Let $G$ be a finite group and $p$ be a prime. Let $H$ be a subgroup of $G$ which contains $N_G(P)$ for some Sylow $p$-subgroup $P$ of $G$. Suppose $P \subseteq H^g$ for some $g \in G$. Prove that $g \in H$.

With the assumptions I can prove that $N_G(H)=H$. What should I do then?

Thanks in advanced.

  • $\begingroup$ DO you understand what does the question mean? $\endgroup$ – user87543 Jan 7 '14 at 2:33
  • $\begingroup$ No. Can you explain to me? $\endgroup$ – user111636 Jan 7 '14 at 2:49

Consider $P\trianglelefteq N_G(P)\subseteq H\subseteq G$. Observe that $P$ is a sylow $p$ subgroup of $H$ and by hypothesis $P\subseteq H^{g}$, which implies $P^{g^{-1}}\subseteq H$. Thus both $P, P^{g^{-1}}$ are sylow $p$ subgroups of $H$ and hence they are conjugate, i.e there exists an $h\in H$ such that $P=P^{g^{-1}h}$. Thus $g^{-1}h\in N_G(P)\subseteq H$, which implies $g^{-1}\in H$. Thus $g\in H$.

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  • $\begingroup$ Spacing seems to improve if you use {}^{g}H: compare $P\subseteq ^{g}H$ to $P\subseteq {}^g H$. $\endgroup$ – Pedro Tamaroff Jan 7 '14 at 5:55
  • $\begingroup$ @Pedro, thanks, that is nice to know. $\endgroup$ – user114539 Jan 7 '14 at 5:56
  • $\begingroup$ To change $\;H^g\;$ for $\;^gH\;$ seems to make things less clear, and I can't see what's the advantage of this notation, im particular when used with the subset sign: it looks like $\;\subset\;$ is being raised to $\;g\;$...pretty weird. $\endgroup$ – DonAntonio Jan 7 '14 at 5:59
  • $\begingroup$ @DonAntonio, ahh i did not even notice, somehow i am used to writing conjugation that way, but i am sure the idea is clear. $\endgroup$ – user114539 Jan 7 '14 at 6:01
  • $\begingroup$ The idea is clear, the notation much less. Just compare between $\;^gP\;$ or even $\;{}^{g}P\;$ and $\;P^g\;$ ...on the left, $\;g\;$ appears very far away. $\endgroup$ – DonAntonio Jan 7 '14 at 6:03

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