16 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
and
respectively. In particular, this gives a way to construct these magmas:
Suppose that \(M\) is a magma such that
and
hold. Then the magma obeys 1323.
Trivial.
So now we would like to construct magmas satisfying Equation 2 and Equation 3. We need some bijections:
Let \(G\) be a countably infinite abelian group of order \(2\). Then there exists a bijection \(\phi _a: G \to \mathbb {Q}^\times \) for each \(a \in G \backslash \{ 0\} \) such that \(\phi _a(0) = 1\) and \(\phi _a(a+b) = -\phi _a(b)\) 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.
Let \(G\) be a countably infinite abelian group of order \(2\), and let \(\phi _a\) be as in the previous lemma. Let \(N\) be the set of pairs \((x,a)\) with \(x \in \mathbb {Q}^\times \) and \(a \in G \backslash \{ 0\} \), and let \(M = G \uplus N\) be the disjoint union of \(G\) and \(N\). Suppose that we have an operation \(\diamond : M \times M \to M\) obeying the following axioms:
(i) We have
\begin{equation} \label{op-0} a \diamond b = a+b \end{equation}4for \(a,b \in G\).
(ii) We have
\begin{equation} \label{op-1} (x,a) \diamond b = (\phi _a(b) x, a) \end{equation}5\begin{equation} \label{op-2} b \diamond (x,a) = (x / \phi _a(b), a) \end{equation}6and
\begin{equation} \label{op-3} (\phi _a(b) x,a) \diamond (x,a) = a+b \end{equation}7for \(x \in \mathbb {Q}^\times \), \(b \in G\), and \(a \in G \backslash \{ 0\} \).
(iii) If \(a,b,0 \in G\) are distinct and \((x,a) \diamond (y,b) = (z,c)\) for some \(x,y,z \in \mathbb {Q}^\times \) and \(c \in G\), then \((y,b) \diamond (z,c) = (\phi _a(b) x,a)\).
Then Equation 2, Equation 3 and hence Equation 1 holds.
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 Equation 2 is satisfied.
Now we verify Equation 3. If \(y\) is a square, then the claim already follows from Equation 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 Equation 6 that
and then from Equation 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 Equation 7 we have
and then by Equation 5
but from construction we have \(\phi _a(a+c)=-\phi _a(c) = \phi _a(b)\phi _a(c)\) 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 \(G \backslash \{ 0\} \). But this follows from (iii).
A partial solution is a finite family \({\mathcal F}\) of tuples \((x,y,z,a,b,c) \in (\mathbb {Q}^\times )^3 \times G^6\) with \(a,b,c,0\) distinct, such that the tuples \((\phi _a(b)^n x,a, \phi _b(c)^n y,b)\), \((\phi _b(c)^n y,b,\phi _c(a)^n z,c)\), \((\phi _c(a)^n z,c,\phi _a(b)^{n+1} x, a)\) for \((x,y,z,a,b,c) \in {\mathcal F}\) and \(n \geq 0\) are all distinct.
Let \({\mathcal F}\) be a parital solution. Then if one defines a partial operation \(\diamond \) on \(M\) by requiring axioms (i), (ii), imposing the additional operations
for all \(n \geq 0\) and \((x,y,z,a,b,c) \in {\mathcal F}\), and no other operations, then this is a well-defined partial operation that obeys axiom (iii) whenever it is defined.
Routine.
If \(\diamond \) is defined by a partial solution, and \((x,a) \diamond (y,b)\) is undefined for some \(x,y \in \mathbb {Q}^\times \) and distinct \(a,b \in G \backslash \{ 0\} \), then it is possible to extend the partial solution so that \((x,a) \diamond (y,b)\) is now defined.
We select a \(c \in G \backslash \{ 0\} \) that has not previously been used by the partial solution, let \(z \in \mathbb {Q}^\times \) be arbitrary, (e.g., one could take \(z=1\) if desired), and add \((x,y,z,a,b,c)\) to \({\mathcal F}\), thus we now also assign
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. The second and third equations in this family cannot collide with previous assignments because \(c\) is novel. 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 the first infinite sequence does not collide with any previously assigned value of \(\diamond \) either.
Every partial solution can be extended to a complete solution that obeys Equation 2 and Equation 3, and hence 1323.
Apply the usual greedy algorithm.
There exists a 1323 magma which does not obey the 2744 equation \(R_y L_{Sy} L_y x = x\).
It suffices to produce a partial solution in which \(L_y\) is not injective. Pick distinct \(a, b, b', c\) with \(\phi _a(b)\), \(\phi _a(b')\) having no prime factors in common in the numerator or denominator. We take the partial solution consisting of \((1,1,1,a,b,c)\) and \((1,1,1,a,b',c)\), that is to say we impose the conditions
and
One can check that no collisions occur; on the other hand, as \(L_{(1,a)} (1,b) = L_{(1,a)} (1,b') = (1,c)\), we already have a violation of left injectivity. Completing this seed to a full magma using Corollary 16.7, we obtain the claim.