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 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?
A group of order $ap^{n}$
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abstract-algebra
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0Thanks for asking, there is an typo in the original question:( – 2012-12-28
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0yes, that is a hint but I dont know how to use the information in the problem. – 2012-12-28
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1It seems like $G$ is a red herring, insofar as you can reduce it to $P\lhd H$ where $H$ has the same property that $|H|=ap^n$ with $(a,p)=(a,p-1)=1$ – 2012-12-28
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0what do you mean by red herring? – 2012-12-28
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0A "red herring" is an English term for a confusing bit of information that can lead you astray. Not sure the origin of the term. – 2012-12-28
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0Thank you:), I have learnt something else except algebra – 2012-12-28
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0Ah, it originates with leading hunting dogs astray by giving them the strong scent of a "red herring," diverting them from their original goal. – 2012-12-28
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0By the way, can you explain to me why I can reduce it to P is normal in H, and H has the same order like G? – 2012-12-28
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1$H=N(P)$ in the original problem, then $P\lhd H$, and this $H$ has the same GCD properties as $G$. – 2012-12-28
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0But how do you know the order of N(P) is the same with G? – 2012-12-28
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0In the answer below, it has been proved that N(P)=P, namely the order of N(p) is $p^{n}$, is that right ? – 2012-12-28
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0You don't know it is the same order, but you know that $p^n$ divides it, and you know it divides $ap^n$, so its order is of the form $bp^n$ where $b|a$ so $(b,p)=(b,p-1)=1$ – 2012-12-28
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0For 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
1 Answers
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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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1then we can get $a^{'}=1,N_{G}(P)=P$,and P is abelian, so P is contianed in the center of its normalizer $N(P)$ – 2012-12-28
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0I have another question is that, suppose that we know that P is only a Sylow subgroup then I want to know that p divides $[N(p):C(P)] $ if and only if P is nonabelian – 2012-12-28
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0I think you better open a new thread for the new question. – 2012-12-28
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0ok, I will try first , then I will post it – 2012-12-28