Suppose $aHbH$ is a left coset. Show that $H$ is normal in $G$. Let $G$ be a groupand $H\le G$ such that for each $a,b \in G$, the set-product $aHbH$ equals a left coset. Show that $H\unlhd G$.
If $aHbH = abH$, then I can show that. But here it's showing it is equal to a left coset.  

$aHbH$ is a left coset containing $ab$. Since any two left cosets are either disjoint or equal, this coset is $abH$. Thus for $g\in G$ we have $gHg^{-1}H=gg^{-1}H=H$, hence since $ghg^{-1}$ is in this coset for any $h\in H$ it follows that $H$ is normal. 
