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We know that any torsion-free group can be imbedded in a vector space over $\mathbb Q$. Now the question is:

if there is a maximal independent subset with $n$ elements of our torsion-free group $G$,$X$, then the group can be imbbeded in a vector space of dimention $n$.

I start with the fact that, we can establish an injective from $G$ to $\mathbb V$, a vector space over $\mathbb Q$ , namely $f$. This function is bijective form $G$ to $f$($G$), so $f$($X$) is independent set in $\mathbb V$. How can I extend this former set into a basis for $\mathbb V$ ?

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    @Arturo: Thanks alot.2011-05-13

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You are talking about abelian groups here, right?

Then take the tensor product $\widehat{G}=G\otimes_\mathbb{Z}\mathbb{Q}$ - the so-called divisible hull of $G$. It is a $\mathbb{Q}$-vector space of dimension $n$ if and only if $G$ contains a maximal $\mathbb{Z}$-linearly independent set with $n$ elements. The proof for this fact is just elementary calculation using the fact that every element of $\widehat{G}$ is a fraction of the form $\frac{g}{n}$, $g\in G$, $n\in\mathbb{N}$.