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If $G$ is a finite group, and $H$, $K$ are proper subgroups of $G$, then it is not necessary that $HK=KH$. But, these two subsets have same size. The question I would like to ask, then, is

If $HK\neq KH$, then what can we say about the size of $HK\cap KH$?

(Note that $H\cup K\subseteq HK\cap KH$.)

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What kind of things would you like to see? Estimates on the size, divisibility conditions, other things? –  Marc van Leeuwen May 8 '13 at 5:09
    
One can give equality/inequality/divisibility conditions etc., for the size of $HK\cap KH$. The first natural inequality (as noted in question) is $|KH\cap KH|\geq |H\cup K| \geq max(|H|,|K|)$. –  RDK May 8 '13 at 5:29
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In a dihedral group of order twice a prime, if $HK\neq KH$, then $HK \cap KH = H \cup K$, so that bound might be best possible. Also the size (3) need not divide the order of the group, nor be divisible by anything too reasonable. –  Jack Schmidt May 8 '13 at 5:38
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Cool question!${}$ –  Alexander Gruber May 8 '13 at 18:22
    
I doubt that you can say anything in general, but I'm pretty sure that there is at least one paper on (modern) additive combinatorics which estimates $|HK \cap KH|$ for nice groups $G$. –  Martin Brandenburg May 10 '13 at 21:58
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It would useful to connect $M=HK\cap KH$ with $N=H\cap K$. Evidently, $M$ is a (disjoint) union of double cosets $NxN, x\in M$. By [M.Hall, The Theory of Groups, Theorem 1.7.1] $$ |NxN|=\frac{|N|^2}{|N\cap x^{-1}Nx|}. $$ So, for example, if $p||N|$ ($p$ is prime) then $p||M|$.

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