Do the quasiverbal subgroups always exist? Let’s define a group quasiword as an element of $F_\infty \times P(F_\infty)$. Suppose $Q \subset F_\infty \times P(F_\infty)$ is a set of quasiwords. Define a quasivariety described by $Q$ as a class of all groups $G$, such that $\forall (w, A) \in Q, h \in Hom(F_\infty, G), (h(A) = \{e\} \to h(w) = e)$. One can easily see, that all group varieties actually are quasivarieties.
Now, suppose $\mathfrak{U}_Q$ is a quasivariety described by $Q$. And $G$ is a group. Does there always exist the unique minimal normal subgroup $V_Q(G)$, such that $\frac{G}{V_Q(G)} \in \mathfrak{U}_Q$ (lets call such subgroup quasiverbal). In the particular case, when $\mathfrak{U}_Q$ is a variety one can easily see that such subgroup always exist and is exactly the verbal subgroup, corresponding to that variety. However, whats about the general case?
I tried to prove the existence of quasiverbal subgroups using Zorn’s lemma. However, it was not a success, as I failed to prove, that in $\{H \triangleleft G| \frac{G}{H} \in \mathfrak{U}_Q\}$ any descending chain has a lower bound.
 A: I’m not positive what you mean by $P(F_{\infty})$ (permutations on $F_{\infty}$?). But it seems clear from your definition that given a class $\mathfrak{U}_{Q}$,


*

*If $G\in\mathfrak{U}_{Q}$ and $H\leq G$, then $H\in\mathfrak{U}_Q$;

*If $\{G_i\}_{i\in I}$ is a family of groups in $\mathfrak{U}_Q$, then $\prod G_i\in\mathfrak{U}_Q$.

*If $G\in\mathfrak{U}_Q$ and $G\cong K$, then $K\in\mathfrak{U}_Q$.


(I’m not quite sure yet if the class is closed under reduced products, and hence would match the standard use of quasivariety; maybe that is what you are using, just with a definition somewhat different than I’m used to).
In particular, suppose that $N_i$ is the collection of all normal subgroups of $G$ such that $G/N_i\in\mathfrak{U}_Q$. Clearly the family is nonempty, since $G/G$ lies in $\mathfrak{U}_Q$. Consider the canonical map $G\to \prod_i (G/N_i)$. The kernel is exactly $\cap N_i$. The map induces an embedding of $G/(\cap N_i)$ into $\prod_i(G/N_i)$. But the latter is in $\mathfrak{U}_Q$ (product of groups in $\mathfrak{U}_Q$), and hence the $G/(\cap N_i)$ is (isomorphic to) a subgroup of a group in $\mathfrak{U}_Q$, and hence lies in $\mathfrak{U}_Q$ itself. Clearly, $N=\cap_i N_i$ is the least normal subgroup with the property that the quotient of $G$ lies in $\mathfrak{U}_Q$, and hence is your “quasiverbal subgroup”.
