Let $G$ be a group, $H\unlhd G$, prove that the commutator subgroup $H'$ of $H$ is a normal subgroup in $G$.

Question

Let $G$ be a group, $H$, a normal subgroup of $G$, prove that the commutator subgroup $H'$ of $H$ is a normal subgroup in $G$.

My ideas:

I want to prove it like this: $gH'g^{-1} = H'$ $\forall g \in G$ but I don't know how to continue, because $g$ may not lie in $H$ and $ghg^{-1}h^{-1}$ it's not a commutator.

Answer 1

Hint: $g[h_1,h_2]g^{-1}=[gh_1g^{-1}, gh_2g^{-1}]$. Why is this a commutator of $H$?