There are a lot of similar questions concerning this fact, and I looked through them all but the answers tend to take larger steps and I'm having a hard time following them.
Suppose that $1\to L \rightarrow H \rightarrow R\to 1$ is a short exact sequence of groups that splits via some $\epsilon: R \rightarrow H$. I want to show that in this case $H$ is the semidirect product of $R$ and $L$.
To do this I consider the $R$-action on $H$ which is just the conjugation $h \mapsto r h r^{-1}$. (Do I understand correctly that this we can only do since we have an injective $\epsilon: R \rightarrow H$?) As $L$ is normal in $H$ it is preserved by every such conjugation and we have for each $r\in R$ a group isomorphism $l \mapsto r l r^{-1}$. In turn this means that we have defined a group homomorphism $\alpha: R \rightarrow Aut(L)$. So I see that this means that we can define the semidirect product of $L$ and $R$ along this map $\alpha$, $L \rtimes_\alpha R$.
However, now I would need to show that this semidirect product is isomorphic to $H$. To do this I would like to write out explicitly the group isomorphism. The answers I've seen on this site didn't spell this out carefully. What is the group isomorphism that demonstrates that $H \equiv L \rtimes_\alpha R$?