Showing that the rank of $(G/H)/(G/H)_t$ is $n - m$ This is an exercise that appears in Hungerford's Algebra (exercise II.2.5) which I've been struggling with (here, $G_t$ denotes the torsion subgroup of $G$):

If $G$ is a finitely generated Abelian group such that $G/G_t$ has rank $n$, and $H$ is a subgroup of $G$ such that $H/H_t$ has rank $m$, then $m \leq n$ and $(G/H)/(G/H)_t$ has rank $n - m$.

The same question appears twice on this site, one with no answers and another with an answer which is far too advanced, given the placement of this exercise on this book (directly after the characterization of all finitely generated Abelian groups, with no mention of the construction of tensor products, exact sequences, etc).
I've managed to prove $m \leq n$ by exhibiting an explicit monomorphism from $H/H_t$ into $G/G_t$, which is easy: simply define $\iota: H/H_t \to G/G_t$ by $h H_t \mapsto hG_t$, but I have been unsuccessful with the second part of the question, which asks us to find the rank of $(G/H)/(G/H)_t$.
I've tried setting up an isomorphism between $(G/H)/(G/H)_t$ and $(G/G_t)/\iota(H/H_t)$, which would be sufficient, but that hasn't gotten me anywhere at all. How should I go about this problem?
Links to the two other posts which contain this question:

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

*$(G/H)/(G/H)_t$ has rank $m-n$ where $G$ is finitely generated abelian and the rank of $G/G_t$ and $H/H_t$ is $m$ and $n$ respectively
 A: Much thanks to @Jack Schmidt for the helpful ideas. This solution uses the quotient groups suggested in his comments to effectively avoid explicitly formulating anything related to vector spaces or exact sequences (but it is essentially the same/similar argument). For clarity, we will use additive notation.
Let $p$ be a prime which does not divide $\lvert{G_t}\rvert$ and $\lvert (G/H)_t \rvert$. Then $pG_t = G_t$, so $G/pG \cong \mathbb{Z}_p \oplus \cdots \oplus \mathbb{Z}_p$, a direct sum of $n$ copies of $\mathbb{Z}_p$. In particular, $\lvert G/pG \rvert = p^n$. Similarly, $H/pH$ is the direct sum of $m$ copies of $\mathbb{Z}_p$, with $\lvert H/pH \rvert = p^m$, and $(G/H)/p(G/H)$ is the direct sum of some finite number of copies of $\mathbb{Z}_p$, equal to the rank of $(G/H)/(G/H)_t$. Thus, to prove our theorem, it suffices to show that $\lvert (G/H)/p(G/H) \rvert = p^{n-m}$.
Define the maps $$\iota: H/pH \to G/pG, \quad h + pH \mapsto h + pG,$$ $$\pi: G/pG \to (G/H)/p(G/H), \quad g + pH \mapsto (g+H) + p(G/H).$$ It is easy to verify that both maps are well-defined; for $\iota$, $h - h' \in pH$ implies $h - h' \in pG$, whereas for $\pi$, $g - g' \in pG$ implies $(g - g') + H \in p(G/H)$. It follows then from the definitions that $\iota, \pi$ are both homomorphisms, with $\iota$ a monomorphism and $\pi$ an epimorphism.
We will show that $\iota(H/pH) = \ker\pi$. For any $h \in H$, we have $\pi(h + pG) = (h + H) + p(G/H) = p(G/H)$, so $\iota(H/pH) \subseteq \ker\pi$. On the other hand, suppose $g + pG \in \ker\pi$, so that $g + H \in p(G/H)$. This implies $g - pg' = h$ for some $g' \in G$ and $h \in H$, so that $g - h \in pG$ and thus $g + pG \in \iota(H/pH)$. Thus, $\ker\pi \subseteq \iota(H/pH)$ and $\iota(H/pH) = \ker\pi$.
In particular, this implies $(G/pG)/\iota(H/pH) = (G/pG)/\ker\pi \cong (G/H)/p(G/H)$; hence, we conclude that $$ \lvert (G/H)/p(G/H) \rvert = \lvert G/pG \rvert / \lvert \iota(H/pH) \rvert = \lvert G/pG \rvert / \lvert H/pH \rvert = p^{n-m},$$ as desired.
