Prove that $G = \prod_{p\hspace{1mm} prime}\frac{\mathbb{Z}}{p\mathbb{Z}}$ is not isomorphic to $\frac{G}{torsion(G)} \bigoplus torsion(G)$. Let $G$ be the group  $ \prod_{p\hspace{1mm} prime}\frac{\mathbb{Z}}{p\mathbb{Z}}$, i.e. the cartesian product of the factor groups of $\mathbb{Z}$ modulo every prime $p$. (so essentially a group of sequences of integers modulo ascending primes). let $torsion(G)$ be the torsion group of $G$, i.e. $\{g \in G| \exists n \in \mathbb{N} s.t. g^n = id_G \}$. Then I want to show that $G$ is not isomorphic to  $\frac{G}{torsion(G)} \bigoplus torsion(G)$, where $\bigoplus$ denotes the direct sum. 
Now it would appear that this is a problem that calls for some of the more abstract areas of mathematics (My hunch is Zorn's lemma...). Here's what I got so far. Because distinct primes are always relatively prime, I can write  $ \prod_{p\hspace{1mm} prime}\frac{\mathbb{Z}}{p\mathbb{Z}} = \frac{\mathbb{Z}}{\prod_{p \hspace{1mm}prime}p\mathbb{Z}}$. This group doesn't have a maximum element under the "divisible" order relation right? A.k.a. no element $x$ that is divisible by every prime $p$. But $\frac{G}{torsion(G)}$ may have, by Zorn's lemma, perhaps?
This problem is very mysterious to me...
 A: Let $\mathcal{P}$ be the set of primes. We will show that


*

*the torsion subgroup $torsion(G)$ is $\bigoplus_{p \in \mathcal{P}} \mathbb{Z}/p\mathbb{Z}$;

*the quotient $G/torsion(G)$ is a divisible abelian group; and

*the only divisible element of $G$ is $0$.


Since $G/torsion(G) \ne 0$, we will have a contradiction.
For 1, let $(a_{p})_{p \in \mathcal{P}} \in G = \prod_{p \in \mathcal{P}} \mathbb{Z}/p\mathbb{Z}$ be an element, and suppose that $n (a_{p})_{p \in \mathcal{P}} = 0$. If $p \in \mathcal{P}$ is a prime for which $a_{p} \ne 0$, then $p$ divides $n$; thus there are finitely many such $p$; thus $(a_{p})_{p \in \mathcal{P}} \in \bigoplus_{p \in \mathcal{P}} \mathbb{Z}/p\mathbb{Z}$.
For 2, let $\alpha \in G/torsion(G)$ be an element and let $n \in \mathbb{Z}_{\ge 1}$ be a positive integer; we want to find $\beta \in G/torsion(G)$ such that $n\beta = \alpha$. Choose $(a_{p})_{p \in \mathcal{P}} \in G$ whose image in $G/torsion(G)$ is $\alpha$. We can without loss of generality alter finitely many coordinates of $(a_{p})_{p \in \mathcal{P}}$, so we may assume that $a_{p} = 0$ for all $p \in \mathcal{P}$ such that $p$ divides $n$. Then we may find $(b_{p})_{p \in \mathcal{P}} \in G$ such that $n(b_{p})_{p \in \mathcal{P}} = (a_{p})_{p \in \mathcal{P}}$, working coordinate-wise. The image of $(b_{p})_{p \in \mathcal{P}}$ in $G/torsion(G)$ is our desired $\beta$. (See also this question and this question.)
For 3, let $(a_{p})_{p \in \mathcal{P}} \in G$ be a nonzero element; choose $p \in \mathcal{P}$ such that $a_{p} \ne 0$; then there does not exist $(b_{p})_{p \in \mathcal{P}} \in G$ such that $p(b_{p})_{p \in \mathcal{P}} = (a_{p})_{p \in \mathcal{P}}$.
