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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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$G$ is imbedded in the subspace of $\mathbb{V}$ spanned by $f(X)$. Since the subspace is spanned by a linear independent set of $n$ vectors, it has dimension $n$. – Jiangwei Xue May 12 '11 at 10:08
@Jiangwei: That is true $f$($X$) spans the subspace, but it could be part of a another independent set in vector space. – Babak S. May 12 '11 at 12:01
@Babak: I do not understand your last comment, nor what "this former set" in your last sentence refers to. Any independent set of a vector space can be extended to a basis for $V$. – Arturo Magidin May 12 '11 at 17:19
@Arturo: Your last sentence of above comment was exactly what I intended. As you said, f (X ) can be extended to a basis for V and it is of dimention n . Maximality of X seemes to make f (X ) be the desired basis for V . in fact, maximality help it to be fit as an independent spaning set in V . If someone suggest the subspace spanned by f (X ) as a solution for this problem, so why X should necessarily be a maximal independent set in G .The former set was f (X ) and thanks for your advices. – Babak S. May 13 '11 at 8:56
@Babak: Sorry, but I don't understand what you are trying to say. Maximality of $X$ means that $f(X)$ is a basis for the span of $f(G)$, but you are not requiring the span of $f(G)$ to be $V$, so it is false that $f(X)$ will always be a basis for $V$. And you keep talking about "this problem", but I don't understand what "this problem" is. – Arturo Magidin May 13 '11 at 16:54
up vote 6 down vote accepted

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 $\frac{g}{n}$, $g\in G$, $n\in\mathbb{N}$.

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