
#1
Jul1911, 11:47 AM

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 [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? 



#2
Jul1911, 12:16 PM

Mentor
P: 16,700

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 [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 [tex]\{(0,0)\},<(1,0)>,<(0,1)>,<(0,2)>,<(1,1)>,<(1,2)>[/tex] so all the subgroups are just products of the above groups. 


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