According to Wikipedia, for a prime $p$,
The Prüfer p-group may be represented as a subgroup of the circle group, $U(1)$, as the set of $p^n$th roots of unity as $n$ ranges over all non-negative integers:
$\mathbb{Z}(p^\infty)=\{\exp(2\pi i m/p^n) \mid m\in \mathbf{Z}^+,\,n\in \mathbf{Z}^+\}=G.$
Alternatively, the Prüfer p-group may be seen as the Sylow $p-$subgroup of $Q/Z$, consisting of those elements whose order is a power of $p$:
$\mathbb{Z}(p^\infty)=\mathbb{Z}[1/p]/\mathbb{Z}=H$
I have a very hard time to prove that the group $G$ and $H$ are isomorphic. Any help is highly appreciated.
Ask yourself: what do elements of each look like?
You should think $e^{2\pi i\color{Red}{\square}}$ in $G$, and $\color{Red}{\square}+\Bbb Z$ in $H$, with both $\color{Red}{\square}$s looking exactly the same...