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This is part of Exercise 4.3.5 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.

Here is the previous part:

An abelian $p$-group has finitely many elements of each order iff it satisfies ${\rm min}$

The Details:

A group $G$ satisfies ${\rm min}$ if every set $S$ of subgroups of $G$ has at least one minimal element (with respect to $H\le K$ for $H,K\in S$).

A group $G$ is a $p$-group, for prime $p$, if each element of $G$ has order a power of $p$.

On page 12, ibid.,

A torsion group (or periodic group) is a group all of whose elements have finite order.

The Question.

[Using the previous part], characterise abelian groups which have only finitely many elements of each order (including $\infty$).

Thoughts:

I think part of the solution has to do with torsion groups. The case when the group has elements of infinite order might be handled separately.${}^\dagger$

Cheating a little, I looked through Rotman's "An Introduction to the Theory of Groups (Fourth Edition)" and found this:

Theorem 10.7 (Primary Decomposition). Every torsion group $G$ is a direct sum of [abelian $p$-groups].

This is something to keep in mind as a guide only. I don't think it would be constructive to cite it in a solution to the exercise (without proof).

A torsion group can have infinitely many elements of a given order; for example, consider

$$\bigoplus_{i=1}^\infty\Bbb Z_2.$$

The question is trivial for finite groups.

I think I could answer this myself if I had enough time. Therefore, I would prefer hints over full solutions.

$\dagger$ If $g\in G$ has infinite order, then so does $g^i$ for each $i\in \Bbb N\setminus \{0\}$, since otherwise, if $\lvert g^i\rvert=n$, then $\lvert g\rvert$ divides $in$, a contradiction. Thus there cannot be finitely many elements of infinite order (other than zero of them).

Please help :)

Shaun
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    The Primary Decomposition theorem is easy: just let $P$ be the set of all $p$-power elements, and notice that for any prime it is a subgroup. Then $G$ is the product of the $P$s over all primes $p$. – David A. Craven Aug 21 '21 at 18:35
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    Moding out by the torsion and using the observation you made you can conclude the group is torsion. Any torsion abelian group is a direct sum (product/restricted product/weak product) of its $p$-parts: the proof is identical to that used in the case of finitely generated abelian groups. That reduces the problem to the $p$-parts, and you just finished characterising the abelian $p$-groups that have only finitely many elements of any given order. – Arturo Magidin Aug 21 '21 at 19:06
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    But: since there may be infinitely many primes $p$ involved, the group need not have min any more. – David A. Craven Aug 21 '21 at 20:06
  • (An example of such a group is $\mathbb Q/\mathbb Z$, no?) – David A. Craven Aug 21 '21 at 20:26
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    @DavidA.Craven: Or more simply, $\oplus_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z}$. ($\mathbb{Q}/\mathbb{Z}$ being the sum of $\mathbb{Z}_{p^{\infty}}$). Once you have reduced to $p$-groups, you then just need to restrict to finitely many nontrivial factors. – Arturo Magidin Aug 21 '21 at 20:30
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    @ArturoMagidin But Q/Z is one of the groups under consideration, as it satisfies the hypothesis. So we cannot restrict to finitely many factors. – David A. Craven Aug 21 '21 at 20:53
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    @DavidA.Craven: I got a little confused and I was less than clear: if you want to describe all abelian groups that satisfy min, then you end up with "finitely many factors" (only finitely many elements of any given order, plus must be a $\pi$-group for some finite set of primes; since a torsionfree element yields an infinite descending chain of subgroups too). For the groups under consideration here, given the context in Robinson, I expect he wants to have it be described as "torsion groups whose $p$-part satisfy min, for each prime $p$". – Arturo Magidin Aug 21 '21 at 22:46
  • @ArturoMagidin I agree that that's the only obvious conclusion, but I don't like it, and can't help but wonder if there's some nice alternative that the author was driving at. – David A. Craven Aug 21 '21 at 22:54

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