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If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ is an injective group homomorphism, then is $\mathbb{Z}^m/\operatorname{im}(f)$ a finite abelian group?

I think yes. Let the homomorphism be given by a $m\times m$ matrix. Then, by existence of smith normal form over $\mathbb{Z}$, I think $\mathbb{Z}^m/\operatorname{im}(f)$ would be a direct product of cyclic groups of orders given by the entries in the diagonal of the smith normal form. Is this right? If so, how do we prove it? Any hints? Thanks beforehand.

user26857
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vidyarthi
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    Hint: show that $f$ induces a linear map $\overline{f}: \mathbb{Q}^m \rightarrow \mathbb{Q}^m$ which is injective. Conclude that $\mathbb{Z}^m/\mathrm{Im},f$ is a torsion abelian group, and thus a finite abelian group. – Aphelli Dec 20 '20 at 18:08
  • @Mindlack could you please elaborate, like every module homomorphism is $\mathbb{Z}$-linear right? So does it invariably imply $\mathbb{Q}$-linearity? And is smith normal form of no use here? – vidyarthi Dec 20 '20 at 18:13
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    @Mindlack torsion abelian groups need not be finite, consider for example the group of $n$th roots of unity for all $n \in \mathbb{N}$ – Noah Solomon Dec 20 '20 at 18:29
  • Torsion subgroups of $\mathbb{Z}^n$ are finite, though. – Max Dec 20 '20 at 18:57
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    @Max What you wrote is technically true, since the only torsion subroup of $\mathbb{Z}^n$ is the zero subgroup. I think you meant "torsion quotients of $\mathbb{Z}^n$ :). – Pierre-Guy Plamondon Dec 20 '20 at 19:12
  • @Noah Solomon: I’m aware of that. The thing is, this torsion group is by definition finitely generated. For the OP: yes, you get automatically $\mathbb{Q}$-linearity when you go from $\mathbb{Z}$ to $\mathbb{Q}$. I’m not sure what you call smith normal form though – if it’s what I think it is, it’s just an explicit reformulation of my abstract argument. – Aphelli Dec 20 '20 at 19:34
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    @Pierre-GuyPlamondon Yes. Sorry/thanks :) – Max Dec 20 '20 at 20:32
  • @Mindlack thanks for the clarification, makes sense. I guess I'm not just not entrenched enough in this material to follow your argument without that part made explicit. – Noah Solomon Dec 20 '20 at 21:58
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    This result has been proved many times on this site. Use search! – user26857 Dec 21 '20 at 00:24

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The Smith normal form works fine (and with $f(x)=Ax$ it proves that $|\Bbb{Z}^n/Im(f)| = |\det(A)|$) but it is easier to find a non-zero matrix $B\in M_n(\Bbb{Z})$ such that $AB=d\ I$.

reuns
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