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According to wikipedia, the Pontryagin dual of a Prüfer group is isomorphic to a group of p-adic integers.

Where can I find a proof for it on the internet?

Minimus Heximus
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    For direct limits of topological groups (here finite), see https://math.stackexchange.com/questions/2865 – Watson Nov 23 '18 at 15:02

2 Answers2

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The Pruefer group ${\mathbb Z}_{p^{\infty}}$ is the direct limit of the sequence $${\mathbb Z}/p{\mathbb Z}\hookrightarrow {\mathbb Z}/p^2{\mathbb Z}\hookrightarrow ...$$ Applying $(-)^{\wedge} = \text{Hom}_{\text{cnt}}(-,{\mathbb S}^1)$ shows that $\left({\mathbb Z}_{p^\infty}\right)^{\wedge}$ is the inverse limit of the Pontryagin duals of ${\mathbb Z}/p^k {\mathbb Z}$. Finite groups are self-dual, so you end up with the $p$-adic integers.

This is a bit sloppy, since, as Alex pointed out, one has to be careful in which categories to talk about direct and inverse limits. See the comment below.

Addendum

Unfolding the definitions, you get the following explicit pairing: If $x := \sum\limits_{n=0}^{\infty} a_n p^n\in{\mathbb Z}_p$ is a $p$-adic integer and $y\in{\mathbb Z}[\frac{1}{p}]/{\mathbb Z}={\mathbb Z}_{p^{\infty}}$ is an element of the Pruefer $p$-group, then $\langle x,y\rangle := xy\in {\mathbb Z}_{p^\infty}\subset{\mathbb Q}/{\mathbb Z}\subset{\mathbb S}^1$ is the value of the character ${\mathbb Z}_{p^\infty}\to{\mathbb S}^1$ associated to $x$ when evaluated at $y$. Here the product $xy$ makes sense since multiples of $y$ by sufficiently high $p$-powers vanish in ${\mathbb Q}/{\mathbb Z}$.

Hanno
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    Silly question. Is it obvious that the $\text{Hom}\text{cont}(\varinjlim A_j,A)=\varprojlim \text{Hom}\text{cont}(A_j,A)$ as topological abelian groups? – Alex Youcis Nov 24 '14 at 08:58
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    @AlexYoucis Not a silly question at all. Pontryagin duality is an anti-equivalence of categories, hence it exchanges direct and inverse limits in the category $\textbf{LCHAG}$ of locally compact Hausdorff abelian topological groups. The point one has to check is that the (discrete) abelian group direct limit used to present the Pruefer group is in fact a direct limit in $\textbf{LCHAG}$ (this is because discrete abelian groups are coreflective in $\textbf{LCHAG}$ through the 'discretization' functor), and that inverse limits of compact groups can be computed as in $\textbf{Top}$. – Hanno Nov 24 '14 at 09:06
  • Interesting! Thanks for the explanation! – Alex Youcis Nov 24 '14 at 09:25
  • Thanks. Good proof. Yet I'd like to see a more elementary proof so that the structure of homomorphisms $f:\Bbb Z_{p^\infty}\to \Bbb T$ will be known. – Minimus Heximus Nov 24 '14 at 14:58
  • @user795571: I added an explicit description - is this more what you where looking for? – Hanno Nov 24 '14 at 15:40
  • Thank you. But what is ${\mathbb Z}[\frac{1}{p}]$? – Minimus Heximus Nov 24 '14 at 16:29
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    @user795571: The localization of ${\mathbb Z}$ at the prime $p$, realized as the rational numbers with denominator a $p$-th power. – Hanno Nov 24 '14 at 16:31
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Since $\mathbb{Z}(p^\infty)$ is torsion, the image of a character $\mathbb{Z}(p^\infty)\to\mathbb{T}=\mathbb{R}/\mathbb{Z}$ is contained in the torsion part of $\mathbb{T}$, which is $\mathbb{Q}/\mathbb{Z}$. Since $\mathbb{Z}_{p^\infty}$ is a $p$-group, the image is further contained in the $p$-component, which is exactly $\mathbb{Z}_{p^\infty}$. So $$ \operatorname{Hom}(\mathbb{Z}(p^\infty),\mathbb{T})= \operatorname{Hom}(\mathbb{Z}(p^\infty),\mathbb{Z}(p^\infty)). $$ The topology on this group is still the one of pointwise convergence.

Now write $\mathbb{Z}(p^\infty)= \operatorname{colim}\limits_{n\in\mathbb{N}}\mathbb{Z}(p^n)$, so $$ \operatorname{Hom}(\mathbb{Z}(p^\infty),\mathbb{Z}(p^\infty))\cong \lim_{n\in\mathbb{N}}\operatorname{Hom}(\mathbb{Z}(p^n),\mathbb{Z}(p^\infty))\cong \lim_{n\in\mathbb{N}}\operatorname{Hom}(\mathbb{Z}(p^n),\mathbb{Z}(p^n)) $$ and it's just a matter of checking that this is the required isomorphism.

egreg
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