If $G$ is a finite group and $H \leq G$ is a subgroup, then $G/H = \frac{G}{H}$.
Is there an easy way to compute $K \backslash G / H$, for $K \leq G$ also a subgroup?
If $G$ is a finite group and $H \leq G$ is a subgroup, then $G/H = \frac{G}{H}$. Is there an easy way to compute $K \backslash G / H$, for $K \leq G$ also a subgroup? 




Take $K=H$ and consider the diagonal action of $G$ on $\Omega\times\Omega$, where $\Omega$ is the set of the right cosets $H\backslash G$. Let the number of orbits of this action be $d$. Then $d$ is the number of double cosets $H\backslash G/H$: If we denote the orbits by $\Omega_1,\Omega_2,\ldots, \Omega_d$, let $g_1,\ldots,g_d$ be representatives of the respective orbits. Then the map sending $(Hg,Hg^{\prime})$ to $Hg^{\prime}g^{1}H$ is easily shown to be a bijection between the set of orbits and double cosets. More details can be found under the topic of Schur bases. 


$BN$
pair. What kind of examples have you looked at? – Jim Humphreys Jun 20 '12 at 19:21$G$
together with the orders of its subgroups$K$
and$H$
must determine the number of double cosets. This is presumably false, though I don't have a small counterexample at hand; but that would need to be sorted out first of all. – Jim Humphreys Jun 20 '12 at 22:24