Double-coset counts need not divide group order
= Double-coset counts need not divide group order
For the order-eight <dihedral group> generated by $r$ of order four and $s$ of order two with $srs=r^{-1}$, let $H=K=\langle s\rangle$. There are exactly three <double cosets>: $H$, $Hr^2H$, and $HrH$, of sizes two, two, and four. Their number does not divide eight. This differs from the number of ordinary <cosets>, which divides group order by <Lagrange's theorem>.