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let $G$ be a free abelian group of rank $n$ with basis $B=\{b_1,\cdots,b_n\}$ then $G$ must be torsion-free. to prove this let $g=\sum{m_ib_i}\not = 0$ an element of $G$. Suppose there exists $q\in \mathbb Z$ such that $qg=0$ then $\sum{qm_ib_i}=0$, so for each $i=1..n$, $qm_i=0$. Now since $g\not = 0$ then $q=0$. Is this correct?

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    But $g$ is made out of basis elements. Say $g = \sum m_i b_i$. Then since $g\neq 0$, at least one $m_j\neq 0$. Then $0 =qg = \sum (qm_i)b_i$ and $qm_j\neq 0$. So, you've written $0$ in a way where not all coefficients are $0$. Since using all coefficients $0$ is another way, you've violated uniqueness.2012-03-07

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Your proof is correct. However, I would add one more step to "since $g \neq 0$ then $q = 0$":

Since $g \neq 0$ and $b_1,\ldots,b_n$ are linearly independent $m_i \neq 0$ for some $i$. Because $\mathbb Z$ has no zero divisors $qm_i = 0$ implies $q = 0$.

EDIT: It is not necessary to restrict to the finite rank case. Your basis may have any cardinality, however, since $g$ can be expressed as a finite linear combination of basis elements the proof works the same in that case.