= Solution
Every $g\in G$ lies in $HgK$, since both <subgroups> contain the identity. If two <double cosets> intersect, choose
$$
h_1gk_1=h_2g'k_2.
$$
Then $g'=h_2^{-1}h_1gk_1k_2^{-1}\in HgK$, so every element of $Hg'K$ lies in $HgK$. Reversing the argument gives equality. Thus \b[the double cosets partition the group].
Unlike single <cosets>, <double cosets> can have different cardinalities. For finite $G$, decompose $HgK$ into left <cosets> of $K$, or apply the <orbit-stabilizer theorem> to the action of $H$ on $gK$. Its <stabiliser subgroup> is $H\cap gKg^{-1}$, giving
$$
\boxed{|HgK|=\frac{|H||K|}{|H\cap gKg^{-1}|}.}
$$
This formula explains the counterexamples in the following parts.
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