Multiplication table of a commutative quasigroup is a symmetric latin square. Is the converse also true?

Question

Or can we find an example where the Latin square is symmetric but its corresponding quasi group is not commutative?

Answer 1

A binary structure $(A,\cdot)$ is a commutative quasigroup if and only if its Cayley table is a symmetric latin square.

The fact that each element of $A$ occurs exactly once in each row and column of the multiplication (operation) table corresponds to division and cancellation.

The symmetry of the table corresponds to the commutative operation: let $x,y\in A$, the element of the table with indexes $(x,y)$ equals $(y,x)$ if and only if $y\cdot x=x\cdot y$.