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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 $\pmod{p}$. –  user641 May 1 '12 at 22:27
    
@SteveD Thanks. Then I guess use "All the orbits must partition $X$".But are there any way that we could know the order of $X$? –  Simonaster May 1 '12 at 23:15
    
@SteveD It seems that there is a formula that $|X|$=$|G|$/$|K|$, but actually I didn't know a lot about that formula. Sigh.. –  Simonaster May 1 '12 at 23:19
    
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?). –  user641 May 2 '12 at 0:31
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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. –  Brett Frankel May 2 '12 at 0:35
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1 Answer 1

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 $H \leq K$.

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