Operation is Right Operation iff Anticommutative with Left Cancellable Element
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Theorem
Let $\struct {S, \circ}$ be a semigroup.
Then:
- $\circ$ is the right operation
- $\circ$ is anticommutative and has a left cancellable element.
Proof
Sufficient Condition
Let $\circ$ be the right operation.
Then from Right Operation is Anticommutative we have that $\circ$ is anticommutative.
Let $x \in S$ be arbitrary.
Let $y, z \in S$ such that:
- $x \circ z = x \circ y$
Then:
\(\ds x \circ z\) | \(=\) | \(\ds z\) | Definition of Right Operation | |||||||||||
\(\ds x \circ y\) | \(=\) | \(\ds y\) | Definition of Right Operation | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds y\) |
That is, $x$ is a left cancellable element for all $x \in S$.
Thus:
- $\circ$ is anticommutative and has a left cancellable element.
$\Box$
Necessary Condition
Let $\circ$ be anticommutative and have a left cancellable element $z$.
As $\struct {S, \circ}$ is a semigroup it follows from Semigroup Axiom $\text S 1$: Associativity that $\circ$ is associative.
Hence from Associative and Anticommutative:
- $\forall x, y, z \in S: z \circ x \circ y = z \circ y$
As $z$ is left cancellable:
- $\forall x, y \in S: x \circ y = y$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.17$