# If H is a p-group, the order of any H-orbit is a power of p.

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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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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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 $p$.
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