8
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Does your counterexample still work with $\mathbb{F}_2^\omega\times\mathbb{F}_3^{\omega_1}$ and $\mathbb{F}_2^{\omega_1}\times\mathbb{F}_3^{\omega}$, if we assume Luzin's hypothesis, that is, that $2^\omega=2^{\omega_1}$? This would still be the size of the continuum $|\mathbb{R}|$, even though $\omega$ and $\omega_1$ have different cardinalities. If so, this would show that it is at least consistent with ZFC that the answer is no, since Luzin's hypothesis is known to be relatively consistent with ZFC. $\endgroup$ – Joel David Hamkins Nov 15 '14 at 19:39
  • 2
    $\begingroup$ A simpler countexample under Luzin's hypothesis than in my first comment would be: $\mathbb{F}_2^\omega$ versus $\mathbb{F}_2^{\omega_1}$, which both have size continuum. $\endgroup$ – Joel David Hamkins Nov 15 '14 at 19:51
  • 1
    $\begingroup$ How about the following proposed counterexample: $\mathbb{F}_2^\omega$ versus direct sum of $\omega_1$ copies of this? (Or one could even use countable support instead of finite support in the direct sum.) These both have size continuum, without any extra set-theoretic assumption. $\endgroup$ – Joel David Hamkins Nov 15 '14 at 20:04
  • 2
    $\begingroup$ This hypothesis "have at most countably many finite Hausdorff quotients" for a profinite group, is the same as being metrizable, and is also equivalent to be either finite or homeomorphic to a Cantor set. $\endgroup$ – YCor Nov 16 '14 at 2:23
  • 1
    $\begingroup$ @QiaochuYuan: The fact that a profinite group with only finitely many quotients of each finite index is determined by the list of isomorphism types of those quotients is indeed a standard fact in profinite group theory. I've definitely seen it in John Wilson's book, and I'm pretty sure it's in Ribes-Zalesskii as well. $\endgroup$ – Colin Reid Nov 17 '14 at 1:24
13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is great! Thanks for this beautiful counterexample :) $\endgroup$ – jacob Nov 15 '14 at 20:34
  • $\begingroup$ I don't know the hat notation used in this context to define G. Would someone add a few words in an explanatory comment or edit, please? (Even if the words are "profinite completion", precisely what is being completed profinitely?) $\endgroup$ – The Masked Avenger Nov 15 '14 at 20:41
  • $\begingroup$ @The Masked Avenger: $\mathbb{Z}$ is being completed profinitely. $\widehat{\mathbb{Z}}$ is the profinite completion of the integers, otherwise known as the profinite integers. By the Chinese remainder theorem it may also be written as a product $\prod_p \mathbb{Z}_p$ where $\mathbb{Z}_p$ denotes the $p$-adic integers. $\endgroup$ – Qiaochu Yuan Nov 15 '14 at 20:44
  • $\begingroup$ And you are then raising that to a countable direct power to build G. (I think.) Thanks for the additional clarity, and thanks for teaching me today. $\endgroup$ – The Masked Avenger Nov 15 '14 at 20:48
1
$\begingroup$

A more geometric example is as follows: let $k$ be a countable algebraically closed field of characteristic $p$, and pick a positive integer $n$. Let $G_n = \pi_1(\mathbb{A}^n_k)$, the etale fundamental group of the $n$-dimensional affine space over $k$. By Raynaud's solution of the Abhyankar conjecture, $G_1$ admits as finite quotients (precisely) all finite groups with no prime-to-$p$ quotient. Therefore, the same holds for all $G_n$. However, the cohomological dimension of $G_n$ equals $n$, and hence $G_n\not\cong G_m$ for $n\neq m$.

$\endgroup$
  • 2
    $\begingroup$ I find this use of the word 'geometric' amusing. $\endgroup$ – HJRW Mar 6 '17 at 15:01
  • $\begingroup$ The use of "geometric" for algebraic features interpretable in algebraic geometry is very common. $\endgroup$ – YCor Mar 6 '17 at 15:31
  • $\begingroup$ @YCor -- I know. It's especially amusing in characteristic $p$. $\endgroup$ – HJRW Mar 6 '17 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.