Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [profinite-groups]

For questions regarding profinite groups and their properties.

Profinite groups are topological groups that can be "assembled" from finite groups. Precisely, a profinite group is a Hausdorff, compact, and totally disconnected topological group.

If given the discrete topology, every finite group is profinite, and the Galois theory of infinite degree field extensions gives rise to profinite Galois groups. Products and closed subgroups of profinite groups are also profinite.

Reference: Profinite group.

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Profinite topology of a Group

Let $G$ be a group. Consider now the set of all (left for instance) cosets in $G$ of subgroups of finite index. This set is a base for a topology in $G$. I found somewhere that if $G$ is residually finite then $G$ is a compact space in this topology…
W4cc0
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Cogroup structures on the profinite completion of the integers

Let $\mathsf{ProFinGrp}$ be the category of profinite groups (with continuous homomorphisms). This is equivalent to the Pro-category of $\mathsf{FinGrp}$. Notice that $\widehat{\mathbb{Z}} = \lim_{n>0} \mathbb{Z}/n\mathbb{Z}$ is a cogroup object,…
Martin Brandenburg
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What is the automorphism group of the field of all constructible numbers?

Let $\Omega\subseteq \mathbb{C}$ be the field of all constructible numbers (i.e. $\Omega$ is the smallest subfield of $\mathbb{C}$ which is closed under taking square roots). What is known about the automorphismgroup of the field $\Omega$?
user 59363
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Is $\Bbb Z_p^2$ a Galois group over $\Bbb Q$?

$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\ab}{\mathrm{ab}} \newcommand{\Gal}{\mathrm{Gal}} \newcommand{\prolim}{\varprojlim} $ Fix a…
Watson
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Continuous function space on a profinite group as a direct limit

I would greatly appreciate any help with the following problem. If there are existing references related to this, kindly provide them. If not, any help in this matter would be highly valued. Problem: Let the profinite group $G$ be the inverse limit…
John
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Fast algorithm for computing $\sum_{m=1}^{n} (n \bmod m)/m!$

Let $$S_n = \sum_{m=1}^{n} \frac{n \bmod m}{m!}$$ be the $n$-th partial sum of a series with remainder $$R_{n}=\sum_{m=n+1}^{\infty}\frac{n \bmod m}{m!}=n\cdot e - \frac{\lfloor e\cdot n\rfloor}{(n-1)!}.$$ This series occurs when embedding the…
Jackson Walters
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Restriction of an irreducible representation to a normal subgroup with cyclic quotient

Let $G$ be a finite group, $H$ a normal subgroup of $G$ such that $G/H$ is cyclic. Let $V$ be an irreducible representation over an algebraically closed field $k$, with $char(k)=0$. Is it true that the restriction of $V$ to $H$ is a direct sum of…
Emiliano Ambrosi
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Does Abelianization Commute with Profinite Completion?

Let $G$ be a group, let $\widehat{G}$ be its profinite completion, and let $G^{\text{ab}}$ be its abelianization. Is is true that abelianization commutes with profinite completion, in the sense that $(\widehat{G})^{\text{ab}} =…
Ashvin Swaminathan
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Field $K$ with $\operatorname{Gal}(\overline{K}/K)\simeq\widehat{F_2}$

Is there a field K such that $\operatorname{Gal}(\overline{K}/K)$ is the profinite free group with two generators? For one generator I know that for all the $\mathbb{F}_p$ we have…
Gabriel Soranzo
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Haar measure on a profinite group is the inverse limit of the counting measure on its quotients?

I've heard this a few times now, though I've never seen a precise result. I guess the precise statement would be close to: Let $N_i$ be a basis normal subgroup neighborhoods of the identity in a profinite group $G$, $\pi_i : G \to G / N_i = G_i$ the…
Elle Najt
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Do the maps need to be surjective in the definition of a profinite group?

Let $G_i, f_{ij}$ be an inverse system of topological groups where each $G_i$ is finite in the discrete topology. A profinite group is defined to be an inverse limit of such an inverse system. However, my professor seemed to assume that the…
Hihihihi
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Nontrivial examples of pro-$p$ groups

I only know a few examples of pro-$p$ groups. Of course the $p$-adics $\mathbf{Z}_p$, and any finite $p$-group. Congruence subgroups of $\text{GL}_n(\mathbf{Z}_p)$: e.g. $\Gamma_1:=\{g\in \text{SL}_n(\mathbf{Z}_p) : g\equiv \text{id} \, (\text{mod…
Ehsaan
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Profinite completion is complete.

Let $G$ be any group, and $\widehat{G}$ its profinite completion. Is it true that $\widehat{\widehat{G}}=\widehat{G}$, i.e. is it true that $\widehat{G}$ is (canonically isomorphic to) its own profinite completion? It seems that it should follow…
Servaes
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A group is profinite if and only if it is a Stone space

A profinite group is an inverse limit of an inverse system of discrete finite groups. Alternatively, a profinite group is a topological group that is also a Stone space. Under the second, axiomatic definition it's obvious that a profinite group $G$…
Alex Byard
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Galois group of $\mathbb{Q}(\underset{n\geq 1}{\bigcup}\mu_n)/\mathbb{Q}$

I was trying to formalise the fact that $G_{\mu_\infty}:=Gal(\mathbb{Q}(\underset{n\geq 1}{\bigcup}\mu_n)/\mathbb{Q})\simeq \underset{n}{\varprojlim}\ (\mathbb{Z}/n\mathbb{Z})^{\times}=\hat{\mathbb{Z}}^{\times}$. When I was done I tried finding…
Jason V
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