Associative and Anticommutative
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Theorem
Let $\circ$ be a binary operation on a set $S$.
Let $\circ$ be both associative and anticommutative.
Then:
- $\forall x, y, z \in S: x \circ y \circ z = x \circ z$
Proof
Let $\circ$ be both associative and anticommutative.
Then from Associative Idempotent Anticommutative:
- $\forall x, z \in S: x \circ z \circ x = x$
and $\circ$ is idempotent.
Consider $x \circ y \circ z \circ x \circ z$.
We have:
\(\ds x \circ y \circ z \circ x \circ z\) | \(=\) | \(\ds x \circ \paren {y \circ z} \circ x \circ z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \circ z\) |
Also:
\(\ds x \circ y \circ z \circ x \circ z\) | \(=\) | \(\ds x \circ y \circ \paren {z \circ x \circ z}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \circ y \circ z\) |
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.17 \ \text{(c)}$