Call an algebra $A$ finitely approximable if $\forall a, b \in A$ with $a \neq b$ there is a congruence $\theta_{ab} \in \text{Con}(V)$ with $(a, b) \notin \theta_{ab}$ and $A / \theta_{ab}$ is finite.
Proposition: $A$ is finitely approximable iff $A$ is a subdirect product of finite algebras.
Proof. Assume $A$ is a subdirect product of finite algebras. Then there is a family $\{A_i \mid i \in I\}$ of finite algebras and an embedding $f \colon A \to \prod_{i \in I} A_i$ with $\pi_i \circ f$ being a surjection for all $i \in I$. Take arbitrary $a, b \in A$ with $a \neq b$. Since $f$ is an embedding we have $f(a) \neq f(b)$, so $\exists i \in I$ such that $\pi_i(f(a)) \neq \pi_i(f(b))$, that is $(f(a), f(b)) \notin \text{ker}(\pi_i \circ f)$. Also we have $$A/\text{ker}(\pi_i \circ f) \cong (\pi_i \circ f) (A_i) = A_i$$ with $A_i$ being finite. Hence $A$ is finitely approximable.
Now assume $A$ is finitely approximable. For any pair $a, b \in A$ with $a \neq b$ let $f_{ab} \colon A \to A/ \theta_{ab}$ be the canonical epimorphism. We define a map $f \colon A \to \prod_{a, b \in A, a \neq b} A/\theta_{ab}$ as follows $f(x) = (f_{ab}(x))$ for $a, b \in A, a \neq b$. Clearly, $f$ is a homomorphism. We have $\pi_{ab} \circ f = f_{ab}$ is a surjection. Also if $a \neq b$ then $f_{ab}(a) \neq f_{ab}(b)$ and hence $f(a) \neq f(b)$. So $A$ is a subdirect product of finite algebras.
Call a variety of algebras $V$
- finitely approximable if every member of $V$ is;
- residually finite if every subdirectly irreducible member of $V$ is finite;
- satisfying the finite model property if $V \models t \approx s \Leftrightarrow (\forall A \in V)[|A| < \infty \rightarrow A \models t \approx s]$, that is, $t \approx s$ holds in $V$ if and only if it holds in all finite members of $V$.
Denote by $V_{SI}$ the class of all subdirectly irreducible members of $V$. Birkhoff's theorem implies that $\forall A \in V$ there is an isomorphism $A \cong B \leqslant \prod_{i \in I} A_i$, where $\{A_i \mid i \in I\} \subseteq V_{SI}$. Hence if $V$ is residually finite then $V$ is finitely approximable. Conversely, if $V$ is finitely approximable take any $A \in V_{SI}$. We know that $A$ is a subdirect product of finite algebras, but since $A$ is subdirectly irreducible it must be isomorphic to one of the algebras in this product. In particular, $A$ must be finite. Thus, $V$ is residually finite iff $V$ is finitely approximable.
If $V$ is finitely approximable then if some identity holds in all finite members of $V$ it also holds in the subalgebra of the direct product of (some of) these algebras and hence in every algebra of $V$. That means if $V$ is finitely approximable then $V$ satisfies the finite model property.
I'm interested in whether the converse implication is also true. However, I'm unable to show this. Here are my thoughts.
Let $F(X)$ be the $V$-free algebra generated by $X$ with $|X| > 1$. Assume that in $F(X)$ we have $$a = t(x_1, \dots, x_n) \neq s(x_1, \dots, x_n) = b.$$ Then $V \not\models t \approx s$, so there is a finite $A \in V$ with $A \not\models t \approx s$, say $$t(a_1, \dots, a_n) \neq s(a_1, \dots, a_n)$$ with $a_1, \dots, a_n \in A$. Now consider the homomorphism $f \colon F(X) \to A$ extending the map $x_1 \mapsto a_1, \dots, x_n \mapsto a_n$. We have $f(a) \neq f(b)$ and hence $(a, b) \notin \text{ker}(f)$. But also the quotient $F(X)/\text{ker}(f) \cong \text{im}(f) \subseteq A$ is finite. This shows that $F(X)$ is finitely approximable. In particular, $F(X)$ is a subdirect product of finite algebras.
Now we know that $(\forall A \in V)(\exists X)$ with $A = \text{im}(f)$ for some homomorphism $f \colon F(X) \to A$. I've tried to "transfer" the property of $F(X)$ being a subdirect product of finite algebras to $A$ using $f$, but I didn't succeeded. I think I'm missing something obvious here.
Question 0. Are my arguments correct?
Question 1. Is the converse true, that is, if $V$ satisfies the finite model property, then $V$ is finitely approximable (residually finite)? Can you give me a hint?
The paper
Olshanskii, A. Ju. Varieties of finitely approximable groups. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 33 1969 915–927
proves that a variety ${\mathcal V}$ of groups is finitely approximable if and only if it is generated by a finite group whose Sylow subgroups are abelian. I will write this as ${\mathcal V}={\mathcal V}(G)$ for a finite group $G$ whose Sylow subgroups are abelian.
On the other hand, the finite model property for ${\mathcal V}$ is just the property that ${\mathcal V}$ is generated as a variety by its finite members. That is, ${\mathcal V}$ is the least variety containing the class ${\mathcal V}_{\textrm{fin}}$ of finite members of ${\mathcal V}$.
Putting things together, if $G$ is a finite group with at least one nonabelian Sylow subgroup, then ${\mathcal V}(G)$ is not finitely approximable (according to Olshanskii's paper), yet it will have the finite model property (because it is generated by one of its finite members).