# Find subgroups of finitely generated abelian groups

by spicychicken
Tags: abelian, finitely, generated, groups, subgroups
 P: 3 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 $G=Z_2\oplus Z$ are easy...you only have $0\oplus nZ$ and $Z_2\oplus nZ$ for $n\geq 0.$ But if you have a different group, say $G=Z_6\oplus Z_4$, it's possible the subgroups aren't of the form $\oplus$ 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?
 Mentor P: 18,345 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&source...5OYWYg&cad=rja Also keep in mind that if G and H are groups such that gcd(|G|,|H|)=1, then $Sub(G\times H)\cong Sub(G)\times Sub(H)$ So in your example $$Sub(\mathbb{Z}_6\times \mathbb{Z}_4)\cong Sub(\mathbb{Z}_3)\times Sub(\mathbb{Z}_2\times \mathbb{Z}_4)$$ so you only need to find the subgroups of $\mathbb{Z}_2\times \mathbb{Z}_4$. The cyclic subgroups of this group are $$\{(0,0)\},<(1,0)>,<(0,1)>,<(0,2)>,<(1,1)>,<(1,2)>$$ so all the subgroups are just products of the above groups.

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