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Question Let $G$ be a countably infinite reduced abelian $p$-group. Is it always possible to write it has an infinite direct sums of non-trivial groups? If it is not true, how far is $G$ from being an infinite direct sum?

On one hand, by Prüfer's second theorem, the answer is true if $G$ has no elements of infinite height. On the other hand, if $G$ has an element of infinite heigth and $G\cong\bigoplus_{i\geq 1} L_i$, at least one of the $L_i$ has an element of infinite height (and is decomposable).

Now, any reduced $G$ with an element of infinite height can be factored has $G= G_1\times H_1$ with $H_1$ cyclic. In fact, on can take $H_1=C_{p^n}$ whenever $G$ has an element of order $p$ and height $n-1$. Since $G_1$ contains an element of infinite height, we can repeat this procedure. That is, for all $j\geq 1$ we have a decomposition $$G=G_j\oplus\bigoplus_{i=1}^jH_i.$$ Let $K$ be the intersection of the decreasing sequence $(G_i)_{i\geq 1}$. Then $G$ contains the direct sum $K\oplus\bigoplus_{i\geq1}H_i$. If the $H_i$ have bounded orders, $\bigoplus_{i\geq1}H_i$ is a direct factor of $G$ and we are done. We can assume that this is the case as soon as $G$ has infinitely many elements of order $p$ and of bounded height.

We also have an obvious map $\varphi\colon G\to\prod_{i\geq 1}H_i$ with kernel $K$. So one can hope that $G$ sits somewhere between $K\oplus\bigoplus_{i\geq1}H_i$ and $K\oplus\prod_{i\geq1}H_i$. However, in general it is not true that $\varphi(G)=\bigoplus_{i\geq1}H_i$ and it is unclear if $1\to K\to G\to\varphi(G)\to 1$ splits.


I claim that morally it is enough to understand what happens for the following classical example of a reduced group with an element of infinite height: $$A=\langle x_0,x_1,\dots\ |\ px_0=0,p^ix_i=x_0,[x_i,x_j]=0\rangle$$.

Indeed, if $G$ is reduced and has an element of infinite height, it has an element $a$ of infinite height such that if $p^nb=a$ with $n\geq 1$, then $b$ has finite height (as otherwise one can find a divisible subgroup). Intuitively, this $a$ will be our $x_0$. For any $b$ with $pb=a$ there exists $c$ with $pc=a$ and $h(c)>h(b)=k$. Hence $b-c$ has order $p$ and height $h(b)$. Such an element gives us a direct cyclic summand $C_{p^{k+1}}$. If for a given $k$ we have infinitely many $b_i$ with $pb_i=a$ and $h(b_i)=k$, then $G$ contains an infinite direct sum of copies $C_{p^{k+1}}$, which is a direct summand (as the orders are bounded). Hence, we can assume that for every $k$ there is only finitely many $b$ with $pb=a$ and $h(b)=k$. The group $A$ is the prototipical example of such a behaviour.

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Let $G$ have an element of infinite height. By Kaplansky's Theorem (Fuchs (2015) p 300) there are infinitely many $k\in\mathbb N$ such that $G$ has an element $b$ with indicator $(0,\dots, k,\infty)$.

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  • $\begingroup$ Could you please explain why this is enough to conclude? As far as I understand, such a $b$ gives us a cyclic direct factor $C_{p^{k+1}}$, but I don't see how this is enough to obtain an infinite direct sum. $\endgroup$
    – PHL
    May 12, 2022 at 6:24
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    $\begingroup$ You are right, this is not enough. In fact a countable reduced abelian $p$--group is not necessarily an infinite direct sum. Counterexample: Pruefer's group $G=\langle a_i\colon i\in \mathbb N\rangle$ with $pa_0=0= p^ia_{i-1}$ for all $i$. Then $G$ is not $\Sigma$--cyclic but $G/\langle a_0\rangle$ is. $\endgroup$ May 12, 2022 at 8:40

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