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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 prevariety 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 prevarieties.

Now, suppose $\mathfrak{U}_Q$ is a prevariety described by $Q$. And $G$ is a group. Lets call the upper quasiverbal subgroup associated with the set of quasiwords $Q$ the unique minimal normal subgroup $\overline{V_Q(G)}$, such that $\frac{G}{\overline{V_Q(G)}} \in \mathfrak{U}_Q$ (the proof that it exists can be found here: Do the quasiverbal subgroups always exist?)

Now let’s define the lower quasiverbal subgroup of group $G$ associated with $Q$ as $\underline{V_Q(G)} = \langle \{h(w)| h \in Hom(F_\infty, G), h(A) = \{e\} \} \rangle$

Is it always true, that $\overline{V_Q(G)} = \underline{V_Q(G)}$? If $\mathfrak{U}_Q$ is a variety, they indeed are. But whats about the general case?

It is quite obvious, that both $\overline{V_Q(G)}$ and $\underline{V_Q(G)}$ are characteristic, and that $\underline{V_Q(G)} \leq \overline{V_Q(G)}$. However, that is clearly not enough.

Chain Markov
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1 Answers1

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Actually, the answer is no, and there are infinitely many counterexamples to this statement:

Suppose, $C_{p^n}$ is the primary cyclic group of order $p^n$, where $n \geq 2$, and $Q = \{(x, x^p)\}$. Then it is not hard to see, that in this case $\overline{V_Q(C_{p^n})} = C_{p^n}$, and $\underline{V_Q(C^{p^n})} \cong C_p$.

Chain Markov
  • 16,012