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 ?

  • $\begingroup$ 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. $\endgroup$ – Milo Brandt Jul 6 '15 at 0:29
  • $\begingroup$ Certainly, it's a counterexample. There's also no reason to expect a Latin square to have the associativity property. $\endgroup$ – Robert Israel Jul 6 '15 at 0:31
  • 2
    $\begingroup$ 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! $\endgroup$ – preferred_anon Jul 6 '15 at 0:40

Taken from http://science.kennesaw.edu/~sellerme/sfehtml/classes/math4361/chapter4section1outline.pdf

Let $G=\{1,2,3,4,5\}$ with multiplication table

\begin{array}{|c||c|c|c|c|c|} \hline *&1&2&3&4&5\\ \hline \hline 1&1&2&3&4&5\\ \hline 2&2&1&4&5&3\\ \hline 3&3&4&5&2&1\\ \hline 4&4&5&1&3&2\\ \hline 5&5&3&2&1&4\\ \hline \end{array}

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.


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