroups on Set given Count of Commutative Algebra Loops

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Theorem

Let $S$ be a finite set of cardinality $n$.

Let $e \in S$.

Let there be $m$ commutative operations $\oplus$ on $S$ such that $\struct {S, \oplus}$ is an algebra loop whose identity is $e$.


Then there are $n! m$ commutative binary operations $\otimes$ on $S$ such that $\struct {S, \otimes}$ is a quasigroup.


Proof

Consider a commutative algebra loop $\struct {S, \oplus}$ whose identity is $e$.

Consider the row of the Cayley table of $S$ which is headed by $e$.

There are $n!$ permutations of the elements of this row.

Each of these corresponds to a different commutative operation on $S$, as the corresponding column headed by $e$ is constrained to be the same as the row.

The result follows by the Product Rule for Counting.

$\blacksquare$


Examples

Order 3

Let $S$ have exactly $3$ elements.

There are $6$ quasigroups $\struct {S, \otimes}$ on $S$ such that $\otimes$ is a commutative operation.


Order 4

Let $S$ have exactly $4$ elements.

There are $96$ quasigroups $\struct {S, \otimes}$ on $S$ such that $\otimes$ is a commutative operation.


Sources

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