Is the semidirect product of normal complementary subgroups a direct product. If $G$ is a group with $K$ and $N$ as normal complementary subgroups of $G$, then we can form $G \cong K \rtimes_\varphi N$ where $\varphi:N \to Aut(K)$ is the usual conjugation. But someone told me that $G \cong K \times N$ also. Is this true? It suffices to prove that $\varphi$ is trivial, but why is this the case?
(This question came up in an attempt at a different problem. In the specific case that this relates to, I also know that $K$ and $N$ are simple groups, but I don't know if this is relevant.)    

This is a consequence of the fact that if two normal subgroups intersect trivially, then elements of the first group commute with elements of the second group, so conjugation does nothing. This is proved by showing the commutator of the two elements is contained in the intersection of the subgroups, which means it is the identity.
