Questions tagged [verbal-subgroups]

A verbal subgroup of a group $G$, generated by the set of words $A \subset F_\infty$ ($F_\infty$ is a free group of countable rank) is a subgroup $V_A(G) = \langle \{h(w): w \in A, h \text{ is a homomorphism from } F_\infty \text{ to }G \} \rangle$. A verbal subgroup is always a characteristic one. To be used with the tag [group-theory].

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Burnside groups with GAP system [closed]

My question is related to Burnside groups $B(n, 3)$ in the GAP system. I'm interested in ways to represent Burnside groups $B(n, 3)$ in GAP. The obvious representation using relations (see example for ...
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How to mention about group words when you haven't said about free groups and show regarding a marginal set computation

Let $F$ be a free group on a countably infinite set $\{x_1, x_2, \ldots \}$ and let $W$ be a nonempty subset of $F$. If $w = w(x_1, \ldots , x_n) \in W$ and $g_1, \ldots, g_n$ are elements of a group $...
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Meaning of Marginal sets and related descriptions

Let $F$ be a free group on a countably infinite set $\{x_1, x_2, \ldots \}$ and let $W$ be a nonempty subset of $F$. If $w = w(x_1, \ldots , x_n) \in W$ and $g_1, \ldots, g_n$ are elements of a group $...
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Show that the group of all complex $2^n$th roots of unity, $n=0,1,2,...$, has a fully-invariant subgroup which is not verbal

This is the last part of Exercise 2.3.3 of Robinson's "A Course in the Theory of Groups (Second Edition)". The following are the other parts. Verbal subgroups of the group of all $2^n$th ...
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Verbal Subgroups of $\mathbb{Z}^2$

So I want to try to show that all the verbal subgroups of an abelian group $G$ are of the form $G(X^n), n \geq 1$. I want to start with $\mathbb{Z}$ and $\mathbb{Z}^n$. So I worked with $\mathbb{Z}$, ...
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Do there exist characteristic subgroups, that are not quasiverbal?

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 $G$ ...
Chain Markov's user avatar
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Verbal subgroups of the group of all $2^n$th roots of unity.

Hi: Let $G$ be the multiplicative group of all complex $2^n$th roots of unity, $n=0,1,2,...$ Prove that $1$ and $G$ are the only verbal subgroups of $G$. Let $F$ be a free group on a countably finite ...
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Is there some sort of formula for $[F_n : V_{\{x^4\}}(F_n)]$?

Suppose $F_n$ is a free group of rank $n$. It is a rather well known fact, that $b_4(n) = [F_n : V_{\{x^4\}}(F_n)]$ is finite for all $n \in \mathbb{N}$. Is there a some sort of formula for $b_4(n)$? ...
Chain Markov's user avatar
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Is finite verbal subgroup equivalent to finite index of marginal subgroup?

There is a well known fact: If $G$ is a finitely generated group. Then $|G’| < \infty$ iff $[G:Z(G)]<\infty$. Suppose $\mathfrak{U}$ is a group variety. Let’s denote the corresponding verbal ...
Chain Markov's user avatar
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Is there a formula for $[F_n : V_{\{x^3\}}(F_n)]$?

Suppose $F_n$ is a free group of rank $n$. It is a rather well known fact, that $b_3(n) = [F_n : V_{\{x^3\}}(F_n)]$ is finite for all $n \in \mathbb{N}$. Is there a some sort of formula for $b_3(n)$? ...
Chain Markov's user avatar
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Are upper quasiverbal and lower quasiverbal subgroups always the same subgroup?

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 ...
Chain Markov's user avatar
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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 ...
Chain Markov's user avatar
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Do all maximal verbal series have the same length?

Suppose, $G$ is a finite group. Let’s call a series of subgroups of $G$ $\{H_k\}_{k=1}^n$ a verbal series of length $n$, iff $H_1 = E$, $H_n = G$ and $\forall k < n$ $H_k$ is a verbal subgroup of $...
Chain Markov's user avatar
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Does there exist an Artinian verbally simple group, which is not characteristically simple?

Does there exist an Artinian verbally simple group, which is not characteristically simple? A characteristically simple group is a group without non-trivial proper characteristic subgroups, a verbally ...
Chain Markov's user avatar
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Does there exist a verbally simple group, which is not characteristically simple?

Does there exist a verbally simple group, which is not characteristically simple? A characteristically simple group is a group without non-trivial proper characteristic subgroups, a verbally simple ...
Chain Markov's user avatar
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Large counterexamples to "Non-isomorphic finite groups have verbal subgroups of different order"

In this question, it was conjectured that for every pair of non-isomorphic finite groups $G$ and $H$, there exists some word $\omega$ such that $|V_{\omega}(G)|\ne|V_{\omega}(H)|$, i.e. their ...
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Does $D_4$ have a verbal subgroup of order 4?

Does $D_4$ have a verbal subgroup of order 4? How did this question arise: In the comments $Q_8$ ad $D_4$ were pointed to be a possible counterexample to this question: Is it true, that for any two ...
Chain Markov's user avatar
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Does there exist some sort of classification of finite verbally simple groups?

Let’s call a group verbally simple if it does not have any non-trivial verbal subgroup. Does there exist some sort of classification of finite verbally simple groups? $G^n$, with $G$ being a finite ...
Chain Markov's user avatar
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Is it true, that for any two non-isomorphic finite groups $G$ and $H$ there exists such a group word $w$, that $|V_w(G)| \neq |V_w(H)|$?

Is it true, that for any two non-isomorphic finite groups $G$ and $H$ there exists such a group word $w$, that $|V_w(G)| \neq |V_w(H)|$? Here $V_w(G)$ stands for the verbal subgroup of $H$, generated ...
Chain Markov's user avatar
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Free groups: normal supplements of the commutator subgroup

Let $F$ be a free group and let $V$ be another verbal subgroup of $F$ such that $$ F = [F,F] V. $$ Is it true that $V=F?$ More generally, if $N$ is a normal (or even characteristic) subgroup of $F$ ...
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Alternating subgroup of symmetric group--is it considered a generalization of verbal?

The alternating subgroup ($A_n$) of the symmetric group($S_n$), is, I believe, not a verbal subgroup. But it is generated by elements of the form $xy$, where $x$ and $y$ have the same cycle structure. ...
Richard Peterson's user avatar
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On a n-Engel Verbal Subgroup be perfect

Given a Group $G$, the $n$-th Engel Word defined by $[x,\ _0{x}]=x: \ [x,_{ \ n}y]=[[x,_{\ {n-1}}y],y]$ in the group $G$ consists by substituting group elements for the determinates. The Group ...
Kalawa's user avatar
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When Verbal Subgroups are propers

Let $w$ be a group-word, and let $G$ be a group. The verbal subgroup $w(G)$ of $G$ determined by $w$ is the subgroup generated by the set consisting of values $w(g_1, \ldots, g_n)$, where $g_1, \ldots,...
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