 Solution

A subgroup $H$ is a normal subgroup of $G$ if $gH{g}^{-1}=H$ for every $g\in G$, equivalently if $gH=Hg$ for every $g$. In the symmetric group $S_3$, the subgroup $A_3=\{e,(123),(132)\}$ is normal: conjugation relabels a three-cycle as a three-cycle, so preserves this set. By contrast, $\{e,(12)\}$ is not normal because conjugating $(12)$ by $(23)$ gives $(13)$ outside the subgroup.

If $H$ is normal, define a product on its left cosets by $(gH)(kH)=(gk)H$. To check that this quotient group operation is well-defined, take different representatives $g{h}_1$ and $k{h}_2$, with $h_1,h_2\in H$. Their product isNormality puts the last two factors in $H$, so its coset is exactly $gkH$. Associativity follows from that in $G$, the identity coset is $H$, and the inverse of $gH$ is $g^{-1}H$. Thus all the group axioms hold, and the projection $g\mapsto gH$ is a group homomorphism.