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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, namely $\mathbb{Z}$ and the trivial group. What you are looking for are free abelian groups, that is a huge difference. – m_l Apr 20 '12 at 14:22
And $Z_4$ isn't a free abelian group anyway. – Chris Eagle Apr 20 '12 at 14:25
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. – wxu Apr 20 '12 at 14:31
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. – Gunbuddy Apr 20 '12 at 14:33
@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. – Arturo Magidin Apr 20 '12 at 14:37

A free group in what category/variety?

Let $\mathcal{C}$ be a class of groups. We say that a group $F\in\mathcal{C}$ is a free $\mathcal{C}$-group if there is a subset $X\subseteq F$ such that for every group $G\in\mathcal{C}$ and every function $f\colon X\to G$, there is a unique group homomorphism $\mathfrak{f}\colon F\to G$ such that $\mathfrak{f}(x)=f(x)$ for all $x\in X$.

When we talk about "free groups", with no qualifications, we are generally talking about the "absolutely free groups", which are the ones where $\mathcal{C}$ is the class of all groups. Another common class is when $\mathcal{C}$ is the class of all abelian groups, in which case the "free $\mathcal{C}$-group" are the "free abelian groups."

It is easy to see that $\mathbb{Z}_4$ cannot be a free group or a free abelian group. To see this: if $X\subseteq\mathbb{Z}_4$ is not empty, with $x\in X$, then no map that sends $x$ to $1\in\mathbb{Z}$ can be extended to a group homomorphism from $\mathbb{Z}_4$ to $\mathbb{Z}$ (all such homomorphism map everything to $0$). And if $X=\varnothing$, then there would have to be a unique group map from $\mathbb{Z}_4$ into any abelian group, and this is not the case: there are, for example, four different homomorphisms $\mathbb{Z}_4\to\mathbb{Z}_4$.

For the same reason, since it is not a free abelian group, it cannot be an absolutely free group either.

Those are the two most reasonable interpretations of the question.

That said, there is a "nice" class of groups for which $\mathbb{Z}_4$ is a $\mathcal{C}$-free group: the class of abelian groups of exponent $4$. Take $X=\{1\}$. If $G$ is any abelian group of exponent $4$, and $f\colon\{1\}\to G$ is any function, then this determines a unique homomorphism $\mathbb{Z}_4\to G$, which works because $4f(1)=0$ in $G$.

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Many examples can be easily dismissed by the fact that free abelian groups are torsion-free. So, if your group has non-trivial torsion subgroup, it cannot be free abelian. This immediately yields that finite groups are not free abelian (except for the trivial group).

More generally, a finitely generated free abelian group is isomorphic to $\mathbb{Z}^n$ for some natural $n$. In this case your question can be decided by checking whether your group is isomorphic to some $\mathbb{Z}^n$.

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