TITLE: Any Subgroup containing commutator subgroup is normal.
QUESTION [16 upvotes]: I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as

If $H$ is a subgroup containing commutator subgroup then $H$ is
  normal.

I.e. we have to show that $\forall g\in G$ such that $gHg^{-1}=H$ with fact that $G'\subset H$
It is for elements in $G'$ to show condition for normality. 
But how to do for elements not in $G'$ but in $H$, that is in $H\setminus G'$?

REPLY [13 votes]: $G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
isomorphism theorem for groups, $H$ is normal in $G$.