21
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 $\...

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 $\Gamma \subset \mathrm{SL}_n({\mathbb Z})$ ...

18
votes

Accepted

### Avoiding countable subgroups of general uncountable groups

Counterexample for Problem 1: According to this answer Saharon Shelah constructed a "Jónsson group" of order $\aleph_1,$ i.e., an uncountable group $G$ in which every proper subgroup is countable. ...

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 ...

14
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 ...

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 ...

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.

12
votes

Accepted

### A hyperbolic group with a small profinite completion

It's a famous open question whether every word-hyperbolic group is residually finite. Kapovich--Wise showed that this is equivalent to asking whether every non-trivial word-hyperbolic group has non-...

12
votes

Accepted

### Is every closed subgroup of $\text{GL}_n(K[[x]])$ finitely generated?

This question has a negative answer is many respects. Firstly, there are simple constructions in the commutative case. Namely, the additive group $K[\![x]\!]$ is an infinite dimensional $\mathbb F_p$...

Community wiki

12
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 ...

11
votes

Accepted

### Union of conjugates of a closed subgroup of a compact group

The usual example is $G = {\rm SU}_2({\bf C})$ and $H$ the diagonal subgroup.
Every unitary matrix is diagonalizable, and thus contained in a
conjugate of $H$.

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 ...

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 ...

10
votes

Accepted

### Finite generation and profinite completion

No. Take $G=\oplus_{p} C_p$, where the sum is over all primes $p$ and $C_p$ is the cyclic groups of order $p$. Then $G$ is clearly locally finite and infinite and therefore not finitely generated. ...

10
votes

### Avoiding countable subgroups of general uncountable groups

Here's another counterexample for Problem 1 not using the fancy (non-explicit) construction of Shelah.
Fix an odd prime $p$. Let $A=\mathbf{F}_p((t))$ be the ring of Laurent series over the field $\...

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 ...

9
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 ...

8
votes

Accepted

### Avoiding countable subgroups of a group homeomorphic to the Cantor space

Yes, there exists.
Since $G$ is profinite, we may write it as $G=\lim_{n\in\mathbb{N}} G_n$, where $G_n$ are finite and the projections $\pi_n:G\to G_n$ are onto.
Since $H$ is countable, we may write ...

8
votes

Accepted

### Applications of Lubotzky's linearity theorem?

According to Alex himself, this theorem is practically useless. It does not mean that it can't be applied, for instance when you have a group with assumptions that it has many quotients in some ...

Community wiki

8
votes

### Subgroups of hyperbolic groups

If there's a non-residually finite hyperbolic group, then the answer is no. By a result of Kapovich-Wise, in this case there is a hyperbolic group $H$ whose profinite completion is trivial. Then ...

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[...

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(...

7
votes

Accepted

### Do free profinite groups satisfy Howson's theorem?

The following example shows that the answer is no.
Let $p$ and $q$ be two different primes. First I want to construct two generated profinite group $G=A\ltimes H$ isomorphic to semi direct product of ...

7
votes

Accepted

### Is there a left-orderable profinite group?

There is no such ordering, which is compatible with the profinite topology in the following way: If $x<y$, then there are small neighbourhoods $U, V$ of $x, y$, such that $u<v$ for all $u\in U, ...

7
votes

Accepted

### Faithful representations of free pro-p groups

It will only happen if $m=1$. See this paper:
http://mlarsen.math.indiana.edu/~larsen/papers/2gen.pdf
Indeed, the pro-$p$ groups that are linear over local fields of characteristic $0$ are just the ...

7
votes

Accepted

### Generators of Sylow subgroups

I don't think so. Let $p$ and $q$ be primes with $q|p-1$. Then there are $q$ inequivalent $1$-dimensional modules for $C_q$ over ${\mathbb F}_p$, If we take the semidirect product of the direct sum of ...

7
votes

Accepted

### Inverse limit of $\Bbb Q/q\Bbb Z$ isomorphic to finite adeles?

For 1 you can indeed use duality, but you can also give an explicit isomorphism:
First for each rational number $q$ it is easy to see that $\mathbb{A}^{f}/(q \hat{\mathbb{Z}}) \simeq \mathbb{Q}/(q\...

7
votes

Accepted

### Dessins d'enfants and absolute Galois group

For the sake of getting this off the unanswered stack...
Please refer to the following reference:
Guillot, P. An elementary approach to dessins d'enfants and the Grothendieck-Teichmüller group, ...

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 ...

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 ...

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