Homomorphic images of $p$-adic integers

Question

Let $J_p$ be the additive group of $p$-adic integers for a fixed prime $p$. I would like to know if the structure of its homomorphic images is known. In particular, if $J_p/N$ is a homomorphic image with finite exponent, is $J_p/N$ also finite?

Answer 1

The structure of quotients of $\mathbf{Z}_p$ (as abstract group) is surprisingly restricted:

1) one has $\mathbf{Z}_p$ itself

2) every proper quotient $\mathbf{Z}_p/H$ decomposes canonically as direct product $T\times A$, where $T$ is its torsion subgroup, which is cyclic of $p$-power order, and its divisible subgroup $A$. Every divisible abelian group $A$ of cardinal $\le 2^{\aleph_0}$ can be obtained this way. $T$ is simply the Hausdorff quotient, namely the quotient by the closure of $H$. We have $A=0$ $\Leftrightarrow$ $H$ is closed $\Leftrightarrow$ $H$ is open.

A corollary is the answer to your second question: the only quotients of finite exponent are the finite ones.

---

To show this, start with the observation that for every prime $\ell\neq p$, $\mathbf{Z}_p$ is $\ell$-divisible (multiplication by $\ell$ is surjective), and this passes to its quotients.

Start now when $H$ is dense. Since $p\mathbf{Z}_p$ is open, we have $p\mathbf{Z}_p+H=\mathbf{Z}_p$. This means that $p(\mathbf{Z}_p/H)=\mathbf{Z}_p/H$. Hence $\mathbf{Z}_p/H$ is $\ell$-divisible for every prime $\ell$, and hence divisible. The structure of divisible abelian groups is simple: a direct sum of indecomposable ones, and the indecomposable ones are $\mathbf{Q}$ and the quasi-cyclic (or Prüfer) groups $\mathbf{Q}/\mathbf{Z}_{(p)}$. In particular the divisible abelian groups of cardinal $\le\kappa$, for any infinite cardinal $\kappa$, are the quotients of $\mathbf{Q}^{(\kappa)}$, a $\mathbf{Q}$-vector space of dimension $\kappa$. Now $\mathbf{Q}_p/\mathbf{Z}$ being torsion-free, divisible of cardinal $2^{\aleph_0}$, it is isomorphic to $\mathbf{Q}^{(2^{\aleph_0})}$. So we obtain all others as quotients.

When $H$ is not necessarily dense (but nonzero), $\mathbf{Z}_p/H$ has the subgroup of finite index $\bar{H}/H$, which is divisible. Since every divisible subgroup of an abelian group is a direct summand, we obtain the required decomposition. Here the direct factor is the Hausdorff quotient, which is cyclic of $p$-power order.

---

Added: Here's a corollary: every torsion-free quotient $Q$ of $\hat{\mathbf{Z}}$ of cardinal $<2^{\aleph_0}$ is divisible (i.e., isomorphic to some vector space over $\mathbf{Q}$).

Indeed, let $I$ be the image of $\bigoplus_p\mathbf{Z}_p$. By the above, $I$ is divisible. Since $\hat{\mathbf{Z}}/(\bigoplus_p\mathbf{Z}_p)$ is divisible, $Q/I$ is also divisible.