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$\begingroup$

How can I prove a given abelian group; such as $\mathbb{Z}_4$ with addition mod 4, is not a free group?

Should I consider all the subsets of the given group and prove any of them cannot be a basis? But this approach will give me a lot of sub groups to consider. Is there any way to prove that multi-element subset of a group cannot be a basis, if all the elements in the subset individually cannot be a basis of the group?

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    Be careful: There are only two abelian groups that are [free groups](http://en.wikipedia.org/wiki/Free_group), namely $\mathbb{Z}$ and the trivial group. What you are looking for are [free abelian groups](http://en.wikipedia.org/wiki/Free_abelian_group), that is a huge difference.2012-04-20
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    And $Z_4$ isn't a free abelian group anyway.2012-04-20
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    If you can find a nonzero element $x$ in an abelian group, such that $nx=0$ for some positive $n$, then this abelian group is not a free abelian group.2012-04-20
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    Yes. I want to show that it is not free. That is why I wanted to consider all the subsets of that and prove each of those cannot be a basis.2012-04-20
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    @Gunbuddy: Is your definition of "free **abelian** group" (*please* don't drop the 'abelian'!) that "it has a basis"? No group with torsion can have a basis: if $g\in A$ is a nonzero torsion element, and $\mathcal{B}=\{b_i\}$ were a basis, then expressing $g$ in terms of the $b_i$ and then adding $g$ to itself enough times to get $0$ would give you a nontrivial relation among the $b_i$, which is impossible.2012-04-20

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