# How to show $\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}\cong \mathbb{Z}/n\mathbb{Z}$

I'm trying to do exercise 1.9 from the following PDF: http://websites.math.leidenuniv.nl/algebra/Lenstra-Profinite.pdf

I'm primarily interested in part (a) showing the titular isomorphic relation and part (c) describing the open/closed subgroups of $\hat{\mathbb{Z}}$.

For part (a) I cannot get far by trying to come up with an isomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}$. A hint here would be appreciated.

For part (c) I know the fact that every closed subgroup will be profinite (i.e. compact, Hausdorff, and totally disconnected), but is this a complete description of the closed subgroups of $\hat{\mathbb{Z}}$?

$\newcommand{\ZZ}{\mathbb{Z}}$ The elements of $\hat{\ZZ}$ have very explicit descriptions in terms of compatible sequences $(x_i)_{i\in\mathbb{N}}$ where each $x_i\in\ZZ/i\ZZ$.
There is a very natural map $\hat{\ZZ}\rightarrow\ZZ/n\ZZ$ given by "projection onto the $n$th coordinate". This is a homomorphism because the elements $(x_i)$ are compatible sequences. It is a good exercise to show that this is surjective, and that its kernel is $n\hat{\ZZ}$.
The closed subgroups however need not be open and in general are far more numerous. Since $\hat{\ZZ}$ is abelian, every closed subgroup $H\le\hat{\ZZ}$ is normal and defines a quotient, which is finite iff $H$ is open. It's easy to see that the finite quotients of $\hat{\ZZ}$ are precisely the finite cyclic groups (you can prove this directly or use the universal property of profinite completions). However, there are many more non-open closed subgroups. For example, let $\pi$ be a set of prime numbers, and let $\hat{\ZZ}(\pi)$ be the set of compatible sequences $(x_i)$ where $i$ ranges over only natural numbers which are divisible only by the primes in $\pi$. Then projection onto such coordinates gives you a surjective map $$\hat{\ZZ}\rightarrow\hat{\ZZ}(\pi)$$ whose kernel is a closed but not open subgroup of infinite index. The group $\hat{\ZZ}(\pi)$ is the pro-$\pi$ completion of $\ZZ$. The kernel of the map is the product $\prod_{p\notin\pi}\ZZ_p$ of the Sylow-$p$ subgroups for $p\notin\pi$ (note that $\hat{\ZZ} = \prod_p\ZZ_p$, where $\ZZ_p$ is the additive group of the $p$-adic integers).
I believe that all closed subgroups of $\hat{\ZZ}$ can be obtained as intersections and joins of the closed subgroups described above, though there are details to be worked out.
• You can use the fact that the Pontryagin dual of $\hat{\mathbb{Z}}$ is $\mathbb{Q}/\mathbb{Z}$, and a classification of the subgroups of $\mathbb{Q}$ containing $\mathbb{Z}$ is available; their orthogonal in the duality are all the closed subgroups of $\hat{\mathbb{Z}}$. – egreg Dec 10 '15 at 0:16