This is a generalization of the following question by John Wiltshire-Gordon.
Consider an inductive family of finite groups: $$ G_0 \hookrightarrow G_1 \hookrightarrow \ldots \hookrightarrow G_i \hookrightarrow \ldots $$
We may view each group as a subgroup of the next. Let $G$ be the directed union of the $G_i$: $$ G := \bigcup_{i=0}^\infty \ G_i $$
Now, $G$ is a countable group. Suppose we're asked a question like, "What proportion of the elements of $G$ are commutators?", or "What proportion of pairs $(g,h) \in G^2$ satisfy $g^2h^2=1$ but $g^2 \neq 1$?" We try to make sense of such questions in terms of density on $G^n$, which is not defined for all subsets. For $E \subseteq G^n$, let $$ d(E) := \lim_{i \to \infty} \frac{|E \cap G_i^n|}{|G_i^n|}, $$ if the limit exists.
My question is: If $E$ is a subset of $G^n$ that is first-order definable in the language of groups, does the density of $E$ necessarily exist? John's question covers the (still unresolved) special case of when $E$ is defined by an atomic formula.
A first-order definable subset $E \subseteq G^n$ is the set of $n$-tuples of group elements where a particular first-order formula in $n$ free variables is true. Such a formula is a finite string of symbols and may involve multiplication, inversion, the identity element, equality, logical connectives ($\wedge$, $\vee$, $\neg$, $\implies$), quatifiers ($\exists$ and $\forall$), and parentheses. For example, the following first-order formula defines the set of commutators $g \in G$:
$$ (\exists x)(\exists y)(g = xyx^{-1}y^{-1}). $$
On the other hand, if you try to define the commutator subgroup by such a formula, you run into trouble. You might want to say, "$g$ is a commutator or $g$ is a product of two commutators or $g$ is a product of three commutators or ...," but infinite disjunctions are not allowed.
Please note that the counterexample to a similar question in a comment by Vipul Naik here is not a counterexample to this question. In the inductive family $$ A_3 \hookrightarrow S_3 \hookrightarrow \ldots \hookrightarrow A_{2i-1} \hookrightarrow S_{2i-1} \hookrightarrow A_{2i+1} \hookrightarrow S_{2i+1} \hookrightarrow \ldots, $$ the groups alternate between having all the elements as commutators and half the elements as commutators. However, in the directed union, which is $A_\infty$, all the elements are commutators. The upshot is that quantifiers in a first-order formula may demand that you "look ahead" in your inductive family.
