= Solution
A <subgroup> $H$ is a <normal subgroup> of $G$ if $gHg^{-1}=H$ 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 $gh_1$ and $kh_2$, with $h_1,h_2\in H$. Their product is
$$
gh_1kh_2=gk(k^{-1}h_1k)h_2.
$$
Normality 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>.
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