 Solution

Assume $H=\ker \theta$ for a group homomorphism $\theta :G\to K$. For $h\in H$ and $g\in G$,Thus $gH{g}^{-1}\subseteq H$. Apply the same inclusion with $g^{-1}$ to obtain the reverse inclusion, giving $gH{g}^{-1}=H$. Hence every kernel is a normal subgroup, proving the reverse implication and the required equivalence.