Questions tagged [profinite-groups]

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An infinite profinite group such that any $p$-adic representation has finite image

Fix a prime $ p $. We call an infinite profinite group $ G $ a Fontaine-Mazur group (with respect to $ p $) if every continuous homomorphism $ G\to {\rm GL}_n(\overline{\mathbb{Q}}_p) $ has finite ...
Nobody's user avatar
  • 817
3 votes
0 answers
285 views

Non-abelian group cohomology, additional information

Let $G$ be a (profinite) group, and let $M$ be a non-abelian $G$-module. We know how to construct reasonably $H^0(G,M)$ and $H^1(G,M)$ and it turns out that $H^1(G,M)$ is just a pointed set and not ...
Or Shahar's user avatar
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5 votes
0 answers
147 views

Can there be non-isomorphic fundamental groups of equivalent Galois categories?

It is known that if $(C, F)$ is a Galois category then there exists an equivalence $C \cong \pi_1(C, F)-FinSets$ between $C$ and the category of finite sets with continuous actions of $\pi_1(C, F) := ...
Dat Minh Ha's user avatar
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4 votes
0 answers
136 views

Proving that the fundamental group of a finite Galois category is profinite

This is sort of a cross-posting of this question of mine over on math.SE. Suppose that I am given a finite Galois category $(\mathcal{G}, F)$, i.e. a Galois category in the sense of SGA 1. One of the ...
Dat Minh Ha's user avatar
  • 1,472
2 votes
1 answer
349 views

Existence of regular semisimple elements in linear group over local field

Let $ L $ be a finite extension of $p$-adic numbers $ \mathbb{Q}_p $. Let $ \text{GL}_{n}(L) $ denote the general linear group $ \text{GL}_{n}(L) $ over $L$ equipped with the topology induced from the ...
Nobody's user avatar
  • 817
2 votes
0 answers
149 views

Artin map and profinite completion of the idèles

One way to formulate local class field theory is by saying that the local Artin map induces an isomorphism from the profinite completion of $K^\times$ to $\operatorname{Gal}(K^\text{ab}/K)$, which ...
Antoine Labelle's user avatar
1 vote
0 answers
128 views

Direct limit of groups rings of finite quotients of a profinite group

Background: Let $G$ be a profinite group, for $M \leqslant N$ open normal subgroups we have the projection map $p_{M,N} \colon G/M \to G/N$ which induces a transfer map on rational group rings $$ p_{M,...
David Barnes's user avatar
2 votes
1 answer
338 views

Difference between definitions of continuous action, profinite case

My setting is the following : let $G$ be a topological group and $X$ be a topological space. I have the head filled with two possible definitions for a continuous action of $G$ on $X$. The first could ...
Nataniel Marquis's user avatar
0 votes
1 answer
116 views

Examples of non-proper profinite HNN extensions

We define a profinite HNN extension as the profinite completion of the abstract HNN extension. In the abstract case, the homomorphim of the base group to the HNN extension is always a monomorphism. ...
Mattheus Pereira's user avatar
1 vote
0 answers
190 views

Fields such that every finite Galois extension is solvable

What are the fields such that every finite Galois extension is solvable? We have algebraically closed fields, real closed fields, p-adic fields. Anything else? A more pointed question after comments: ...
Buckeye's user avatar
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1 vote
1 answer
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Free partners in semi-direct products

Let $G = N \rtimes K$ be a semi-direct product of groups and suppose that $K$ is a finite group. Call the set $\mathcal{F} = \{ \alpha \in G \mid \langle \alpha, K \rangle = \langle \alpha \rangle \...
Doryan Temmerman's user avatar
5 votes
0 answers
218 views

When is the profinite completion of a Noetherian group ring also Noetherian?

Let $G$ be a group, and let $\mathbb{Z}[G]$ denote its group ring. Its profinite completion is the inverse limit over all ideals of finite index. By Benjamin Steinberg's answer here, this profinite ...
stupid_question_bot's user avatar
4 votes
2 answers
329 views

Topology on the hom space between profinite groups

$\DeclareMathOperator\Hom{Hom}$Let $G,H$ be profinite groups. Let $\Hom(G,H)$ be the set of continuous group homomorphisms, equipped with the compact-open topology. I'd like to understand the ...
stupid_question_bot's user avatar
6 votes
1 answer
456 views

Irreducible representations of product of profinite groups

It is a standard fact in the representation theory of finite groups that for $G,H$ finite groups, all of the irreducible representations of $G \times H$ are the external tensor product of irreps of $G$...
Martin Skilleter's user avatar
4 votes
1 answer
262 views

Confusion with self-dual representations of $\mathrm{GL}_n$ over a $p$-adic field

The following surely is kind of a trivial question, but it keeps me confused. It concerns a detail in Lust and Stevens' paper "On depth zero L-packets for classical groups" London Math. Soc. ...
Suzet's user avatar
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Projective limit of copies of same group w.r.t. some fixed endomorphism

In our study of automorphism groups of transcendental field extensions, we have encountered the situation where we have a group $F$ together with an endomorphism $\alpha \colon F \to F$, resulting in ...
Tom De Medts's user avatar
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2 votes
0 answers
91 views

Normal subgroups of prosupersolvable groups

Let $G$ be a finite supersolvable group, and if $p$ is the biggest prime dividing $\vert G \vert$. Then $G$ has normal subgroups of order every possible power of $p$. Analogous statement in case of ...
Shri's user avatar
  • 273
2 votes
0 answers
134 views

Profinite projective and free modules

I am studying cohomology of profinite groups and the following question came to my mind: suppose we have $G$ a pro-$p$ group which is Poincaré Dual of dimension $d$. This means that $\mathbb{Z}_p$ as ...
N.B.'s user avatar
  • 757
6 votes
1 answer
240 views

Top cohomology of profinite Poincaré duality group

The paper "Cohomology of p-adic analytic groups" by Symonds and Weigel is considered one of the main references for continuous cohomology of profinite groups. There is a passage I do not ...
N.B.'s user avatar
  • 757
3 votes
1 answer
211 views

Absolute Galois group with unique closed non-open subgroup

Is there an absolute Galois group that is not a subgroup of $\hat{\mathbb{Z}}$ and that has one and only one closed non-open subgroup?
klmntn's user avatar
  • 53
8 votes
3 answers
705 views

Definition of a profinite category

When studying objects like profinite groups, profinite spaces and profinite rings, I have noticed that some properties just remain the same. For example they will always be inductive limits of some ...
Keen's user avatar
  • 201
1 vote
0 answers
70 views

Pro-p completion of a quotient of $U/w(U)$ is virtually nilpotent for a finitely generated free group $U$

Let $w$ be a word of a free group. Assume that $H/\overline{w(H)}$ is virtually nilpotent for every finitely generated pro-$p$ group $H$. Let $U$ be a finitely generated free group and $T$ the maximal ...
Lucas's user avatar
  • 289
3 votes
0 answers
204 views

Finitely generated subgroups of the absolute Galois group

Consider the absolute Galois group $\operatorname{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$. It seems to me that, in general, providing an "explicit" description of the elements of this ...
Carl-Fredrik Nyberg Brodda's user avatar
4 votes
0 answers
102 views

Bound of word width in compact $p$-adic analytic group

A theorem proved by A. Jaikin-Zapirain in On the verbal width of finitely generated pro-$p$ groups says that: If $G$ is a compact $p$-adic analytic group, then every word $w$ of a free group $F$ has ...
Lucas's user avatar
  • 289
2 votes
1 answer
229 views

Fast algorithm for computing $\sum_m (n \mod m)/m!$

I'm interested in quickly computing an embedding of the profinite integers $\widehat{\mathbb{Z}}$ into the unit interval $\left[0,1\right]$. $\widehat{\mathbb{Z}}$ can be represented as compatible ...
Jackson Walters's user avatar
4 votes
0 answers
189 views

Rational cohomology cohomology of $p$-adic analytic groups

It is a result of Lazard that given $G$ a compact $p$-adic analytic group then we have an isomorphism \begin{equation} H^*(G; \mathbb{Q}_p) \cong H^*(T_eG; \mathbb{Q}_p) \end{equation} where $T_eG$ is ...
N.B.'s user avatar
  • 757
4 votes
2 answers
246 views

Why is every nilpotent-by-finite finitely generated pro-p-group always $p$-adic analytic

I'm studying the paper "On the verbal width of finitely generated pro-p groups" by Andrei Jaikin-Zapirain (link at ProjectEuclid) and I cannot see a claim made in a proof. I don't know if ...
Lucas's user avatar
  • 289
0 votes
0 answers
71 views

Mapping property of $p$-Sylow groups of profinite groups

Let $G$ be an abelian profinite groups. Then we have the Sylow group decomposition $$G\cong \prod_p G_p.$$ In the case of finite groups, we have $ \prod_p G_p\cong \bigoplus_p G_p$ and thus $$\text{...
curious math guy's user avatar
2 votes
0 answers
249 views

kernel and cokernel of corestriction map in cohomology of a profinite group

Let $G$ be a profinite group, $N$ a normal open subgroup and $A$ a discrete $G$-module. We have a corestriction map $cor: H^1(N, A)_{G/N} \to H^1(G, A)$. Are there any results on the kernel and ...
Ahmed Matar's user avatar
2 votes
0 answers
76 views

On exactness of associating smooth representation-functor $(\,)^\infty$

Let $G$ be a locally profinite group, e.g. reductive group over $\mathbb{Q}_p$. For a (abstract) representation $(\pi,V)$ of $G$ and $K\subset G$ compact open subgroup denote by $V^K\subset V$ the $\...
KKD's user avatar
  • 463
4 votes
0 answers
167 views

Countability of conjugacy classes in profinite groups

In the MOF question [1] it was asked if $G$ is a second-countable profinite group with uncountably many subgroups, does it follow that it has uncountably many closed subgroups modulo conjugacy? A ...
Scott Balchin's user avatar
3 votes
1 answer
383 views

Spaces of closed subgroups of a profinite group up to conjugacy

$\DeclareMathOperator{\Sub}{\operatorname{Sub}}$ Let $G$ be a profinite group and consider the space $\Sub(G)$ of closed subgroups of $G$ equipped with the profinite topology. That is, we have $G = \...
Scott Balchin's user avatar
4 votes
0 answers
102 views

What do the eigenvalues of a random element of $\mathbb Z_\ell[\Gamma]$ look like?

Let $\Gamma = \varprojlim \Gamma_n$ be a profinite group with $\Gamma_n$ finite quotients. For concreteness, let us fix $\Gamma_n = \operatorname{PGL}_2(\mathbb Z/\ell^n)$ so $\Gamma = \operatorname{...
Asvin's user avatar
  • 7,648
3 votes
0 answers
97 views

Colimits in cohomology of profinite arithmetic groups

Let $G\subset \operatorname{GL}_n$ be a linear algebraic group over $\mathbb{Q}$ and let $\Gamma\subset G\cap \operatorname{GL}_n(\mathbb{Z})$ be an arithmeric subgroup without torsion. Using a result ...
Jrodri26's user avatar
  • 123
5 votes
0 answers
156 views

Rational cohomology of p-adic general linear groups

I wanted to compute the cohomology ring $H^*(GL_n(\mathbb{Z}_p); \mathbb{Q}_p)$ (with $p$ fixed prime as usual). I found some incomplete notes stating that the computation should go as follows. First ...
N.B.'s user avatar
  • 757
1 vote
1 answer
90 views

Infinite pro-$p$ group of finite solvable length and finite coclass

I was reading about infinite pro-$p$ groups of finite coclass from the book "The Structure of Groups of Prime Power Order" by Leedham-Green and McKay. I asked this question in math....
usermath's user avatar
  • 243
4 votes
0 answers
276 views

Galois representation with infinite image but finite image everywhere locally

Fix a prime $l$. Let $\phi:\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_n(\mathbb{Q}_l)$ be a semisimple continuous representation. Assume $\phi$ has finite image when restricted to $\mathrm{...
user avatar
5 votes
1 answer
513 views

Sylow subgroups of abelian profinite groups

If $G$ is a finite abelian group, then we have a decomposition $$G\cong \prod_{p} G(p)$$ where $G(p)$ is the $p$-Sylow subgroup of $G$. This product makes sense as for all but finitely many primes $p$,...
curious math guy's user avatar
2 votes
0 answers
159 views

Outer Galois representations and Tate modules of Jacobian varieties

Let $X$ be a proper smooth curve over a field $k$. Then we have an exact sequence of profinite groups \begin{equation*} 1 \to \pi_1(X_{\overline k}) \to \pi_1(X) \to G_k \to 1, \end{equation*} ...
Aoi Koshigaya's user avatar
8 votes
0 answers
216 views

Finitely generated commutative rings with the same profinite completion

Let $R_1$ and $R_2$ be two finitely generated commutative rings. Assume that their profinite completions are isomorphic: $\widehat{R_1}\cong \widehat{R_2}$. Suppose that $R_1$ is a domain. Does ...
Andrei Jaikin's user avatar
3 votes
0 answers
107 views

Regular growth of ranks in Iwasawa tower

$\newcommand{\rank}{\operatorname{rank}}$Let $G=H \times K$ be a torsion free pro-$p$, $p$-adic Lie group. Let $H =\mathbb{Z}_p$, the ring of $p$-adic integers and $K$ is a non-commutative torsion ...
user100603's user avatar
8 votes
1 answer
466 views

Commutator subgroup of the absolute Galois group - a closed subgroup

Let $K$ be a finite extension of $\mathbb{Q}$. Is it possible that the commutator subgroup of the absolute Galois group of $K$ (considered as an abstract group) is a closed subgroup? This property ...
user avatar
1 vote
0 answers
138 views

How a profinite group can be obtained from its normal open subgroups?

Let $\Delta$ be a set, each element of which is a profinite group (2 distinct elements of $\Delta$ may be isomorphic). Under what conditions on $\Delta$, there exists a profinite group $G$ which has $\...
Meisam Soleimani Malekan's user avatar
1 vote
1 answer
307 views

Is there a free profinite abelian group on a profinite set?

Let $\mathit{Profinite}_{\mathrm{Ab}}$ be the category of profinite abelian groups, and let $\mathit{Profinite}_{\mathrm{Set}}$ be the category of profinite sets. Does the forgetful functor $$\mathit{...
Profinite Questioner's user avatar
4 votes
1 answer
330 views

Measure of subsets of profinite groups

Let $G$ be an infinite profinite group, so $$G=\lim_{\longleftarrow}G/N$$ where $N$ runs through the open normal subgroups. I have two questions: Is $G$ of Haar measure zero in the compact group $\...
Meisam Soleimani Malekan's user avatar
3 votes
1 answer
814 views

Continuous function defined by measurable sets

Is the following slightly generalization of Corollary 20.17 in Hewitt and Ross Book (page 296) correct? Let $A$ be a subset of a profinite group $G$ ( compact, Hausdorff, totally disconnected ...
Alireza Abdollahi's user avatar
5 votes
2 answers
526 views

Mackey theory in the setting of locally profinite groups

$\DeclareMathOperator\Hom{Hom}$Let $R$ be a commutative ring (not necessarily unital). Let $G$ be a finite group, and let $H_1, H_2$ be subgroups of $G$. Recall the following standard result [1, Thm. ...
Pedro A. Matos's user avatar
12 votes
1 answer
778 views

Does a (nice) centerless group always have a centerless profinite completion?

This is an extension of a question I asked here on Math.SE Assume that I have a finitely generated residually finite centerless group $G$. Is it true that the profinite completion $\hat{G}$ also has ...
Santana Afton's user avatar
1 vote
2 answers
281 views

Meaning of epimorphism from full Galois group to some group

My problem has two parts: let $\;G:=\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)\;$ be the full Galois group of the rationals and $\;K\;$ be some finite group, then: (1) Does having an epimorphism (...
Ninet's user avatar
  • 11
5 votes
0 answers
212 views

Applications of one of Serre's Theorems

This theorem is due to Serre: Let $G$ be a profinite group, $p$ prime. Assume that $G$ has no element of order $p$ and let $H \leq G$ be an open subgroup. Then $cd_p(G) = cd_p(H)$. Where $cd_p(...
Mariah's user avatar
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