Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm currently reading Munkres-Algebraic topology text, and in his review chapter of abelian groups, he gives the classification theorem for finitely generated abelian groups. He ommited some important proofs from this chapter, so I'll be glad if you'll be able to help me complete the missing proofs:

Lemma: Let $F$ be a free abelian group. If $ R$ is a subgroup of $F$, then $R$ is also a free abelian group. If $F$ has rank n, then $R$ has rank $r \leq n $. Furthermore, there is a basis $e_1 , ..., e_n $ for $F $ and integers $t_1 , ..., t_k $ with $ t_i>1$ such that: (1) $t_1 e_1 ,...,t_ke_k, e_{k+1},...,e_r $ is a basis for $R$. (2) $t_1 |t_2|...|t_k $. The integers $t_1,...,t_k $ are uniquely determined by $F$ and $R$, although the basis $e_1 ,...,e_n$ is not.

I really need your help with proving the bold part of this lemma.

Hope you'll be able to help,

Thanks a lot!

share|improve this question
    
In my edition of "Elements Of Algebraic Topology" everything is there (§11, Computability of Homology Groups) –  Blah Feb 24 '12 at 13:09
    
Dear Jack: You can look at this answer. –  Pierre-Yves Gaillard Feb 24 '12 at 15:10
    
Thanks a lot ! I totally missed it ! It's indeed in chapter 11... Thanks a lot and sorry ! –  joshua Feb 24 '12 at 16:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.