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 &lt;a&gt;\oplus&lt;b&gt; 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)\},&lt;(1,0)&gt;,&lt;(0,1)&gt;,&lt;(0,2)&gt;,&lt;(1,1)&gt;,&lt;(1,2)&gt;
so all the subgroups are just products of the above groups.