Cayley Table for Commutative Operation is Symmetrical about Main Diagonal

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Definition

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


Then:

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

if and only if:

$\circ$ 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.

$\blacksquare$


Sources