Papers
Topics
Authors
Recent
Detailed Answer
Quick Answer
Concise responses based on abstracts only
Detailed Answer
Well-researched responses based on abstracts and relevant paper content.
Custom Instructions Pro
Preferences or requirements that you'd like Emergent Mind to consider when generating responses
Gemini 2.5 Flash
Gemini 2.5 Flash 62 tok/s
Gemini 2.5 Pro 48 tok/s Pro
GPT-5 Medium 14 tok/s Pro
GPT-5 High 13 tok/s Pro
GPT-4o 93 tok/s Pro
Kimi K2 213 tok/s Pro
GPT OSS 120B 458 tok/s Pro
Claude Sonnet 4 38 tok/s Pro
2000 character limit reached

The Congruence Subgroup Problem for low rank Free and Free Metabelian groups (1608.04151v2)

Published 14 Aug 2016 in math.GR

Abstract: The congruence subgroup problem for a finitely generated group $\Gamma$ asks whether $\widehat{Aut\left(\Gamma\right)}\to Aut(\hat{\Gamma})$ is injective, or more generally, what is its kernel $C\left(\Gamma\right)$? Here $\hat{X}$ denotes the profinite completion of $X$. In this paper we first give two new short proofs of two known results (for $\Gamma=F_{2}$ and $\Phi_{2}$) and a new result for $\Gamma=\Phi_{3}$: 1. $C\left(F_{2}\right)=\left{ e\right}$ when $F_{2}$ is the free group on two generators. 2. $C\left(\Phi_{2}\right)=\hat{F}{\omega}$ when $\Phi{n}$ is the free metabelian group on $n$ generators, and $\hat{F}{\omega}$ is the free profinite group on $\aleph{0}$ generators. 3. $C\left(\Phi_{3}\right)$ contains $\hat{F}{\omega}$. Results 2. and 3. should be contrasted with an upcoming result of the first author showing that $C\left(\Phi{n}\right)$ is abelian for $n\geq4$.

Summary

We haven't generated a summary for this paper yet.

List To Do Tasks Checklist Streamline Icon: https://streamlinehq.com

Collections

Sign up for free to add this paper to one or more collections.

Lightbulb On Streamline Icon: https://streamlinehq.com

Continue Learning

We haven't generated follow-up questions for this paper yet.