Let $G$ be a group, $H$ be a subgroup of $G$. Define
$G/H := \{~gH~|~g \in G~\}$
$H\backslash G := \{~Hg~|~g \in G~\}$.
I want to show $G/H=H\backslash G$ implies $H\unlhd G$:
Assume $G/H=H\backslash G$. Then $\forall g\in G, g\in gH$, where $gH = Hg'$ for some $g'\in G$. So $g\in Hg'$ for some $g'\in G$. Hence $g=hg'$ for some $h\in H$. So $gH = Hg' = Hh^{-1}g = Hg$, and $H$ is therefore normal in $G$.
I feel very strange about this, could anyone help me check whether it is correct?
Your argument is ok but perhaps this one is simpler:
The left cosets form a partition of $G$. So do the right cosets.
If $gH$ is a right coset, then it must be $Hg$ because that's the only one that contains $g$.