# Definition of Rank of a Finitely Generated Abelian Group

I'm a little confused about the definition of the rank of a finitely-generated abelian group $$G$$, which I've read is the size of a largest linearly independent set in $$G$$. I'm looking at a proof that proves that a certain abelian group has rank $$n$$ by exhibiting a linearly independent set of $$n$$ elements and showing that it's maximal. But how is this enough? Just because that particular independent set can't be extended doesn't on its own stop there from being some totally different linearly independent set with more elements.

I'm led to infer that all maximal independent sets in a finitely-generated abelian group have the same number of elements, but this isn't at all obvious to me. Why is it true? (I'd like to be able to see this without using tensors---I'm aware the arguments are more succinct from that perspective but I haven't learned enough about tensors yet.)

Edit: Just realized I accidentally assumed the groups were free - I meant to assume just that they’re finitely generated.

• Sure, it should be a lemma that all the maximal linearly independent sets has the same size. That shouldn't be to hard to prove. An alternative approach is to consider $G/T$ where $T$ is the group of torsion element. $G/T$ is torsion free and finitely generated, so actually $\mathbb{Z}^n$ for some $n$. That $n$ is the rank. – Knaus Sep 8 '20 at 20:40
• @Knaus How would you prove the lemma? I’m coming up blank. – Nick A. Sep 8 '20 at 20:57

There are two kind of ranks of a finitely generated Abelian group $$G$$. The $$\Bbb{Z}$$-rank is the smallest number of cyclic factors in a representation of $$G$$ as a direct product of cyclic groups, the $$\Bbb{Q}$$-rank is the number (=smallest number) of infinite cyclic factors in such a representation.
Edit For the free Abelian groups the ranks are the same and are equal to the number of elements in a maximal independent set. The proof is that if $$X$$ is a maximal independent set of a free Abelian group $$G$$ then $$X$$ must generate a finite index subgroup $$H$$ of $$G$$ (if the index is not finite then $$X$$ is not maximal). The growth function $$f_H(n)$$ of $$H$$ is $$\approx n^{|X|}$$. It must be equivalent to the growth function of $$G$$ which is $$n^{rank(G)}$$. So $$rank(G)=|X|$$
• My definition of independence is exactly the same as for vector spaces, but with coefficients taken from $\mathbb{Z}$. – Nick A. Sep 8 '20 at 21:20