Questions tagged [profinite-groups]

For questions regarding profinite groups and their properties.

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Citations to published literature for the profiniteness of Inn(G) for G a profinite group

A recent question asked whether the inner automorphism group of a profinite group is profinite, and received an answer which was mathematically unimpeachable: yes because (1) the center is always a ...
D.N. Yetter's user avatar
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Action of profinite groups on primes

Let $G = \varprojlim G_i$ be a profinite group, $A$ a commutative ring with a $G$-action. Suppose furthermore that $A = \bigcup A^{U_i}$, where $U_i$ range over the open normal subgroups of $G$. Is it ...
Emory Sun's user avatar
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The semidirect product of profinite groups as a surjective inverse system of finite groups

Let $G$ be a second countable profinite group. Then $G$ can be written as a denumerable projective limit $\varprojlim_{i}(\cdots \to G_i\to G_{i-1}\to \cdots )$, where the $G_i$'s are finite and the ...
Nobody's user avatar
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Calculating the Hausdorff Dimension of specific profinite groups

I'm trying to understand how to calculate the Hausdorff dimension of some profinite groups and came across the following paper (here) which defines the Hausdorff dimension for a closed subgroup $H \...
Flo's user avatar
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Closed subgroup of automorphisms of a rooted tree

I am currently working with the automorphisms of a rooted tree $T$, and I have a question regarding $\text{Aut}(T)$ as a profinite group. Recall that we can make $\text{Aut}(T)$ into a profinite group ...
Alice in Wonderland's user avatar
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Smooth versus classical induction for an open subgroup of a profinite group

Let $G$ be a locally profinite group and $H \leq G$ an subgroup. If $(\sigma, W)$ is a smooth representation of $G$ then its classical induction is the pair $(\Sigma, \text{IND}_H^G(W))$ where $$\text{...
Tom Adams's user avatar
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Injective homomorphism between profinite groups

Given an inclusion $P\subset \Gamma$, it induces a homomorphism $\phi:\hat P\to \hat\Gamma$ on profinite completions. Then is that true that $\phi$ is injective iff given any normal subgroup of finite ...
Danny's user avatar
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Convergence in Profinite Exponentation

Unfortunately, I have some difficulties understanding the following proof from Profinite Groups (https://link.springer.com/book/10.1007/978-3-662-04097-3) on page 123. Let $G$ be a profinite group and ...
JackYo's user avatar
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Does profinite completion preserve injectivity?

Let $G$ be an abelian group. Let $\widehat{G}$ be a profinite completion of $G$. Profinite completion means a inverse limit of $G$ by a system given by homomorphisms $G/N\to G/M$ where $N$ and $M$ are ...
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local fields are locally profinite

Let $F$ be a local field with ring of integer $\mathfrak{o}$ and maximal ideal $\mathfrak{p}$. It is clear that $$\mathfrak{o}\supseteq \mathfrak{p}\supseteq \mathfrak{p}^2\supseteq \dots $$ is a ...
Mario's user avatar
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locally profinite and profinite group

Let $G$ be a locally profinite group (so a topological group such that every open neighbourhood of the identity in $G$ contains a compact open subgroup of $G$). It is well know that $$G \text{ is ...
Mario's user avatar
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Orbit of closed set under compact group action

Let $X$ be a Hausdorff totally disconnected space and $G$ be a compact (or even profinite) group acting continuously on $X$. Is it true that the orbit of any closed subset $Y$ under $G$ is closed in $...
Henrique Augusto Souza's user avatar
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Induced module is abelian

Let $G$ a profinite group and $H\subset G$ a normal open subgroup such that $G/H$ is abelian. Let $M$ be a discrete $H$-module. We say that the $H$-module is abelian if the image of the action map $\...
Mario's user avatar
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Infinite kernels of profinite integers when considered as endomorphisms of $\mathbb{Q}/\mathbb{Z}$

Considering the profinite integers as the ring of endomorphisms of $\mathbb{Q}/\mathbb{Z}$, I am looking for examples where the kernel of a profinite integer is an infinite subgroup of $\mathbb{Q}/\...
Orr Sela's user avatar
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Finding open subgroups of profinite groups from description as an inverse limit

For the profinite group $\widehat{\mathbb Z} := \varprojlim_n \mathbb Z/n\mathbb Z$ (which ofcourse actually has a ring structure as well), you can show its open subgroups are precisely the groups $n \...
Sverre's user avatar
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Lifting homomorphism in profinite group

Let $G$ be a profinite group and all its finite quotients be $G_i$. Let $\phi_i : G \longrightarrow G_i$ be standard map and $\phi_{i,j}: G_j \longrightarrow G_i$ be the standard surjective maps ...
Shri's user avatar
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Finite covers of $X$ and continuous profinite group actions

I have a question that is related to the one here, but I found the answer there unsatisfactory and kind of confusing. For context, we are considering the action of the fundamental group $\pi_1(X,x)$ ...
Vasting's user avatar
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5 votes
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A conjugacy class of an element in a profinite group is either finite or uncountably infinite?

Let $g$ be an element of a profinite group $G$. Is it true that the conjugacy class $g^G=\{hgh^{-1}~|~h\in G\}$ of $g$ in $G$ is finite or uncountably infinite? I know that profinite groups are always ...
Nobody's user avatar
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Maps from a finitely generated pro-p group to $\mathbb F_p$ factors through Frattini quotient

Let $G$ be a finitely generated pro-p group, these notes (p.99, Corollary 5.4.21) claims that all maps from $G$ to $\mathbb F_p$ factor through the Frattini quotient ($G/\Phi(G)$), where $\Phi(G)$ is ...
Kat's user avatar
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Uniform open subgroups of $p$-adic Lie group

A pro-$p$ group $G$ is called uniform if it is finitely generated, torsion-free with $G/\overline{G^p}$ abelian. Let $\Gamma$ be a $p$-adic Lie group. Then it can be shown that $\Gamma$ has a uniform ...
Yijun Yuan's user avatar
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Continuous homomorphisms from $\widehat{\mathbb Z}$ to $\overline{\mathbb F}_p^\times$

I would like to compute the continuous homomorphisms from $\widehat{\mathbb Z}$ to $\overline{\mathbb F}_p^\times$, where $\widehat{\mathbb Z} = \varprojlim_N \mathbb Z/N\mathbb Z$ has the profinite ...
gimothytowers's user avatar
2 votes
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A profinite group is the inverse limit of all quotients by open normal subgroups

My definition of a profinite group is it's the inverse limit of a system of finite groups, $G=\varprojlim_{i\in I} G_i$. But I can't see why $G=\varprojlim_{j\in J} G/N_j$, where $\{N_j\}_{j\in J}$ is ...
Kat's user avatar
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Normal closure of closed subgroup is closed in a f.g. profinite group?

Is there an example of a topologically finitely generated profinite group $G$ and a closed subgroup $H$ such that we simultaneously have: The profinite normal closure $L$ of $H$ is open in $G$. The ...
Henrique Augusto Souza's user avatar
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Closed ideals of ring of integral Adeles

If $\mathbb{A}_\mathbb{Z} = \mathbb{R} \times \hat{\mathbb{Z}}$, where $\hat{\mathbb{Z}} = \prod_{p \ prime} \mathbb{Z}_p$ denotes the profinite integers, is there anything that can be said about its ...
Pedro Lourenço's user avatar
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1 answer
164 views

Is it true that every residually finite group is a profinite group under some topology?

I know that a profinite group is residually finite. I am interested in the converse, which seems true. My reasoning is as follows. Please let me know if I am making some mistakes. Let $G$ be a ...
Hakuna Matata's user avatar
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55 views

Definition of profinite group

Consider the following definition: A profinite group is a topological group that is compact, Hausdorff, and admits a neighborhood basis of $1$ made of normal subgroups. My question is: it is ...
Luigi Traino's user avatar
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Artin's theorem in the infinite dimensional case

I am talking about the theorem from Galois theory, specifically to Milne's course notes on field theory proposition 7.10 (which can be easily found online). In the proof of this proposition, we have ...
DevVorb's user avatar
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Automorphism group of the functor $P:G\text{-FinSet} \to \text{Set}$

I am trying to understand the proof of the following theorem from Qiaochu Yuan's answer Theorem: Let $G$ be a group and let $P : G\text{-FinSet} \to \text{Set}$ be the forgetful functor from the ...
Kat's user avatar
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0 answers
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Decomposition as inverse limit

We know that if $G$ is a profinite group and $H$ a closed subgroup of $G$ then we can write $$H = \varprojlim H/N$$ where $N$ varies on the open normal subgroups of $H$. Now, if $K$ is a closed ...
Greg's user avatar
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Does profinite integer multiplication have integer "eigenvalues" in general?

Suppose $a \in \widehat{\Bbb{Z}}$, say $a = (\overline{a_1}, \overline{a_2}, \overline{a_3}, \overline{a_4}, \dots)$. I'm wondering if for any $b \in \widehat{\Bbb{Z}}$ whether there exists a number $...
Daniel Donnelly's user avatar
1 vote
1 answer
42 views

Is a profinite free product of pro-$p$ groups a pro-$p$ group?

I have a question. Maybe it be trivial, but I'm cannot conclude nothing yet. Suppose that we have a free profinite product $$G = G_1 \amalg \cdots \amalg G_n.$$ By definition in Ribes-Zalesskii $G$ is ...
Greg's user avatar
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0 answers
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Finiteness of $n$th Galois cohomology$H^n(U, \mathbb{F}_p)$ for open subgroups $U$ of a pro-$p$-group $G$

I am reading a book named 'cohomology of number fields' by Neukirch, Schmidt, Wingberg. Let $G$ be a pro-$p$-group. Suppose the $p$-cohomological dimension $cd_p G=n<\infty$. Suppose $H^n(G, \...
MiRi_NaE's user avatar
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8 votes
2 answers
351 views

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: ...
John's user avatar
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1 vote
1 answer
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Non-discrete LCAG's whose closed non-trivial subgroups are open

Let G be a locally compact abelian group and suppose that for any non-trivial closed subgroup H of G, H is open. It also makes sense to suppose that G is non-discrete (since in that case this ...
Pedro Lourenço's user avatar
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Limit of Natural Isomorphisms of Factors of a Diagram

Diagrams $F:\mathcal{C}\to\mathcal{D}$ are objects of the functor category $\mathcal{D}^\mathcal{C}$, when $\mathcal{C}$ is small. A particular case of a small category is the category associated to a ...
Alex Byard's user avatar
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0 answers
50 views

Convolution of Double Coset Indicator Function in Hecke Algebra of Locally Profinite Group

Let $G$ be a locally profinite group (i.e. a topological group that is locally compact Hausdorff and totally disconnected or, equivalently, a Hausdorff topological group s.t. $1 \in G$ has a ...
Tragomix's user avatar
2 votes
1 answer
119 views

Continuous action by profinite completion of a group [closed]

I'm working on a problem involving diagram dependence of a specific group action. It turns out that taking the profinite completion of the group will allow me to speak from a point of universality in ...
Alex Byard's user avatar
2 votes
1 answer
551 views

Subgroups of the $p$-adics that miss $\mathbb{Z}$

I am trying to understand the complementation of $\mathbb{Z}$ in the $p$-adics. The motivation for this question is that it will help me understand the situation in $\text{Tak}(\mathbb{Z}_p)$, where $\...
Alex Byard's user avatar
2 votes
1 answer
147 views

Is every closed subgroup of a profinite group contained in a maximal subgroup?

I recently asked this question about whether or not every subgroup of a profinite group is contained in a maximal subgroup. The answer is no, and a counterexample is given by $\mathbb{Z}<\mathbb{Z}...
Alex Byard's user avatar
3 votes
1 answer
59 views

Is every subgroup of a profinite group contained in a maximal subgroup?

I recently asked this question about whether or not profinite groups admit maximal subgroups: And indeed, profinite groups admit subgroups of finite index, so taking any minimum index subgroup ...
Alex Byard's user avatar
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0 answers
70 views

Proving object cannot be profinite without Stone topology

A profinite group $G$ arises as an inverse limit $G=\varprojlim G_\lambda$ of finite discrete groups $G_\lambda$. There is a nice property which states that a group $G$ is profinite if and only if it ...
Alex Byard's user avatar
3 votes
1 answer
125 views

Do profinite groups admit maximal subgroups

I've been looking into profinite groups, their topological subgroup lattices, etc. I asked the question does every profinite group admit maximal subgroups? I can't find an example of a profinite group ...
Alex Byard's user avatar
6 votes
1 answer
101 views

Basic closed subsets of Stone topological group presented as inverse limit, inner automorphism group of profinite group

Take a profinite group $G=\varprojlim G_\alpha$. We know that the inner automorphism group $\text{Inn}(G)$ of $G$ is profinite since $\text{Inn}(G)=G/Z(G)$, and the quotient of a profinite group by a ...
Alex Byard's user avatar
3 votes
1 answer
145 views

Inner automorphism group of a profinite group is profinite

I'm continuing to investigate my professor's statement regarding profinite groups and Stone topological groups. See this post for more information. A profinite group $G$ is an inverse limit of an ...
Alex Byard's user avatar
6 votes
1 answer
155 views

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 ...
Alex Byard's user avatar
4 votes
1 answer
112 views

Sufficient condition for a profinite group to be topologically finitely generated

Is it true that if $G = \varprojlim G_i$ is a profinite group such that every $G_i$ has a generator set $S_i$ whose cardinality is uniformly bounded for all $i$, then $G$ is (topologically) finitely ...
Santiago Radi's user avatar
2 votes
0 answers
50 views

The first Galois cohomology commutes with projective limits

I am reading Serre's paper "Sur les groupes de congruence des variétés abéliennes" (here is the link to this paper: https://www.mathnet.ru/links/016949238724700ec2209f00e507a40f/im3061.pdf). ...
Khainq's user avatar
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A field has a unique algebraic extension of each degree if and only if its absolute Galois group is the profinite completion of integers

I would like to know how to prove that a field $K$ which has a unique algebraic extension of each degree has its absolute Galois group isomorphic to the profinite completion of the integers $\hat{\...
subobject_classifier's user avatar
3 votes
0 answers
63 views

Profinite completion motivation

I have started galois theory recently, and curiosity quickly leads one to the subject of profinite groups. Although I have yet to be comfortable using these, I get what they are and we define them as ...
DevVorb's user avatar
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4 votes
1 answer
87 views

Is FC-center of a topologically finitely generated profinite group closed?

Let $ G $ be a group. A element $ g\in G $ is an FC-element if it has only finitely many conjugates in $ G $. The set $ \Delta(G) $ of FC-elements of $ G $ is a characteristic subgroup of $ G $, and ...
stupid boy's user avatar

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