Solution
= Solution
Given $h\in H$, choose $g\in G$ with $\phi(g)=h$. For $k\in K$,
$$
h\phi(k)h^{-1}=\phi(gkg^{-1})\in\phi(K),
$$
because $K$ is normal. Thus $\phi(K)\triangleleft H$.
= Solution
Given $h\in H$, choose $g\in G$ with $\phi(g)=h$. For $k\in K$,
$$
h\phi(k)h^{-1}=\phi(gkg^{-1})\in\phi(K),
$$
because $K$ is normal. Thus $\phi(K)\triangleleft H$.