First we show that every finite model of Definition 2.56 is trivial. Write \(y^2 := y \diamond y\) and \(y^3 := y^2 \diamond y\). For any \(y,z\), introduce the functions \({f_y: x \mapsto y^3 \diamond x}\) and \({g_{yz}: x \mapsto x \diamond (y^2 \diamond z)}\). Definition 2.56 says that \(g_{yz}(f_y(x))=x\), hence by finiteness \(g_{yz}=f_y^{-1}\), showing that \(g_{yz}\) does not depend on the value of \(z\). Since

it follows that \(f_y(y^2 \diamond z)=f_y(y^3)\) which by injectivity of \(f_y\) implies that \(z\mapsto y^2 \diamond z\) is a constant function (with \(y\) fixed). Substituting \(y^2\) for \(y\) shows that the same is true for \(z\mapsto (y^2 \diamond y^2) \diamond z\), and since

we conclude that \(f_y\) is also a constant function. But this function is already known to be injective, thus there do not exist distinct elements in its domain, showing that the model must be trivial.

To construct an infinite model, consider the magma of positive integers \(\mathbb {Z}^+\) with the operation \(x \diamond y\) defined by

Then \(y \diamond y = 2^y\) and \((y \diamond y) \diamond y = 1\) for all \(y\). If \(x\neq 1\) we have that

and since \((y \diamond y) \diamond z\) is a power of two for all \(y, z\) it follows that

The case \(x=1\) requires a further argument: observe that \(w = (y \diamond y) \diamond z\) evaluates to one unless \(z = 2^y\), in which case it evaluates to \(2^{2^y}\) (which is greater than or equal to four). In particular, \(w\) never takes the value two. Thus

concluding our proof that this magma is a model of Definition 2.56

Using the \(y^2\) and \(y^3\) notation as before, the law reads

In particular, for any \(y\), the map \(T_y \colon x \mapsto y^3 \diamond x\) is injective, hence bijective in a finite model \(G\). In particular we can find a function \(f : G \to G\) such that \(T_y f(y) = y^3\) for all \(y\) Applying Equation 5 with \(x = f(y)\), we conclude

and thus \(y \diamond z\) is independent of \(z\) by injectivity of \(T_y\). Thus, the left-hand side of Equation 5 does not depend on \(x\), and so the model is trivial. This shows there are no non-trivial finite models.

To establish an infinite model, use \(\mathbb {N}\) with \(x \diamond y\) defined by requiring

and

for \(x \neq 3^y\), and

for \(z \neq 3^y 5^x\). Finally set

for \(z \neq 3^y, 2^{3^y}\). All other assignments of \(\diamond \) may be made arbitrarily. It is then a routine matter to establish Equation 5.

From Definition 2.52 we see that the map \(w \mapsto y \diamond w\) is onto, hence injective in a finite model. Using this injectivity four times in Definition 2.52, we see that \(z \diamond y\) does not depend on \(z\), hence the expression \(x \diamond (z \diamond y)\) does not depend on \(x\). By Definition 2.52 again, this means that \(x\) does not depend on \(x\), which is absurd in a non-trivial model.

For a finite magma \(M\), consider the set \(S = \{ x \diamond y | x, y \in M\} \). Now \(f_z : x \mapsto z \diamond x\) and \(g_z : x \mapsto x \diamond z\). They both map \(S\) to \(S\), and due to the hypothesis \(g_z \diamond f_z\) is the identity on \(S\), so because \(S\) is finite \(f_z\) and \(g_z\) must be inverse bijections on it, and therefore they commute.

Consider \(\mathbb {N}\), with \(x \diamond y\) defined as \(x \oplus y\) (bitwise XOR) if \(x\) and \(y\) are even, \(y+2\) if only \(y\) is even, \(x \dot- 2\) if only \(x\) is even, and \(0\) if both are odd. Note that the range of the operation is the set of even naturals. Definition 2.44 is satisfied, because for even \(z\) we get \(z \oplus (x \diamond y) \oplus z = x \diamond y\) and for odd \(z\) we get \((x \diamond y) + 2 \dot- 2 = x \diamond y\). Setting \(x = y = z = 1\), Definition 2.41 isn’t satisfied.

If neither side of the law is a single variable then the zero law \(x \diamond y = 0\) will work, so one can assume the law takes the form \(x = f(x,y)\). Consider a finite field \(F\) with the operation \(x \diamond y := ax+by\) for some coefficients \(a,b \in F\). Then the law becomes a pair of equations \(P(a,b)=0\), \(Q(a,b)=1\) in the coefficients for some polynomials \(P,Q\) with integer coefficients, which one can verify to not divide each other (they have the same degree, and do not have the same set of non-zero monomials). From Bezout’s theorem, this equation has a solution in some field, and hence by the Lefschetz principle it has a solution in a finite field.

Call a point \(x \in X\) periodic if \(f^n(x) = x\) for some \(n{\gt}0\). Because the forward orbit \(x, f(x), f^2(x), \dots \) in a finite space \(X\) must repeat, we see that for each \(x\) there exists an \(n\) such that \(f^n(x)\) is periodic. Let \(n_x\) be the first \(n\) for which this occurs. If \(n_x {\gt} 1\), then \(g(x), f(g(x)), f(f(g(x))\) cannot be periodic (as this would imply \(f(x)\) periodic), and then we can see that \(n_{g(x)} = n_x + 1\). Hence the maximal value of \(n_x\) is at most \(1\), which implies that \(f(x)\) is periodic for every \(x\). Thus there is an \(n {\gt} 1\) such that \(f^n = f\). Since \(f = f^2 \circ g\), we have \(f^n = f^{n-2} \circ f \circ f = f^{n-2} \circ (f^2 \circ g) \circ f = f^n \circ g \circ f = f \circ g \circ f\), as desired.

By hypothesis, \(f^2(x)\) uniquely determines \(f(x)\). This prevents \(n_x = 2\) for any \(x\) (because if \(n{\gt}2\) is such that \(f^{n+2}(x) = f^2(x)\), then \(f^{n+1}(x) \neq f(x)\)), and hence also prevents \(n_x {\gt} 2\) for any \(x\). So again we have \(f(x)\) periodic for all \(x\), so \(f^n = f\) for some \(n{\gt}1\). Then \(f = f^n = f \circ (g \circ f^2) \circ f^{n-2} = f \circ g \circ f^n = f \circ g \circ f\) as required.