# Direct sum of Prüfer groups and $\mathbb Q/\mathbb Z$

It can be easily shown that, the Prüfer $p$-group $\mathbb Z(p^\infty)$ is isomorphic to multiplicative group $$R_p=\{e^{2\pi ik/p^n}|k\in\mathbb Z,n\geq0\}$$ Now I want to prove that:

$$\frac{\mathbb Q}{\mathbb Z}\cong\sum_p \mathbb Z(p^\infty)$$

What I did is to consider the following map: $$\frac{\overbrace{p_1^{a_1}p_2^{a_2}...}^{k_i}}{p_i^{a_i}}+\mathbb Z\to \big(e^{2\pi ik_1/p_1^{a_1}},e^{2\pi ik_2/p_2^{a_2}},...\big)$$ Do you know another map? :-) Thanks.

• Your question is equivalent to asking for automorphisms of $\mathbb Q/\mathbb Z$. Can you give examples of these? – Mariano Suárez-Álvarez Oct 13 '12 at 6:46
• @MarianoSuárez-Alvarez: Honestly, I didin't think about this point of view. You say my map is not a practical map? – mrs Oct 13 '12 at 7:04
• Your map looks fine: the point was that once you have one isomorphism between $G$ and $H$, asking for another is equivalent to asking for automorphisms of either $G$ or $H$, because given one of those we just compose it on to the isomorphism we already have to get a new one. Hence the question: can you think of any automorphisms of $\mathbb{Q}/\mathbb{Z}$? – Kevin Carlson Oct 13 '12 at 7:49
• Look up profinite completions and Pontryagin duality. – darko Oct 13 '12 at 8:04
• Thanks @MarianoSuárez-Alvarez. – mrs Oct 13 '12 at 9:38

As pointed out in the comments above, asking for a different map is equivalent to asking for an automorphism of $\mathbb{Q} / \mathbb{Z}$. I show here the construction of a map going the other way.
The $\mathbb{Z}$-module embeddings $\mathbb{Z} [1/p] / \mathbb{Z} \rightarrow \mathbb{Q} / \mathbb{Z}$ induce a map $\alpha : \oplus_{p \in P} \mathbb{Z} [1/p] / \mathbb{Z} \rightarrow \mathbb{Q} / \mathbb{Z}$, where $P$ is the set of prime integers. We show it is an isomorphism.
$\alpha$ is surjective: for any integers $a, p, q$ with $p$ and $q$ relatively prime, there are integers $n$ and $m$ such that $\frac{a}{pq} = \frac{n}{p} + \frac{m}{q}$. Thus, for a rational number $A = \frac{a}{\prod_{p} p^{k_p}}$ with $k_p = 1$ for all but finitely many primes, we can write $A$ as $\sum_{p} A_p$, where $A_p = a_p / p^{k_p}$, for integers $\{ a_p \}_{p \in P}$. Then $\overline{A} \in \mathbb{Q} / \mathbb{Z}$ can be written as $\alpha \left( ( \overline{A_p} )_{p \in P } \right)$.
$\alpha$ is injective: take integers $\{ a_p \}_{p \in P}, \{ k_p \}_{p \in P}$ with all but finitely many $a_p$ equal to $0$, such that $\alpha \left( ( \overline{a_p / p^{k_p} })_{p \in P} \right) = \overline{0}$. Without loss of generality, $a_p$ can be taken such that $\sum_{p \in P} a_p / p^{k_p} = 0$. Clearing the demoninators, we have $\sum_{p \in P} a_p \prod_{q \neq p} q^{k_q} = 0$, from which we can see that $p^{k_p} | a_p \forall p \in P$. Therefore $( \overline{a_p / p^{k_p} })_{p \in P} = 0$.