Does anyone know of a non-trivial algebraic structure satisfying these four identities?

Question

Does anyone know of a non-trivial (i.e. cardinality $\geq 2)$ algebraic structure $(X,+,-)$ satisfying the following identities?

1. $(x+a)-a=x$

2. $(x-a)+a=x$

3. $(x+y)+a = (x+a)+(y+a)$

4. $(x-y)+a = (x+a)-(y+a)$

Remark. The Abelian group of order $2$ doesn't satisfy the last two conditions.

Motivation. I think its cool that if $X$ is such an algebraic structure, then for every $a \in X$, the functions $$x \mapsto x+a, \qquad x \mapsto x-a$$

are automorphism of $X$. This mean that if $a \in X$ and $f \in \mathrm{Aut}(X)$, then $f+a \in \mathrm{Aut}(X)$ and $f-a \in \mathrm{Aut}(X).$

Answer 1

Instead of $+$ and $-$, let us use $\tilde{+}$ and $\tilde{-}$ as the algebraic operations we wish to invent.

Let $\mathcal{D} = \{\; a + b \epsilon : a, b \in \mathbb{R} \;\}$ be the the set of dual numbers over $\mathbb{R}$. i.e. the algebra extending $\mathbb{R}$ by adjoining one new element $\epsilon$ with the property $\epsilon^2 = 0$. Now define binary operations $\tilde{+}$ and $\tilde{-}$ on $\mathcal{D}$ by

$$\begin{align} ( a + b\epsilon)\;\tilde{+}\;(c + d\epsilon ) &\;\stackrel{def}{=} (a + b\epsilon) + (c + d\epsilon)\epsilon = a + (b + c)\epsilon\\ ( a + b\epsilon)\;\tilde{-}\; (c + d\epsilon ) &\;\stackrel{def}{=} (a + b\epsilon) - (c + d\epsilon)\epsilon = a + (b - c)\epsilon \end{align}$$

It is easy to check

1. $(x \;\tilde{+}\; a) \;\tilde{-}\; a = (x + a\epsilon) - a\epsilon = x$.
2. $(x \;\tilde{-}\; a) \;\tilde{+}\; a = (x - a\epsilon) + a\epsilon = x$.
3. $(x \;\tilde{+}\; y) \;\tilde{+}\; a = ( x + y\epsilon) + a\epsilon = ( x + a\epsilon ) + (y + a\epsilon)\epsilon = (x \;\tilde{+}\; a) \;\tilde{+}\; (y \;\tilde{+}\; a)$
4. $(x \;\tilde{-}\; y) \;\tilde{+}\; a = (x - y\epsilon) + a \epsilon = ( x + a\epsilon ) - ( y + a\epsilon)\epsilon = ( x \;\tilde{+}\; a ) \;\tilde{-}\; ( y \;\tilde{+} a )$

As a result, $( \mathcal{D}, \tilde{+}, \tilde{-} )$ is a non-trivial example for the algebra structure you are seeking.

Answer 2

Below I will demonstrate that, if $+$ is associative, then $x+a = x$ and $x-a = x$ for all $x$ and $a$. This is enough, I think, to qualify as a "trivial" algebraic structure, even though the underlying set can be as large as you like.

Beginning from (3):

\begin{align*} (x+y)+a &= (x+a) + (y+a)\\ (x+y)+a &= ((x+a)+y) + a & \text{by associativity}\\ x+y &= (x+a)+y & \text{by (1)}\\ x &= x+a & \text{by (1)} \end{align*}

Further, using (2) we find that $x = (x+a) - a = x - a$.

Answer 3

Well I get one stupid example where $X$ is any set and $+$ and $-$ act trivially. What I mean by trivially is:

$x+a=x-a=x$ for all $x, a \in X$. Which is clearly not interesting.

Answer 4

A structure with these properties is called an involuntary quandle, or kei. They arise as important algebraic invariants of knots. See here.