# Finite-by-torsion-free abelian groups (or compact abelian groups with finitely many components)

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).

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

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.
• @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 $H$. – Jeremy Rickard Jan 12 at 14:01
• ...and you get $\text{Tor}$ on the right since $\text{Hom}(-,\mathbb Q/\mathbb Z)$ is exact. Fine. – მამუკა ჯიბლაძე Jan 12 at 19:34