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?



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 $H$ 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.


