13 Equation 1323

In this chapter we study magmas that obey equation 1323,

for all \(x,y\). Using the squaring operator \(Sy := y \diamond y\) and the left and right multiplication operators \(L_y x := y \diamond x\) and \(R_y x = x \diamond y\), this law can be written as

Among other things, this implies that \(L_{Sy}\) is injective and \(L_y\) is surjective, hence \(L_{Sy}\) is invertible. It is convenient to impose an additional axiom

that is to say \(R_{Sy}\) is the inverse of \(L_{Sy}\), then the law 1 simplifies to

So now we would like to construct a magma satisfying 2 and 3. Let \(G\) be a countably infinite abelian group of order \(2\), and work on a carrier \(G \uplus N\), where \(N\) is the set of pairs \((x,a)\) with \(x \in \mathbb {Q}^\times \) and \(a \in G \backslash \{ 0\} \). For each \(a \in G \backslash \{ 0\} \), we let \(\phi _a: G \to \mathbb {Q}^\times \) be a bijection such that \(\phi _a(0) = 1\) and \(\phi _a(a+b) = -1/\phi _a(b)\) for all \(b \in G\), so in particular \(\phi _a(a)=-1\); such a bijection can be easily constructed from the axiom of choice and a greedy algorithm, defining \(\phi _a\) one pair \(\{ b,a+b\} \) at a time. We then partially define an operation \(\diamond \) on \((G \times N) \times (G \times N)\) by setting

for \(a,b \in G\), and

and

for \(x \in \mathbb {Q}^\times \), \(b \in G\), and \(a \in G \backslash \{ 0\} \). This defines \(\diamond \) except for \((x,a) \diamond (y,b)\) with \(x,y \in \mathbb {Q}^\times \) and \(a,b,0\) distinct.

With these rules, \(0\) is a unit, and the squaring operator is given by \(Sa = 0\) and \(S(x,a) = a\), so the set of squares is \(G\). If \(a \in G\) is a square number, we have

and

so 3 is satisfied.

We can also verify some cases of 3. If \(y\) is a square, then the claim already follows from 2 since \(Sy=0\) is a unit. Otherwise, for \(y \in \mathbb {Q}^\times \) and \(b \in G \backslash \{ 0\} \), we have to show that

for \(a \in G\) and

for \(x \in \mathbb {Q}^\times \) and \(a \in G \backslash \{ 0\} \). In the first case, we have from 6 that

and then from 7 we have

as required. In the second case, if \(a=b\), then by the bijective nature of \(\phi _a\), we can write \(x = \phi _a(c) y\) for some \(c \in G\), then from 7 we have

and then by 5

but from construction we have \(\phi _a(a+c)=-1/\phi _a(c) = \phi _a(b)/\phi _a(c)\) and hence \(\phi _a(a+c) y = \phi _a(b) x\) as required.

It remains to handle the case when \(a,b\) are distinct elements of \(G \backslash \{ 0\} \). Here we will set \((x,a) \diamond (y,b)\) to some \((z,c)\) with \(c\) distinct from \(a,b,0\) to be determined. The axiom to verify is then

• Axiom: If \((x,a) \diamond (y,b) = (z,c)\), then \((y,b) \diamond (z,c) = (\phi _a(b) x,a)\).

To do this we perform a greedy algorithm. Suppose that \((x,a) \diamond (y,b)\) is undefined for some \(x,y \in \mathbb {Q}^\times \) and distinct \(a,b \in G \backslash \{ 0\} \). We select a \(c \in G \backslash \{ 0\} \) that has not previously been used by the greedy algorithm, and set

for some arbitrary \(z \in \mathbb {Q}^\times \) (e.g., one could take \(z=1\) if desired). To verify the axiom, we then also need to make an infinite number of other assignments, namely

for all natural numbers \(n\). As the \(a,b,c\) are distinct, the \(\phi _a(b), \phi _b(c), \phi _c(a)\) are not equal to \(\pm 1\), and the members of this infinite sequence do not collide with each other. If we arrange matters so that \(\phi _b(c)\) involves primes in the numerator or denominator that do not appear in any previous \(\phi _a(b), \phi _b(c), \phi _c(a), x, y, z\) used by the greedy algorithm, or the current \(x,y,z\), then we also see that this infinite sequence does not collide with any previously assigned value of \(\diamond \) either. If we then add all these assignments to the multiplication table, we maintain the desired axiom. Iterating this a countable number of times, we obtain a magma that obeys 2 and 3, and hence 1323.

A very similar construction gives