# “Inverse” homomorphism of a group monomorphism

Assume $\varphi:H\rightarrow G$ is a finite group monomorphism. Is it true that there exists a homomorphism $$\psi:G\rightarrow H$$ such that $$\psi\circ\varphi=id_H?$$

• $\varphi \colon \mathbb{Z}/(2) \to \mathbb{Z}/(4)$ given by $x + 2\mathbb{Z} \mapsto 2x + 4\mathbb{Z}$. – Daniel Fischer Nov 24 '17 at 15:03

Consider $G$ any simple group of non-prime order $n$ and let $p|n$ be a prime. Then $G$ has $\mathbb{Z}_p$ as a proper subgroup (Cauchy's theorem). Thus we have a monomorphism (the inclusion)
$$f:\mathbb{Z}_p\to G$$
This monomorphism does not have the one-sided inverse, because there's only one homomorphism $g:G\to\mathbb{Z}_p$, namely the trivial one.
Indeed, if $g$ is non-trivial, then $\ker(g)$ is a proper normal subgroup of $G$. Since $G$ is simple then $\ker(g)$ has to be trivial and thus $g$ has to be a monomorphism. This is impossible since $p < n$.