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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!

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  • $\begingroup$ 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. $\endgroup$
    – reuns
    Dec 12 '18 at 17:58
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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.

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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.

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  • $\begingroup$ Beware the commutators are not a subgroup. They only *generate $G,G]$. $\endgroup$
    – Bernard
    Dec 12 '18 at 18:37
  • $\begingroup$ For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem $\endgroup$
    – leibnewtz
    Dec 12 '18 at 22:40

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