Let $\varphi : G \rightarrow G/N$. Prove/Dis-prove that there exists a right inverse of $\varphi$ that is *homomorphic*. Statement: Let $G$ be a finite group, $N$ be a normal subgroup of $G$ and let $\varphi: G \rightarrow G/N$ be the cannonical map. Prove/Dis-prove that there exists a right inverse of $\varphi$ that is homomorphic.
Testing the statement with $C_n$ and $D_n$, we see that there is a right inverse which is an homomorphism, for every quotient map.
How does one think about the statement for a general group.
Any hints/ideas are highly appreciated.
 A: 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.
A: Hint: Consider $\{1,x^2\}\lhd D_4$ where $x$ is rotation by $+90^\circ$.
A: For $\bar\varphi: G/N\to G$ to be such an inverse (namely $\varphi\bar\varphi=Id_{G/N}$), it must be $\bar\varphi(Na)\in Na, \forall a\in G$. If we further require it to be a homomorphism, then $\bar\varphi(G/N)\le G$. Therefore, the sought map $\bar\varphi$ picks one element out of each coset in a way that the set of such elements is a subgroup of $G$. Take e.g. $G:=\Bbb Z/4\Bbb Z=\lbrace0,1,2,3\rbrace$ and $N:=\lbrace0,2\rbrace$; thence, $G/N=\lbrace\lbrace0,2\rbrace,\lbrace1,3\rbrace\rbrace$, but neither $\lbrace 0,1\rbrace$ nor $\lbrace 0,3\rbrace$ is a subgroup of $G$: in this case, the canonical map $\varphi$ hasn't got any right inverse which is also a homomorphism.
