**Question: Let $G,H$ be profinite groups of cardinality $|\mathbb{R}|$, with the same finite quotients (here I only consider quotients by normal, *open* subgroups). Then are $G$ and $H$ isomorphic?**


Background: First, note that one needs some sort of cardinality assumption, as otherwise one could take $G=\mathbb{F}\_2^I$ and $H=\mathbb{F}\_2^J$ where $I$ and $J$ are infinite of distinct cardinality. Moreover, it is NOT sufficient to just require that $G$ and $H$ have the same cardinality, as one could take $G=\mathbb{F}\_2^I\times \mathbb{F}\_3^J$ and $H=\mathbb{F}\_2^J\times\mathbb{F}\_3^I$.


As for motivation, if we take $C$ to be an affine hyperbolic curve over the algebraic closure of a finite field, then it is known which subgroups occur as finite quotients of the etale fundamental group. Given this phrasing, I assume that this is not enough information to ecover the actual etale fundamental group, and so I wanted an explicit counterexample.


**EDIT: Due to the comments, I think a better question is as follows: Rather than assuming $G$ and $H$ have cardinality $\mathbb{R}$, let me assume instead that they have at most countably many homomorphisms to any finite group.** 


This avoids (at least) some counterexamples assuming Luzin' hypothesis.


Let me add that the answer is yes if $G$ and $H$ are topologically finitely generated. This is well know, but as its quick I include a proof: 


Note that for each positive integer $n$, $G$ has only finitely many homomorphisms to fintie groups of order at most $n$. Let $G(n)$ be the kernel of all of these, and $G\_n$ the corresponding finite quotient. Then $G$ is the inverse limit of the $G\_n$. To see this note that the natural map $G\to\varprojlim\_n G\_n$ is clearly injective, and as it is surjective on every quotient it must also be surjective by compactness.


So it is enough to see that $G\_n\cong H\_n$. But by assumption, $G$ surjects onto $H\_n$, and such a surjection must factor through $G\_n$, so $G\_n$ surjects onto $H\_n$. Likewise $H\_n$ surjects onto $G\_n$ and since these are finite groups they must be isomorphic. 



Are profinite groups of cardinality $|\mathbb{R}|$ determined by their finite quotients?
Without the extra finiteness conditions, but also not relying on any set-theoretic assumptions, take $G = \widehat{\mathbb{Z}}^{\mathbb{N}}$ and $H = G \times A$ for a nontrivial finite abelian group $A$. Both $G$ and $H$ have every possible finite abelian quotient, but only $H$ has torsion. 


