Zobrazeno 1 - 6
of 6
pro vyhledávání: '"20E18, 20E06"'
Publikováno v:
International Mathematics Research Notices, Volume 2023, Issue 24, December 2023, Pages 21320--21345
Can one detect free products of groups via their profinite completions? We answer positively among virtually free groups. More precisely, we prove that a subgroup of a finitely generated virtually free group $G$ is a free factor if and only if its cl
Externí odkaz:
http://arxiv.org/abs/2207.00912
Let G be a finitely generated infinite pro-p group acting on a pro-p tree such that the restriction of the action to some open subgroup is free. Then we prove that G splits as a pro-p amalgamated product or as a pro-p HNN-extension over an edge stabi
Externí odkaz:
http://arxiv.org/abs/1103.2955
Autor:
Lorensen, Karl
Assume $G$ is a polycyclic group and $\phi:G\to G$ an endomorphism. Let $G\ast_{\phi}$ be the ascending HNN extension of $G$ with respect to $\phi$; that is, $G\ast_{\phi}$ is given by the presentation $$G\ast_{\phi}= < G, t \ |\ t^{-1}gt = \phi(g)\
Externí odkaz:
http://arxiv.org/abs/1009.2645
Autor:
Lorensen, Karl
Publikováno v:
J. Pure Appl. Algebra 214 (2010), 6-14
For any prime $p$ and group $G$, denote the pro-$p$ completion of $G$ by $\hat{G}^p$. Let $\mathcal{C}$ be the class of all groups $G$ such that, for each natural number $n$ and prime number $p$, $H^n(\hat{G^p},\mathbb Z/p)\cong H^n(G, \mathbb Z/p)$,
Externí odkaz:
http://arxiv.org/abs/0809.3046
Can one detect free products of groups via their profinite completions? We answer positively among virtually free groups. More precisely, we prove that a subgroup of a finitely generated virtually free group $G$ is a free factor if and only if its cl
Externí odkaz:
https://explore.openaire.eu/search/publication?articleId=doi_dedup___::d30967dbbfe5e21dc546bb4a5df09cb9
Autor:
Karl Lorensen
Publikováno v:
Journal of Pure and Applied Algebra. (1):6-14
For any prime $p$ and group $G$, denote the pro-$p$ completion of $G$ by $\hat{G}^p$. Let $\mathcal{C}$ be the class of all groups $G$ such that, for each natural number $n$ and prime number $p$, $H^n(\hat{G^p},\mathbb Z/p)\cong H^n(G, \mathbb Z/p)$,
Externí odkaz:
https://explore.openaire.eu/search/publication?articleId=doi_dedup___::1a6aa963ebf7309e029735137c2d17f2
