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Let G be a group of order $ap^{n}$ where p is prime and (a,p)=(a,p-1)=1.Suppose that some Sylow p-subgroup $\, P is cyclic. Prove that P is contained in the center of its normalizer N(P).

we can consider the homomorphism $N(P)\rightarrow Aut(P)$,then the kernel is the centralizer of P, but I haave no idea of what is going on, can anyone help me with this?

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    For the answer below, ask there. I didn't write it, nor did I have an answer, I just noted the red herring.2012-12-28

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The homomorphism $\,\phi: N_G(P)\to Aut(P)\,$ is

$\phi(n)(x):=x^n:=n^{-1}xn\Longrightarrow \ker\phi=C_G(P)\Longrightarrow N_G(P)/C_G(P)\cong K\leq Aut(P)$

But since $\,P\,$ is abelian we get that $\,P\leq\ker\phi\,$ , whence we have that $\,N_G(P)/P\cong H\leq Aut(P)\,$

But $\,|Aut(P)|=p^{n-1}(p-1)\,\,,\,\,\left|N_G(P)/P\right|=a'\,\,,\,\,a'\mid a\Longrightarrow $ ...and $\,(a,p)=(a,p-1)=1\,$ , so...

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    ok, I will try first , then I will post it2012-12-28