Is an abelian group characterized by its localizations? Let $G$ and $H$ be countable abelian groups. Assume that for every prime number $p$ there is an isomorphism $G\otimes_{\mathbb Z} \mathbb Z[\frac{1}{p}]\cong H\otimes_{\mathbb Z} \mathbb Z[\frac{1}{p}]$. Does it follow that $G$ and $H$ are isomorphic as abelian groups?
(Note that this is certainly true for finitely generated groups. Moreover it also holds if all isomorphisms $G\otimes_{\mathbb Z} \mathbb Z[\frac{1}{p}]\cong H\otimes_{\mathbb Z} \mathbb Z[\frac{1}{p}]$ are induced by one fixed group homomorphism $G\to H$. In both cases it is enough to consider two different prime numbers.)
 A: The answer to the question is yes, and it only needs two different prime numbers $p,q$. Moreover, the proof is valid for arbitrary abelian groups $G$ and $H$ (the compability of the isomorphisms isn't needed either). It uses Hurkyl's approach and goes as follows. 
For any abelian group $A$ and $n\in \mathbb N$, we use the notation $A_n:=A\otimes_{\mathbb Z}\mathbb Z[1/n]$. Suppose that there are group isomorphisms $\varphi_p:G_p\to H_p$ and $\varphi_q:G_q\to H_q$. Consider the short exact sequence
\begin{array}{ccccccccc}
 0&\to &\mathbb Z&\xrightarrow{\iota_p\oplus -\iota_q} & \mathbb Z[1/p]\oplus \mathbb Z[1/q] &\xrightarrow{\kappa_p+\kappa_q} & \mathbb Z[1/(pq)]  &\to &0,\\
\end{array}
where all maps $\iota_p,\iota_q,\kappa_p$ and $\kappa_q$ are the canonical embeddings.
Since $\operatorname{Tor}(G,\mathbb Z[1/(pq)])\cong \operatorname{Tor}(H,\mathbb Z[1/(pq)])\cong0$ tensoring with $G$ and $H$ yields the following commutative diagram with exact rows
\begin{array}{ccccccccc}
 0&\to &G&\xrightarrow{} &G_p\oplus G_q &\xrightarrow{} &G_{pq}&\to &0\\
  & & \downarrow{\psi}& & \downarrow{\varphi_p\oplus\varphi_q}& & \downarrow{\tilde{\varphi}_p+\tilde{\varphi}_q} \\
 0&\to &H&\xrightarrow{}&H_p\oplus H_q &\xrightarrow{}&H_{pq}&\to &0
\end{array}
where $\tilde{\varphi}_p$ and $\tilde{\varphi}_q$ are the canonical extensions of $\varphi_p$ and $\varphi_q$, respectively. We now show that $\psi$ is an automorphism. Set $G(p^{\infty}):=G\otimes_{\mathbb Z}\mathbb Z(p^{\infty})$, where $\mathbb Z(p^{\infty})$ denotes the $p$-Prüfer group, and define $H(p^{\infty})$ analogously. Furthermore, write $G(p)$ and $H(p)$ for the $p$-torsion subgroups of $G$ and $H$, respectively. We obtain a commutative diagram with exact rows
\begin{array}{ccccccccc}
 0&\to &G/G(p)&\xrightarrow{} &G_p&\xrightarrow{} &G(p^{\infty})&\to &0\\
  & & \downarrow{\overline{\psi}}& & \downarrow{\varphi_p}& & \downarrow{\overline{\varphi}_p} \\
0&\to &H/H(p)&\xrightarrow{} &H_p&\xrightarrow{} &H(p^{\infty})&\to &0\\
\end{array}
where $\overline{\psi}$ and $\overline{\varphi}_p$ are the induced morphisms. Since $\varphi_p$ is an isomorphism, the Snake lemma yields an isomorphism $\operatorname{coker}(\overline{\psi})\cong \operatorname{ker}(\overline{\varphi}_p)$. It also follows that $\overline{\psi}$ is injective. However,  $\operatorname{ker}(\overline{\varphi}_p)$ is a $p$-group and consequently $\overline{\psi}$ is an isomorphism. Performing the same argument for $q$ also shows that $\psi:G\to H$ restricts to an isomorphism $G(p)\to H(p)$. Finally, the diagram
\begin{array}{ccccccccc}
 0&\to &G(p)&\xrightarrow{} &G&\xrightarrow{} &G/G(p)&\to &0\\
  & & \downarrow{\psi}& & \downarrow{\psi}& & \downarrow{\overline{\psi}} \\
0&\to &H(p)&\xrightarrow{} &H&\xrightarrow{} &H/H(p)&\to &0\\
\end{array}
together with an application of the Five Lemma yields that $\psi:G\to H$ is an isomorphism of abelian groups.
