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This comes from the proof of the third Sylow Theorem in Michael Artin's "Algebra".

Let S be the set of Sylow p-groups in a given group G of order $p^em$. Let H be any Sylow group. If we decompose S into orbits for the operation of conjugation by H, then to establish s=1 modulo p, we only must show no element of S is fixed, since if H is a p-group, the order of any H-orbit is a power of p.

I'm having trouble with this last claim, I can't see why it is true. I don't know how much superfluous information I gave because I really don't understand the inference. Can anyone help?

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3 Answers 3

up vote 3 down vote accepted

For any group $\,G\,$ acting on any set $\,X\,$ , we have that for any $\,x\in X\,$ :

$$|\mathcal Orb(x)|=[G:G_x]\;,\;\;G_x:=\{g\in G\;;\;gx=x\}$$

Since the index of any subgroup of $\,G\,$ divides the order of $\,G\,$, the claim follows.

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The order of any orbit in any permutation action of any finite group equals the index of the stabilizer of any point in that orbit. In particular, it divides the order of the group.

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I need to learn to type faster. –  Andreas Blass Apr 24 '13 at 23:57

Aren't we forgetting something? You also have to show that $H$ is the only element of the set $S$ that is fixed, in other words the orbit of $H$ has size 1 and the rest has an orbit size a power of $p$ (as pointed out in the previous answers!) strictly greater than 1. This goes as follows. $H$ fixes an element $K \in S$ iff $H \subseteq N_G(K)$ (the normalizer of $K$ in $G$). Assume the latter. $K$ is a Sylow $p$-subgroup and certainly $K$ is normal in $N_G(K)$, hence its only Sylow $p$-subgroup. Since $H$ is a $p$-subgroup of $N_G(K)$, according to a more general lemma, it must lie in some Sylow $p$-subgroup of $N_G(K)$ for which there is only one choice, $K$. Hence $H \subseteq K$ and this implies $H=K$ because they have the same order. To go short, if $H \neq K$, then the orbit of $K$ is a non-trivial power of $p$. You can now conclude that $|S| \equiv 1$ mod $p$, where the $1$ comes from the orbit of $H$ and all the other orbit sizes are divisible by $p$.

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By the way, it is not too hard to show that the orbit size of $K$ actually equals $[H:H\cap K]$ ($= [K:H\cap K]$) –  Nicky Hekster Apr 25 '13 at 15:08

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