If $H$ is a subgroup of $G$ and contains all commutators $aba^{-1}b^{-1}$ with $a, b \in G$, then $H$ is normal.

Question

I've stalled out on coming up with ideas to prove this.

Thought One:

Using $hb = aba^{-1}$ and $a^{-1}h = ba^{-1}b^{-1}$ somehow. However I can't assume $hb \in H$, I think.

Thought Two:

Using $bh = baba^{-1}b^{-1}$ and $ha^{- 1}= aba^{-1}b^{-1}a^{-1}$ somehow. But I can't assume those are in $H$ either.

I can't think of a plan to start solving this. Hints are welcome.

Answer 1

Let $C$ be the commutator subgroup $G/C$ is commutative, Let $p:G\rightarrow G/C, h\in H, g\in G$, $p(ghg^{-1})=p(h)$ implies there exists $u\in C, ghg^{-1}=uh$, since $u\in H$ we deduce that $H$ is normal.

If you do not know the notion of group quotient, use this: $ghg^{-1}=h(h^{-1}ghg^{-1})$ since $h^{-1}ghg^{-1}\in C\subset H$, we deduce that $ghg^{-1}\in H$.

Answer 2

You want to show that the subgroup $H$ is normal in $G$. That is for any $h \in H$, for every $g \in G$ $$ghg^{-1} \in H.$$

You already know that $h^{-1}ghg^{-1} \in H$ because every commutator is in $H$ that means that if $$h^{-1}ghg^{-1} = h'$$ then $ghg^{-1} = hh'$ such that $hh' \in H$. Thus we have proved that $gHg^{-1} = H$ for every $g \in G$, so that $H$ is a normal subgroup.