Idempotent Non-Trivial Quasigroup is Not a Loop

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Let $\left({S, \circ}\right)$ be an idempotent quasigroup whose underlying set $S$ comprises more than one element.

Then $\left({S, \circ}\right)$ is not an algebra loop, that is, it has no identity element.


Aiming for a contradiction, suppose $\left({S, \circ}\right)$ has an identity element $e$.

Then by Identity Element is Idempotent:

$e \circ e = e$

Consider $e' \in S$ where $e' \ne e$.

Since $e$ is an identity element:

$e' \circ e = e'$

Also, by assumption, $\circ$ is idempotent, so:

$e' \circ e' = e'$

Then by the definition of a quasigroup, the left regular representation:

$\lambda_{e'}: S \to S, x \mapsto e' \circ x$

is a permutation, and in particular, an injection.

Hence $e' = e$, which contradicts our assumption that $e' \ne e$.

So $\left({S, \circ}\right)$ has no identity element and is not an algebra loop.