Let $H\triangleleft G$. Prove that $G/H$ is abelian iff $ [G, G] \subseteq H$ 
The commutator of two elements $a, b \in G$ is defined as
$[a, b] = aba^{−1}b^{−1}$.


Let $[G, G] =\langle [a, b] | a, b \in G\rangle $ be the generated subgroup of all commutators of the elements of $G$.


Let $H\triangleleft G$.   Prove that  $G/H$ is abelian iff $ [G, G] \subseteq H$.

I have previously proved $[G,G] \triangleleft G$ and $G/[G,G]$ is abelian
The following solution is provided, but there are some things I don't understand. I feel need the intermediate steps  they are not provinding to fully understand it. Can someone shed some light?
Solution:
If $G/H$ is abelian for every $a, b \in G$ it follows that $abH = baH$,that is $aba^{−1}b^{−1} \in H$.

(1)  How do they get this?

so for every  $a, b \in G$ we have $[a, b] = aba^{−1}b^{−1} \in H$,  but now the subgroup
$[G, G]$ generated by the elements of the form $[a, b]$ is contained in $H$.

(2) They proved one commutator is in $H$, how does it extent to the whole generated subgroup?

Viceversa, if $ [G, G]  \subseteq H$ then for every $a, b \in G$ we have
$(abH)(a^{−1}b^{−1}H) = H = 1_{G/H}$;

(3) I would write $(abH)(a^{−1}b^{−1}H) =[a,b]H$, why does this equals H?

So $abH = (a^{−1}b^{−1}H)^{−1} = baH$.
 A: $(1)$ It is well known that we have $g_1H=g_2H$ if and only if $g_1^{-1}g_2\in H$. So if $G/H$ is Abelian then for any $a,b\in G$ we have $a^{-1}b^{-1}H=b^{-1}a^{-1}H$ and hence $(b^{-1}a^{-1})^{-1}(a^{-1}b^{-1})=aba^{-1}b^{-1}=[a,b]\in H$. So all the commutators are in $H$.
$(2)$ Since $a,b$ were arbitrary, we actually proved that all the commutators are in $H$. And since $[G,G]$ is by definition the smallest subgroup of $G$ which contains all the commutators this implies $[G,G]\leq H$.
$(3)$ Another standard result about cosets is that we have $gH=H$ if and only if $g\in H$. So since by assumption $[a,b]\in H$ we have $[a,b]H=H$.
A: (1) Consider $abH=baH$. It is the same as saying
$$\{abh\mid h\in H\}=\{bah'\mid h'\in H\}.$$
So for any $k\in H$, $abk=bak'$ for some $k'\in H$. But then $abkk'^{-1}=ba$, i.e., $a^{-1}b^{-1}ab=k'k^{-1}$, which is equivalent to saying $[a,b]\in H$ since $k'k^{-1}\in H$. (Why?)
(2) Note that $a,b\in G$ are arbitrary.
(3) See (1). Since $[a,b]\in [G,G]\subseteq H$, $[a,b]H=H$.
