classification of free abelian groups I want to know whether there is a precise classification of free abelian groups? For example, Which of the following statements is true?
1-Abelian group $G$ is free iff be an infinite cyclic group.
2-If $G$ be a free abelian group then $G$ is isomorphic by product of infinite groups.
3-If $G$ be a free abelian group then $G$ is isomorphic by $\Bbb Z^n$ for some $n\in\Bbb N$.
 A: None of the above statements are true.  They can all be answered by using what a free abelian group is: an arbitrary direct sum of copies of $\mathbb{Z}$: $\bigoplus_{i \in I} \mathbb{Z}$.  (Here $I$ is any set, called an "index set".)  This is one of the two common definitions of a free abelian group.  If you are given the other definition -- via a universal mapping property -- then directly after being given this other definition you should be given the equivalence with the former one.  Now:
1) It is easy to see that $\bigoplus_{i \in I} \mathbb{Z}$ is infinite cyclic iff $I$ has 
exactly one element.  It could have zero elements (yielding the trivial group) or more than one element, e.g. $\mathbb{Z} \oplus \mathbb{Z}$.  So this is false.
2) This one is false because of the use of the word "product" rather than "sum".  Basic set theory shows that if $I$ is infinite, $\prod_{i \in I} \mathbb{Z}$ is uncountably infinite, whereas if $I$ is countably infinite, $\bigoplus_{i \in I} \mathbb{Z}$ is countably infinite.  It is also infinitely generated.  So not every free abelian group is isomorphic to a direct product of 
copies of $\mathbb{Z}$. 
(What is also true, but significantly harder to show, is that when $I$ is countably infinite, $\prod_{i \in I} \mathbb{Z}$ is not a free abelian group.)
3) This is false because free abelian groups can be infinitely generated.  (Maybe you were only told about finitely generated free abelian groups?)
A: Moreover, every  free abelian group is a direct sum (not product!) of copies (maybe infinite number) of $\Bbb Z$ (see L.Fuchs, Infinite Abelian Groups, v.1).
