 Solution

Every $g\in G$ lies in $HgK$, since both subgroups contain the identity. If two double cosets intersect, chooseThen $g^{\prime }=h_2^{-1}{h}_{1}g{k}_1{k}_2^{-1}\in HgK$, so every element of $H{g}^{\prime }K$ lies in $HgK$. Reversing the argument gives equality. Thus 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 $H\cap gK{g}^{-1}$, givingThis formula explains the counterexamples in the following parts.