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 $x\ne 1$ 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 $T_y:x\mapsto y^3\diamond x$ 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 $x\ne 3^y$, and

for $z\ne 3^y{5}^x$. Finally set

for $z\ne 3^y,2^{3^y}$. 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 $n>0$. 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 $n_x>1$, 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 $n>1$ 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 $n>2$ is such that $f^{n+2}(x)=f^2(x)$, then $f^{n+1}(x)\ne f(x)$), and hence also prevents $n_x>2$ for any $x$. So again we have $f(x)$ periodic for all $x$, so $f^n=f$ for some $n>1$. 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.

By the pigeonhole principle, there exists $n\ge 1$, $m\ge 0$ such that $f^{m+n}=f^m$, which implies on iteration that $f^{m+mn}=f^m$ and hence $f^{2mn}=f^{mn}$, giving the claim.

Write $Sx:=x\diamond$, $fx:=x\diamond Sx$ and $Cx=Sx\diamond x$, then we have $x\diamond y=y\diamond fx$, and our task is to show $x\diamond y=Cx\diamond y$ and $x\diamond y=x\diamond Cy$ for finite magmas, giving the four open implications.

Note that $x\diamond y=y\diamond fx=fx\diamond fy$, hence $f$ is a homomorphism. By Lemma 3.9, there is $n\ge 1$ with $f^n=f^{2n}$. From 3342 we have $Sx=Sfx$ and $Cx=Sx\diamond x=f^{n}Sx\diamond f^{n}x=f^{n}Sx\diamond f^{2n}x=f^{2n-1}x\diamond f^{n}Sx=f^{n-1}x\diamond Sx=f(f^{n-1}x)=f^{n}x$. Then

and a similar argument gives $x\diamond Cy=x\diamond y$.

We write 1167 as

hence $L_y$ is invertible and $L_{z\diamond Sy}{L}_y=I$. In particular $L_{z\diamond Sy}Sy=y$, hence squaring is injective, hence surjective. So 1167 can be rewritten as $L_{S^{-1}y}{L}_{z\diamond y}=I$, hence $L_{z\diamond y}$ is independent of $z$. In particular

by 1167, hence the right-hand side of 1096 is independent of $z$. Replacing $z$ with $y$, the claim now follows from 1167.

1133 asserts that $L_y{L}_{y\diamond (z\diamond y)}=I$, hence $L_y$ is invertible and $L_{y\diamond (z\diamond y)}=L_y^{-1}$ does not depend on $z$. Setting $z=y\diamond Sy$ we see from 1133 that $y\diamond (z\diamond y)=y$ and hence we have left-involution: $L_y=L_y^{-1}$. Replacing $y$ with $L_{z}y$ in 1133 we now get $L_{L_{z}y}{L}_{L_{z}y\diamond y}=I$, which since $L_{L_{z}y}$ is its own inverse gives

In particular

hence as $L_{R_y^{2}z}$ is its own inverse

Thus squaring is surjective, hence invertible.

Next, by 6 we have $L_{z\diamond Sy}=L_w$ where $w=R_{Sy}^{2}z$. By 6 and the fact that $L_w$ is its own inverse, we have

hence $Sw=L_{w}w=Sy$. As $S$ is injective, we have $L_w=L_y$, and then $L_y{L}_{z\diamond Sy}=L_y^2=I$, giving 1167.

Write $\tilde{C}x=x\diamond Sx$, then 1441 asserts that $R_{\tilde{C}x}{R}_{y}x=x$. Setting

we get $x=S\tilde{C}x$, so by finiteness $\tilde{C}Sx=x$. Replacing $x$ with $Sx$ in 1441 we conclude $R_x{R}_{y}Sx=Sx$ which is 4067.

Clearly 1443 implies 1441, hence 4067 by the previous implication, so that 3055 simplifies to $x=Sx\diamond x$. Meanwhile, 1443 also implies $x=S(x\diamond (x\diamond z))$. On taking square roots and using $S\tilde{C}=x$, we obtain $x\diamond (x\diamond z)=\tilde{C}x$. Applying this with $z=Sx$ we conclude $L_x\tilde{C}x=\tilde{C}x$, hence on replacing $x$ with $Sx$ we get $L_{Sx}x=x$ as required.

This is very similar to the previous proof. Write $Cx=Sx\diamond x$, then 1681 asserts $R_{Cx}{L}_{y}x=x$. Again setting $y=Sx$ yields $x=SCx$ hence $x=CSx$, and on replacing $x$ with $Sx$ in 1681 one obtains 3877.

For the second part, we simplify to $x=x\diamond Sx$, while 1701 gives $x=S((z\diamond x)\diamond x)$ hence $(z\diamond x)\diamond x=Cx$, so $R_{x}Cx=Cx$, so $R_{Sx}x=x$ as required. □