Theorem concerning free abelian groups

[LHS1]

I spend much time to study a theorem - Let G be a nonzero free abelian group of finite rank n, and let K be a nonzero subgroup of G. Then K is free abelian of rank s smaller or equal to n. Furthermore, there exists a basis (x1,x2,...,xn) for G and positive integers d1,d2,...,ds where di divides d(i+1) for i=1,...s-1, such that (d1x1,d2x2,...dsxs) is a basis for K.(Theorem 4.19 on page 253, Fifth edition, A First Course In Abstract Algebra, by John B. Fraleigh) and I still do not understand the proof at all. Could anyone help me by explaining the proof in more detail ,elaborate and illustrate the theorem by examples.

 

[morphism]

Perhaps you could be specific about what you're having trouble with? (Also, not all of us have that book, so we don't know what kind of proof the author is presenting.)

Maybe it will help if you conceptually think of a free abelian group of rank n as a "vector space" over the integers of dimension n. Can you think of an analogous theorem for vector spaces?

 

[mathwonk]

try my free notes on my website for math 8000 and math 844. these results are as just suggested, analogous to linear vector space results. the simplest proof is to diagonalize an integer matrix.

 

[mathwonk]

the first part of the theorem is easily proved by induction on n, where the case n=1 follows from the fact that the integers are a principal ideal domain.

then for the second part just define a map Z^s-->Z^n with image the given subgroup, then diagonalize the matrix for this map. the integers di are the entries on the diagonal after this process.

 

[Symmetryholic]

It is hard to understand for me too. I read the proof of the above problem several times, I barely grasp the main idea though.

The toughest part is to find a concrete example.
Below is my example.

Let G be a free abelian group generated by $$\{x_{1}, x_{2}, x_{3}\}$$. Let H be a subgroup generated by $$\{2x_{1}, 3x_{2}, 5x_{3}\}$$. To the best of my knowledge, there is an equivalent basis of G such that the basis of a subgroup H has the divisor property.

Could anyone show me the link (free abelian group or vector space) of the available procedure or algorithm to find the equivalent basis of G that has the divisor property?

Thanks.