Calculate the Pontragin dual $\text{Hom}_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})$. I'm calculating the $\text{Ext}^1_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z}, \mathbb{Z})$. In particular, $\text{Ext}^1_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z}, \mathbb{Z}) \cong \text{Hom}_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})$ the Pontragin dual. However how should I see what this group is? I only know that for finite abelian groups on the first position, this dual is isomorphic to it self. But no idea for infinitely generated groups. 
 A: Throughout $\Bbb Q_p$ and $\Bbb Z_p$ denote the $p$-adic numbers and $p$-adic integers.
Observe that
$$\Bbb Q/\Bbb Z\cong\bigoplus_p \Bbb Q_p/\Bbb Z_p.$$
Thus
$$\textrm{Hom}(\Bbb Q/\Bbb Z,\Bbb Q/\Bbb Z)
\cong\prod_p \textrm{Hom}(\Bbb Q_p/\Bbb Z_p,\Bbb Q/\Bbb Z).$$
Any homomorphism from $\Bbb Q_p/\Bbb Z_p$ to $\Bbb Q/\Bbb Z$
has image within $\Bbb Q_p/\Bbb Z_p$ and so
$$\textrm{Hom}(\Bbb Q/\Bbb Z,\Bbb Q/\Bbb Z)
\cong\prod_p \textrm{Hom}(\Bbb Q_p/\Bbb Z_p,\Bbb Q_p/\Bbb Z_p).$$
Each endomorphism of $\Bbb Q_p/\Bbb Z_p$ is induced by multiplication
by an element of $\Bbb Z_p$. Therefore
$$\textrm{Hom}(\Bbb Q/\Bbb Z,\Bbb Q/\Bbb Z)
\cong\prod_p \Bbb Z_p.$$
This group is often denoted as $\hat {\Bbb Z}$, the profinite
completion of $\Bbb Z$. It's also the absolute Galois group of each finite field.
A: Slightly different argument: Note that $\mathrm{Hom}_\mathbb{Z}(\varinjlim_i A_i, B) \simeq \varprojlim_i\mathrm{Hom}_\mathbb{Z}(A_i, B)$ (this is just restating of the universal property for $\varinjlim:$ the left hand side consists of morphisms $\varinjlim_i A_i \rightarrow B$, the right hand side describes co-cones $A_i \rightarrow B, i \in I, $ of compatible systems of maps).
Then, as $\mathbb{Q}/\mathbb{Z} = \varinjlim_{n | m} \mathbb{Z}[\frac{1}{n}]/\mathbb{Z},$ we have
$$\mathrm{Hom}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})=\mathrm{Hom}_\mathbb{Z}(\varinjlim_{n | m} \mathbb{Z}[\frac{1}{n}]/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \varprojlim_{n | m}\mathrm{Hom}_\mathbb{Z}(\mathbb{Z}[\frac{1}{n}]/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})\simeq \varprojlim_{n | m} \mathbb{Z}/n\mathbb{Z}=\widehat{\mathbb{Z}}.$$
