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

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

Not in general.

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 $p < n$.


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