Models of a certain (weird) equational theory

There is such a theory that was actually going to be the topic of my dissertation. Let $R$ be a root system, let $x,y$ be roots, and let $s_x$ be the reflection in the hyperplane normal to $x$. Then we may define $$x*y = s_x(y)$$ Then there does not exist any $x$ for which $x*x=x$. In fact, for all $x$ we have $$x*x = -x$$ We therefore have $$(x*x)*(x*x) = - (-x) = x$$ Note we have $$x*(-y)=-(x*y)$$ and $$(x*x)*y = x*y$$

We also have $$(x*y)*(x*y)=-(x*y)$$ and $$(x*x)*(y*y) = x*(-y) = -(x*y)$$ so $$(x*x)*(y*y)=(x*y)*(x*y)$$

The two axioms necessary to prove these things are

1. $x*(x*y)=y$

2. $x*(y*z)=(x*y)*(x*z)$

There is a third axiom that ensures that the resulting algebra is a root system, but it is a bit more of a pain to state.

This answer expands on a comment of Captain Lama; if they post an answer of their own I'll delete this one, and I've made it community-wiki so I don't get reputation for their work.

Note that the second condition is an immediate consequence of associativity and commutativity. So any commutative semigroup satisfying $x^4=x$ will satisfy your theory - for example, the group $\mathbb{Z}/3\mathbb{Z}$.

(Of course, there are structures satisfying your theory which are not commutative semigroups, but commutative semigroups are relatively simple things to think about.)

Let me describe how to produce typical models of this equational theory using a different but equivalent language.

First, number the two given axioms:

Axiom $(1)$. $(x*x)*(x*x)=x$
Axiom $(2)$. $(x*y)*(x*y)=(x*x)*(y*y)$

Let $\sigma(x)=x*x$ be the squaring map with respect to $*$, and let $x\odot y=\sigma(x*y)$. Axiom (1) asserts exactly that $\sigma$ is a permutation of exponent $2$, while Axiom (2) asserts exactly that $\sigma$ commutes with $*$. Since $\sigma$ also commutes with itself, it will then commute with $\odot$, which is a composition of $\sigma$ and $*$. Since $x\odot y$ is defined to be $\sigma(x*y)$, and $\sigma$ has exponent $2$, we can recover $*$ from $\sigma$ and $\odot$ by $x*y=\sigma(\sigma(x*y))=\sigma(x\odot y)$.

Altogether, this shows that we can convert between the $*$-language and the $\odot,\sigma$-language using these definitions:

$\sigma(x):=x*x$.

$x\odot y:=\sigma(x*y)$.

$x*y:=\sigma(x\odot y)$.

Now, in order to translate theories, we observe that an algebra $\langle A; *\rangle$ in the language $\{*\}$ satisfies Axioms (1) and (2) iff the corresponding algebra $\langle A; \odot, \sigma\rangle$ in the language $\{\sigma,\odot\}$ satisfies

Axiom $(1)'$. the binary operation of $\langle A; \odot\rangle$ is idempotent, and
Axiom $(2)'$. $\sigma$ is an exponent-2 automorphism of $\langle A; \odot\rangle$.

That is, up to a change of language, a model of the original axioms is simply an idempotent binary algebra equipped with an exponent-$2$ automorphism.

Examples.

It is not hard to characterize the examples where $\sigma$ is trivial (i.e., $\sigma$ is the identity function). Any such algebra is obtained from an idempotent binary algebra $\langle A; \odot\rangle$ by setting $x*y=x\odot y$.

It is not hard to characterize the examples where $\odot$ is trivial (i.e., $\odot$ is one of the projections $x\odot y = x$ for all $x, y$ or $x\odot y = y$ for all $x, y$). In this case, for any set $A$ let $\sigma: A\to A$ be any permutation of exponent $2$ ($\sigma^2(x)=x$). Then $x*y:=\sigma(x)$ or $x*y:=\sigma(y)$ are both operations on $A$ satisfying Axioms (1) and (2). The example in the question statement is of this type.

Let $\mathbb A = \langle A; \odot\rangle$ be any idempotent binary algebra. Let $\mathbb B = \mathbb A\times \mathbb A$. Let $\sigma: \mathbb B\to \mathbb B: (b,c)\mapsto (c,b)$ be the automorphism of switching coordinates. A change of language converts $\langle B; \odot, \sigma\rangle$ into a model of Axioms (1) and (2).

Let $M$ be an $R$-module. Suppose that $r,s\in R$ commute with each other and $s^2=1$. Then $x\odot y:=rx+(1-r)y$ is idempotent and $\sigma(x)=sx$ is an exponent 2 automorphism of $\langle M; \odot\rangle$, so if we equip $M$ with only the operation $x*y=\sigma(x\odot y) = srx+s(1-r)y$, then $\langle M; *\rangle$ will satisfy Axioms (1) and (2).