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Free abelian group proof help

  1. Dec 25, 2006 #1
    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?
    Last edited: Dec 25, 2006
  2. jcsd
  3. Dec 25, 2006 #2


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    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|.
  4. Dec 25, 2006 #3
    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.
  5. Dec 29, 2006 #4


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    what are you trying to prove?
    Last edited: Dec 29, 2006
  6. Dec 30, 2006 #5
    I figured it out, but thanks anyway.
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