Suppose $aHbH$ is a left coset. Show that $H$ is normal in $G$.

Question

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.

Answer 1

$aHbH$ 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.

Answer 2

Suppose for $a,b\in G$ we have $c\in G$ with $(aH)\cdot(bH)=cH$.

In particular, for $g,g^{-1}\in G$ we have $x_g\in G$ such that, $(gH)\cdot (g^{-1}H)=x_gH.$

That is, for any $h_1,h_2\in H$ we have $h_3\in H$ such that, $gh_1g^{-1}h_2=x_gh_3$.

In particular, for $h_2=e=h_1$, the identity element of $G$, the element $x_g$ will be in $H$ : $x_gh_4=gh_1g^{-1}h_2=geg^{-1}e=e$ for some $h_4\in H$. Implies that, $x_g=h_4^{-1}\in H$.

Next for $h_2=e$ and for any arbitrary $h_1$ we also have $gh_1g^{-1}=x_gh_5\in H$ for some $h_5\in H$. So $gHg^{-1}\subseteq H$.

That is $H$ is normal in $G$.