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If $H$ is a $p$-subgroup of $G$ with order of $p^{a}$ and $K$ is a Sylow $p$-subgroup of $G$ with order of $p^{b}$. $X$ is the set of left coset of $K$.

Let $H$ acts on $X$, what is the order of orbits? I have a feeling that it might be related to the prime $p$. But I don't know how to get that.

(I am trying to use this result to establish that $H$ is contained in $K$. )

Thanks!

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    @SteveD Thanks! I got it!2012-05-02

1 Answers 1

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If you want to prove that $H \leq K$, you will need an additional condition. Here is the general statement:

Let $G$ be a finite group, $K\leq G$ a $p$-Sylow subgroup, and $H\leq G$ a $p$-group. If $KH=HK$ then $H \leq K$.

The proof goes as follows:

From $KH=HK$ it follows that $HK$ is a subgroup and therefore $|HK|=\frac{|H||K|}{|H \cap K|}$. The order of $HK$ is a power of $p$, the order of $K$ is the maximal power of $p$ (since it is a Sylow subgroup). Since $K\leq HK$, $|K|=|HK|$. Thus $\frac{|H|}{|H \cap K|}=1$ and $|H \cap K|=|H|$. Therefore (since $H\cap K \leq H$) we have $H \leq K$.