Latin Square is not necessarily Cayley Table of Group

From ProofWiki
Jump to navigation Jump to search

Theorem

While it is true that the Cayley table of a (finite) group is in the form of a Latin square it is not necessarily the case that a Latin square is the Cayley table of a group.


Proof

Proof by Counterexample:

Let $\left({S, \circ}\right)$ be the algebraic structure defined by the following Cayley table:

$\begin{array}{c|ccccc} \circ & e & a & b & c & d \\ \hline e & e & a & b & c & d \\ a & a & e & d & b & c \\ b & b & c & e & d & a \\ c & c & d & a & e & b \\ d & d & b & c & a & e \\ \end{array}$

By inspection it can be seen that the Cayley table for $\left({S, \circ}\right)$ is a Latin square.

However, we have that:

$\left({a \circ b}\right) \circ c = d \circ c = a$
$a \circ \left({b \circ c}\right) = a \circ d = c$

Thus $\circ$ is not an associative operation.

So, by definition, $\left({S, \circ}\right)$ is not a group.

$\blacksquare$


Sources