Prove $aH\cdot bH := (ab)H \Rightarrow Hb=bH$ Background: In an abstract algebra book, the author wants to motivate the definition of "normal subgroup" to give a group structure to a quotient set "$G/H=\{aH\mid a\in G\}$" where $H$ is a subgroup of the group $G$. 
He says "If the operation on the set $G/H$ is defined by $aH\cdot bH= (ab)H,$ then $H\cdot bH=bH$. Therefore we must have $Hb=bH$". 
Could anybody explain why "Therefore we must have $Hb=bH$"? Thank you.
 A: First, we are not defining $aHbH$ as $(ab)H$, we are defining it as
$$aHbH = \{ ah_1bh_2 : h_1, h_2 \in H \}.$$
What we want to show is that this definition yields a sensible product. I.e. $aHbH = (ab)H$. For this to make sense, we would like to do something like
$$ aHbH = a(Hb)H = a(bH)H = (ab)H^2 = (ab)H. \tag{1}$$
For this to work, we need $Hb = bH$. I.e. every right coset is also a left coset.
You should also be able to think of these equalities on the level of sets:


*

*$aHbH = a(Hb)H \iff \{ ah_1bh_2 : h_1, h_2 \in H \} = \{ agh : g \in Hb, h \in H \}$

*$a(Hb)H = a(bH)H \iff \{ agh : g \in Hb, h \in H \} = \{ agh : g \in bH, h \in H \}$

*$a(bH)H = (ab)H^2 \iff \{ abg : g \in H^2 \}$ where $H^2 = \{h_1h_2 : h_1, h_2 \in H\}$
If you set $a = e$ in $(1)$, you obtain $HbH = bH$. I.e.
$$ \{ h_1bh_2 : h_1, h_2 \in H\} = \{ bh : h \in H\}. \tag{2}$$
The hint Wojowu gives is that $Hb \subseteq HbH$ so you get $Hb \subseteq bH$ out of $(2)$. Now you need to show that $bH \subseteq Hb$.
