Let $\varphi : G \rightarrow G/N$. Prove/Dis-prove that there exists a right inverse of $\varphi$ that is *homomorphic*.

Suppose there is, call it $\psi$. Then, by assumption, $\varphi\psi\colon G/N\to G/N$ is the identity map.

Let $K$ be the image of $\psi$ and define $\alpha=\psi\varphi\colon G\to G$. Take $g\in G$; then $$ \varphi(g\alpha(g^{-1}))=gN\cdot g^{-1}N=N $$ because $\varphi\alpha=\varphi\psi\varphi=\varphi$.

Therefore $y=g\alpha(g^{-1})\in N$ and we conclude that $g=y\alpha(g)\in NK$. Hence $G=NK$.

Now look for a group $G$ having a proper normal subgroup $N$ so that, for no proper subgroup $K$, we have $NK=G$.

The minimal such example is the cyclic group with four elements.

Hint: Consider $\{1,x^2\}\lhd D_4$ where $x$ is rotation by $+90^\circ$.