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].
Questions tagged [verbal-subgroups]
23 questions
7
votes
1 answer
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 by the group word $w$.
Initially, the question I…
Chain Markov
- 16,012
6
votes
1 answer
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 corresponding verbal subgroups have unequal order. The…
Santana Afton
- 6,978
5
votes
2 answers
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 $g_1, \ldots, g_n$ are elements of $G$.
I'm tryng to prove a…
Kalawa
- 131
5
votes
1 answer
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 subgroup as a $V_{\mathfrak{U}}(G)$ and the…
Chain Markov
- 16,012
5
votes
1 answer
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 $b_3(n)$? Here $V_Q$ is the verbal subgroup for the…
Chain Markov
- 16,012
5
votes
0 answers
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 simple group is a group without non-trivial…
Chain Markov
- 16,012
5
votes
1 answer
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 simple group, is always verbally simple as it has…
Chain Markov
- 16,012
4
votes
1 answer
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 non-isomorphic finite groups $G$ and $H$ there…
Chain Markov
- 16,012
3
votes
0 answers
Burnside groups with GAP system
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 $B(3,3)$ below) is quite complex, because…
arthurbesse
- 149
3
votes
2 answers
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 set $\{x_1,x_2,...\}$. I must prove that for any…
stf91
- 813
3
votes
0 answers
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 $H_{k+1}$. It is rather obvious, that in this case…
Chain Markov
- 16,012
3
votes
0 answers
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 $F$ which also supplements $[F,F],$
$$
F=[F,F] N
$$
is…
P.H.
- 133
- 5
2
votes
2 answers
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 $\mathbb{Z}$, I know that all the subgroups of $\mathbb{Z}$ are…
Sasha
- 707
- 6
- 14
2
votes
1 answer
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 $G$ generated by $Q$ as $V_Q(G) := \langle\{h(w)|(w, A)…
Chain Markov
- 16,012
2
votes
1 answer
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 of all groups $G$, such that $\forall (w, A) \in…
Chain Markov
- 16,012