Let $G$ be an abelian group and use additive notation. Call $G$ a divisible group if given x in G and a positive integer m we can always find an element y of $G$ such that $my = x$. For example, $\mathbb{Q, R, C}$ and C are all divisible groups. Show that $\mathbb{Z}$ and $\mathbb{Q}^{pos}$ are not divisible. Prove that a non-trivial divisible group cannot be finitely generated.
Edit: I'm inclined to say that for $\mathbb{Z}$ - and probably for $\mathbb{Q}^{pos}$ - that $my = x \rightarrow y = x/m$, and then say $x = 1, m = 2$, and point out that y is no longer in the group $\mathbb{Z}$, so I can't always find a y for which the original conditions are satisfied. Is that a correct interpretation of the question?
Edit: This is the exercise 21.13 of Groups and symmetry, M.A Armstrong
For $\mathbb{Z}$ (under addition), observe that there is no integer $x$ for which $2x=1$. This shows that $\mathbb{Z}$ is not divisible.
For $\mathbb{Q}^{pos}$ (under multiplication), observe that there is no positive rational number $x$ for which $x^2=2$. This shows that $\mathbb{Q}^{pos}$ is not divisible either.
Suppose that $G$ is a divisible abelian group. To show that $G$ cannot be finitely generated unless it is trivial, suppose that $S=\{g_1,g_2,...,g_n\}$ is a finite generating set for $G$. Then, given any $g_i \in S$, one can kill all of the generators except $g_i$ to get a divisible cyclic group. But the only divisible cyclic group is the trivial group (either the generator cannot be "halved" if the cyclic group is infinite, or no nonzero element can be "divided" by the order of the cyclic group if it is finite), so getting rid of $g_i$ from the generating set $S$ will still leave another generating set. This process can be repeated $n$ times, to get the conclusion that the even the empty set (after getting rid of all the generators in $S$) still generates $G$, so $G$ must be trivial.