I asked this question two days ago om Math SE but didn't receive an answer: <https://math.stackexchange.com/questions/1597321/generalization-of-hkh-kk-cap-h>


Suppose we are given subgroups $H,K$ of a finite group $G$. Denote by $\langle H,K\rangle$ the subgroup generated by $H$ and $K$. 


If $H$ normalizes $K$, then $\langle H,K\rangle =HK$ and 
$$(\langle H,K\rangle:H) = (K:H \cap K)\tag{$\ast$}$$
Is there a generalization of $(\ast)$ if $H$ isn't supposed to normalize $K$ ? 


Comment: $(\ast)$ always holds for $HK$ on the left hand side (in place of $\langle H,K\rangle$) by he double coset formula and we have $\langle H,K\rangle =HK\cdots HK$ for a fixed number of factors. Maybe this property could be used in some way ? 



Generalization of $(HK:H)=(K:H\cap K)$
I don't really know what you are looking for, but, for the record, if we know a full set $T$ of the $(H,K)$-double coset representatives in $G = \langle H,K \rangle$, ( so that $G$ is the disjoint union $\bigcup\_{t \in T}(HtK)$), then, for each $t$, we $|HtK| = |t^{-1}HtK| = \frac{|H||K|}{|t^{-1}Ht \cap K|}$. Hence $[G:H] = \sum\_{t \in T} [K: t^{-1}Ht \cap K]$. 


So if we know how $K$ intersects with conjugates of $H$, and we can determine a set of $(H,K)$-double coset representatives in $G$, we can get a formula of sorts.


I don't know if Sylow invented double cosets, but he certainly made use of the formula above when he proved Sylow's theorem(s). 


As I said in comments, it is sometimes difficult to determine the number of $(H,K)$-double cosets if you don't already know $|G|$.


Double coset decompositions are also useful in representation theory, for example in Mackey's formula for the restriction back to $K$ of a module induced from $H$.


