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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 $H\setminus G'$?
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 $G$.
