Is $G$ isomorphic to $\frac{G}{H}\times H$? Let $G$ be a group and $H$ a normal subgroup. Is $G$ necessarily isomorphic to $\frac{G}{H}\times H$?

This only happens when $G$ is a direct product of two groups to begin with (and even in that situation not for all normal subgroups), i.e. when $G$ is not indecomposable. Some counterexamples:


*

*The symmetric group $S_n$ for $n\geq 3$

*The quaternion group $Q_8$

*Any semidirect product of indecomposable groups that is not direct

*Any cyclic group $\mathbb{Z}_{p^k}$ for prime $p$ and $k\geq 2$

*The integers $\mathbb{Z}$

*etc.


The situation you describe is actually quite rare. To see that note that for example this doesn't work even for groups that are products, e.g. $G=\mathbb{Z}_4\oplus\mathbb{Z}_2$. Now take the subgroup $H=\mathbb{Z}_2\oplus\mathbb{Z}_2$ and note that
$$H\oplus G/H\simeq \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\not\simeq G$$
