$H \unlhd G$, $\pi:G \to H$ is a group homomorphism with $\pi(h)=h$. Show that $G \cong H × G/H$

Question

Let $H$ be a normal subgroup of a group $G$ such that there is a group homomorphism $\pi:G \to H$ with $\pi(h)=h$ for all $h \in H$. Prove that $G$ is isomorphic to $H × G/H.$

I don't know how to solve this. Any help is appreciated.

And if you know some abstract algebra book that deals with this kind of problem please mention, thanks.

Answer 1

As $H$ is normal, we have a short exact sequence of groups \begin{equation} 1\xrightarrow{\hspace{1cm}}H\xrightarrow{\hspace{.5cm}\iota\hspace{.5cm}} G\xrightarrow{\hspace{.5cm}p\hspace{.5cm}} G/H\xrightarrow{\hspace{1cm}}1 \end{equation} and our hypothesis is that there exists a map $\pi:G\to H$ such that $\pi\circ\iota=\operatorname{id}_H$. What you are trying to prove is sometimes referred to as "the splitting lemma", you can find a proof in these notes, Theorem $3.2$.