Operation which is Left Distributive over Every Commutative Associative Operation is Right Operation
Theorem
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ have the property that:
- for every arbitrary operation $*$ on $S$ which is both commutative and associative, $\circ$ is left distributive over $*$.
Then $\circ$ is the right operation $\to$:
- $\forall a, b \in S: a \to b = b$
Proof
First recall from Right Operation is Left Distributive over All Operations that the right operation is indeed left distributive over all operations, whether commutative or associative.
Let $*$ be an arbitrary operation on $S$ which is both commutative and associative.
As asserted, let $\circ$ be left distributive over $*$.
Let $c \in S$ be arbitrary.
Consider the constant operation $\sqbrk c$:
- $\forall a, b \in S: a \sqbrk c b := c$
We have from Constant Operation is Commutative and Constant Operation is Associative that $\sqbrk c$ is both commutative and associative.
Hence $\circ$ must be left distributive over $\sqbrk c$.
But then from Condition for Operation to be Left Distributive over Constant Operation:
- $x \circ c = c$
As $c$ is arbitrary:
- $\forall a, b \in S: a \circ b = b$
so $\circ$ has to be the right operation.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.25$