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Let $p$ be a prime, let $G$ be a finite group whose order is divisible by $p$, and assume that $P \leq G$ is a maximal $p$-subgroup (if $Q \leq G$ is a $p$-subgroup and $P \leq G$ then $P=Q$). Prove that every conjugate of $P$ is also a maximal $p$-subgroup. Futhermore, if $P$ is the only maximal $p$-subgroup of $G$, then $P$ is normal in $G$.

I think that the definiton maximal $p$-subgroup is same as Sylow $p$-subgroup. So we can prove the results with the proof similarly Sylow's theorems. Is it true?

Thanks a lot.

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  • $\begingroup$ Yup, maximal = Sylow. $\endgroup$ – Daniel Fischer Dec 11 '13 at 12:51
  • $\begingroup$ Not quite: it should be if $Q$ is a $p$-subgroup with $P \subseteq Q$, then $P=Q$. As user110834 puts it, $P$ is normal by definition! $\endgroup$ – Nicky Hekster Dec 11 '13 at 13:52
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Hint: if $P$ is maximal $p$-subgroup then also any conjugate $P^g$ is a maximal $p$-subgroup.

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  • $\begingroup$ Can you give me more details? $\endgroup$ – user110834 Dec 12 '13 at 12:01
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    $\begingroup$ Sure, let $P$ be a maxiaml $p$-subgroup and pick a $g \in G$. Let $Q$ be a $p$-subgroup containing $P^g$. But $P^g \subseteq Q$ implies $P \subseteq Q^{g^{-1}}$. Moreover, $|Q^{g^{-1}}|= |Q|$. By the maximality of $P$, this yields $P=Q^{g^{-1}}$, so equivalently $P^g=Q$. Hence $P^g$ is also a maximal $p$-subgroup. If $P$ is the only maximal $p$-subgroup can you now show that it must be normal? $\endgroup$ – Nicky Hekster Dec 12 '13 at 13:36
  • $\begingroup$ Ehm ... of course. Thank you very much. $\endgroup$ – user110834 Dec 17 '13 at 10:02

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