# 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:

1. $G<\infty$
2. $G=\infty.$

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.

• – Seirios Sep 25 '15 at 9:52

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 $a\in A$).
1. 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.
2. 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.