# Any Subgroup containing commutator subgroup is normal.

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 $$H\setminus G'$$?

• See this question. Aug 4 '18 at 18:09

If $g\in G$ and $h\in H$, then $ghg^{-1}h^{-1}=h'$, for some $h'\in H$ (since $H$ contains the commutator subgroup). But then $ghg^{-1}=h'h\in H$. Therefore, $gHg^{-1}\subset H$.
$G'$ 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 $G$.