Do there exist characteristic subgroups, that are not quasiverbal?

Question

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 quasiverbal subgroup of a group $G$ generated by $Q$ as $V_Q(G) := \langle\{h(w)|(w, A) \in Q, h \in Hom(F_\infty, G), h(A) = \{e\}\}\rangle$.

It is not hard to see, that all verbal subgroups are quasiverbal. Moreover, all marginal subgroups are also quasiverbal. Indeed, if a subgroup is a marginal subgroup for a collection of words $\{w_i(x_1, ... , x_n)\}_{i \in I}$, then it is quasiverbal for a quasiword $$(g, \{w_i(x_1, ..., x_{j-1}, gx_j, x_{j + 1}, ..., x_n) w_i^{-1}(x_1, ... ,x_n)\}_{i \in I, j \leq n} \cup \{w_i(x_1, ..., x_{j-1}, x_jg, x_{j + 1}, ..., x_n) w_i^{-1}(x_1, ... ,x_n)\}_{i \in I, j \leq n})$$

Also, all quasiverbal subgroups are obviously characteristic.

My question is:

Do there exist characteristic subgroups, that are not quasiverbal?

Personally, I failed to construct any.

Answer 1

Let $S$ be a nonabelian simple group which contains the group $\mathbb Z$ as a subgroup. Let $G = S\times \mathbb Z$. The subgroup $\{e_S\}\times \mathbb Z$ is the center of $G$, so it is characteristic in $G$. Let's argue that it is not quasiverbal.

Assume instead that $Q$ is a set of quasiwords for which $V_G(Q)=\{e_S\}\times \mathbb Z$. A homomorphism $h: F_{\infty}\to G=S\times \mathbb Z$ is a pair $(h_1(x),h_2(x))$ where $h_1: F_{\infty}\to S$ is a homomorphism and $h_2: F_{\infty}\to \mathbb Z$ is a homomorphism. If $V_G(Q)=\{e_S\}\times \mathbb Z$, then there must exist a homomorphism $h = (h_1,h_2)$ such that $h(A)=\{e_G\}$ for all $(w,A)\in Q$, but $h(w)\in (\{e_S\}\times \mathbb Z)-\{e_G\}$. Thus, $h_2: F_{\infty}\to \mathbb Z$ satisfies $h_2(A)=\{e_{\mathbb Z}\}$ for all $(w,A)\in Q$, but $h_2(w)\neq e_{\mathbb Z}$. But now the function $h' = (h_2,h_2): F_{\infty}\to \mathbb Z\times \mathbb Z\subseteq G$ is a homomorphism satisfying $h'(A)=\{e_G\}$ for all $(w,A)\in Q$; yet $h'(w)=(h_2(w),h_2(w))\notin \{e_S\}\times \mathbb Z$, since $(h_2(w),h_2(w))$ has equal coordinates and the second is not the identity element. This shows that $V_G(Q)$ cannot be $\{e_S\}\times \mathbb Z$ after all.