Left Operation is not Commutative

Theorem

Let $S$ be a finite set.

Let $\leftarrow$ denote the left operation on $S$.

Then $\leftarrow$ is not commutative on $S$ unless $S$ is a singleton.

Proof

Let $S$ be a singleton, $S = \set s$, say.

Then:

$s \leftarrow s = s$

and so $\leftarrow$ is trivially commutative on $S$

Otherwise, $\exists s, t \in S$ such that $s \ne t$.

Then:

$s \leftarrow t = s$

but:

$t \leftarrow s = t$

and the result follows by definition of commutative operation.

$\blacksquare$

Also see

• Left Operation is Anticommutative

• Right Operation is not Commutative

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.4$