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 :)