Prove that a finitely generated module need not be finitely as an abelian group. The question is given in the following picture:
The hint is given in the following picture:

But I did not understand what exactly should I do, so could anyone give me a more detailed hint please?
Thanks!
 A: Let $R = \mathbb{Q}$.

Then $R$ is finitely generated as an $R$-module (it's cyclic), but $R$ is not finitely generated as an abelian group.

If you've done the earlier exercise, then you're done. If not, show that for any finitely generated subgroup of $G$ of $\mathbb{Q}$, the elements of $G$ can all be expressed with a common denominator, namely, the LCM of the least positive integer denominators of the generators of $G$. But of course, no single denominator would suffice as a common denominator for all elements of $\mathbb{Q}$.

For the main problem, here's another proof . . .

Let $R$ be any ring with uncountably many elements (e.g., $\mathbb{R}$ or $\mathbb{C}$). Then once again, $R$ is finitely generated as an $R$-module, but can't be finitely generated as an abelian group, since any finitely generated abelian group is at most countably infinite.
A: (a) If $\mathbb{Q}$ is generated by $\frac{a_1}{b_1}\dots\frac{a_n}{b_n}$ with $(a_i,b_i)=1$, then it is generated by $\frac{1}{b}$ where $b$ is the minimum common multiple. It will be isomorphic to $\mathbb{Z}$, which is a contradiction. By example, $\mathbb{Q}$ does not have finite quotients.
(b) As $\mathbb{Q}$ is divisible, the only abelian group that is free and divisible is $0$
And as mentioned by @quasi $\mathbb{Q}$ is generated by $1$ over $\mathbb{Q}$.
