This is my homework question and my solution is;
Let $G$ be a group and $H$ be a subgroup of $G$.
$G'$ is the smallest normal subgroup of $G$ whose quotient $G/G'$ is abelian.
If $N$ is a normal subgroup of $G$ and $G/N$ is abelian, then $G'$ is a subgroup of $N$.
Hence, as $H$ is a subgroup of $G$, $G'≤H≤G$ and then $H$ is a normal subgroup of G.
$ghg^{-1} \in H$ for all $g \in G$ and for all $h \in H$
($[g,h]$ is called a commutator, $G'$ is commutator subgroup of $G$)
$ghg$$^{-1}$=$ghg^{-1}(h^{-1} h)=[g,h]h$ and then $G'≤H$
And I know that $G'$ is a normal subgroup of $G$ and $G/G'$ is abelian.
But I am stuck, I would appreciate it if you help. Thank you.
Best Answer
I think you had it in your post, though a bit messy. Let me order your thoughts.
Proposition If $H$ is a subgroup of a group $G$ with $G' \subseteq H$, then $H$ is normal.
Corollary For any subgroup $H$ of a group $G$, $HG'$ is a normal subgroup.
Proof Since $G'$ is normal, $HG'$ is certainly a subgroup, and of course $G' \subseteq HG'$.
So let's prove the proposition: let $g \in G$ and $h \in H$, then $g^{-1}hg=h(h^{-1}g^{-1}hg)=h[h,g]$ and $[h,g] \in G' \subseteq H$ and of course $h \in H$, so their product is also in $H$.$\square$