# Finite group with cyclic Sylow subgroups

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

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