Let $G$ be a group and let $N$ be a normal subgroup of $G.$ Let $N'$ denote the commutator subgroup of $N.$ Prove that $N'$ is a normal subgroup of $G.$
What I do know is that the commutator subgroup is characteristic. What I am not sure about is whether or not $N'$ is characteristic in $N$ or in $G$. If it is characteristic in $G$, then it is invariant under the conjugation automorphism $gN'g^{-1}$ for every $g \in G$ and I am done. Otherwise it is characteristic in $N$ only, which means I am stuck. I would appreciate your help.
Hint: Let $G$ be a group and $H \leq K$ be two subgroups of $G.$ If $H$ is characteristic in $K$ and $K$ is normal in $G$ then $H$ is normal in $G.$
EDIT: Let $g \in G$ and consider the automorphism $\phi_g:G \to G, x \mapsto gxg^{-1}.$ Since $K$ is normal, $gKg^{-1} = K.$ So $\phi_g|_K$ is an automorphism of $K.$ Since $H$ is a characteristic subgroup of $K, \phi_g|_K(H) = H.$