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

Question

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.

Answer 1

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.