Wikipedia:Reference desk/Archives/Mathematics/2012 October 4

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October 4

Finitely generated abelian groups

I'm stumped on what should be an easy problem. The book defines a free finitely generated abelian group as one with a basis, i.e., a linearly independent generating set. Then an exercise defines a torsion-free group in the usual way, and asks me to prove that a finitely generated abelian group that is torsion-free must be free. I know this follows from the fundamental theorem of finitely generated abelian groups, but I get the impression I'm not supposed to import a big gun to solve this problem.

Can I just start from the fact that I've got a finite generating set, add the fact that there's no torsion, and show that there must be a basis? I know I can't count on the basis being a subset of my generating set, because I've got a simple counter-example: the set {2,3} generates the integers, but neither {2} nor {3} does it alone. Help! Thanks in advance. -GTBacchus^(talk) 20:07, 4 October 2012 (UTC)

Suppose $X=\{x_{1},\ldots ,x_{n}\}$ is a generating set for a torsion-free abelian grop. If $i\neq j$ then we can replace $x_{i}$ by either $-x_{i}$ or $x_{i}+x_{j}$ and we will still have a generating set with $n$ elements.

Suppose that $X$ is not linearly independent, so there is a linear dependence $a_{1}x_{1}+\cdots +a_{n}x_{n}=0$. First, we can assume that each coefficient is non-negative since if $a_{i}<0$ then we can replace $x_{i}$ by $-x_{i}$ and $a_{i}$ by $-a_{i}$.

Now, if at least two of the coefficients are non-zero, say $0<a_{i}\leq a_{j}$ then we can replace $x_{i}$ by $x_{i}+x_{j}$ and since $a_{i}x_{i}+a_{j}x_{j}=a_{i}(x_{i}+x_{j})+(a_{j}-a_{i})x_{j}$, we have a linear dependence for this new generating set with non-negative coefficients and the sum of the coefficents is smaller than before. Hence after doing this finitely many times we must obtain a generating set which has a linear dependence with only one non-zero coefficient, and so the corresponding generator has finite order. But the group is torsion-free so this generator must be trivial, and so the group has a generating set with $n-1$ elements.

It follows that if $X$ is a generating set with the smallest possible cardinality, then $X$ is a basis. 60.234.242.206 (talk) 11:40, 6 October 2012 (UTC)