if $G' <H < G$ then $H$ is normal in $G$.

Question

if $G' <H < G$ then $H$ is normal in $G$. ($G'$ is the commutator subgroup of $G$.)

This is what I do:

because $G' < H$ we have $\frac{H}{G'} \triangleleft \frac{G}{G'}$. because $\frac{G}{G'}$ is abelian then $\frac{\frac{G}{G'}}{\frac{H}{G'}} \approx \frac{G}{H} $ is abelian and it means $\frac{G}{H}$ is abelian.

now we have $Hg_1Hg_2=Hg_2Hg_1 \Rightarrow Hg_1g_2=Hg_2g_1 \Rightarrow (g_2g_1)^{-1}(g_1g_2)=h $ for some $h \in H$.

now I stuck here.I need some help to finish this, I feel that I am in right path.

Thank you very much.

Answer 1

The homomorphism theorems imply that when $N$ is a normal subgroup of $G$, there is a bijection between the subgroups of $G$ containing $N$ and the subgroups of $G/N$ given by $$ H\mapsto H/N $$ In this correspondence, normal subgroups correspond to normal subgroups.

When $N=G'$, the quotient $G/G'$ is abelian, so each of its subgroups is normal. So, since $H/G'$ is normal in $G/G'$, you have that $H$ is normal in $G$.

Answer 2

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 5(d) on p.65 in Herstein's book.
I could not solve this problem, so I referred to "An Introduction to Algebraic Systems" (in Japanese) by Kazuo Matsuzaka.

Let $a\in G$ and $x\in H$.
Since $H$ includes all the commutators of $G$, $axa^{-1}x^{-1}\in H$.
So, $axa^{-1}=hx$ for some $h\in H$.
So, $axa^{-1}=hx\in H$.
So, $H$ is a normal subgroup of $G$.