Groups with finitely many long commutators of maximal order
Fuente:
arXiv
Salvato in:
| Autori principali: | , , |
|---|---|
| Natura: | Preprint |
| Pubblicazione: |
2025
|
| Soggetti: | |
| Accesso online: | |
| Tags: |
Aggiungi Tag
Nessun Tag, puoi essere il primo ad aggiungerne!!
|
| _version_ | 1866908318486757376 |
|---|---|
| author | Heras, Iker de las Di Concilio, Federico Shumyatsky, Pavel |
| author_facet | Heras, Iker de las Di Concilio, Federico Shumyatsky, Pavel |
| contents | Given a group $G$ and elements $x_1,x_2,\dots, x_\ell\in G$, the commutator of the form $[x_1,x_2,\dots, x_\ell]$ is called a commutator of length $\ell$. The present paper deals with groups having only finitely many commutators of length $\ell$ of maximal order. We establish the following results.
Let $G$ be a residually finite group with finitely many commutators of length $\ell$ of maximal order. Then $G$ contains a subgroup $M$ of finite index such that $γ_\ell(M)=1$. Moreover, if $G$ is finitely generated, then $γ_\ell(G)$ is finite.
Let $\ell,m,n,r$ be positive integers and $G$ an $r$-generator group with at most $m$ commutators of length $\ell$ of maximal order $n$. Suppose that either $n$ is a prime power, or $n=p^αq^β$, where $p$ and $q$ are odd primes, or $G$ is nilpotent. Then $γ_\ell(G)$ is finite of $(m,\ell,r)$-bounded order and there is a subgroup $M\le G$ of $(m,\ell,r)$-bounded index such that $γ_\ell(M)=1$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2504_10377 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Groups with finitely many long commutators of maximal order Heras, Iker de las Di Concilio, Federico Shumyatsky, Pavel Group Theory Given a group $G$ and elements $x_1,x_2,\dots, x_\ell\in G$, the commutator of the form $[x_1,x_2,\dots, x_\ell]$ is called a commutator of length $\ell$. The present paper deals with groups having only finitely many commutators of length $\ell$ of maximal order. We establish the following results. Let $G$ be a residually finite group with finitely many commutators of length $\ell$ of maximal order. Then $G$ contains a subgroup $M$ of finite index such that $γ_\ell(M)=1$. Moreover, if $G$ is finitely generated, then $γ_\ell(G)$ is finite. Let $\ell,m,n,r$ be positive integers and $G$ an $r$-generator group with at most $m$ commutators of length $\ell$ of maximal order $n$. Suppose that either $n$ is a prime power, or $n=p^αq^β$, where $p$ and $q$ are odd primes, or $G$ is nilpotent. Then $γ_\ell(G)$ is finite of $(m,\ell,r)$-bounded order and there is a subgroup $M\le G$ of $(m,\ell,r)$-bounded index such that $γ_\ell(M)=1$. |
| title | Groups with finitely many long commutators of maximal order |
| topic | Group Theory |
| url | https://arxiv.org/abs/2504.10377 |