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Let $G$ be a finitely generated abelian group and $H$ be a subgroup. Let subscript $t$ denote the torsion subgroup. If $G/G_t$ is free of rank $n$ and $H/H_t$ is free of rank $m$, it is easy to embed $H/H_t\hookrightarrow G/G_t$ and deduce that $m\le n$. Now the question is that I want to show that $(G/H)/(G/H)_t$ is free of rank $n-m$.

This is harder than it looks and I have not succeeded in finding a proof after many hours.

[EDIT] I'm looking for a group theory proof.

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  • $\begingroup$ Try using the elementary divisor structure theory. $\endgroup$ May 1, 2014 at 14:08
  • $\begingroup$ It is definitely possible to derive this from a known theorem about the basis of a subgroup of a finitely generated free abelian group. $\endgroup$
    – Dan Shved
    May 13, 2014 at 20:30

2 Answers 2

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The rank of $G/G_t$ is the dimension of $G\otimes\mathbb{Q}$ as a vector space.

From the exact sequence $0\to H\to G\to G/H\to 0$, you get the exact sequence $$ 0\to H\otimes\mathbb{Q}\to G\otimes\mathbb{Q}\to (G/H)\otimes\mathbb{Q}\to 0 $$

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  • $\begingroup$ why the rank of $G/G_t$ is the dimension of $G\otimes\mathbb{Q}$ as a vector space ? $\endgroup$
    – WLOG
    May 1, 2014 at 16:25
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    $\begingroup$ @WLOG $G\otimes\mathbb{Q}$ is isomorphic to $(G/G_t)\otimes\mathbb{Q}$ (easy); since $G/G_t\cong\mathbb{Z}^n$ where $n$ is the rank, we're done. $\endgroup$
    – egreg
    May 1, 2014 at 16:27
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The rank is the dimension of G⊗Q because it is isomorphic to (G/Gt)⊗Q as G/Gt=Z^n as n is the rank

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  • $\begingroup$ But why is it this?!? $\endgroup$
    – user1729
    May 20, 2014 at 13:13
  • $\begingroup$ Because it is isomorphic to (G/Gt)⊗Q as G/Gt=Z^n as n is the rank. $\endgroup$
    – user150369
    May 20, 2014 at 13:21
  • $\begingroup$ Then say so in your answer! $\endgroup$
    – user1729
    May 20, 2014 at 13:24
  • $\begingroup$ My apologies, I will edit it. Thank you. $\endgroup$
    – user150369
    May 20, 2014 at 13:27
  • $\begingroup$ Also, this is the same as egreg's answer, unless I am missing some subtlety? $\endgroup$
    – user1729
    May 20, 2014 at 13:38

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