I'm working on a proof for subgroups of free abelian groups and am having trouble with a step (I know other methods, but would like to try and make this one work if possible). The basic idea is let G be a free abelian group with generators [itex](g_1...g_n)[/itex] and let H be a subgroup of G. Assuming a suitable renumbering of the generators so that g1 does not have all zero coefficients in H, we can find a minimal coefficient of g1 in H (with respect to absolute value, and non-zero of course), and then it is easy to show that all other coefficient of g1 in H have to be multiples of this coefficient. Call it [itex]\alpha[/itex]. I realized that might not have been clear. What I mean is let: [itex]h=a_1 g_1 +...+a_n g_n \in H[/itex] then [itex]\alpha[/itex] is minimal of all a1 so that we know there is some element [itex]h_2[/itex] in H such that [itex]h_2=\alpha g_1 +...+a_n g_n \in H[/itex] And all other a1's in an arbitrary h are multiples of [itex]\alpha[/itex]. What I would like to do is show that [itex]\alpha g_1 \in H[/itex] But maybe this isn't necessarily true? Or am I missing something simple to show this?