Left Operation is not Commutative
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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
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.4$