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 $H\setminus G'$?


2 Answers 2


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 $G$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.