= Solution
\b[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 $H=K=\langle s\rangle$. Its <double cosets> are precisely
$$
H=\{e,s\},\quad Hr^2H=\{r^2,r^2s\},\quad
HrH=\{r,r^3,rs,r^3s\}.
$$
These are disjoint and exhaust the eight symmetries. There are three <double cosets>, and $3\nmid8$. 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>.
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