22
votes
Accepted
Does $GL_2(\widehat{\mathbb{Z}})$ contain a dense finitely generated subgroup?
The answer is no. If $GL_2(\widehat{\mathbb Z})$ is topologically finitely generated, then so is the quotient $\widehat {\mathbb Z}^*$ (quotient via the determinant map). The latter has quotient $\...
- 11.1k
20
votes
Accepted
Does a (nice) centerless group always have a centerless profinite completion?
The answer is No in general. Let $n\geq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=\mathrm{SL}_n({\mathbb Z})$. There exists a subgroup $\Gamma \subset \mathrm{SL}_n({\mathbb Z})$ ...
- 11.1k
17
votes
Accepted
Profinite completion of finitely presented groups
Yes. Take the Baumslag-Solitar group
$$G=\mathrm{BS}(2,3)=\langle t,x\mid tx^2t^{-1}=x^3\rangle$$
Then $G$ is finitely presented; the image of $G$ in its profinite completion (i.e., the largest ...
- 60.4k
15
votes
Accepted
The Unit Group of $\mathbb{Z}_p$
This is probably more suitable to MathStackExchange, although there may well be researchers in allied areas that aren't aware of this fact. So in that spirit, here's the standard proof. There is an ...
- 45.8k
13
votes
Is $SL_n(\mathbb{Z}_p)$ virtually torsion free?
$SL_n(\mathbb{Z}_p)$ is virtually torsion free as it is $p$-adic analytic and therefore contains a uniformly powerful open subgroup.
- 5,200
13
votes
Is $\widehat{\mathbb{Z}}[[t]]\cong\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$?
It seems to me that for $p$ prime $\hat{\mathbb Z}[[t]]$ has exactly one continuous ring homomorphism to $\mathbb Z/p\mathbb Z$, while $\hat{\mathbb Z}[[\hat{\mathbb Z}]]$ has exactly $p-1$ such ...
- 54.4k
13
votes
Without choice, can every homomorphism from a profinite group to a finite group be continuous?
There are two common definitions of what it means for a topological group to be profinite, which I’ll distinguish as “formally profinite” and “compact profinite”.
Formally profinite: an inverse limit ...
- 1,906
11
votes
Accepted
Example of reflective subcategory of (Groups) whose reflector doesn't preserve finite products
The answer is no: every such full subcategory is stable under direct products, or equivalently $L$ preserves direct products.
Here all Hom are in the category of groups. The definition of $L$ means ...
- 60.4k
11
votes
Accepted
Are there open subgroups of $SL_2(\widehat{\mathbb{Z}})$ which are $GL_2(\widehat{\mathbb{Z}})$-conjugate, but not $SL_2$-conjugate?
Yes, there are indeed such subgroups. Since the open subgroups are exactly the congruence subgroups, it suffices to find two subgroups of $\mathrm{SL}_2(\mathbf{Z}/N\mathbf{Z})$ which are conjugate in ...
- 20.5k
11
votes
Accepted
Classification of natural endomorphisms on finite groups
It suffices to look at symmetric groups, as these are generated by transpositions. Indeed, let $z \in \widehat{\mathbf Z}$ and suppose it induces a homomorphism $\phi_z$ on $S_n$ for all $n$ given by $...
- 27.3k
10
votes
Is there a residually finite non-elementary hyperbolic group whose profinite completion is boundedly generated?
I'm fairly certain that no example is known. Of course, it's a famous open problem whether every hyperbolic group is residually finite. This turns out to be equivalent to many other questions about ...
- 24.1k
10
votes
Accepted
Sylow subgroups of abelian profinite groups
This is Proposition 2.3.8 of Ribes and Zaleskii - Profinite groups (second edition). (I originally gave references specifically for the finer structure of profinite Abelian groups, but assuming ...
- 11.4k
8
votes
Accepted
Difference between the completed group algebra and the profinite completion of a group ring
It seems to me that these are indeed isomorphic. Namely, if $n$ is an integer and $U$ is a finite index normal subgroup of $G$, then the kernel $I_{n,U}$ of the natural map $\mathbb ZG\to \mathbb Z/n[...
- 37.4k
8
votes
Accepted
What are the LCA groups that are the Pontryagin dual of a locally profinite abelian group?
Totally disconnected LCA groups are (profinite)-by-(discrete) LCA groups. Hence their Pontryagin dual are (compact)-by-(discrete torsion) groups. These are precisely locally elliptic LCA groups.
(I'm ...
- 60.4k
8
votes
Accepted
Commutator subgroup of the absolute Galois group - a closed subgroup
No, the abstract commutator subgroup $[G_K,G_K]$ of the absolute Galois group $G_K$ of a number field $K$ is never closed:
Write $[G,G]$ for the commutator subgroup of $G$ as an abstract group,
and $c(...
- 1,989
8
votes
Definition of a profinite category
There is a notion of pro-object in a general category $\mathbf{C}$, which generalises the usual profinite objects - profinite spaces are the case $\mathbf{C} = \mathbf{FinSet}$, profinite groups are $\...
8
votes
Accepted
Topological generators for $\mathrm{SL}_2(\mathbf{Z}_p)$
$\def\ZZ{\mathbb{Z}}\def\SL{\text{SL}}\def\Id{\text{Id}}$This seems false to me.
Lemma: $e:=\left[ \begin{smallmatrix} 1&p\\0&1 \\ \end{smallmatrix} \right]$ and $f:=\left[ \begin{smallmatrix} ...
- 151k
7
votes
Accepted
Finite Homomorphic images of infinite products of finite solvable groups
That's correct: every finite quotient is solvable. Indeed let $G$ be your product of finite solvable groups. Let $p:G\to F$ be a (possibly non-continuous) surjective homomorphism to a finite group $F$....
- 60.4k
7
votes
Does continuous action by profinite groups mean finite orbits?
To the question in the title: Not in general, e.g the action of (an infinite profinite group) $G$ on itself by left multiplication is continuous with infinite orbits.
However, note that the action of ...
- 11.4k
7
votes
Is the absolute Galois group the same as the automorphism group?
This has already been answered in the comments, but perhaps you can see it more clearly like this.
Take the isomorphism $\mathrm{Gal}(\bar{\mathbb{Q}}|\mathbb{Q}) \cong \varprojlim(K|\mathbb{Q})$ as ...
- 17.5k
7
votes
Every group of totally disconnected type is locally profinite?
"Is there an example of a group of td-type which is not locally profinite?"
No. This was proved by D. van Dantzig in the 1930s:
Van Dantzig, D.: Zur topologischen Algebra. III. Brouwersche und ...
- 4,678
7
votes
Accepted
Linear representation of the free metabelian / 2-step nilpotent profinite groups on 2 generators
The group $B$, the free pro-metabelian group, has the following description, due to Jorge Almeida. I’ll do it for an arbitrary finite set $|X|$ of cadinality at least $2$. Consider $\widehat{\mathbb ...
- 37.4k
7
votes
Definition of a profinite category
There are two natural definitions of a profinite category. You can look at inverse limits of finite categories or you can look at topological categories whose underlying spaces are profinite (call ...
- 37.4k
7
votes
Accepted
Irreducible representations of product of profinite groups
This is not even true for finite groups, in this generality, and not even in characteristic $0$. Consider, for example, the group $Q_8 \times C_3$, where $Q_8$ is the quaternion group and $C_3$ is ...
- 12.8k
7
votes
An infinite profinite group such that any $p$-adic representation has finite image
The group $SL_n({\mathbb F}_p[t])$, for $n \geq 3$ has super rigidity property and hence any representation over characteristic zero has finite image. Since this is dense in $SL_n({\mathbb F}_p[[t]])$,...
- 11.1k
7
votes
An infinite profinite group such that any $p$-adic representation has finite image
Let $G$ be a finitely generated, residually finite group for which every linear representation in char. zero has a finite image. For instance, this holds if $G$ is a torsion group. An example of such ...
- 60.4k
7
votes
Structure of a profinite group as a condensed set with an action of an open subgroup
Yes, that is true. I'm not sure I'm explaining the step that's confusing you, but you are asking about a special case of the statement that the functor $S\mapsto\underline{S}$ from profinite sets to ...
- 19.9k
7
votes
Accepted
Looking for an example of profinite groups
You can take $G=\prod_{k>0}\mathbb{Z}/2^k$ and $H=\bigoplus_{k>0}\mathbb{Z}/2^k$. Then every finitely generated subgroup of $H$ is finite, and $\overline{H}=G$, but the element $(1,1,1,\dotsc)\...
- 55k
7
votes
Accepted
Is the free profinite group (or pro-$p$) torsion-free?
Both the free profinite and free pro-$p$ groups are torsion-free (also free pro-solvable groups or free pro-$\mathcal C$ groups for any collection $\mathcal C$ of groups closed under extensions, ...
- 37.4k
6
votes
Is every compact topological ring a profinite ring?
The earliest reference I could find to the fact that compact Hausdorff rings are profinite (objects in the category of topological rings) is in Johnstone's "Stone Spaces" (VI.4.11 on page 266); in the ...
- 17.5k
Only top scored, non community-wiki answers of a minimum length are eligible