An abelian group is called

freely generatedif it is generated by such that (such an expression is called arelation in) holds for for a finite sum of elements of iff for all (such a relation is calledtrivial).

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 has a unique expression as a sum of the generators with appropriate integer coefficients; if there were two such expressions then would be a nontrivial relation in , 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 be freely generated by a set and also by a set Then

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 can be written as for Group all the latter in an matrix , with entries Without loss in generatlity we may assume that If then there exists a nonzero vector i.e. Clearing denominators in the entries of we may assume that they are all integer. Then is a nontrivial relation among elements of a contradiction establishing the claimed property.

Thus we see that the number of generators in a free generating sets is an invariant of It is called the *rank* of and denoted 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 of a f.g. free abelian group is free, and

Let be freely generated by (so ).

We proceed by induction on When we already proved this as a corollary to 1st Isomorphism Theorem.

So we can consider now By induction, is f.g., i.e. is freely generated by Then otherwise, arguing as in the previous proof, we can find a nontrivial relation between This settles the 2nd claim of the theorem.

If then we are done. So now let us consider expressions for each and define to be the minimal positive with the minimum taken over all Note that each is divisible by Let an satisfy

We claim that is f.g. by First, we observe that there are no nontrivial relations between the latter elements. Indeed, if there was one, say then by induction assumption, and so But the latter gives a nontrivial relation between ‘s, as a contradiction.

It remains to show that every can be written as Consider Then and so as required.

4 February, 2009 at 12:41 |

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.

4 February, 2009 at 12:53 |

Thanks, you caught a typo (fixed now). It must be

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…

4 February, 2009 at 13:20 |

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

Thanks. =)

4 February, 2009 at 13:32 |

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.)

4 February, 2009 at 13:57 |

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