The equational and quasi-equational theory of commutative groups in the signature $*$.

Question

I know that the equational theory of groups in the signature $*$ is axiomatized by the associative law, so does that mean the equational theory of commutative groups in the signature $*$ is axiomatized by the commutative and associative laws? Also, I know that the quasi-equational theory of groups in the signature $*$ is not finitely based, but is the quasi-equational theory of commutative groups in the signature $*$ also not finitely based? If it is, can someone exhibit a finite number of quasi-identities for commutative groups, and if it is not, can someone prove that it is not finitely based?

Answer 1

The quasi-equational theory of the class of commutative groups is axiomatized by

$x*(y*z)\approx (x*y)*z$

$x*y\approx y*x$

$(x*y\approx x*z)\to (y\approx z)$.

To see this, let $\mathcal C$ be the class of commutative groups in the language of $*$, let $\mathcal Q$ be the quasivariety generated by $\mathcal C$, and let $\mathcal Q'$ be the quasivariety defined by the three quasi-identities above. Since any commutative group satisfies the three quasi-identities, ${\mathcal Q}\subseteq {\mathcal Q}'$. Since any $*$-algebra satisfying the three quasi-identities is embeddable in a commutative group, ${\mathcal Q}'\subseteq {\mathcal Q}$. This shows that ${\mathcal Q}={\mathcal Q}'$, so the three quasi-identities axiomatize the quasi-variety generated by $\mathcal C$.

To believe everything I just wrote, you have to believe that any commutative, cancellative $*$-semigroup $S$ is embeddable in a commutative $*$-group. To see this, let $S$ be any commutative cancellative $*$-semigroup.

Case 1. ($S$ has an identity element) Apply the Grothendieck construction to $S$. This produces a commutative $*$-group into which $S$ embeds.

Case 2. ($S$ does not have an identity element) We formally adjoin an identity element to $S$ first and then apply the Case 1 argument to $S^1$.

To believe what I wrote in Case 2, you have to believe that if $S$ is a commutative cancellative semigroup, then $S^1$ will be a commutative cancellative extension of $S$. (This is not very hard to show.)

Answer 2

Here's a partial answer (leaving the quasi-equational theory part open):

It is indeed the case that the associativity and commutativity axioms axiomatize the equational $\{*\}$-theory of abelian groups. However, I don't think this follows trivially from the fact that associativity alone axiomatizes the equational $\{*\}$-theory of groups: in general, even if $T$ axiomatizes the equational $\Sigma$-theory of a class $\mathbb{K}$ of structures in a language containing $\Sigma$ and $\eta$ is an equation in $\Sigma$, we need not have $T\cup\{\eta\}$ axiomatize the equational $\Sigma$-theory of $\mathbb{K}\cap Mod(\{\eta\})$. This is because equational logic does not let us express conditionals.

Instead, we just prove the specific claim directly. Modulo associativity and commutativity, every equation is equivalent to one of the form $$x_1^{n_1}x_2^{n_2}...x_k^{n_k}=x_1^{m_1}x_2^{m_2}...x_k^{m_k}$$ (with implicit $*$s and right-associating parentheses) for some distinct variables $x_1,x_2,...,x_k$ and some natural numbers (including zero) $n_1,n_2,...,n_k,m_1,m_2,...,m_k$. It's easy to see that any such equation not satisfying $n_i=m_i$ for all $i$ can be made to fail in an abelian group (consider mapping each $x_i$ to a different generator of $\mathbb{Z}^k$). On the other hand, if $n_i=m_i$ for every $i$, then the equation in question is trivial.