Here's a question I should know the answer to but don't:

Suppose $1\to F \to G \to G/F \to 1$ is a short exact sequence of abelian groups with $F$ finite and $G/F$ torsion-free. Must the sequence split?

This is not true if you merely assume that $F$ is torsion. A counterexample is given by YCor here: https://mathoverflow.net/a/314536/20598.

Equivalently, suppose $G$ is a compact abelian group with finitely many components. Then does $G_0 \to G \to G/G_0$ split? This is not true without assuming there are finitely many components, as YCor's example shows, and it's also not true for nonabelian groups, as Max's answer here shows: https://math.stackexchange.com/a/954539/23805 (though I think it's true whenever $G/G_0$ is cyclic).

  • $\begingroup$ The answer is yes. I don't remember right now where to find a reference. $\endgroup$ – YCor Jan 12 at 11:55
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    $\begingroup$ In fact, you only need the torsion subgroup has bounded order. See the (currently broken, but I’ll fix it as soon as I’ve posted this) link in mathoverflow.net/questions/60525/… $\endgroup$ – Jeremy Rickard Jan 12 at 12:10
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    $\begingroup$ @JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent". $\endgroup$ – YCor Jan 12 at 12:15
  • $\begingroup$ @YCor I agree, but I decided to stick with the terminology of the reference. $\endgroup$ – Jeremy Rickard Jan 12 at 12:24

Here’s a quick homological proof.

Suppose $F$ is finite and $H$ torsion free. Then $F\cong\text{Hom}(F,\mathbb{Q}/\mathbb{Z})$, so $$\text{Ext}^1(H,F)\cong\text{Ext}^1\left(H,\text{Hom}(F,\mathbb{Q}/\mathbb{Z})\right) \cong\text{Hom}\left(\text{Tor}_1(H,F),\mathbb{Q}/\mathbb{Z}\right), $$ which is zero since torsion free abelian groups are flat.

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    $\begingroup$ Is the last isomorphism obvious? I mean, I could probably easily prove it, but... $\endgroup$ – მამუკა ჯიბლაძე Jan 12 at 13:06
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    $\begingroup$ @მამუკაჯიბლაძე $\text{Hom}(-,\text{Hom}(F,\mathbb{Q}/\mathbb{Z}))\cong\text{Hom}(-\otimes F,\mathbb{Q}/\mathbb{Z})$. I’m just applying the first right derived functor of each side to $H$. $\endgroup$ – Jeremy Rickard Jan 12 at 14:01
  • $\begingroup$ Very slick. Nice. $\endgroup$ – Sean Eberhard Jan 12 at 14:14
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    $\begingroup$ ...and you get $\text{Tor}$ on the right since $\text{Hom}(-,\mathbb Q/\mathbb Z)$ is exact. Fine. $\endgroup$ – მამუკა ჯიბლაძე Jan 12 at 19:34

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