2
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.