Finitely generated free abelian groups

An abelian group A is called freely generated if it is generated by S\subset A such that f=\sum_{s\in S}k_s s=0 (such an expression is called a relation in S) holds for for a finite sum f of elements of S iff k_s=0 for all s\in S (such a relation is called trivial).

Note a direct analogy between a free generating set and a basis of a vectorspace (here of course the coefficients are integers, not field elements.) Indeed, each a\in A has a unique expression a=\sum_{s\in S}k_s s as a sum of the generators with appropriate integer coefficients; if there were two such expressions a=\sum_{s\in S}k_s s=\sum_{s\in S}k'_s s then 0=\sum_{s\in S}(k_s-k'_s) s would be a nontrivial relation in S, contradiction.
Freely generated groups are also called free. Finitely generated (f.g., for short) free abelian groups are quite close in their behavour to finite-dimensional vectorspaces, although there are important differences, too.
The first important property of f.g. free abelian groups is the following:

Let A be freely generated by a set S and also by a set S'. Then |S|=|S'|.

You can compare it to the well-known fact that two bases of a vectorspaces have the same cardinality. The proof is similar, too, albeit with a twist stemming from the fact that not all nonzero integers have multiplicative inverses.
Each s'\in S' can be written as s'=\sum_{s\in S} k_s(s') s, for k_s(s')\in\mathbb{Z}. Group all the latter in an |S'|\times |S| matrix Q, with entries Q_{s',s}=k_s(s'). Without loss in generatlity we may assume that |S|\leq |S'|. If |S|<|S'| then there exists a nonzero vector v\in \mathbb{Q}^{|S'|}\cap Ker(Q^T), i.e. vQ=0. Clearing denominators in the entries of v, we may assume that they are all integer. Then \sum_{t\in S'}v_t t=0 is a nontrivial relation among elements of S', a contradiction establishing the claimed property.

Thus we see that the number of generators in a free generating sets is an invariant of A. It is called the rank of A and denoted rk(A). This notion is parallel to the notion of the dimension of a vectorspace, for obvious reasons.
The next question is to study subgroups of free abelian groups. Namely, the following holds.

A subgroup H of a f.g. free abelian group A is free, and
rk(H)\leq rk(A).

Let A be freely generated by S=\{s_1,\dots,s_n\} (so n=rk(A)).
We proceed by induction on n. When n=1, we already proved this as a corollary to 1st Isomorphism Theorem.
So we can consider now H_1=H\cap\langle s_1,\dots,s_{n-1}\rangle. By induction, H_1 is f.g., i.e. H_1 is freely generated by h_1=\sum_{j=1}^{n-1} k_{1,j} s_j,\dots,h_{m-1}=\sum_{j=1}^{n-1} k_{m-1,j} s_j. Then m\leq n, otherwise, arguing as in the previous proof, we can find a nontrivial relation between s_1,\dots,s_{n-1}. This settles the 2nd claim of the theorem.

If H_1=H then we are done. So now let us consider expressions h=\sum_{j=1}^n k_j(h) s_j for each h\in H, and define r to be the minimal positive k_n(h), with the minimum taken over all h\in H. Note that each k_n(h) is divisible by r. Let an h_m\in H satisfy k_n(h_m)=r.
We claim that H is f.g. by h_1,\dots, h_m. First, we observe that there are no nontrivial relations between the latter elements. Indeed, if there was one, say \sum_j \ell_j h_j=0, then \ell_m\neq 0 by induction assumption, and so \ell_m h_m\in \langle s_1,\dots,s_{n-1}\rangle. But the latter gives a nontrivial relation between s_j‘s, as \ell_m h_m=\ell_m \sum_j k_j(h_m) s_j, a contradiction.
It remains to show that every h\in H can be written as h=\sum_j \ell_j(h) h_j. Consider x=h-\frac{k_n(h)}{r}h_m. Then x\in H_1, and so h=x+\frac{k_n(h)}{r}h_m=\sum_{j=1}^{m-1}\ell_j(x) h_j + \frac{k_n(h)}{r}h_m, as required.

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5 Responses to “Finitely generated free abelian groups”

  1. Stephanie Says:

    Hi sir, can ask about the following proof?
    If |S|<=|S’| then there exists a nonzero vector
    v … i.e. vQ=0.
    Why is it so? I can’t understand it even after thinking about it for a while.

    Also, what does it mean by trivial and non-trivial relation?

    Thank you.

    • Dima Says:

      Thanks, you caught a typo (fixed now). It must be |S|<|S'|.
      Relations, and trivial ones, are defined in the very beginning of the post. And a non-trivial relation is a relation that is not trivial…

  2. Stephanie Says:

    Okay.. But actually what I am asking is why if |S|<|S’| then there exists such a nonzero vector v?

    Thanks. =)

    • Dima Says:

      That’s linear algebra!!! (a proof that two bases of a vectorspace have the same cardinality is very similar.)
      Q has more rows than columns, and thus vQ=0 always has a nonzero solution. (or, if you like, QT has more columns that rows, and (vQ)T=QT vT=0 has a nonzero solution.)

  3. Stephanie Says:

    Oowh.. Now I understand.. Thanks a lot, sir. =)

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