# Rank of $(G/H)/(G/H)_t$ where $G$ is finitely generated abelian and $H$ is a subgroup.

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.

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

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 $$0\to H\otimes\mathbb{Q}\to G\otimes\mathbb{Q}\to (G/H)\otimes\mathbb{Q}\to 0$$

• why the rank of $G/G_t$ is the dimension of $G\otimes\mathbb{Q}$ as a vector space ?
– WLOG
May 1, 2014 at 16:25
• @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. May 1, 2014 at 16:27

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

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