Suppose $\phi:G\rightarrow H$ is a group homomorphism. When is it true that $G\cong$ ker$(\phi)\oplus G/$ker$(\phi)$?
If $G$ and $H$ are abelian and there exists a homomorphism $\varphi:H\rightarrow G$ such that $\phi \varphi=1_{H}$ then this is true.
I seem to remember something about this when I studied exact sequences, but I don't have Dummit and Foote on hand.
I thought there was a case where the short exact sequence always splits like this?
The key phrase is "direct product"; $G\cong\ker(\phi)\oplus G/\ker(\phi)$ happens if and only if there exists some normal subgroup $N\lhd G$ such that $N\cap\ker(\phi)=1$ and $G=N\ker(\phi)$. (In particular, $N\cong G/\ker(\phi)$ here.)
If $N$ is not required to be normal then you get a "semidirect product", which characterises the $\phi\varphi=1_H$ condition you give.