A group in between the commutator subgroup and the original group must be normal

Question

Let $C$ be the commutator subgroup of a group $G$. then by some easy arguments, we know that

$1$. $C$ is normal in $G$

$2$. $G/C$ is abelian

$3$. If $N$ is normal in $G$ and $G/N$ is abelian, then $C ⊂ N$.

My question is, how to prove that if $C ⊂ H ⊂ G$, then $H$ is normal in G?

My attempt is the following:

Since $C ⊂ H$, for any $a, b$ in $G$, $aba^{-1}b^{-1}$ is in $H$, so $ab(ba)^{-1}$ is in $H$. i.e. $abH = baH$, i.e. $(aH)(bH)=(bH)(aH)$, and we have $G/H$ is abelian. But my goal is to get $gH=Hg$ for any $g$in $G$. Do I have to use something like the Correspondence Theorem? Any help is appreicated.

Answer 1

Consider the quotient group $G/C$, it is abelian. In an abelian group very subgroup is normal, then $HC/C = H/C$ is normal in $G/C$.

By the correspondence theorem this means that $H$ is normal in $G$.