Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation

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Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

Let $\circ$ be such that every operation on $S$ which is both commutative and associative is distributive over $\circ$.


Then $\circ$ is either the left operation $\gets$ or the right operation $\to$.


Proof

Recall from:

Every Operation is Distributive over Right Operation

and:

Every Operation is Distributive over Left Operation

that if $\circ$ is either $\gets$ or $\to$, then every operation is distributive over it, whether commutative or associative.


Lemma

Let $S$ be a set.

Let $A \subsetneqq S$ be a proper subset of $S$.

Let $a \in A$ and $b \in \relcomp S A$.

Let $\odot$ be the operation on $S$ defined as:

$\forall x, y \in S: x \odot y = \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}$


Then $\odot$ is commutative and associative.

$\Box$


Hence $\odot$ must be distributive over $\circ$.


First note that we have:

\(\ds \forall x, y, z \in S: \, \) \(\ds x \odot \paren {y \circ z}\) \(=\) \(\ds \begin {cases} a & : \set {x, y \circ z} \subseteq A \\ b & : \set {x, y \circ z} \not \subseteq A \end {cases}\) Definition of $\odot$
\(\ds \forall x, y, z \in S: \, \) \(\ds \paren {x \odot y} \circ \paren {x \odot z}\) \(=\) \(\ds \begin {cases} a \circ a & : \set {x, y, z} \subseteq A \\ a \circ b & : \set {x, y} \subseteq A, \set {x, z} \not \subseteq A \\ b \circ a & : \set {x, y} \not \subseteq A, \set {x, z} \subseteq A \\ b \circ b & : \set {x, y} \not \subseteq A, \set {x, z} \not \subseteq A \end {cases}\) Definition of $\odot$
\(\ds \forall x, y, z \in S: \, \) \(\ds \paren {x \circ y} \odot z\) \(=\) \(\ds \begin {cases} a & : \set {x \circ y, z} \subseteq A \\ b & : \set {x \circ y, z} \not \subseteq A \end {cases}\) Definition of $\odot$
\(\ds \forall x, y, z \in S: \, \) \(\ds \paren {x \odot z} \circ \paren {y \odot z}\) \(=\) \(\ds \begin {cases} a \circ a & : \set {x, y, z} \subseteq A \\ a \circ b & : \set {x, z} \subseteq A, \set {y, z} \not \subseteq A \\ b \circ a & : \set {x, z} \not \subseteq A, \set {y, z} \subseteq A \\ b \circ b & : \set {x, z} \not \subseteq A, \set {y, z} \not \subseteq A \end {cases}\) Definition of $\odot$

But we have that $\odot$ is distributive over $\circ$.

That is:

$\forall x, y, z \in S: x \odot \paren {y \circ z} = \paren {x \odot z} \circ \paren {y \odot z}$

and so as $a$ and $b$ are arbitrary:

$\forall a, b \in S: \set {a \circ a, a \circ b, b \circ a, b \circ b} = \set {a, b}$

This of course also applies to:

$y \circ z \in \set {y, z}$

and:

$x \circ y \in \set {x, y}$



As $a$ and $b$ are arbitrary, the result applies for all $a, b \in S$.


Sources