Does there exist two binary operations $+$ and $*$ on $\{0,1\}$ such that $+$ and $*$ satisfy the same equations, but not the same quasi-equations? If not, are there such binary operations on a finite set of higher cardinality, and if so, what is the smallest possible cardinality? For those who don't know, a quasi-equation is a conditional statement, the antecedent of which is a conjunction of finitely many equations, and the consequent of which is a single equation.
1 Answers
Does there exist two binary operations + and ∗ on {0,1} such that + and ∗ satisfy the same equations, but not the same quasi-equations? If not, are there such binary operations on a finite set of higher cardinality, and if so, what is the smallest possible cardinality?
(Partial) answers:
$2$-element magmas.
There is an easy reason why
two $2$-element magmas
that satisfy the same identities must
satisfy the same quasi-identities,
but it requires some checking.
The reason is that two $2$-element magmas that
satisfy the same identities have to be isomorphic,
hence they have to
satisfy the same quasi-identities.
Let me sketch the argument showing that
two $2$-element magmas that
satisfy the same identities must be isomorphic.
Write operations on $\{0,1\}$ in their Boolean
form using the symbols $\wedge$ (and),
$\vee$ (or), $\neg$ (not).
If $\mathbf{A}=\langle \{0,1\}; m(x,y)\rangle$
and $\mathbf{A}'=\langle \{0,1\}; m'(x,y)\rangle$
are isomorphic magmas, then the isomorphism must be the identity
function $\textrm{id}\colon \{0,1\}\to \{0,1\}$ or the
function $\pi\colon \{0,1\}\to \{0,1\}$ that swaps $0$ and $1$.
The identity function will be an isomorphism from $\mathbf{A}$
to $\mathbf{A}'$ if and only if $m(x,y)$ and $m'(x,y)$
are equal operations on $\{0,1\}$,
while the swapping function $\pi$ will
be an isomorphism if and only if $m(x,y)$
is conjugate to $m'(x,y)$ in the sense that
$m(x,y) = \pi^{-1}(m'(\pi(x),\pi(y)))$.
There are only $16$ binary operations $m(x,y)$
on the set $\{0,1\}$, so only $16$ magmas
with universe $\{0,1\}$.
Up to conjugacy/duality there are only ten magmas
on the set $\{0,1\}$.
There are four $1$-element orbits under the conjugacy action
(i.e., four self-dual magmas) and six $2$-element
orbits under this action (i.e., six pairs of
isomorphic magmas).
Choosing one of each type,
the magma operations can be taken to be:
Any $2$-element magma is isomorphic to one
with universe $\{0,1\}$ equipped with
one of the binary operations on this list.
For any two nonisomorphic magmas from this list, there is an identity that distinguishes them. That is, we may isolate any one of them from the others by some set of identities. For example, I can express that $m(x,y)$ is independent of its second variable/place with the identity $m(x,y)= m(x,z)$. The first type of magma listed above may be isolated from all the others by the identities which express that its operation is independent of both places. The second type are the only ones whose operation is independent of exactly one of its places, so these magmas may be isolated from all others by identities. They can be distinguished from one another by checking which place the operation depends on and which of the identities $m(x,y)=x$ or $m(x,y)=y$ are satisfied. The third type may be isolated from the others by identities expressing that it commutes with itself ($m(m(x,y),m(u,v)) = m(m(x,u),m(y,v))$) and that $p(x,y,z):=m(x,m(y,z))$ is a Maltsev operation (i.e., $p(x,y,y)=x=p(y,y,x)$). We are left with the problem of isolating each of the fourth type of magma by identities. Here it suffices to test whether the magma operation is commutative, $m(x,y)=m(y,x)$, whether $m(x,x)=x$, or whether $m(m(x,y),y)=m(x,y)$. This completes a sketch of why two $2$-element magmas that satisfy the same identities are isomorphic.
$5$-element magmas.
Next I explain why there do exist two $5$-element magmas that satisfy the same identities, but not the same quasi-identities. This relies on ideas from Caroline Shannon's 1979 PhD thesis, Non-Finitely Based Binary Algebras Derived from Lattices.
Let $G = \langle V; E(x,y)\rangle$ be a graph. (I.e., $V\neq \emptyset$ and $E(x,y)$ is binary predicate on $V$ that is irreflexive and symmetric.) Define a binary operation $m(x,y)$ on $V\cup \{\infty\}$ by $m(u,v) = u$ if $u,v\in V$ and either $E(u,v)$ or $u=v$, and by $m(u,v)=\infty$ otherwise. Write $\mathbf{A}_G$ for the magma $\langle V\cup \{\infty\}; m(x,y)\rangle$. Here are some facts about these graph-related magmas.
Let $P_4$ be a path of $4$ vertices and let $Q$ be obtained from $P_4$ by adding an edge that is not a loop and does not create a $4$-cycle. For example, if $P_4$ is the path $a-b-c-d$, let $Q$ be obtained by adding an edge $a-c$. Both $P_4$ and $Q$ have induced paths of three vertices, so the variety generated by either one of the $5$-element magmas $\mathbf{A}_{P_4}$ and $\mathbf{A}_{Q}$ is the same as the variety generated by $\mathbf{A}_{P_3}$. Therefore $\mathbf{A}_{P_4}$ and $\mathbf{A}_{Q}$ satisfy the same identities. If $\mathbf{A}_{P_4}$ and $\mathbf{A}_{Q}$ satisfied the same quasi-identities, then (since they are finite) they would each embed into a power of the other. Since they are simple, they would have to embed into each other. This would force $\mathbf{A}_{P_4}\cong \mathbf{A}_{Q}$. But $\mathbf{A}_{P_4}\not\cong \mathbf{A}_{Q}$, because the underlying graph structure of each can be recovered from the multiplication. Thus, $\mathbf{A}_{P_4}$ and $\mathbf{A}_{Q}$ are $5$-element magmas that satisfy the same identities but not the same quasi-identities.
$4$-element unary algebras.
Let me briefly mention how
to produce $4$-element examples
if you allow me to change from magmas
to unary algebras.
The $8$-element dihedral group $D_4$ acts by rigid motions on a square. This yields a $4$-element faithful $D_4$-set $V$ given by the action of $D_4$ on the vertices of the square and also yields a $4$-element faithful $D_4$-set $E$ given by the action of $D_4$ on the edges of the square. By faithfulness, both $V$ and $E$ generate the variety of all $D_4$-sets, so they satisfy the same identities. $V$ and $E$ are nonisomorphic $D_4$-sets of size $4$, so neither can be embedded into the other. Both $V$ and $E$ are subdirectly irreducible algebras, so neither can be embedded into a power of the other. Since $V$ and $E$ are finite, neither can belong to the quasivariety generated by the other, so there must be a quasi-identity that holds in one of them that fails in the other one. (Since $D_4$ is $2$-generated, this example can be presented in the language of two unary operations.)
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