Extending Herstein's Challenging Exercise to Modules Anybody who has worked through Herstein's Topics in Algebra might remember Exercise 26 of Section 2.5 (in second edition):

If $G$ is an abelian group containing subgroups of order $m$ and $n$, then $G$ contains a subgroup whose order is the least common multiple of $m$ and $n$.

Robert Beals has given a very nice solution in [1]. My question is whether or not we can extend the result of this exercise to modules (and if we can, to which extent). More concretely, 

Suppose $R$ is a commutative ring. If $M$ is an $R$-module that contains submodules of cardinality $m$ and
  $n$, then must $M$ necessarily contain a submodule whose cardinality is the least
  common multiple of $m$ and $n$?

Since abelian groups can be viewed as $\mathbb{Z}$-modules, this generalizes the above situation. Now, I am guessing this fails for arbitrary rings. I would be especially interested to know the conclusion if $R$ is Noetherian. (I think if $R$ is a PID, the conclusion follows, but I have no proof for this. Beals' proof uses purely group-theoretic concepts some of which I cannot generalize to modules).
Thanks for your time :)
[1] Beals, Robert. "On Orders of Subgroups in Abelian Groups: An Elementary Solution of an Exercise of Herstein." The American Mathematical Monthly, Vol. 116, No. 10 (Dec., 2009), pp. 923-926 
 A: I may be missing something, but can't you show this as follows?
To prove the result for  (additive) abelian groups, we can first show that a finite subgroup is the direct sum of its primary components (which are also its Sylow subgroups), and then show that the sum $H = H_1+H_2$ of two finite subgroups of coprime order $m$, $n$ has order $mn$. By combining these two techniques, for any two finite subgroups, we can construct a subgroup of order equal to the least common multiple of their orders: we first decompose both subgroups into primary components, and then recombine the primary components appropriately from the two subgroups.
We can do the same thing with submodules. An $R$-module $M$ of finite order $n$ is also an additive abelian group of finite order $n$, and so is the direct sum of its primary components $M_p$, for the primes $p$ dividing $n$, where $M_p$ can be defined as $M_p = \{m \in M \mid p^am=0 {\ \rm for\ some\ } a \ge 0 \}$. Then $M_p$ is also an $R$-submodule, because $m \in M_p \Rightarrow p^am=0 \Rightarrow p^arm=0$ for any $r \in R$, so $rm \in M_p$.
Secondly, if $M_1$ and $M_2$ are finite $R$-submodules of coprimes orders $m$, $n$, then, as an abelian group, $M_1+M_2$ has order $mn$, but $M_1+M_2$ is also an $R$-submodule.
