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Given an abelian group $G$ (not necessarily finitely generated) And a fixed subgroup $H \le G$, can we always 'factor' $G$ into $$ G \cong H\oplus K $$ for some group $K$? If we assume that $G$ is finitely generated, we can use the structure theorem to deduce this fairly easily, but I am stuck when it comes to non-finitely generated groups.

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  • $\begingroup$ The torsion free and the torsion parts. $\endgroup$ – Bernard Dec 8 '17 at 13:04
  • $\begingroup$ @DietrichBurde I think the premise is that we're given an $H$. At least, that's what it sounds like here ("Given an abelian group $G$ (not necessarily finitely generated) And a subgroup $H \le G$"). $\endgroup$ – Arthur Dec 8 '17 at 13:04
  • $\begingroup$ That is way stronger than just asking that every subgroup is a quotient. And even this is false. $\endgroup$ – MooS Dec 8 '17 at 13:05
  • $\begingroup$ The first two people misunderstood the question. We are GIVEN a fixed $H$. $\endgroup$ – Elie Bergman Dec 8 '17 at 13:06
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Take $\mathbb Z \subset \mathbb Q$. We cannot have $\mathbb Q \cong \mathbb Z \oplus A$, becase the RHS either has rank at least two or is not torsion free.

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  • $\begingroup$ @ElieBergman As I edited my answer to fit your newly formulated question, you should delete that comment, since it looks weird now. $\endgroup$ – MooS Dec 8 '17 at 13:31
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If $G$ is finitely generated, then the fundamental theorem gives a decomposition of $G$ into cyclic groups. This, at least can be easily refuted for infinitely-generated groups:

Show that $\mathbb{Q}^+/\mathbb{Z}^+$ cannot be decomposed into the direct sum of cyclic groups.

Your question does not really ask this (although the title suggests it a bit). In general, for a given subgroup $H$ of $G$, there may or may not be a "complement" $K$ such that $G\cong H\oplus K$.

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This is not true, even for finite groups. Take $G=C_4$ and $H$ a subgroup of order $2$. If $G=H \oplus K$, then $K$ would have order $2$ and $G$ would have exponent $2$, not $4$.

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