For each positive integer $n$, let $\eta_n$ denote the following identity in the language of monoids.
$$(xy)^n = x^n y^n$$
For example, $\eta_2$ is the identity $xyxy = xxyy.$
Question. Is it true that for any subset $Q$ of the prime numbers $\{2,3,5,\ldots\}$, there exists a monoid $M$ such that for all prime numbers $p,$ $M$ satisfies $\eta_p$ iff $p \in Q$? In other words, is $\{\eta_2,\eta_3,\eta_5,\ldots\}$ a logically independent set of sentences? (In the presence of the monoid axioms).
For example (and I may be overlooking something obvious here), but I do not think that $(xy)^5 = x^5 y^5$ follows from just $(xy)^2=x^2 y^2$ and $(xy)^3 = x^3y^3$.
Consider the monoid $M$ given by the presentation $\langle x,y : \{(xy)^p = x^p y^p\}_{p \in Q} \rangle$. In order to show that $(xy)^q \neq x^q y^q$ in $M$ for $q \notin Q$, we use that two words $w,w'$ in $x,y$ become equal in $M$ if and only if $(w,w')$ lies in the equivalence relation generated by the following relation: There are words $a,b$ and $p \in Q$ such that $w=a (xy)^p b$, $w'=a x^p y^p b$. The good thing is that (in contrast to groups) no cancellation occurs. Thus, it becomes a "combinatorial" exercise to show that $(xy)^q$ and $x^q y^q$ are not equal in $M$. If they were, we would have to replace $(xy)^p$ by $x^p y^p$ somewhere in $(xy)^q$, which only works when $p \leq q$. We end up with $(xy)^i x^p y^p (xy)^{q-p-i}$. We can never reach $x^q$ or $y^q$ this way. Of course one really has to prove this, by induction perhaps. Please tell me if this makes any difficulties.