5
$\begingroup$

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).

$\endgroup$
4
  • $\begingroup$ The answer is yes. I don't remember right now where to find a reference. $\endgroup$
    – YCor
    Jan 12 '19 at 11:55
  • 1
    $\begingroup$ 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/… $\endgroup$ Jan 12 '19 at 12:10
  • 1
    $\begingroup$ @JeremyRickard great! "bounder order" (used in the post you link) is awkward, it should be "of bounded exponent". $\endgroup$
    – YCor
    Jan 12 '19 at 12:15
  • $\begingroup$ @YCor I agree, but I decided to stick with the terminology of the reference. $\endgroup$ Jan 12 '19 at 12:24
12
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Is the last isomorphism obvious? I mean, I could probably easily prove it, but... $\endgroup$ Jan 12 '19 at 13:06
  • 2
    $\begingroup$ @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 $H$. $\endgroup$ Jan 12 '19 at 14:01
  • $\begingroup$ Very slick. Nice. $\endgroup$ Jan 12 '19 at 14:14
  • 1
    $\begingroup$ ...and you get $\text{Tor}$ on the right since $\text{Hom}(-,\mathbb Q/\mathbb Z)$ is exact. Fine. $\endgroup$ Jan 12 '19 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.