I wrote the following proof on an exam, I was wondering if it makes sense.
Question: let $H$ be a subgroup of $G$, written in multiplicitive notation. Prove that if the coset multiplication defined by $(aH)(bH) = (ab)H$ is well defined for all $a,b \in G$, then $H$ is a normal subgroup.
Proof: $H$ being a normal subgroup of G means that $gH = Hg$ If we define multiplication of cosets by multiplication of their representative elements, then suppose $H$ is not normal.
ii) since $Ha\ne aH$, then let $Ha = a'H$
iii) $(a'H)(bH) = (a'b)H \ne (ab)H$
So we multiplied $Ha$ and $aH$ against $bH$ and got different results. Since $Ha$ and $aH$ have the same representative elements, this means that multiplication would not be well-defined. Therefore, if $H$ was not normal, then multiplication by representative elements would not be well defined. Therefore $H$ must be a normal subgroup of $G$.