16 Equation 1323

In this chapter we study magmas that obey equation 1323,

x = y ⋄ (((y ⋄ y) ⋄ x) ⋄ y)

for all $x, y$ . Using the squaring operator $S y := y ⋄ y$ and the left and right multiplication operators $L_{y} x := y ⋄ x$ and $R_{y} x = x ⋄ y$ , this law can be written as

L_{y} R_{y} L_{S y} x = x .

In particular, this gives a way to construct these magmas:

Lemma 16.1 Construction of 1323 magmas

Suppose that $M$ is a magma such that

R_{S y} L_{S y} = 1

and

L_{y} R_{y} = R_{S y}

hold. Then the magma obeys 1323.

Proof ▶

So now we would like to construct magmas satisfying Equation 2 and Equation 3. We need some bijections:

Lemma 16.2 Bijections

Let $G$ be a countably infinite abelian torsion group of exponent $2$ . Then there exists a bijection $ϕ_{a} : G \to Q^{\times}$ for each $a \in G ∖ {0}$ such that $ϕ_{a} (0) = 1$ and $ϕ_{a} (a + b) = - ϕ_{a} (b)$ for all $b \in G$ , so in particular $ϕ_{a} (a) = - 1$ .

Proof ▶

Such a bijection can be easily constructed from the axiom of choice and a greedy algorithm, defining $ϕ_{a}$ one pair ${b, a + b}$ at a time.

Lemma 16.3 Building a magma

Let $G$ be a countably infinite abelian torsion group of exponent $2$ , and let $ϕ_{a}$ be as in the previous lemma. Let $N$ be the set of pairs $(x, a)$ with $x \in Q^{\times}$ and $a \in G ∖ {0}$ , and let $M = G ⊎ N$ be the disjoint union of $G$ and $N$ . Suppose that we have an operation $⋄ : M \times M \to M$ obeying the following axioms:

(i) We have

$a ⋄ b = a + b$
4

for $a, b \in G$ .
(ii) We have

$(x, a) ⋄ b = (ϕ_{a} (b) x, a)$
5

$b ⋄ (x, a) = (x / ϕ_{a} (b), a)$
6

and

$(ϕ_{a} (b) x, a) ⋄ (x, a) = a + b$
7

for $x \in Q^{\times}$ , $b \in G$ , and $a \in G ∖ {0}$ .
(iii) If $a, b, 0 \in G$ are distinct and $(x, a) ⋄ (y, b) = (z, c)$ for some $x, y, z \in Q^{\times}$ and $c \in G$ , then $(y, b) ⋄ (z, c) = (ϕ_{a} (b) x, a)$ .

Then Equation 2, Equation 3 and hence Equation 1 holds.

Proof ▶

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

L_{a} b = R_{a} b = a + b

and

L_{a} (x, b) = (x / ϕ_{b} (a), b); R_{a} (x, b) = (x ϕ_{b} (a), b)

so Equation 2 is satisfied.

Now we verify Equation 3. If $y$ is a square, then the claim already follows from Equation 2 since $S y = 0$ is a unit. Otherwise, for $y \in Q^{\times}$ and $b \in G ∖ {0}$ , we have to show that

L_{(y, b)} R_{(y, b)} a = R_{b} a = a + b

for $a \in G$ and

L_{(y, b)} R_{(y, b)} (x, a) = R_{b} (x, a) = (ϕ_{a} (b) x, a)

for $x \in Q^{\times}$ and $a \in G ∖ {0}$ . In the first case, we have from Equation 6 that

R_{(y, b)} a = (y / ϕ_{a} (b), b)

and then from Equation 7 we have

L_{(y, b)} R_{(y, b)} a = a + b

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

R_{(y, b)} (x, a) = a + c

and then by Equation 5

L_{(y, b)} R_{(y, b)} (x, a) = (ϕ_{a} (a + c) y, a)

but from construction we have $ϕ_{a} (a + c) = - ϕ_{a} (c) = ϕ_{a} (b) ϕ_{a} (c)$ and hence $ϕ_{a} (a + c) y = ϕ_{a} (b) x$ as required.

It remains to handle the case when $a, b$ are distinct elements of $G ∖ {0}$ . But this follows from (iii).

Definition 16.4 Partial solution

A partial solution is a finite family $F$ of tuples $(x, y, z, a, b, c) \in (Q^{\times})^{3} \times G^{6}$ with $a, b, c, 0$ distinct, such that the tuples $(ϕ_{a} (b)^{n} x, a, ϕ_{b} (c)^{n} y, b)$ , $(ϕ_{b} (c)^{n} y, b, ϕ_{c} (a)^{n} z, c)$ , $(ϕ_{c} (a)^{n} z, c, ϕ_{a} (b)^{n + 1} x, a)$ for $(x, y, z, a, b, c) \in F$ and $n \geq 0$ are all distinct.

Lemma 16.5 Soundness

Let $F$ be a partial solution. Then if one defines a partial operation $⋄$ on $M$ by requiring axioms (i), (ii), imposing the additional operations

(ϕ_{a} (b)^{n} x, a) ⋄ (ϕ_{b} (c)^{n} y, b) = (ϕ_{c} (a)^{n} z, c)

(ϕ_{b} (c)^{n} y, b) ⋄ (ϕ_{c} (a)^{n} z, c) = (ϕ_{a} (b)^{n + 1} x, a)

(ϕ_{c} (a)^{n} z, c) ⋄ (ϕ_{a} (b)^{n + 1} x, a) = (ϕ_{b} (c)^{n + 1} y, b)

for all $n \geq 0$ and $(x, y, z, a, b, c) \in F$ , and no other operations, then this is a well-defined partial operation that obeys axiom (iii) whenever it is defined.

Proof ▶

Lemma 16.6 Greedy extension

If $⋄$ is defined by a partial solution, and $(x, a) ⋄ (y, b)$ is undefined for some $x, y \in Q^{\times}$ and distinct $a, b \in G ∖ {0}$ , then it is possible to extend the partial solution so that $(x, a) ⋄ (y, b)$ is now defined.

Proof ▶

We select a $c \in G ∖ {0}$ that has not previously been used by the partial solution, let $z \in Q^{\times}$ be arbitrary, (e.g., one could take $z = 1$ if desired), and add $(x, y, z, a, b, c)$ to $F$ , thus we now also assign

(ϕ_{a} (b)^{n} x, a) ⋄ (ϕ_{b} (c)^{n} y, b) = (ϕ_{c} (a)^{n} z, c)

(ϕ_{b} (c)^{n} y, b) ⋄ (ϕ_{c} (a)^{n} z, c) = (ϕ_{a} (b)^{n + 1} x, a)

(ϕ_{c} (a)^{n} z, c) ⋄ (ϕ_{a} (b)^{n + 1} x, a) = (ϕ_{b} (c)^{n + 1} y, b)

for all natural numbers $n$ . As the $a, b, c$ are distinct, the $ϕ_{a} (b), ϕ_{b} (c), ϕ_{c} (a)$ 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 $ϕ_{b} (c)$ involves primes in the numerator or denominator that do not appear in any previous $ϕ_{a} (b), ϕ_{b} (c), ϕ_{c} (a), x, y, z$ 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 $⋄$ either.

Corollary 16.7 Iterated greedy extension

Every partial solution can be extended to a complete solution that obeys Equation 2 and Equation 3, and hence 1323.

Proof ▶

Corollary 16.8 1323 does not imply 2744

There exists a 1323 magma which does not obey the 2744 equation $R_{y} L_{S y} L_{y} x = x$ .

Proof ▶

It suffices to produce a partial solution in which $L_{y}$ is not injective. Pick distinct $a, b, b^{'}, c$ with $ϕ_{a} (b)$ , $ϕ_{a} (b^{'})$ 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

(ϕ_{a} (b)^{n}, a) ⋄ (ϕ_{b} (c)^{n}, b) = (ϕ_{c} (a)^{n}, c)

(ϕ_{b} (c)^{n}, b) ⋄ (ϕ_{c} (a)^{n}, c) = (ϕ_{a} (b)^{n + 1}, a)

(ϕ_{c} (a)^{n}, c) ⋄ (ϕ_{a} (b)^{n + 1}, a) = (ϕ_{b} (c)^{n + 1}, b)

and

(ϕ_{a} (b)^{n}, a) ⋄ (ϕ_{b^{'}} (c)^{n}, b^{'}) = (ϕ_{c} (a)^{n}, c)

(ϕ_{b^{'}} (c)^{n}, b^{'}) ⋄ (ϕ_{c} (a)^{n}, c) = (ϕ_{a} (b^{'})^{n + 1}, a)

(ϕ_{c} (a)^{n}, c) ⋄ (ϕ_{a} (b^{'})^{n + 1}, a) = (ϕ_{b} (c)^{n + 1}, b^{'}) .

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. □