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Suppose I have a group $G$ such that $n_p \not \equiv 1 \pmod{p^2}$, where $n_p = |\operatorname{Syl}_p(G)|$. Consider a $P \in \operatorname{Syl}_p(G)$ acting on $ \operatorname{Syl}_p(G)$ by conjugation. I want to show that $\exists Q \in \operatorname{Syl}_p(G) : |\mathcal{O} (Q)| = p$. I was considering how to apply the orbit-stabilizer theorem, but I do not know how to apply the fact that $n_p \not \equiv 1 \pmod{p^2}$.

Shaun
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user100101212
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    $n_{p} \neq 1 \bmod{p^{2}}$, so $|\mathrm{Syl}_{p}(G) \setminus {P}| \neq 0 \bmod{p^{2}}$. What sort of action does $P$ have on the remaining Sylow $p$-groups? – xxxxxxxxx Sep 30 '19 at 16:32
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    You are essentially re-asking this other question. To be precise, you are asking a question to try to get an answer for the hint I posted there (so as to get the full solution that I purposely did not provide). Please do not post separately when dealing with the same question. – Arturo Magidin Sep 30 '19 at 18:08

1 Answers1

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  1. For $P$ acting on $Syl_p(G)$, each orbit has size equal to some power of $p$. (Use orbit-stabilizer.)
  2. Show that there is only one orbit of size $1$. (Use Sylow's theorems.)
  3. The sizes of the orbits sum up to $n_p$, so conclude that there is some orbit of size $p$.
spin
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