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Find subgroups of finitely generated abelian groups

  1. Jul 19, 2011 #1
    Is there an "easy" method to finding subgroups of finitely generated abelian groups using the First Isomorphism Theorem? I seem to remember something like this but I can't quite get it.

    For example, the subgroups of [itex]G=Z_2\oplus Z[/itex] are easy...you only have [itex]0\oplus nZ[/itex] and [itex]Z_2\oplus nZ[/itex] for [itex]n\geq 0.[/itex]

    But if you have a different group, say [itex]G=Z_6\oplus Z_4[/itex], it's possible the subgroups aren't of the form [itex]<a>\oplus<b>[/itex] correct? Like <(2,2)>.

    How would you describe all the subgroups? I can do it by brute force..I'm looking for an quick easier asnwer if one exists...even in only some situations

    EDIT: maybe this makes more sense if I only need to know subgroups of a specific index?
    Last edited: Jul 19, 2011
  2. jcsd
  3. Jul 19, 2011 #2
    It seems you're looking for the subgroup lattice of finitely generated abelian groups?

    Well, the following article may help: http://www.google.be/url?sa=t&sourc...g=AFQjCNHJPHZy0JvaO0tmsu8F8EfX5OYWYg&cad=rja"

    Also keep in mind that if G and H are groups such that gcd(|G|,|H|)=1, then
    [itex]Sub(G\times H)\cong Sub(G)\times Sub(H)[/itex]

    So in your example
    [tex]Sub(\mathbb{Z}_6\times \mathbb{Z}_4)\cong Sub(\mathbb{Z}_3)\times Sub(\mathbb{Z}_2\times \mathbb{Z}_4)[/tex]

    so you only need to find the subgroups of [itex]\mathbb{Z}_2\times \mathbb{Z}_4[/itex]. The cyclic subgroups of this group are
    so all the subgroups are just products of the above groups.
    Last edited by a moderator: Apr 26, 2017
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