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Let $H,K \le G$ and consider the set of double cosets $ H \backslash G / K$. If we have $H' \le H$, is there a formula expressing $| H' \backslash G / K|$ in terms of $|H \backslash G / K|$, $|H:H'|$ and other similar quantities? It may be assumed that $K \le H$ if necessary.

The motivation for this question is computing the number of cusps of modular forms, as in this question. It seems to me that there should be a way to express the number of double cosets $|\Gamma_1(p) \backslash \Gamma(1) / \Gamma_{\infty}|$ in terms of $|\Gamma_0(p) \backslash \Gamma(1) / \Gamma_{\infty}|$, $|\Gamma_0(p) : \Gamma_1(p)|$ and possibly some quantity involving $\Gamma_{\infty}$. As someone with limited knowledge of group theory, it would be helpful to have a formula which allows for a reduction in this way.

For completeness, the expressions not defined in the link are $\Gamma(1) = \text{SL}_2(\mathbb{Z})$ and $\Gamma_{\infty} = \text{stab}(\infty)$.

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  • $\begingroup$ Does $H$ act freely on $G/K$ in your setting? $\endgroup$ Commented May 28 at 17:35
  • $\begingroup$ @MoisheKohan I think it would. I don't know for sure, but I could try proving it. Is there a general formula if so? $\endgroup$
    – J. S.
    Commented May 28 at 17:38
  • $\begingroup$ If the action is free then the answer will be quite simple (just the product), but in general I do not think there is a simple formula and you would need more information about orders of point-stabilizers. $\endgroup$ Commented May 28 at 17:44
  • $\begingroup$ @MoisheKohan Thank you, that sounds quite promising to me. I will try to prove the result you have mentioned and see if I can generalise it. I think the action is free for $p =2$ and free for $p > 2$ after quotienting by $\pm 1$, which would give the desired result. To ask a silly question, I imagine it is true in general that $|H \backslash G / K| = |K \backslash G / H|$? $\endgroup$
    – J. S.
    Commented May 28 at 17:54
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    $\begingroup$ Yes, use the inversion to construct a bijection of the two quotients. $\endgroup$ Commented May 29 at 15:55

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