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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?

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  • $\begingroup$ 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? $\endgroup$ Commented Jun 20, 2012 at 19:21
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    $\begingroup$ 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 $\frac{1}{|K|\,|H|}\sum i(k)$. $\endgroup$
    – Steve D
    Commented Jun 20, 2012 at 19:25
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    $\begingroup$ 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 $H$. $\endgroup$
    – user6976
    Commented Jun 20, 2012 at 21:04
  • $\begingroup$ 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. $\endgroup$ Commented Jun 20, 2012 at 22:24

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See http://drexel28.wordpress.com/2011/05/02/double-cosets/

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    $\begingroup$ 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. $\endgroup$
    – Matt Young
    Commented Jun 20, 2012 at 19:10
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    $\begingroup$ The point of the link is that it summarizes all the "easy" enumeration facts about double cosets. $\endgroup$
    – Igor Rivin
    Commented Jun 20, 2012 at 19:46
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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.

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    $\begingroup$ And, of course, one can play the same game with $K\backslash G\times H\backslash G$... $\endgroup$
    – HJRW
    Commented Jun 21, 2012 at 11:58

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