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
5
votes
Accepted
Subgroup of p-adic units
More is true: the generating set $\{f_p:\text{$p$ is a prime}\}$ is dense in $\smash{\widehat{\mathbb Z}}^\times$. To see this, it suffices to show that, for any finite set of primes $S$, the ...
- 99.2k
4
votes
Accepted
Profinite groups with isomorphic proper, dense subgroups are isomorphic
Let $G$ be a compact group and $H$ a dense subgroup. I claim that $H$, as topological group, determines $G$. For simplicity, let me assume that $G$ is metrizable.
Note that a sequence $(h_n)$ in $H$ ...
- 60.4k
4
votes
Topological generators for the Sylow pro-$p$ subgroup of $\mathrm{SL}_2(\mathbf{Z}_p)$
If $x$ and $y$ generate a group then so do $x$ and $xy$, so take $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} ...
- 137k
2
votes
Accepted
If $F$ is a prosoluble subgroup of a free profinite product $\amalg G_i$ and $F \cap G_i^g$ is pro-$p$, is also $F$ pro-$p$?
Let $I = \{1,2\}$ and let $G_1 = G_2$ be groups of order $p=2$. Their free product is $G = \langle \delta, \varepsilon |\, \delta^2= \varepsilon^2=1\rangle$, which is $\langle \tau, \varepsilon |\, \...
- 81
2
votes
Stone-topological/profinite equivalence for quandles
Here is how the proof works for monoids. This doesn't really answer the question so I made this communitywiki, but maybe the OP will find it useful.
Let $X$ be a compact Hausdorff space. Call an ...
2
votes
Agemo-of-agemo inclusions for p-groups
Well, unfortunately it's not true :(
I just ran the calculation on the free $2$-nilpotent group on two generators and class $8$, for which the groups $\mho_1(\mho_2(G))$ and $\mho_2(\mho_1(G))$ are ...
- 2,489
2
votes
Accepted
Open conjugacy classes in a second countable profinite group
Yes, there exist profinite groups $G$ with a conjugacy class of empty interior and consisting of elements of finite order, generating $G$ as an abstract group.
Let $H$ be a nonabelian finite simple ...
- 60.4k
2
votes
Accepted
Confusion with self-dual representations of $\mathrm{GL}_n$ over a $p$-adic field
You are right, there is an error in Lust-Stevens, and any unramified character provides a counter-example. Here is a way to fix their statement: $\rho=\mathrm{cInd}_{KF^\times}^G\widetilde\tau$ is ...
- 1,897
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