# Weird duality between integers and p-adic integers

The integers are, up to isomorphism, the unique infinite discrete abelian group that is isomorphic to all its non-trivial subgroups. This can be seen easily by Pontryagin Duality: It's equivalent to saying that $$S^1$$ is the only infinite compact abelian group isomorphic to all its proper quotients by closed subgroups. Indeed, if G is an infinite compact abelian group with such property, then if $$\chi$$ is any non trivial character of G, we get an isomorphism of topological groups $$G/ker(\chi) \cong Im(\chi)$$ (compacity of $$G/ker(\chi)$$ guarantees it's a homeomorphism). The property imposed on G implies that G is isomorphic to a compact subgroup of $$S^1$$, and since G is infinite, we get $$G \cong S^1$$.

One can also show, in a similar spirit, that the only infinite discrete abelian groups isomorphic to all their proper quotients are the Prüfer p-groups $$\mathbb{Z}(p^{\infty})$$, although this takes a bit more work. By Pontryagin duality, this implies that the only compact infinite abelian groups isomorphic to all their closed non-trivial subgroups are the p-adic integers $$\mathbb{Z}_p$$.

So we can look at the property P of a topological (Hausdorff) abelian group being isomorphic to all closed non-trivial subgroups, and if we restrict ourselves to the categories of infinite discrete and infinite compact abelian groups, we get two different answers in each case (up to isomorphism): For discrete groups, only the integers satisfy P, and for compact groups only the p-adic integers satisfy it.

At the same time, it's well known that the two mentioned categories are in correspondence by Pontryagin duality, so with that in mind, it almost feels like there is a "duality" between the integers and the family of groups of p-adic integers. Is there anything actually interesting happening from a categorical point of view at all, or am I overthinking it?

Edit: I just noticed that we don't even need to purposely restrict ourselves to compact or discrete groups. Indeed, if G is any locally compact abelian abelian group isomorphic to all its non-trivial closed subgroups, then it will in particular have to be monothetic. As such, it's known that it's either compact or isomorphic to the integers. So the weird role played by the integers and the p-adic integers technically extends to locally compact abelian groups.

• Minor point: in your first paragraph, does the argument work if the original infinite abelian group is not finitely generated? Commented Jul 19, 2023 at 22:32
• I'm confident it works yes. What really matters is that we consider abelian groups with the discrete topology, which will always guarantee that their duals are compact. The compacity is the most important part of that argument. Commented Jul 19, 2023 at 22:46
• What is the Pontryagin dual of the group $\Bbb{Z}^\omega$ (countably infinite sequences of integers under pointwise addition)? Commented Jul 20, 2023 at 18:43
• I see. Well unfortunately I don't think I can figure out anything besides this: it is a torsion free abelian group of rank $\omega$, so it embeds into $\mathbb{Q}^{(\omega) }$, which implies that its Pontryagin dual is isomorphic to a quotient of $\widehat{\mathbb{Q}} ^{\omega}$ by a closed subgroup. Maybe studying the topology of $\widehat{\mathbb{Q}} ^{\omega}$ can give you some insight, I don't know, sorry. Commented Jul 20, 2023 at 21:17
• Thanks for those remarks - I will work on it. Commented Jul 20, 2023 at 21:19

You are describing similar properties between $$\mathbf Z$$ among discrete abelian groups and every $$\mathbf Z_p$$ among compact abelian groups.

All I want to do here is record some other ways they appear in analogous roles.

1. $$S^1 \cong \mathbf R/\mathbf Z$$ and $$\mathbf Z(p^\infty) \cong \mathbf Q_p/\mathbf Z_p$$ as topological groups: $$\mathbf Z$$ is discrete and cocompact in $$\mathbf R$$, while $$\mathbf Z_p$$ is compact and open (= codiscrete) in $$\mathbf Q_p$$, with $$\mathbf R$$ and each $$\mathbf Q_p$$ being self-dual groups.

2. In $$\mathbf R^n$$, a lattice can be defined concretely as the $$\mathbf Z$$-span of a basis and conceptually as a discrete cocompact subgroup. In $$\mathbf Q_p^n$$, a lattice can be defined concretely as the $$\mathbf Z_p$$-span of a basis and conceptually as a compact open subgroup.

• Excuse me, what does Co compact and Co discrete mean? Commented Jul 24, 2023 at 23:32
• The "co" refers to "COmplementary" properties: sine and cosine, homology and cohomology, adjoint and coadjoint representations, domain and codomain, kernel and cokernel, and so on. (But in category theory we have covariant and contravariant functors rather than variant and covariant functors. ¯_ (ツ)_/¯ ). Sometimes the complementary properties are related to a vector space and its dual space (homology/cohomology and adjoint/coadjoint) while other times it is related to reversing arrows in a diagram of maps (which happens with duals), such as domain/codomain and kernel/cokernel.
– KCd
Commented Jul 25, 2023 at 18:52
• Now I'll address your specific questions about cocompact and codiscrete. In a short exact sequence of abelian groups $1 \to A \to B \to C \to 1$, $A$ is a kernel and $C \cong B/A$ is a cokernel and quotient: a property P of a quotient $B/A$ is called a "co"-property of $A$ as a subgroup of $B$, so when these are topological groups, to say $A$ is cocompact in $B$ means $B/A$ is compact, and to say $A$ is codiscrete in $B$ is to say $B/A$ is discrete (equivalently, $A$ is open in $B$).
– KCd
Commented Jul 25, 2023 at 18:52