# Latin square property sufficient?

So I know that for any group table, Every row must contain distinct group elements and the same holds for every column for a group table. And this property is called the Latin Square property. However, Every time I read a book about Abstract Algebra, They say that the latin square property is necessarily but not sufficient for a table to form a group. However, I have not seen any counter example for that.

So I was wondering if there exists a counter example for a table that has the Latin square property but is not a group ? And what property will it violates from the 4 group axioms ?

I mean I know for sure that closure is not violated.

When I was thinking of examples, The easiest I could think of is to construct a table which has no identity element as follows

$$\begin{array}{c|lcr} & e & a & b \\ \hline e & e & b & a \\ a & a & e & b \\ b & b & a & e \end{array}$$

Is that a valid counter example ?

• It's worthy of note that the structure where we only need the latin square property is called a quasigroup. The Wikipedia page doesn't list any examples, but the table therein says there are $4$ quasigroups of order $3$ which are not groups - and yours is such a quasigroup. – Milo Brandt Jul 6 '15 at 0:29
• Certainly, it's a counterexample. There's also no reason to expect a Latin square to have the associativity property. – Robert Israel Jul 6 '15 at 0:31
• Your counterexample is actually the best possible in some sense - you don't have associativity or an identity, so inverses don't make sense either. So you just have closure! – preferred_anon Jul 6 '15 at 0:40

Let $G=\{1,2,3,4,5\}$ with multiplication table
It is easy to see that the bottom right $5\times5$ array is a Latin square. However, we have $$2*(3*4)=2*2=1$$ and $$(2*3)*4=4*4=3$$ So this is an example where the associative property is not met.