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.
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:
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 $4$.