# 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.

## 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

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): Exercise $1.4: \ 6$ - 1964: Walter Ledermann:
*Introduction to the Theory of Finite Groups*(5th ed.) ... (previous) ... (next): $\S 5$: The Multiplication Table - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 2$ - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.4$ - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{II}$: Exercise $\text{A}$