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?
a free abelian group is torsion-free
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group-theory
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0What's your definition of free abelian group? – 2012-03-07
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0a "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 $g=m_1b_1+\cdots+m_nb_n$ – 2012-03-07
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0Then uniqueness is key. If there is an element which is torsion, say $q g=0$, then $0$ has 2 expressions: $qg$ and $0$. – 2012-03-07
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0you 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 – 2012-03-07
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1But $g$ 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