# Number of double cosets

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?

-
I doubt that there is any "easy" method for this in general, since the choices for the two subgroups can vary arbitrarily. Even when the subgroups are equal, it doesn't look straightforward: for example, you might be computing the "Weyl group" order for a finite group with a $BN$-pair. What kind of examples have you looked at? –  Jim Humphreys Jun 20 '12 at 19:21
Probably the best you can do is the Cauchy-Frobenius formula, with $K$ acting on the cosets of $H$. If $i(k)=| \lbrace g\in G\ |\ k^g\in H\rbrace |$, then the number of double cosets is $\frac{1}{|K|\,|H|}\sum i(k)$. –  Steve D Jun 20 '12 at 19:25
An easy way cannot exist. Note that if $G=KH$ (which happens quite often), then the number of double cosets is 1. So there is no relation between the number of double cosets and the indices of $K$ and $H$. –  Mark Sapir Jun 20 '12 at 21:04
To reinforce Mark's sensible comment, I'd add that the question itself takes for granted that the order of $G$ 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

The link gives a formula for the number of elements in a given double coset $KgH$ (note it depends on $g$) while the OP seems to want the number of double cosets. –  Matt Young Jun 20 '12 at 19:10
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.
And, of course, one can play the same game with $K\backslash G\times H\backslash G$... –  HJRW Jun 21 '12 at 11:58