Structure Induced by Permutation on Algebra Loop is not necessarily Algebra Loop
Theorem
Let $\struct {S, \circ}$ be an algebra loop.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
- $\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is not necessarily also an algebra loop.
Proof
Consider the Cayley table of the algebra loop on $S = \set {e, a, b}$:
- $\begin{array}{r|rrr}
\circ & e & a & b \\ \hline e & e & a & b \\ a & a & b & e \\ b & b & e & a \\ \end{array}$
Consider the permutation on $S$:
Let $\sigma$ denote the permutation on $S$ defined as:
\(\ds \map \sigma e\) | \(=\) | \(\ds a\) | ||||||||||||
\(\ds \map \sigma a\) | \(=\) | \(\ds e\) | ||||||||||||
\(\ds \map \sigma b\) | \(=\) | \(\ds b\) |
Then the Cayley table of the structure induced by $\sigma$ on $\circ$ is seen to be:
- $\begin{array}{r|rrr}
\circ & e & a & b \\ \hline e & a & e & b \\ a & e & b & a \\ b & b & a & e \\ \end{array}$
It is apparent by inspection that this is the Cayley table of a quasigroup.
However, there is no identity element.
Hence by definition $\struct {S, \circ_\sigma}$ is not an algebra loop.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(a)}$