No finitely generated abelian group is divisible My question might be ridiculously easy. But I want prove 

No finitely generated abelian group is divisible.

Let $G$ be a finitely generated abelian group. By definition, group $G$ is divisible if for any $g\in G$ and natural number $n$ there is $h\in G$ such that $g=h^n$.
  There 2 case: 


*

*$G<\infty$   

*$G=\infty.$ 


In the first case I tried to use the fundamental theorem of finetely generated abelian group. By this theorem,  finitely generated abelian group is  finite group iff its free rank is zero. At this point I am now stuck.
In the second  case I can use the following statement "No finite abelian group is divisible." But I cannot prove this state. 
 A: Proof with no structure: let $G$ be a f.g. divisible abelian group. Assuming by contradiction that $G\neq 0$, it has a maximal (proper) subgroup $H$; then $G/H$ is a simple abelian group, hence cyclic of prime order, hence is not divisible, contradiction since being divisible passes to quotients.
The same shows that a nonzero f.g. module over any commutative ring $A$ that is not a field, is never divisible (where divisible means that $m\mapsto am$ is surjective for every nonzero $a\in A$).
A: *

*For the finite case, show that in a group $G$ of cardinal $n$, $g\in G$ non-trivial, you cannot find a $h\in G$ such that $h^n=g$. Whence a finite group $G$ cannot be divisible.

*If you have the fundamental theorem for finitely generated abelian group. Show that in $G=\mathbb{Z}^r$ the element $(1,0,...,0)$ cannot be obtained as a proper power of an element of $G$. 
Now, in general, $G=Tor(G)\oplus \mathbb{Z}^r$. show that the number of elements $g\in G$ such that  $g^k=(0,1,0,...,0)$ for some $k$ is finite. Deduce from this that the group $G$ cannot be divisible.
