= Solution
A <subgroup> $N$ is a <normal subgroup> of $G$ if $gNg^{-1}=N$ for every $g\in G$. Define multiplication on its left <cosets> by
$$
(gN)(hN)=ghN.
$$
To check that this is well-defined, replace $g,h$ by $gn_1,hn_2$ with $n_1,n_2\in N$. Then
$$
(gn_1)(hn_2)=gh(h^{-1}n_1h)n_2\in ghN,
$$
because normality gives $h^{-1}n_1h\in N$. Thus the product depends only on the two <cosets>. Associativity follows from that of $G$, the identity is $N$, and the inverse of $gN$ is $g^{-1}N$. These are all the <group axioms>, so \b[$G/N$ is a <group> under coset multiplication]. The natural projection $g\mapsto gN$ is a surjective <group homomorphism> with kernel $N$.
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