# Quotient group $G/H$ is abelian iff $[G,G] \subseteq H$

So I've been working on this problem for a while. I just can't seem to find the solution. Also in the previous subquestion, I showed that $$G$$ is abelian iff $$[G,G] = {e}$$. Where $$e$$ is the neutral element of $$G$$ and $$[G,G]$$ is the commutator group of $$G$$.

Let $$H\triangleleft G$$, Show that $$G/H$$ is abelian iif $$[G,G] \subseteq H$$.

I've proved this direction

$$\Rightarrow$$) If $$G/H$$ is abelian, $$\forall x,y \in G$$,

$$xHyH = xyH = yxH = yHxH$$.

Therefore have that $$G$$ is abelian. As shown earlier, $$[G,G] = {e}$$ if $$G$$ is abelian. So $$[G,G] = {e} \subseteq H$$.

I can't seem to figure out the other direction...

$$\Leftarrow$$)

I only have that for $$x,y \in G$$,

$$x^{-1}y^{-1}xy \in H$$ and that $$H\triangleleft G$$. It does not seem to lead me anywhere.

Thanks for the help!

• If $\phi$ is a (surjective) morphism $G \to A$ with $A$ abelian then $[G,G] \subset \ker(\phi)$. Conversely if $[G,G] \subset \ker(\phi)$ then $\phi : G \to G/[G,G]\to A$ so $A$ is abelian. Dec 12, 2018 at 17:58

Recall that $$[G,G]=\langle xyx^{-1}y^{-1} | x,y \in G\rangle$$.

If $$G/H$$ is abelian, this means $$xyH=yxH$$, for all $$x,y \in G$$. Hence $$xyx^{-1}y^{-1} \in H$$, so $$[G,G] \subseteq H$$. Be careful here, we don't know that $$G$$ is abelian as you suggest.

On the other hand, if $$[G,G] \subseteq H$$, then for all $$xyx^{-1}y^{-1} \in [G,G]$$ we have $$xyx^{-1}y^{-1}H=H$$. But this means $$xyH=yxH$$, so $$G/H$$ is abelian.

• Beware the commutators are not a subgroup. They only *generate $G,G]$. Dec 12, 2018 at 18:37
• For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem Dec 12, 2018 at 22:40
• Just a little supplement: Since $H$ is normal in $G$, we know $Hxy=xyH=yxH=Hyx$, from which we can derive that $xyx^{-1}y^{-1} \in H$. Dec 13, 2022 at 15:03

Let :

$$\pi : G \to G / H$$

Then :

$$\forall x, y \in G, \pi([x,y]) = [\pi(x), \pi(y)]$$

Yet since $$[G,G] \subset H$$ then $$\pi([G,G]) = \{e\}$$. Hence we have :

$$\forall x, y, [\pi(x), \pi(y)] = e$$

So : $$G/H$$ is abelian.