# Factoring an Abelian group

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.

• The torsion free and the torsion parts. – Bernard Dec 8 '17 at 13:04
• @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 $H \le G$"). – Arthur Dec 8 '17 at 13:04
• That is way stronger than just asking that every subgroup is a quotient. And even this is false. – MooS Dec 8 '17 at 13:05
• The first two people misunderstood the question. We are GIVEN a fixed $H$. – Elie Bergman Dec 8 '17 at 13:06

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.
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 $G\cong H\oplus K$.
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 $4$.