# For $G$ profinite abelian, do we have $G \cong \varprojlim G/G^n$?

For $G$ a profinite abelian group, do we have $G \cong \varprojlim G/G^n$ (where the limit is taken over all natural numbers ordered by divisibility)? (If it's wrong, is there any nice class of abelian profinite groups for which it holds?)

Here's what I did so far:

It's easy to show that this holds for any group of finite expotent, in particular finite groups. It's also not difficult to see for $\mathbb Z_p$. (From this, the statement follows for topologically finite generated $p$-groups and products thereof, see below)

Since $G/G^n \cong \mathbb Z /(n) \otimes_{\mathbb Z}G$ and $\mathbb Z/(n)$ is a finitely presented module over $\mathbb Z$, the functor $\mathbb Z/(n) \otimes_{\mathbb Z} -$ commutes with arbitrary products, also products commute with projective limits, furthermore any abelian profinite group is a direct products of pro-$p$ groups, thus it would be enough to show this for abelian pro-$p$ groups. Now any abelian pro-$p$ group is a module over $\mathbb Z_p$, thus for any $k$ coprime to $p$, as $k$ is invertible in $\mathbb Z_p$, we have $G^k=G$, from this it follows that for a pro-$p$ group $G$, we may replace the limit by $\varprojlim G/G^{p^n}$, where the limit is taken over all natural numbers with the usual order. If we consider $G$ as a $\mathbb Z$-module, then the above limit is the completion with respect to the ideal $(p)$, thus the question may be reduced to the statement:

Is a pro-$p$ group a $(p)$-adically complete module?

Maybe this formulation of the question allows some kind of topological argument, but I can't seem to relate $(p)$-adic topology and profinite topology.

• If $G = \lim_k \mathbb{Z}/2^k \mathbb{Z}$. Won't $\lim_{n \to \infty} G/G^{n}$ contain $\lim_k \mathbb{Z}/3^k \mathbb{Z}$ ? – reuns Aug 17 '17 at 12:17
• @reuns No, in this case $G = \mathbb{Z}_p$ and if $k$ is coprime to $p$, then $G^k=G$, as $k \in \mathbb{Z}_p^{\times}$, thus only powers of $p$ contribute to the limit and the isomorphism follows from the well-known fact that $\mathbb{Z} _p /p^k \mathbb{Z}_p \cong \mathbb{Z}/p^k\mathbb{Z}$ – MatheinBoulomenos Aug 17 '17 at 12:21

## 1 Answer

Consider the two sets:

1. the set $A(C)$ of cones with apex $C$ over the diagram of continuous finite quotients of $G$
2. the set $B(C)$ of cones with apex $C$ over the diagram of quotients of the form $G/G^n$

Elements in $A(C)$ are cones factoring through $G$, and hence extend to map to each quotient $G/G^n$. And in the other direction, we can extend an element of $B(C)$ to a finite quotient $G/H$ using $C\to G/G^{|G/H|}\to G/H$. It is easy to see these maps are mutually inverse and natural in $C$.

But $A$ and $B$ are the hom functors for the limit objects, so by Yoneda's lemma the limits are isomorphic.

• I think that for $A(C)$ you want to take cones over the diagram of continuous finite quotients of $G$, as $G$ is not necessarily the limit of its abstract finite quotients. – Jeremy Rickard Aug 18 '17 at 7:36
• @JeremyRickard: good point, I've specified "continuous" now. (And more generally one needs to check that all the maps involved in the proof are continuous, though I think that is not too hard to see.) – Dap Aug 19 '17 at 10:19