$ \DeclareMathOperator{\bN}{\mathbb N} \DeclareMathOperator{\bZ}{\mathbb Z} $Just playing around with profinite completions. As a basic example (ignoring the trivial case of finite groups or $\bZ$) I chose to compute the profinite completion of the $p$-adic integers, an infinite group I am more familiar with than others.

And here we go. The profinite completion of an arbitrary group $G$ is the projective limit over the system of quotients $G/N$ for $N\trianglelefteq G$ of finite index with the natural projection maps. Let us consider the case of the $p$-adic integers $\bZ_p$ and consider an arbitrary (normal) subgroup $N$ of finite index (I actually do not think the latter matters that much except for excluding $N=0$ which, OTOH, is important now that I think about it...). Certainly $N\ne0$ as the $\bZ_p$ is an infinite group. Hence we find $x\in N$ of minimal valuation $n=\nu_p(x)$ (recall that the valuation of $x\in\bZ_p\setminus\{0\}$ is defined as the minimal $n\in\bN$ such that $\pi_n(x)\equiv0\mod p^n$ with $\pi_n\colon\bZ_p\to\bZ/p^n\bZ$ as usual). As $n$ is minimal as such we have $\nu_p(y)\ge n$ for all $y\in N$ and hence $N\subseteq p^n \bZ_p$. From this argument we conclude that the quotients $\bZ_p/p^n\bZ_p\cong\bZ/p^n\bZ$ form a cofinal subsystem and hence it suffices to compute $\varprojlim\bZ_p/p^n\bZ_p\cong\varprojlim\bZ/p^n\bZ$. But this is simply $\bZ_p$ and we are done.

Is the given argument correct? if not, where does it fail? I am not completely sure about the $N\subseteq p^n\bZ_p$-part implying the cofinal-part (even though I used this argument before and it appears to be correct there).
Bonus: Is there a more sophisticated reason to consider only normal subgroups of finite index besides excluding the quotient $G/1\cong G$ (which renders the computation trivial immediately)?

Thanks in advance!


$\newcommand\Z{\mathbb Z}$The argument you gave is correct, and if fact shows something a little more than you claim. You chose $x\in N$ of minimal valuation $n:=\nu_p(x)$. This means that $x=up^n$ where $u\in\Z_p^\times$ is a unit, so not only is $N\subset p^n\Z_p$, but in fact, $N=p^n\Z_p$. That is, these don't just form a cofinal system of finite quotients; there are exactly all finite quotients.

For you bonus question, you certainly want to exclude $G$ itself since otherwise your completion will just give back the original group, i.e. $G\cong\varprojlim\limits_{\text{all }N\triangleleft G} G/N$ for completely formal reasons. Considering finite quotients has the benefit that the group you get in the end will be compact, since it will be a closed subset of the compact space $$\prod_{\substack{N\triangleleft G\\ [G:N]<\infty}}G/N,$$ where each $G/N$ is a finite space with discrete topology. Compact groups are especially nice, vaguely because compact spaces are nice and more concretely e.g. because you can do Fourier analysis on them. You can also make sense of the "order" of a profinite group (given as a supernatural number, e.g. the order of $\Z_p$ is $p^\infty$) and talk about Lagrange's theorem and Sylow $p$-subgroups and the like from finite group theory. Hopefully this helps to motivate profinite completions.

As a side note, it is sometimes useful to form the inverse limit of only some finite quotients; one could take the limit of only the cyclic quotients (and talk of procyclic groups), of only the quotients with $p$-power order (and talk of pro-$p$ groups), etc.

Finally, I don't think computing the profinite completion of $\Z$ is as trivial as you suggest. Let $\widehat\Z$ denote this completion. Can you show that $\widehat\Z\cong\prod_p\Z_p$?

Edit: I made blunder in the first paragraph. It is not the case that all finite index subgroups are of the form $p^n\mathbb Z_p$ (when I wrote "in fact, $N=p^n\Z_p$" I somehow had switched to thinking about ideals in my mind). So where does this leave the claim about cofinality?

We would like to say that the quotients $\Z_p/p^n\Z_p$ form a cofinal sequence of finite quotients of $\Z_p$, i.e. if $\Z_p/N$ is a finite quotient, then our projective system includes a morphism $\Z_p/p^n\Z_p\twoheadrightarrow\Z_p/N$, i.e. $p^n\Z_p\subset N$ for some $n$. This is the reverse inclusion of what you showed. Unfortunately, I have not been able to convince myself one way or the other about this system actually being cofinal, but I figured I should include this edit anyways just to make it clear that what I initially said was incorrect (See edit 2).

As a general note, one thing one might do to avoid these issues is to define the profinite completion of a topological group $G$ to be the inverse limit of its finite-index, open subgroups. Then is the same definition when $G$ is finite, and has the added benefit that it is obvious that the profinite completion of a profinite group is itself under this definition.

Edit 2: Thanks to @TorstenSchoeneberg in the comments for pointing out how to actually answer your question. If $N$ is a subgroup of finite index $n$, then $nx\in N$ for any $x\in\Z_p$, so $N\supset n\Z_p=p^{v_p(n)}\Z_p$ and we have cofinality.

  • $\begingroup$ Could you elaborate on the first part? I'm aware of an argument showing this for closed subgroups; I don't see right now how this generalizes to arbitrary subgroups. For the latter: I know that the profinite completion of $\mathbb Z$ is $\hat{\mathbb Z}$ (and, in fact, I know how to construct an isomorphism to $\prod_p\mathbb Z_p$) but it is more of definition than a computation (of course, depending on your POV). $\endgroup$
    – mrtaurho
    Aug 9 at 23:14
  • $\begingroup$ After a little more thought, the first part of my post is incorrect. I was thinking about ideals and not arbitrary subgroups when I wrote it. This complicates things. To get the cofinality statement you want, you need that no matter how small $N$ is (i.e. no matter how big $\mathbb Z_p/N$ is), there's always some $p^n\mathbb Z_p$ that's smaller (i.e. some surjection $\mathbb Z_p/p^n\mathbb Z_p\to\mathbb Z_p/N$). This is the opposite direction from what you showed. See my (forthcoming) edit for the resolution. $\endgroup$
    – Niven
    Aug 9 at 23:52
  • $\begingroup$ I appreciate your edit! That being said, it points out a serious flaw in my thinking (which I am also not able to fix right away; but I will think about this). However, if I am not mistaken we have open $+$ finite index $\implies$ closed in which case I can handle the rest. Is this restriction common enough? Also, why does it suffice to only consider cyclic quotient? I am still not seeing this part. $\endgroup$
    – mrtaurho
    Aug 10 at 0:22
  • 2
    $\begingroup$ If $N$ is a subgroup of $\mathbb Z_p$ of finite index $n$, then $nx \in N$ for all $x \in \mathbb Z_p$, i.e. $N$ contains the open subgroup $n \mathbb Z_p = p^{v_p(n)} \mathbb Z_p$ and $n$ is a power of $p$. There are no non-open, finite index subgroups of $\mathbb Z_p$. $\endgroup$ Aug 10 at 2:47
  • 3
    $\begingroup$ Can I just point out that the profinite completion of a profinite group is not, in general, itself. It is a deep theorem of Nikolov-Segal that this is true for finitely generated profinite groups. This is because for them every finite-index subgroup is open. This depends on the Classification of Finite Simple Groups, and I think there's still no CFSG-free proof. There are easy examples on non-finitely generated profinite groups with subgroups that are of finite index but not open. $\endgroup$ Aug 11 at 23:58

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