This question asked me to show that the normalizer $$N(H) = \{g \in G \mid gHg^{-1} = H \} \subset G$$ Is a subgroup containing $H$ and that $H$ is a normal subgroup of $N(H)$. I've shown that $N(H) < G$ and $H \subset N(H)$ I'm just having some difficulty parsing everything for this last part.

My first instinct is to say this is almost trivial since the definition of a normal subgroup $N < G$ is $\forall g \in G$ $$ \{gng^{-1} \mid n \in N \} = gNg^{-1} = N$$ And we see that the normalizer of $H$, $N(H)$, is the largest subgroup of $G$ that contains $H$ where $H$ is normal. So given that, could I just say that by definition of the normalizer we can see that $H$ fulfills the requirements for a normal subgroup?