Identifying all the quasigroups of order $3$ up to isomorphism

Question

According to this, there are $5$ non-isomorphic quasigroups of order $3$. I have been able to find $4$ of them:

• the cyclic group of order $3$
• a commutative quasigroup with $3$ idempotent elements
• a commutative quasigroup with no idempotents
• a noncommutative quasigroup with $1$ idempotent (i.e. subtraction $\bmod 3$)

Can someone help me find the fifth one?

Answer 1

All $5$ different quasigroups are listed with multiplication tables in figure $1$ on page $4$ in the article Classification results in quasigroup and loop theory by Sorge, Colton, Mccasland and Meier.

Answer 2

If you have access to a recent version of Maple, you can generate these (or for other, small orders) using the built-in Magma package:

> with( Magma ):
> L := Enumerate( 3, quasigroup, output = list );
                [1    2    3]  [1    2    3]  [1    3    2]  [1    3    2]  [2    1    3]
                [           ]  [           ]  [           ]  [           ]  [           ]
          L := [[2    3    1], [3    1    2], [2    1    3], [3    2    1], [1    3    2]]
                [           ]  [           ]  [           ]  [           ]  [           ]
                [3    1    2]  [2    3    1]  [3    2    1]  [2    1    3]  [3    2    1]

> map( IsGroup, L );
                       [true, false, false, false, false]

> map( IsCommutative, L );
                        [true, false, false, true, true]

> map( CountIdempotents, L );
                              [1, 1, 1, 3, 0]

The output produces the quasigroups as matrices representing their multiplication tables, with the underlying set taken to be $\{1, 2,\ldots, n\}$; here, with $n=3$. I've assigned the list of Cayley tables to the name L in order to use it in the subsequent property explorations.

Answer 3

The remaining quasigroup structure is hard to see because it is nearly identical to one you've already identified: For any quasigroup $(S, \,\cdot\,)$, we can canonically define another quasigroup $(S, \odot)$ by $$s \odot t := t \cdot s .$$ The first three quasigroup operations $(S, \,\cdot\,)$ in the question statement are commutative and so in each case $(S, \odot) = (S, \,\cdot\,)$. But for the fourth operation, $(\Bbb Z / 3 \Bbb Z, -)$, our construction gives the quasigroup operation $$a \ominus b := b - a ,$$ and the quasigroups $(\Bbb Z / \Bbb 3 Z, -)$ and $(\Bbb Z / \Bbb 3 Z, \ominus)$ are not isomorphic (the former has a right identity but the latter does not).

Remark If a bijection $\phi: (S, \,\cdot\,) \to (T, \star)$ of quasigroups satisfies $$\phi(a - b) = \phi(b) \ominus \phi(a)$$ for all $s, s' \in S$, we call $\phi$ a (quasigroup) anti-isomorphism. By construction, for any quasigroup $(S, \,\cdot\,)$, the identity map $S \to S$ is an anti-isomorphism $(S, \,\cdot\,) \to (S, \odot)$.

Up there are five quasigroups of order $3$ up to isomorphism, but only four quasigroups of order $3$ up to isomorphism and anti-isomorphism.

See this answer to a closely related question for more.