Latin Square is not necessarily Cayley Table of Group

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 $\struct {S, \circ}$ 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 $\struct {S, \circ}$ is a Latin square.

However, we have that:

$\paren {a \circ b} \circ c = d \circ c = a$

$a \circ \paren {b \circ c} = a \circ d = c$

Thus $\circ$ is not an associative operation.

So, by definition, $\struct {S, \circ}$ is not a group.

$\blacksquare$

Sources

• 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 5$: The Multiplication Table