Operation Induced by Permutation on Semigroup is not necessarily Associative
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Theorem
Let $\struct {S, \circ}$ be a semigroup.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\circ_\sigma$ be the operation on $S$ induced by $\sigma$:
- $\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\circ_\sigma$ is not necessarily associative on $S$.
Proof
Let $S = \set {a, b, c}$.
Let $\circ$ denote the right operation on $S$:
- $\forall x, y \in S: x \to y = y$
From Structure under Right Operation is Semigroup, $\struct {S, \circ}$ is a semigroup.
Hence we have:
- $a \circ \paren {b \circ c} = c = \paren {a \circ b} \circ c$
Let $\sigma$ denote the permutation on $S$ defined as:
| \(\ds \map \sigma a\) | \(=\) | \(\ds b\) | ||||||||||||
| \(\ds \map \sigma b\) | \(=\) | \(\ds c\) | ||||||||||||
| \(\ds \map \sigma c\) | \(=\) | \(\ds a\) |
We have:
| \(\ds a \circ_\sigma \paren {b \circ_\sigma c}\) | \(=\) | \(\ds a \circ_\sigma \map \sigma {b \circ c}\) | Definition of Operation Induced by Permutation | |||||||||||
| \(\ds \) | \(=\) | \(\ds a \circ_\sigma \map \sigma c\) | Definition of Right Operation | |||||||||||
| \(\ds \) | \(=\) | \(\ds a \circ_\sigma a\) | Definition of $\sigma$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sigma {a \circ a}\) | Definition of Operation Induced by Permutation | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sigma a\) | Definition of Right Operation | |||||||||||
| \(\ds \) | \(=\) | \(\ds b\) | Definition of $\sigma$ |
Then:
| \(\ds \paren {a \circ_\sigma b} \circ_\sigma c\) | \(=\) | \(\ds \map \sigma {a \circ b} \circ_\sigma c\) | Definition of Operation Induced by Permutation | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sigma b \circ_\sigma c\) | Definition of Right Operation | |||||||||||
| \(\ds \) | \(=\) | \(\ds c \circ_\sigma c\) | Definition of $\sigma$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sigma {c \circ c}\) | Definition of Operation Induced by Permutation | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sigma c\) | Definition of Right Operation | |||||||||||
| \(\ds \) | \(=\) | \(\ds a\) | Definition of $\sigma$ |
So $a \circ_\sigma \paren {b \circ_\sigma c} \ne \paren {a \circ_\sigma b} \circ_\sigma c$ and the result follows.
$\blacksquare$