In recent days I am studying semidirect product of groups and I have come up with the following question which has already been answered here (From semidirect to direct product of groups), but I can't understand its solution from the comments, can you please clarify me with step by step solutions?
Let $G = N \rtimes_{\varphi} H$. If there exists a homomorphism $f: G \rightarrow N$ which is the identity on $N$, then is it true that $G$ is the direct product of $N$ and $H$?
Since the question concerns equality rather than isomorphism, it seems to me that, with the standard convention for identifying subgroups of $N \rtimes_\phi H$ with $N$ and $H$, the answer to the question is no.
Let $H=N$ be any nonabelian group, and let $\phi$ be the action of $H$ on $N$ by conjugation.
Then there is homomorphism from $G = N \rtimes_\phi H$ to $N$ which is the identity on $N$ and has kernel $\{(h,h^{-1}) : h \in H \}$, but $G$ is not the direct product of its subgroups $\{(h,1): h \in H \}$ and $\{(1,n): n \in N\}$, which are the subgroups that are customarily identified with $H$ and $N$ in the semidirect product.
Of course we do have $G \cong N \times H$, but that was not the question.