 Solution

False. Realize the symmetries of a square as a dihedral group of order eight, generated by $r=(1234)$ and $s=(24)$, with $r^4=s^2=e$ and $srs=r^{-1}$. Take $H=K=⟨s⟩$. Its double cosets are preciselyThese are disjoint and exhaust the eight symmetries. There are three double cosets, and $3\nmid 8$. This proves that double-coset counts need not divide group order. Specifying the order-eight realization avoids competing conventions for the notation of a dihedral group.