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This is Exercise 4.3.8 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

(NB: Modules aren't covered in Robinson's book yet.)

The Details:

Let $p$ be prime. A $p$-group is a group all of whose elements have a $p$ power order.

On page 12 of Robinson's book,

A torsion group [. . .] is a group all of whose elements have finite order.

On page 94, ibid.,

An element $g$ of an abelian group $G$ is said to be divisible in $G$ by a positive integer $m$ if $g=mg_1$ for some $g_1$ in $G$. [. . .]

An abelian group $G$ is said to be divisible of each element is divisible by every positive integer.

On page 106, ibid.,

A subgroup $H$ of an abelian group $G$ is called pure if

$$nG\cap H=nH$$

for all integers $n\ge 0$.

On page 107, ibid.,

Let $G$ be an abelian torsion group. A subgroup $B$ is called a basic subgroup if it is pure in $G$, it is the direct sum of cyclic groups, and $G/B$ is divisible.

On page 108, ibid.,

An additively written group is called bounded if its elements have boundedly finite orders: [. . .] multiplicative groups with this property are said to have finite exponent.

The Question:

An abelian $p$-group has a bounded basic subgroup if and only if it is the direct sum of a divisible group and a bounded group.

Thoughts:

Let $G$ be an abelian $p$-group.

There's a lot going on here. Divisible groups familiar to me:

Of course, so are abelian groups, $p$-groups, cyclic groups, bounded groups (in their multiplicative form), pure subgroups, torsion groups, and direct sums.

I think $G$ has to be infinite.

It might be easier to consider when $G$ is finitely generated, since then we could use the fundamental theorem of finitely generated abelian groups to get $G$ as a direct sum of some groups; I'm not sure how to take it from there. However, Theorem 4.2.9, page 103, ibid., states,

A finitely generated abelian group $G$ is finite if it is a torsion group.

This is a question I think I could answer myself. However, I'm coming towards the end of my degree so I don't have enough time to spend on reading textbooks thoroughly - but I think I would lose momentum in Robinson's book if I left it any longer. The kind of answer I'm hoping for is a full solution.

Please help :)

Shaun
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    One direction should be clear: since divisible groups are not bounded, if $G=D\oplus B$ with $B$ bounded and $D$ divisible, then $B$ is basic (and bounded). Conversely, if $G$ has a basic bounded subgroup $B$, and $D$ is any divisible subgroup of $G$, then $B\cap D$ must be trivial. I would take a maximal divisible subgroup of $G$ and try to show that $G$ is the sum of $B$ and such a subgroup. – Arturo Magidin Sep 09 '21 at 19:22
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    Scratch the second part... No divisible subgroup does not mean intersection with any divisible subgroup is trivial. As any proper nontrivial subgroup of the Prufer group shows. – Arturo Magidin Sep 09 '21 at 20:30
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    Okay, if $B$ is basic, $D$ divisible, and $x\in D\cap B$, write $B$ as a direct sum of cyclic groups; then $x$ lies in a finite sum $C_{p^{a_1}}\oplus\cdots \oplus C_{p^{a_k}}$ which is a direct summand of $B$. Since $B$ is pure and $x\in nG$ for all $n\gt 0$, then $x\in nB$ for all $n\gt 0$, and in particular $x\in p^{a_1+\cdots+a_k}B$, But this implies that $x$ is in fact trivial, so $B\cap D$ is indeed trivial. – Arturo Magidin Sep 09 '21 at 21:01

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