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!

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

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