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

= 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 $$a_1x_1 + \cdots + a_nx_n = 0$$. 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 $$a_ix_i + a_jx_j = a_i(x_i + x_j) + (a_j - a_i)x_j$$, 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)