## 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:

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. =)