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 $S$.

$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$

Then $\struct {S, \circ_\sigma}$ is not necessarily also an algebra loop.

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:

$\ds \map \sigma e$

$=$

$\ds a$

$\ds \map \sigma a$

$=$

$\ds e$

$\ds \map \sigma b$

$=$

$\ds b$

$\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.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(a)}$