\b[False.] Use the same <subgroups> of $S_3$. The <double coset> $HeH=H$ has two elements, while $H(13)H$ has four. Thus \b[even a fixed pair of subgroups can give different double-coset sizes]. Their disjoint union is all six elements of $S_3$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 7D / b / ii / Solution

 Solution (source code)