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).(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Free abelian group proof help

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