Subgroups of a finitely generated abelian group

[bubchi89]

I have a homework problem which asks to prove that the subgroups of a finitely generated abelian group are finitely generated.

The hint in the book says to prove it by induction on the size of X where the group G = <X>. It also says to consider the quotient group G/a_n+1 (with a_n+1 in X) in the induction step.

I've spent a lot of time on this problem, but I've pretty much made no progress. I know that every element of G is a product (I'm thinking of it as a multiplicative group but I don't think it matters) of the elements in X. For the quotient group, I know that it is finitely generated. That's... pretty much it. Also, I can't use rings which seem to be the easiest way to prove this problem because we haven't covered it. However, I have covered three fundamental Isomorphism Theorems as well as a Correspondence Theorem. I'll be honest though, my understanding of these important theorems is weak, so if the inductive proof relies on an understanding of them then it's no surprise I haven't gotten anywhere. I don't really see why isomorphisms would be applicable which is why I haven't considered it very thoroughly. I have a hard time pursuing a path that doesn't seem fruitful at first glance (bad habit), so I've mostly been going over the same ideas in my head which haven't led me anywhere. For example, the Correspondence Theorem says that there is a bijection from a subgroup S to its quotient group S/K with K <= S. Any help would be great!

 

[Office_Shredder]

Can you do the case where G is generated by one element?

Then the inductive step: If the result is true for when G is generated by up to n elements, suppose that G is generated by n+1 elements a₁,...,a_n+1. Let H be a subgroup of G. What can we say about H by examining G/<a_n+1>?

 

[bubchi89]

Can you do the case where G is generated by one element?

Then the inductive step: If the result is true for when G is generated by up to n elements, suppose that G is generated by n+1 elements a₁,...,a_n+1. Let H be a subgroup of G. What can we say about H by examining G/<a_n+1>?

Ah yes, I forgot to mention (I had been up for 7 hours past my bedtime trying the problem) that I had done the base case. And by done I mean I cited a theorem I never proved correctly, but assume we did in class at one point: every subgroup of a cyclic subgroup is itself cyclic. Unfortunately I don't have time to prove it in this pset either. As for the induction step, I don't know. My understanding of quotient groups is that they partition the set into cosets... that's about it. The Correspondence theorem I know says that if H contains <a_n+1> then there is a bijection of each subgroup to a subgroup in that quotient group, but that isn't helpful. In particular, I don't see how the new G is related to the old G so the invocation of the induction hypothesis is unclear to me too.

 

[bubchi89]

I should give a better answer. In the case where H is a subgroup that contains elements who are a product of <n+1 elements in the generator, then we can invoke the induction hypothesis. So I guess I sort of do see how the IH plays in. However, I have no idea how the quotient group helps me in the case when H contains elements who are a product of all n+1 generators. That's when I tried using the fact that the quotient group is finitely generated by the cosets of <a_n+1> and coefficients a₁ through a_n. But I didn't reach any conclusions from that

Mm, again I missed a thought I should explicate. The subgroup must contain some power of a_n+1 so that element lies in some member of the quotient group. We can also consider the cycle generated by that element, but I don't see how that helps either

Okay, if we consider each element to be a product, the "prefix" (my own choice of notation) of the products are the n generators in the induction hypothesis. Ignoring the a_n+1 element in the product, we know every element can be generated. I don't think I can say that, because this is true, we can simply tack on an a_n+1 to each generator and be done. Can we? I think the set generated by such a generating set would be a superset of the subgroup S we are considering. To be clear, if S had a set of n generators in the induction hypothesis, I can multiply each of them by a_n+1. The set created by such a generating set is bound to have every power of a_n+1 multiplied by the old S. This seems really iffy though, like I've incorrectly invoked the induction hypothesis. Furthermore, finitely generated doesn't say superset, it says exactly equals.

 

[Ali_Kuscu]

Hi.
if G = <a> then Every subgroup of G is Cyclic so generated by one element (finitely generated).
if the hypothesis is true for abelian groups that generated by less than n elements (n>=2) and G = <a1,...,an> then let H = <a1,...,an-1> and let A < G.
Now let R = {dn | d1*a1+...+dn*an is an element of A where di are in Z}.
Since 0*a1+...+0*an = 0 is in A so 0 is in R.
If R \cap Z+ = Empty then R \cap Z- = Empty so R = {0} then A < H so by inductive hypothesis A is finitely generated.
If R \cap Z+ is not Empty then there is a least element in it (say dn)
then let v = d1*a1+...+dn*an is an element of A
Now the claim is that A = <v> + (A \cap H).
Now since <v> < A and (A \cap H) < A
we get <v> + (A \cap H) < A.
Now if \alpha \in A then \alpha = c1*a1+...+cn*an now observe that cn is divisible by dn (let cn = q*dn + r where 0<= r < dn then \alpha-q*v = r1*a1+...+r*an \in A so if r > 0 then r \in R \cap Z+ but r < dn this contradicts with the choice of dn, so r = 0).
Now \alpha-q*v = r1*a1+...+r(n-1)*a(n-1) \in A\capH
so \alpha \in <v> + (A \cap H).
Now since (A \cap H) < H by induction hypothesis it is finitely generated.
Let (A \cap H) = <g1,...,gk>. Then claim is A = <v,g1,...,gk>.
Since v \in A and gi \in A we get <v,g1,...,gk> < A.
If \alpha \in A, \alpha = q*v + h where h \in (A \cap H) Now since (A \cap H) = <g1,...,gk>
we get \alpha = q*v + s1*g1+ ... + sk*gk so A < <v,g1,...,gk>.
So A = <v,g1,...,gk> (A is finitely generated.)!