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We say that an abelian group $G$ is mixed if it has elements $ \neq 0$, that are of finite order (torsion elements), as well as elements of infinite order (torsion-free elements). We denote the torsion subgroup of $G$ as $T(G)$. We call a mixed group "splitting" if it can be written as a direct sum $G = T(G) \oplus H$, where $H$ is a torsion-free subgroup of $G$.

In the given exercise, we have a group $G=\underset{p\in\mathbb{P}}{\prod}\mathbb{Z}/p\mathbb{Z}$ the product of $\mathbb{Z}/p\mathbb{Z}$ over all prime numbers $p$. We can demonstrate without difficulty that the torsion subgroup $T(G)$ is equal to the direct sum of $\mathbb{Z}/p\mathbb{Z}$ for all prime numbers $p$ : $T(G)=\underset{p\in\mathbb{P}}{\bigoplus}\mathbb{Z}/p\mathbb{Z}$

Assume that $G$ is splitting, and we have $G=\bigoplus_p\mathbb{Z}/p\mathbb{Z}\oplus H$, where $H$ is a torsion-free subgroup of $G$.

I tried to find a contradiction, but I failed. I would like your help with this problem.

Daniel Donnelly
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John123
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  • Which kind of group product is $\prod_{p \in \mathbb{P}} \mathbb{Z}/p\mathbb{Z}$? The element $(1,1,1,\ldots) \in \bigoplus_{p \in \mathbb{P}} \mathbb{Z}/p\mathbb{Z}$ has infinite order, so is not in $T(G)$. – aschepler Jul 10 '23 at 15:25
  • This is discussed in the Mixed Groups chapter of Fuchs' book Abelian Groups (see Example 1.2) and here https://math.stackexchange.com/questions/10007/the-torsion-submodule-of-prod-mathbbz-p-is-not-a-direct-summand-of-prod?rq=1 – kabenyuk Jul 10 '23 at 15:28
  • kabenyuk, in Example 1.2, proves that $A_1=\langle T_1,b_1,\dots,b_n,\dot\rangle$ is not splitting, not that $G=\prod_{p\in\mathbb{P}}\mathbb{Z}/p\mathbb{Z}$ is not splitting,??? – John123 Jul 10 '23 at 16:23
  • This is certainly true, but it is easy to see that if $G=T(G)\oplus H$, then since $T(G)<A_1$ we have $T(A_1)=T(G)$ and $A_1=T(A_1)\oplus(H\cap A_1)$. Contradiction. – kabenyuk Jul 10 '23 at 18:04
  • kabenyuk, I could not understand his reasoning, especially when he shows that $p_nt_n+p_ng_n=p_nb_n=b_0-a_n=(t_0+a_n)+g_0 \Longrightarrow p_nt_n=t_0-a_n$. I couldn't grasp where the second implication came from. That's why I was trying to find another way. – John123 Jul 10 '23 at 21:17
  • If you want to reason differently. Try to prove that the factor group $G/T$ is divisible, but for any element $g\in G$, $g\neq1$, there exists an integer $n$ such that the equation $nx=g$ has no solutions. Hence it follows easily that $G$ is not splitting. – kabenyuk Jul 11 '23 at 04:54
  • Kabenyuk, thank you for your help. That was really helpful. However, if it is not a bother, could you explain to me in that example how Fuchs arrived at that implication? – John123 Jul 11 '23 at 11:50
  • Starting from $p_nt_n+p_ng_n=p_nb_n=b_0-a_n=(t_0-a_n)+g_0$, we have:

    $p_nt_n+p_ng_n=b_0-a_n$

    Rearranging, we get:

    $p_nt_n = (b_0-a_n) - p_ng_n$

    Using the expression $b_0-a_n=(t_0-a_n)+g_0$, we substitute it into the equation: $p_nt_n = [(t_0-a_n)+g_0] - p_ng_n$. Expanding, we have: $p_nt_n = t_0 - a_n + g_0 - p_ng_n$ At this point, we can further rearrange the terms as: $p_nt_n = t_0 - a_n - p_ng_n + g_0$. Since $g_n$ belongs to the torsion-free subgroup $G$, we can rewrite $-p_ng_n + g_0$ as $g_n'$, where $g_n' \in G$. Thus, the equation becomes: $p_nt_n = t_0 - a_n + g_n'$

     – John123 Jul 11 '23 at 12:18
  • How can I show that $g_n'=0$ ? – John123 Jul 11 '23 at 12:21

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