Operation on Structure of Cardinality 4+ where Every Permutation is Automorphism is Left or Right Operation

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be a set whose cardinality is at least $4$.

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 either the left operation or the right operation.


Proof

From Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent, we have that $\circ$ is idempotent:

$\forall a \in S: a \circ a = a$


Aiming for a contradiction, suppose $\circ$ is such that:

$x \circ y = z$

for some distinct $x, y, z \in S$.

As $S$ has cardinality of at least $4$, there exists $w \in S$ such that $w \ne x$, $w \ne y$ and $w \ne z$.

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

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

Then:

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

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


Hence we have shown that:

$\forall x, y \in S: x \circ y = x \lor x \circ y = y$

$\Box$


Aiming for a contradiction, suppose $\circ$ is neither the left operation nor the right operation.

From the above, that means there exist $w, x, y, z \in S$ such that:

\(\ds x \circ y\) \(=\) \(\ds x\)
\(\ds w \circ z\) \(=\) \(\ds z\)


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

\(\ds \map f x\) \(=\) \(\ds w\)
\(\ds \map f y\) \(=\) \(\ds z\)
\(\ds \map f z\) \(=\) \(\ds x\)
\(\ds \map f w\) \(=\) \(\ds y\)


Then we have:

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

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

Hence $\circ$ is either the left operation or the right operation.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Operation_on_Structure_of_Cardinality_4%2B_where_Every_Permutation_is_Automorphism_is_Left_or_Right_Operation&oldid=544272"