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I am developing a sort of standard representation for profinite quandles. This involves profinite groups a lot, actually. In one part of my construction the filtered diagram used to construct a profinite group becomes important.

Suppose we have two profinite groups $G_1$ and $G_2$, with proper, dense subgroups $\Gamma_1<G_1$ and $\Gamma_2<G_2$ with a topological isomorphism $\Gamma_1\cong\Gamma_2$. Does this imply that $G_1\cong G_2$? Note that this is certainly not true for topological spaces, but I hope the unique form of compactness offered by profinite things helps the argument some. If this is not true, is there a known counterexample?

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    $\begingroup$ Is the isomorphism between $\Gamma_i$ supposed to be topological? It's not true if you only assume an isomorphism between abstract groups $\endgroup$
    – Wojowu
    Commented Jul 6, 2023 at 20:22
  • $\begingroup$ @Wojowu Yes, it is meant to be a topological isomorphism. I'll edit the question. $\endgroup$
    – Alex Byard
    Commented Jul 6, 2023 at 20:26
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    $\begingroup$ If $\Gamma$ is dense in a compact group $G$, I think one obtains the topological group $G$ from the topological group $\Gamma$ by a completion argument. $\endgroup$
    – YCor
    Commented Jul 6, 2023 at 22:18
  • $\begingroup$ @YCor So if a compact group $G$ is determined by a dense subgroup, then two compact groups containing that subgroup ought to be isomorphic. Do you know the argument for your claim, or where I might find it? $\endgroup$
    – Alex Byard
    Commented Jul 7, 2023 at 1:47

1 Answer 1

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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 $H$ converges in $G$ if and if $h_n^{-1}h_m\to 1$ when $n,m\to\infty$, and two such sequences $(h_n)$, $(h'_n)$ have the same limit if and only if $(h_n^{-1}h'_n)$ converges to 1. Thus one can define $G_H$ as set of Cauchy sequences modulo this relation, which is an equivalence relation, endow it with the group law (noting that being Cauchy passes to products and that the product factors through the equivalence relation). In $G_H$, a sequence $(h_n^k)_k$ of Cauchy sequences converges to $1$ if and only if for every neighborhood $V$ of $1$ there exists $k_0$ such that for every $k\ge k_0$ there exists $n_k$ such that $h_n^k\in V$ for all $n\ge n_k$. Thus there exists at most one metrizable group topology on $G_H$ for which these are the converging sequences to $1$.

From this it follows that any two compact [metrizable] groups with isomorphic dense subgroups (isomorphic as topological groups) are isomorphic as topological groups.

The general case (no metrizability) can be done by a suitable use of filters. Also compactness is not fully used; local compactness is enough, and also it should be enough to suppose some kind of completability, but I haven't seriously checked and I'm not sure what uniformity is best to use.

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  • $\begingroup$ This is what I was looking for, thank you. Would you in addition be able to explain how to use filters to prove this in general, or provide a reference? $\endgroup$
    – Alex Byard
    Commented Jul 7, 2023 at 5:29
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    $\begingroup$ I don't have a reference in mind. I'll maybe write things down if I have time, but it's a good exercise to do it. $\endgroup$
    – YCor
    Commented Jul 7, 2023 at 7:59
  • $\begingroup$ Okay, if you have time to write down the filter argument I'd really appreciate it. The reason is that I'd like to work with uncountable inverse limits. I've accepted your answer. $\endgroup$
    – Alex Byard
    Commented Jul 7, 2023 at 17:08
  • $\begingroup$ Alternatively, if you could briefly indicate how a filter proof would go, I can work it out on my own. I don't see how to generalize your argument, as it seems to rely heavily on having a metric. $\endgroup$
    – Alex Byard
    Commented Jul 7, 2023 at 19:05
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    $\begingroup$ If you have an ultrafilter on a compact space, it is supported at a single point. This applies to an ultrafilter on any dense subset, and conversely given any dense subset, every point of the ambient compact space is the limit point of an ultrafilter on the dense subset. Then one needs to define the topology, the product. I didn't rely that deeply on a metric: just to use that the topology is determined by converging sequences. $\endgroup$
    – YCor
    Commented Jul 7, 2023 at 21:30

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