= Solution
For a <normal subgroup> $H$, the <quotient group> consists of its <cosets> with multiplication
$$
\boxed{G/H=\{gH:g\in G\},\qquad(gH)(kH)=(gk)H.}
$$
To check that this is well-defined, replace $g,k$ by $gh_1,kh_2$ with $h_1,h_2\in H$. Their product is
$$
gh_1kh_2=gk(k^{-1}h_1k)h_2\in gkH,
$$
since normality puts $k^{-1}h_1k$ in $H$. Hence the product does not depend on representatives. <Associativity> follows from that in $G$, the <identity element> is $H=eH$, and $(gH)^{-1}=g^{-1}H$. These verify the <group axioms>.
<Lagrange's theorem> counts the equal-sized cosets:
$$
\boxed{|G|=|G/H|\,|H|,\qquad |H|\mid |G|.}
$$
Finally, if $|G|=p$ is <prime>, choose $g\ne e$. Its <cyclic subgroup> has order greater than one dividing $p$, hence equals $G$. All elements are powers of $g$, so $g^jg^k=g^{j+k}=g^kg^j$. Thus
$$
\boxed{G\cong C_p\text{ and }G\text{ is abelian}.}
$$
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