# Free abelian group proof help

gonzo
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 $(g_1...g_n)$ 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 $\alpha$.

I realized that might not have been clear. What I mean is let:

$h=a_1 g_1 +...+a_n g_n \in H$

then $\alpha$ is minimal of all a1 so that we know there is some element $h_2$ in H such that

$h_2=\alpha g_1 +...+a_n g_n \in H$

And all other a1's in an arbitrary h are multiples of $\alpha$.

What I would like to do is show that $\alpha g_1 \in H$

But maybe this isn't necessarily true? Or am I missing something simple to show this?

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## Answers and Replies

Staff Emeritus
Gold Member
Hrm. I still don't understand exactly what you want to say. But maybe a bad example will be useful!

Consider the free abelian group on 2 generators, G = Z².

Let H = { (m, n) | m + n is even }

H is a subgroup of Z². The smallest scalar multiple of the generator (1, 0) that lies in H is 2(1, 0). Alas, 1(1, 0) + 1(0, 1) lies in H, and |1| < |2|.

gonzo
I guess I wasn't clear enough, sorry. However, even though you seem to have misunderstood what I meant your example does show why I was having trouble proving my result since it is apparently not true, so thanks.

I was looking for a simplification of a proof, but I guess that was wishful thinking.