1
$\begingroup$

From what I can gather, table of each group present latin square. But I wonder, how can I give an example of latin square which isn't a result of the operations on group?

$\endgroup$
3
$\begingroup$

\begin{array} & &2 &1 &3 \\ &1 &3 &2 \\ &3 &2 &1 \\ \end{array}

This cannot represent a group because there is no identity element. $1*1=2$, $2*2=3$, and $3*3=1$.

$\endgroup$
1
$\begingroup$

The answer is yes; there are just a few (up to relabeling) $3\times 3$ latin squares. One is the group multiplication table for the cyclic group of order 3. The other will be the multiplication table of what is called a quasigroup. These are algebraic objects interesting in their own right and even have their own representation theory. In this case, the idea is to try and make a latin square where no element acts as an identity element. You will probably get a right (or left) identity, though.

$\endgroup$

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.