Testwiki: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 $X=\{x_1,\dots ,x_n\}$ is a generating set for a torsion-free abelian grop. If $i\ne j$ 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 $0<a_i\le a_j$ 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)