Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent

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Theorem

Let $S$ be a set whose cardinality is at least $3$.

Let $\struct {S, \circ}$ be an algebraic structure on $S$ such that every permutation on $S$ is an automorphism on $\struct {S, \circ}$.


Then $\circ$ is an idempotent operation.


Proof

Aiming for a contradiction, suppose $\circ$ is not idempotent.

Then there exists $x \in S$ such that:

$\exists y \in S: x \circ x = y$

where $x \ne y$.


Because there are at least $3$ distinct elements of $S$ by hypothesis:

$\exists z \in S: z \ne x, z \ne y$


Let $f: S \to S$ be a permutation on $S$ such that:

$\map f x = x$
$\map f y = z$


We have:

\(\ds z\) \(=\) \(\ds \map f y\) Definition of $f$
\(\ds \) \(=\) \(\ds \map f {x \circ x}\) by hypothesis: $x \circ x = y$
\(\ds \) \(=\) \(\ds \map f x \circ \map f x\) by hypothesis: $f$ is an automorphism
\(\ds \) \(=\) \(\ds x \circ x\) Definition of $f$
\(\ds \) \(=\) \(\ds y\) by hypothesis: $x \circ x = y$

This contradicts our assertion that $x$ and $z$ are distinct.

From Proof by Contradiction it follows that our assumption that $\circ$ is not idempotent must have been false.

Hence $\circ$ is idempotent.

$\blacksquare$


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