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Let $G$ be an abelian group and $H \subseteq G$ a subgroup with the following property: $$ H = \{ g \in G \mid \exists n\in \mathbb N^{+} \: ng \in H \}.$$

Does the short exact sequence $0 \to H \to G \to G/H\to 0$ split? Or in other words: Do we have $G \cong H \times G/H$?

Eric Wofsey
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principal-ideal-domain
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  • Are you defining $H$ by that property? If so, doesn't $G$ itself fit the definition (with $n=1$)? Or do you mean that $G/H$ is torsion? – peter a g Oct 31 '16 at 12:39
  • @peterag Like Peter Franek said: It's not a definition, it's a property. If you have already a subgroup $H$ you can check if it satisfies the property. – principal-ideal-domain Oct 31 '16 at 12:42
  • @peterag That does not satisfy the property mentioned. – Tobias Kildetoft Oct 31 '16 at 12:44
  • @TobiasKildetoft, and everybody, right.... OK. Deleting (one of) my previous comment(s)! – peter a g Oct 31 '16 at 12:44
  • Sorry if I continue to be stupid - the property can be rephrased as "$G/H$ has NO torsion". Correct? – peter a g Oct 31 '16 at 13:18
  • @peterag Yes, you are right. That's a shorter and more elegant way to say it. Thank you for that nice characterization! – principal-ideal-domain Oct 31 '16 at 14:10
  • Then I think it is false: see, for example,@Siddharth Venkatesh's comment at http://math.stackexchange.com/questions/769677/why-mathbbq-is-not-a-projective-mathbbz-module. The $H$ here would be the kernel of the map $A\to B$ there. Agree? – peter a g Oct 31 '16 at 14:40

1 Answers1

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No. Your condition is equivalent to asking that $G/H$ be torsion-free (since an element $g\in G$ such that $ng\in H$ for some $n$ is just an element whose image in $G/H$ is torsion). So your question is then whether if $K$ is a torsion-free abelian group, then every short exact sequence $0\to H\to G\to K\to 0$ splits (i.e., $K$ is projective).

This is false: let $K$ be any torsion-free abelian group which is not free (e.g., $K=\mathbb{Q}$). Let $G$ be a free group with a surjective homomorphism $f:G\to K$ and let $H=\ker f$. Then the sequence $0\to H\to G\to K\to 0$ does not split, since if it did then $K$ would be isomorphic to a subgroup of $G$ and hence free.

Eric Wofsey
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