Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation
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:
and:
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:
- $x \circ y \in \set {x, y}$
This needs considerable tedious hard slog to complete it. In particular: We now examine cases, which promises to be tedious. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
As $a$ and $b$ are arbitrary, the result applies for all $a, b \in S$.
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.26 \ \text {(c)}$