An abelian $p$-group has finitely many elements of each order iff it satisfies ${\rm min}$ 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.
(NB: I have left out the modules tag for a reason: the tools available here are entirely group theoretic.)
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 $p$.
Let $n\in\Bbb N$. Then $G[n]$ is the subgroup of a group $G$ of all elements $g\in G$ such that $ng=0$.
The Question:

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

Thoughts:
Let $p$ be prime. Suppose $G$ is an abelian $p$-group.

$(\Leftarrow)$
Suppose $G$ satisfies ${\rm min}$. Suppose, further, that $G$ has infinitely many elements of order $p^k$ for some $k\in\Bbb N$. I think it would help to consider $G[p^k]$; I'm not sure why.
I don't know where to go from here.

$(\Rightarrow)$
Suppose $G$ has finitely many elements of each order.
What do I do next?

I'm aware that it is possible to have an infinite group with finitely many elements of each order; for example, consider
$$\bigoplus_{p\text{ prime}}\Bbb Z_p.$$
The question is trivial if $G$ is finite.

This appears to be a question I could answer myself with more time. Thus hints are preferred over full solutions.

Please help :)
 A: Let $\Omega_i(G)$ denote the subgroup generated by all elements of order $p^i$. Recall that an elementary abelian group is simply a direct product of (possibly infinitely many) copies of the cyclic group $C_p$*. Its rank is the number of copies of $C_p$. Notice that raising to the power $p$ is a map from $\Omega_i(G)$ to $\Omega_{i-1}(G)$ for each $i$. In particular, it induces a homomorphism from the quotient groups $\Omega_{i+1}/\Omega_i(G)\to \Omega_i(G)/\Omega_{i-1}(G)$, and this homomorphism must be injective. Since all elements of this quotient have order $p$, it is an elementary abelian $p$-group, so we are interested in its rank. In particular, the rank of $\Omega_i/\Omega_{i-1}(G)$ is at most the rank of $\Omega_1(G)/\Omega_0(G)=\Omega_1(G)$. For $i\in \mathbb N$, write $r(G,i)$ for the rank of $\Omega_i(G)/\Omega_{i-1}(G)$ (which could be $\infty$), a weakly descending sequence for any fixed $G$, i.e., $r(G,i)\geq r(G,j)$ for all $i\leq j$. This applies for all subgroups of $G$ as well, and of course $r(G,i)\geq r(H,i)$ for any subgroup $H$ of $G$.
Thus if there are infinitely many elements of order $p^i$ for some $i$ then there are infinitely many of order $p$. These look like an infinite-dimensional $\mathbb{F}_p$-vector space. Take the set of all infinite subgroups of $\Omega_1(G)$. Certainly there are infinitely many of them, and they cannot have a minimal element, just by removing one basis element at a time. Thus $G$ cannot have min.
Conversely, let
$$H_1>H_2>H_3>\cdots$$
be an infinite descending chain of subgroups of $G$, but that $G$ has only finitely many elements of order $p$ (and hence of order $p^i$ for all $i$). For each $i$, let $f(i)$ denote the smallest order of an element of $H_i\setminus H_{i+1}$, which exists since all elements of $G$ have finite order. Notice that $f:\mathbb N\to\mathbb N$ cannot take the same value $m$ infinitely often, as then there would be infinitely many elements of order $p^m$ in $G$.
Note that $r(H_i,j)\geq r(H_{i+i},j)$ for all $i$ and $j$. If $f(1)=m_1$, then we see that $r(H_1,m_1)>r(H_2,m_1)$, which means that the latter is strictly less than $r$, and indeed therefore $r(H_i,m)<r$ for all $i\geq 2$ and $m\geq m_1$. We will now repeat the process, but we have to be a little careful.
Since the sequence $f(i)$ diverges, we can produce a weakly increasing subsequence, with indices $i_1,i_2,\dots$, and values $m_j=f(i_j)$. We have
$$ r\geq r(H_{i_1},m_1)>r(H_{i_1+1},m_1)\geq r(H_{i_2},m_2)>r(H_{i_2+1},m_2)\geq r(H_{i_3},m_3)>\cdots.$$
Such a sequence cannot have more than $r$ strict inequalities, and so we obtain a contradiction.
*Here direct product means direct sum in the categorical sense, sometimes a restricted direct product. How about: this is an abelian group all of whose elements have order $p$.
