Find subgroups of finitely generated abelian groups

[spicychicken]

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 $<a>\oplus<b>$ 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?

 

[micromass]

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
$Sub(G\times H)\cong Sub(G)\times Sub(H)$

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.