Where does the following argument break down?
Let $G$ be a group, with $H \triangleleft G$, $K \leq G$ and $H \cap K = 1$. Taking the semidirect product, we have $G \cong H \rtimes K$. Identifying $K$ with $1 \times K$, we consider conjugation of $K$ with any element in $H \rtimes K$. Pick any element $(h, k)$, then $(h, k)^{-1} \cdot (1, K) \cdot (h, k) = (1, K)$ Therefore, we conclude that $K$ must be normal in $G$ as well. But this is clearly wrong.