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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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What's your definition of free abelian group? –  Jason DeVito Mar 7 '12 at 15:13
    
a "free abelian group" $G$ on a basis $B=\{b_1,\cdots,b_n\}$ (this is the finite case): for each $g\in G$ there is a unique sequence of integers $m_1,\cdots,m_n$ such that $g=m_1b_1+\cdots+m_nb_n$ –  palio Mar 7 '12 at 15:19
    
Then uniqueness is key. If there is an element which is torsion, say $q g=0$, then $0$ has 2 expressions: $qg$ and $0$. –  Jason DeVito Mar 7 '12 at 15:21
    
you mean $qg$ and $0g$ both are $0$ so there is two integers $q$ and $0$ but here $g$ need not be a basis element. so we can't conclude i think unless we do all the work above –  palio Mar 7 '12 at 15:26
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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. –  Jason DeVito Mar 7 '12 at 15:28

1 Answer 1

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.

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