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

Question

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.

Answer 1

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

Answer 2

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