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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.11 \ \text {(c)}$