Logo
ScienceStack
Table of Contents
Fetching ...
Sign In

Groups of profinite type and profinite rigidity

Tamar Bar-On, Nikolay Nikolov

Abstract

We say that a group $G$ is of \textit{profinite type} if it can be realized as a Galois group of some field extension. Using Krull's theory, this is equivalent to the ability of $G$ to be equipped with a profinite topology. We also say that a group of profinite type is \textit{profinitely rigid} if it admits a unique profinite topology. In this paper we study when abelian groups and some group extensions are of profinite type or profinitely rigid. We also discuss the connection between the properties of profinite type and profinite rigidity to the injectivity and surjectivity of the cohomology comparison maps, which were studied by Sury and other authors.

Groups of profinite type and profinite rigidity

Abstract

We say that a group is of \textit{profinite type} if it can be realized as a Galois group of some field extension. Using Krull's theory, this is equivalent to the ability of to be equipped with a profinite topology. We also say that a group of profinite type is \textit{profinitely rigid} if it admits a unique profinite topology. In this paper we study when abelian groups and some group extensions are of profinite type or profinitely rigid. We also discuss the connection between the properties of profinite type and profinite rigidity to the injectivity and surjectivity of the cohomology comparison maps, which were studied by Sury and other authors.
Paper Structure (4 sections, 19 theorems, 5 equations)

This paper contains 4 sections, 19 theorems, 5 equations.

Table of Contents

  1. Introduction
  2. Abelian groups of profinite type
  3. Profinite rigidity
  4. Connections with cohomological goodness

Key Result

Proposition 1

Let $G$ be an abelian group of exponent $n$. Let $n=\prod_{i} p_i^{t_i}$ be a factorization of $n$ as a product of prime powers. By kaplansky2018infinite$G$ is isomorphic to for some cardinals $\textbf{m}_{j_i}$. $G$ is of profinite type if and only if for every $j_i$, either $\mathbf{m}_{j_i}$ is finite, or there exists some cardinal $\mathbf{n}_{j_i}$ such that $2^{\mathbf{n}_{j_i}}=\mathbf{m}_

Theorems & Definitions (42)

  • Proposition 1
  • Remark 2
  • proof
  • Lemma 3
  • proof
  • Proposition 4
  • proof
  • Example 5
  • Corollary 6
  • Proposition 7
  • ...and 32 more