# 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 $n\in\Bbb N$.

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Here are three points about free abelian groups (See Rotman's book): 1) Two free abelian groups $$F=\sum_{x\in X}<x>,~G=\sum_{x\in Y}<y>$$ are isomorphic iff $|X|=|Y|$. This may help you to find desired facts for 1 above. 2) An abelian group is free group iff it has the projective property. 3) If $A\xrightarrow{f}B\xrightarrow{g}C\xrightarrow{h}D$ is exact of free groups then $B\cong im f\oplus ker h$. – Babak S. Mar 19 '13 at 10:14

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?)

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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).

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This is just the definition of a free abelian group: it is a free module over the ring $\mathbb{Z}$. – user641 Mar 19 '13 at 9:38
@ Steve D: Of course. – Boris Novikov Mar 19 '13 at 9:43
@SteveD free abelian groups can also be defined as those groups in the essential image of the left adjoint to the forgetful functor $Ab\to Set$. Then the criterion above becomes a theorem. Some might say it is more natural to define freeness in terms of a left adjoint. – Ittay Weiss Mar 19 '13 at 10:16