Structure of $\Bbb Q_p/\Bbb Q$ The group $\Bbb A_\Bbb Q/\Bbb Q$ of adeles mod rationals is:


*

*Isomorphic to the solenoid

*Isomorphic to the group $\Bbb A_\Bbb Z/\Bbb Z$

*Dual to the rational numbers


This raises the question of the structure of the individual groups $\Bbb Q_p / \Bbb Q$ of p-adic numbers mod rationals, as well as $\Bbb Z_p/\Bbb Z$.
Is there any known way to decompose these groups? Or are they isomorphic to other known structures, or are their Pontryagin duals known?
 A: I don't know a simpler description of $\mathbb{Z}_p/\mathbb{Z}$ than as $\mathbb{Z}_p/\mathbb{Z}$. Its Pontryagin dual is a particular subgroup of the Pontryagin dual of $\mathbb{Z}_p$, so let's try to identify that first. Pontryagin duality sends cofiltered limits to filtered colimits, so it sends $\mathbb{Z}_p$, which is the cofiltered limit of the diagram
$$\dots \to \mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$$
to the filtered colimit of the diagram
$$\dots \leftarrow \mathbb{Z}/p^2\mathbb{Z} \leftarrow \mathbb{Z}/p\mathbb{Z}.$$
But if we want to think of $\mathbb{Z}_p/\mathbb{Z}$ as a discrete group (and we do; it doesn't inherit a compact Hausdorff topology from the usual one on $\mathbb{Z}_p$ because $\mathbb{Z}$ is not a closed subgroup) then we need to take this filtered colimit in compact Hausdorf abelian groups. The group we get this way is the Bohr compactification of the Prüfer $p$-group; I don't know a simpler description of it than this. 
This compactification comes with a natural surjection to $S^1$ dual to the natural inclusion $\mathbb{Z} \to \mathbb{Z}_p$, and the Pontryagin dual of $\mathbb{Z}_p/\mathbb{Z}$ is the kernel of this surjection. 
Neither of the questions you ask tell you much about the situation with the adeles. What makes the adeles special is that $\mathbb{Q}$ sits inside them as a discrete subgroup, so the quotient $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}$ retains an interesting (compact Hausdorff, I think?) topology from the usual topology on $\mathbb{A}_{\mathbb{Q}}$. This isn't the situation with either $\mathbb{Q}_p$ or $\mathbb{Z}_p$: the subgroups $\mathbb{Q}$ and $\mathbb{Z}$ are both dense, and in particular not closed, so with the quotient topologies the quotient groups are not even Hausdorff. The only reasonable topology I see to use is the discrete topology, and as discrete groups $\mathbb{Q}_p/\mathbb{Q}$ is boring and $\mathbb{Z}_p/\mathbb{Z}$ is just $\mathbb{Z}_p/\mathbb{Z}$. 
A: You might be interested in the fact that $\mathbb{Z}_p/\mathbb{Z}\cong\mathrm{Ext}(\mathbb{Z}[1/p],\mathbb{Z})$.
This may be seen by considering the short exact sequence $$0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}[1/p]\rightarrow\mathbb{Z}[p^\infty]\rightarrow0$$ and examining the resulting 6-term exact sequence after applying $\mathrm{Hom}(-,\mathbb{Z})$.
One also has that $\mathbb{Z}_p/\mathbb{Z}$ is isomorphic to $\varprojlim^1(\cdots\rightarrow\mathbb{Z}\rightarrow\mathbb{Z})$, where the maps are multiplication by $p$. This can be proved directly by examining a tower of short exact sequences as above and examining the resulting exact sequence; one can also observe directly that it must be isomorphic to the $\mathrm{Ext}$ group above.
