# 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 $n\in\Bbb N$.

• 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 $B\cong im f\oplus ker h$. – Mikasa 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?)

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).

• 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