Is power-associativity finitely axiomatizable?

Question

Let the signature under consideration be $(S,*)$, which is the signature of magmas. Power-associativity can be axiomatized by an infinite set of equations. Can it also be axiomatized by a finite set of equations, or if not, a finite set of first order axioms?

Answer 1

I believe that the following is a proof that no finite set of axioms can axiomatize power-associativity, however I will caution that this is not my area of expertise and I have only proven this today for myself.

First we shall inductively define some notation (notice that this is only notation and so can in principle be replaced by the original symbols, but it is necessary to use it for the proof)

Let $x^1=x$ and for all $n\geq1$, $x^{n+1}=x*x^n$

Now we give an axiom schema characterizing power-associativity. $\forall n,m\geq1$

$$x^{n+1}x^m=x^nx^{m+1}$$

It is clear that power-associativity implies this, and a simple induction argument shows that this implies power-associativity.

Now assume for a contradiction that we have some finite axiomatization of power-associativity, notice that this is supposed to prove exactly the same statements as our infinte axiomatizion and so the finitely many axioms must be theorems of the infinite axiom scheme.

However any proof must only use finitely many axioms and so we must therefore be able to prove the finitely many new axioms (and by extension axiomatize power-associativity) using only finitely many of the axioms in our original axiom schema.

Now each axiom has $2$ numbers $n$ and $m$. Choose some prime $p>2$ greater than $n+1,m+1$ for all of the axioms.

Now by adding any missing axioms we see that the finite axiom scheme given by $\forall1\leq n,m < p-1$

$$x^{n+1}x^m=x^nx^{m+1}$$

axiomatizes power-associativity.

However it is possible to construct a finite model which is not power-associative and yet satisfies these axioms, which is a contradiction and so power-associativity is not finitely axiomatizable. We now give such a model to finish the proof.

(For the purpose of making things clearer, in the following $kp\bmod p=p$)

Let $S=\{x_1,x_2,...x_p\}$ and define $*$ as follows:

$\forall i,j\neq p$, $x_i*x_j=x_{i+j\bmod p}$

$\forall i$, $x_i*x_p=x_p*x_i=x_p$

It is fairly routine (although tedious) to check that this satisfies all of the axioms, however for example $x_1^1*x_1^p=x_p$ but $x_1^2*x_1^{p-1}=x_1$. Thus it is not power-associative.

$\square$

Hopefully this is all correct and is what you were looking for, let me know if you have any questions.

Answer 2

This may not be exactly what you're looking for, but you could introduce a new binary predicate $a\uparrow x$, intuitively meaning "$x$ is a power of $a$", and axioms

$$ a\uparrow a \\ a\uparrow x \land a \uparrow y \to a\uparrow(x*y) \\ a\uparrow x \land a \uparrow y \land a \uparrow z \to (x*y)*z = x*(y*z)$$

This doesn't guarantee that only powers of $a$ will satisfy $a\uparrow x$, but at least it will be the case that (1) every model of these axioms becomes a power-associative magma when you forget the interpretation of $\uparrow$, and (2) every power-associative magma can be extended with an interpretation of $\uparrow$ that makes it a model of the axioms.

This would allow you to apply at least some first-order model theory while having a finite set of axioms.