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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 ?

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  • $\begingroup$ I think that for any pair of positive integers $m,n >1$ there is a cyclic group $H$ and $K$ of respective orders $m,n$ with $\langle H,K \rangle$ arbitrarily large. This is clear when $m = n =2$, for example. $\endgroup$
    – Geoff Robinson
    Commented Jan 4, 2016 at 20:49
  • $\begingroup$ Isn't the generalisation just your comment?? $\endgroup$
    – eric
    Commented Jan 4, 2016 at 21:00
  • $\begingroup$ @Geoff Robinson: Maybe one should expect a sum on the RHS ? $\endgroup$
    – user63850
    Commented Jan 4, 2016 at 21:02
  • $\begingroup$ I think the question is too vague in its present form. It has to depend on more that just $|H|$ and $|K|$, but when $H$ and $K$ are cyclic of the same prime order $p$ but $H$ and $K$ are not conjugate in $\langle H,K \rangle$, all $(H,K)$ double cosets have size $p^{2}$ ( so you have no chance of even knowing the number of double cosets unless you can already determine $|\langle H,K \rangle |$. $\endgroup$
    – Geoff Robinson
    Commented Jan 4, 2016 at 21:15
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    $\begingroup$ As you might have already noticed, you don't need $H$ to normalize $K$ for (*) to hold, only for the two groups to permute setwise. $\endgroup$
    – Russ Woodroofe
    Commented Jan 4, 2016 at 21:47

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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 $H$.

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  • $\begingroup$ Thank you Geoff, that is exactly what I was looking for. $\endgroup$
    – user63850
    Commented Jan 7, 2016 at 18:08

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