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Definition: A finite group is said to satisfy condition $\mathcal{C}_p$ if every subgroup of a Sylow $p$-subgroup is normal in $N_G(P)$.

Let G be a finite group with cyclic Sylow subgroups. I want to show that $G$ satisfies $\mathcal{C}_p$.

Following a similar argument posted as an answer on my previous question.

Let $P$ be a Sylow $p$-subgroup of $G$ and let $H \leq P$. I need to show that $H \unlhd N_G(P)$. Since $P$ is cylic, it is abelian, and so $H \unlhd P$. Thus $P \leq N_G(H)$. Let $x \in N_G(P)$. Then $H^x \unlhd P$, and so $P \leq N_G(H^x)$. So for any $p\in P$, we have that $H^xp = H^x$. Thus $xpx^{-1} \in N_H(H)$. This implies that $xPx^{-1} \in N_G(H)$. By Sylow's Theorem, there exists some $y \in N_G(H)$ such that $P^y = P^{x^{-1}}$. Hence $yx \in N_G(P)$. Now $H^{yx} = H^x$, as $y \in N_G(H)$. I'm unable to show that $x \in N_G(H)$. I know there's some Theorems (Burnside) on cyclic Sylow groups but I'm not sure if they are to be used here

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    $\begingroup$ Every subgroup of a cyclic $p$-group $P$ is fixed by all the automorphisms of $P$. $\endgroup$ – ancientmathematician May 15 '17 at 15:11
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    $\begingroup$ Probably the definition should say "...if every subgroup of every Sylow $p$-subgroup ...". $\endgroup$ – Derek Holt May 15 '17 at 15:12
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    $\begingroup$ Every subgroup $H$ of a finite cyclic group $P$ is characteristic in $P$ (because $H$ is the unique subgroup of $P$ with order $|H|$). What do you know about characteristic subgroups of normal subgroups? $\endgroup$ – Bungo May 15 '17 at 15:36
  • $\begingroup$ @Bungo, Thank you I might see how it follows. $H$ char $P$ and $P \unlhd N_G(P)$, then $H \unlhd N_G(P)$? $\endgroup$ – R Maharaj May 15 '17 at 15:42
  • $\begingroup$ @RMaharaj Yes, that's right. $\endgroup$ – Bungo May 15 '17 at 16:43

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