= Solution
For a <subgroup> $N\leq H$ and $h\in H$, a left <coset> is $hN=\{hn:n\in N\}$ and a right <coset> is $Nh=\{nh:n\in N\}$. A <normal subgroup> satisfies $hNh^{-1}=N$ for every $h\in H$, equivalently $hN=Nh$.
If $N$ is normal, the <quotient group> is the set $H/N$ of <cosets>, with multiplication
$$
\boxed{(hN)(kN)=(hk)N.}
$$
To check independence of representatives, replace $h,k$ by $hn_1,kn_2$. Then
$$
(hn_1)(kn_2)=hk(k^{-1}n_1k)n_2\in hkN,
$$
because normality puts $k^{-1}n_1k$ in $N$. Thus the operation is well defined. Associativity follows from that in $H$, the identity is $N$, and the inverse of $hN$ is $h^{-1}N$. This proves the group axioms for $H/N$ and explains the role of <coset multiplication is well-defined exactly for normal subgroups>.
Back to article page
