Count of Commutative Quasigroups on Set given Count of Commutative Algebra Loops/Examples/Order 4

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

Let $S$ have exactly $4$ elements.

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

Proof

Let $e \in S$ be a distinguished element of $S$.

Let $N$ be the number of quasigroups $\struct {S, \otimes}$ on $S$ such that $\otimes$ is a commutative operation.

From Count of Commutative Quasigroups on Set given Count of Commutative Algebra Loops we have that:

$N = n! m$

where:

$n$ is the cardinality of $S$

$m$ is the number of commutative operations $\oplus$ on $S$ such that $\struct {S, \oplus}$ is an algebra loop whose identity is $e$.

In this case, we have:

$n = 4$ by hypothesis

$m = 4$ from Algebra Loops of Order 4

Hence:

$N = 4! \times 4 = 96$

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(c)}$