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I feel like Im overlooking some simple fact on this one, so any hints would be appreciated.

I saw this solution, but I'm wondering if there's a way to do it without short exact sequences.

Thanks in advance.

Sebastiano
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Bears
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  • have you proven that subgroups of free abelian groups are free and that free abelian groups have well defined ranks? – hunter Aug 08 '20 at 22:41
  • I haven't.

    We did some stuff with finitely generated abelian groups, but nothing with free abelian groups.

     – Bears Aug 08 '20 at 22:44
  • If this makes sense to you: a short exact sequence $0\to A \to B \to C \to 0$ splits if $C$ is free, in that case $B\cong A\oplus C$ (I think direct sum is more appropriate here, though the difference doesn't matter for these groups). – Justin Young Aug 08 '20 at 22:54

1 Answers1

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This is simple enough to be proved explicitly.

We pick any $v \in \Bbb Z^2$ such that $f(v) = 1$ and we define a homomorphism $h:\Bbb Z \times \ker(f) \rightarrow \Bbb Z^2$ by $h(a, x) = av + x$.

$h$ is surjective: if $y$ is any vector in $\Bbb Z^2$, then $h(f(y), y - f(y)v) = y$.

$h$ is injective: if $h(a, x) = 0$, then we have $av + x = 0$ and applying $f$ gives $a = 0$, which then implies $x = 0$.

WhatsUp
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  • Interesting.

    Can you share what kinds of considerations you had at the front of your mind when you came up with this answer? Did it "just occur" to you, or is there some crucial set of ideas that this question primed you to think about which made your solution come to you?

     – Bears Aug 08 '20 at 22:51
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    In fact, I just translated the answer using short exact sequences (see comment of @JustinYoung above) to this explicit version. This might be a motivation for you to learn the more abstract method. – WhatsUp Aug 08 '20 at 23:09
  • That's definitely motivation for me to learn the more abstract method. Thanks for sharing your thoughts. – Bears Aug 08 '20 at 23:12