2
$\begingroup$

Here is a very simple claim: if $ G $ is profinite group isomorphic to its completion $ \hat{G} $. Then the natural map $ \phi: G \to \hat{G} $ is an isomorphism. I'm not sure how to show this. The universal property of ($ \hat{G}, \phi $) seems to only give that $ \phi $ is injective. I know that $ G $ is isomorphic to the inverse limit taken over all $ G/N$ with $ N $ open normal subgroups (of finite index), and $ \hat{G} $ is isomorphic to the inverse limit taken over all $ G/K$ taken over all $ K $ normal subgroups of finite index. But it's not so clear how to use this, since the isomorphism is not explicitly given. Help is appreciated.

This claim is taken from these notes, page 6.

$\endgroup$
  • 1
    $\begingroup$ I would assume that the way to prove this the contrapositive: prove that if $\phi$ fails to be an isomorphism, it must fail very badly so that $\hat{G}$ is somehow "larger" than $G$ in a way that prevents them from being isomorphic. For instance, it seems plausible to me that if $\phi$ is not an isomorphism, then $\hat{G}$ must have greater cardinality than $G$. (I think I can prove this in the case that $G$ is abelian, at least.) $\endgroup$ – Eric Wofsey Jul 2 '18 at 2:31
  • $\begingroup$ @EricWofsey thank you for your comment. That is interesting. I was reading some online notes when the author simply sidesteps this saying that it follows from the universal property of $ (\hat{G}, \phi) $. Maybe there's a way to use it that I'm simply not following. $\endgroup$ – user564167 Jul 2 '18 at 3:51
  • 1
    $\begingroup$ If an author claims this just follows from the universal property, then I think they are simply mistaken. (And, for that matter, it is not clear to me that the result is even true.) $\endgroup$ – Eric Wofsey Jul 2 '18 at 4:00
  • $\begingroup$ @EricWofsey he was using this claim to show that if $ G $ is isomorphic to its profinite completion, then every finite index subgroup of $ G $ is open. Is there a way to see this without showing that $ \phi $ is an isomorphism? Or must this statement also include the fact that $ G $ is isomorphic to $ \hat{G} $ via $ \phi $? $\endgroup$ – user564167 Jul 2 '18 at 4:11
  • $\begingroup$ I have included the reference in my question. $\endgroup$ – user564167 Jul 2 '18 at 4:15
0
$\begingroup$

This appears to simply be an error in the notes. Certainly, the claim does not follow formally from the universal property as the text seems to imply. Consider the following alternate statement, which has the same formal categorical structure (just replace the forgetful functor adjunction between groups and profinite groups with the forgetful functor adjunction between groups and sets): if a group $G$ is isomorphic to the free group $F(G)$ on its underlying set, then the canonical map $G\to F(G)$ is an isomorphism. This statement is false, though (a free group on any infinite set is a counterexample).

I do not know whether the claim in question is true or not. If it is true, I would expect the proof to involve a hands-on analysis of how $\phi$ can fail to be an isomorphism rather than any formal abstract nonsense.

$\endgroup$

Your Answer

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