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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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    $\begingroup$ In your last paragraph, those quotients need not be isomorphic. Take G=Z, H=2Z. Then you have (Z/2Z)/(Z/2Z) = Z/Z vs (Z/0Z)/(2Z/0Z) = Z/2Z $\endgroup$ Jul 16, 2021 at 18:30
  • $\begingroup$ Ah, I miss the times when any of my arbitrary conjectures supporting a convenient solution would all magically turn out to be true. $\endgroup$
    – Emory Sun
    Jul 16, 2021 at 18:45
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    $\begingroup$ There is another tensor proof that might be easier to do before tensors. Since $G_t$ is finitely generated, its order is divisible by only finitely many primes. Take a different prime $p$. Then $p G_t = G_t$, and so $G/G_t$ is similarly to $G/pG$. The former is a direct sum of copies of Z, the latter is a direct sum of copies of Z/pZ. Same number. The latter is a vector space, so dimensions work like you want ranks to work. $\endgroup$ Jul 16, 2021 at 19:04
  • $\begingroup$ I don't think I had learned tensor products or Z/pZ vector spaces by the time I was taught section II of Hungerford, but here's hoping things have gotten better :-) $\endgroup$ Jul 16, 2021 at 19:07
  • $\begingroup$ Thank you for your helpful comments; I think I managed to find a solution. $\endgroup$
    – Emory Sun
    Jul 16, 2021 at 20:43

1 Answer 1

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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.

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  • $\begingroup$ Nice work! I think there is on error in the last part of the second to last sentence of your second paragraph (same error in my comment). In order for (G/H)/(G/H)_t = Z^k and (G/H)/p(G/H) = (Z/pZ)^k for the same k, you want p(G/H)_t = (G/H)_t. So not only do we need p not dividing |G_t|, but also not dividing |(G/H)_t|. As before G=Z, H=2Z, p=2 shows the problem. $\endgroup$ Jul 16, 2021 at 21:09
  • $\begingroup$ Thank you for the correction. I had assumed that $p \not\mid \lvert G_t \rvert$ automatically implied $p \not\mid \lvert (G/H)_t \rvert$ for some reason. $\endgroup$
    – Emory Sun
    Jul 17, 2021 at 2:16

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