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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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    The idea is the orbit-stabilizer theorem, which immediately implies the order of any orbit divides the order of the group (here, $H$). Since $H$ is a p-group, it has a very restricted set of numbers which divide $|H|$! Now think about everything $\pmod{p}$.2012-05-01
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    @SteveD Thanks. Then I guess use "All the orbits must partition $X$".But are there any way that we could know the order of $X$?2012-05-01
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    @SteveD It seems that there is a formula that $|X|$=$|G|$/$|K|$, but actually I didn't know a lot about that formula. Sigh..2012-05-01
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    Yes, the cosets partition $|G|$, so they are disjoint. Each has the same number of elements - $|K|$. So there are $|G|/|K|$ of them, this is easy (and is usually called Lagrange's theorem!). What's most important for your actual question, however, is that $|G|/|K|$ is *not* divisible by $p$ (why?).2012-05-02
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    Note that what you are trying to establish, namely that $H\leq K$, is *false*. The most you can guarantee is that $H$ is contained in one of the conjugates of $K$, for example when $H$ is also a Sylow $p$-subgroup.2012-05-02
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    @SteveD Thanks! I got it!2012-05-02

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