Cayley Table for Commutative Operation is Symmetrical about Main Diagonal

Definition
Let $\struct {S, \circ}$ be an algebraic structure.

Then:
 * the Cayley table for $\struct {S, \circ}$ is symmetrical about the main diagonal


 * $\circ$ is a commutative operation.
 * $\circ$ is a commutative operation.

Proof
Let $\mathcal C$ denote the Cayley table for $\struct {S, \circ}$.

Let $\sqbrk c_{a b}$ denote the entry of $\mathcal C$ corresponding to the element $a \circ b$ of $S$.

That is, $\sqbrk c_{a b}$ is the entry in the row headed by $a$ and the column headed by $b$.

Necessary Condition
Let $\circ$ be a commutative operation.

Then by definition:
 * $\forall a, b \in S: a \circ b = b \circ a$

Thus:
 * $\sqbrk c_{a b} = \sqbrk c_{b a}$

and so:
 * the entry in the row headed by $a$ and the column headed by $b$

is the same as:
 * the entry in the row headed by $b$ and the column headed by $a$.

This applies to all $a, b \in \S$.

Hence for all $a \in S$, the row headed by $a$ is the same as the column headed by $a$.

It follows that $\mathcal C$ is symmetrical about the main diagonal.

Sufficient Condition
Let $\mathcal C$ is symmetrical about the main diagonal.

Then:
 * $\forall a, b \in S: \sqbrk c_{a b} = \sqbrk c_{b a}$

That is, by the definition of the Cayley table:
 * $\forall a, b \in S: a \circ b = b \circ a$

Thus, by definition, $\circ$ is a commutative operation.