High School What are the properties of a quasigroup table?

[LagrangeEuler]

TL;DR

Quasigroup need invertibility. If in every row and every column any element shows only once is this then quasigroup?

I have a problem regarding what table is quasigroup. If we have multiplication table and that table is of the sudoku type is that quasigroup? See picture quasigroup.png. Is only (3) quasigroup?

 

[fresh_42]

Please tell us what a quasigroup is for you. The definition on Wikipedia doesn't require a neutral element, which makes the existence statement about inverse elements in your question meaningless.

So which definition do you use?

 

[LagrangeEuler]

Wikipedia definition. What different definition you have also?

 

[fresh_42]

A quasigroup (Q, ∗) is a set, Q, with a binary operation, ∗, (that is, a magma), obeying the Latin square property. This states that, for each a and b in Q, there exist unique elements x and y in Q such that both
a ∗ x = b,
y ∗ a = b
hold.

And how do you prove the existence of inverse elements then?

 

[LagrangeEuler]

I do not know? So you think wikipedia definition is completely wrong? It is true. How do you say what is inverse element if you do not have neutral element. So this definition that you right is like latin square in any column every elements you have only once. Right?

 

[fresh_42]

No, I do not think it is wrong. I only expect that those less standard terms might be used slightly differently depending on the author and their purpose, which is why I asked. The answer to your question directly follows from the definition, so it is natural to ask for that definition.

Group properties and the solution of equations like the ones above are closely related. As to your question:

If the magma table has at least one "e" in every row, then all elements have a right inverse.
If the magma table has at least one "e" in every column, then all elements have a left inverse.
If there is only one "e", then the inverses are unique.

Other entries are irrelevant for the question about inverses. Of course in a group table, all elements occur exactly once in every row and every column. Given that, it still has to be proven whether associativity holds.
Commutativity means a symmetric group table.

Solvability of $ax=b$ and $xb=a$ means, that the $a-$row has an entry $b$, and the $b-$column has an entry $a$.