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Thank you~

Show that if $G$ is a finite nilpotent group, then every Sylow subgroup is normal in $G$.

I know that the normalizer of any proper subgroup of a nilpotent group contains this subgroup properly. So I think maybe I can prove the normality of the Sylow subgroup of $G$, say $P$, by showing that $N(P)=N(N(P))$, thus showing that $N(P)=G$. (Here, $N(P)$ represents the normalizer of $P$ in $G$.) But I don't know how to complete this step.

I'll appreciate your help. Many thanks.

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You can find a complete solution at,fng.html for instance. – lhf Feb 28 '11 at 12:04
Please make the body of your message self-contained; the title is meant as indexing. You wouldn't ask a reader to go look at the spine of the book for content, would you? – Arturo Magidin Feb 28 '11 at 14:51
up vote 3 down vote accepted

Let $G$ be any finite group, and let $P$ be a Sylow $p$-subgroup of $G$, and $N(P)$ the normalizer in $G$ of $P$.

Note that $P$ is a Sylow $p$-subgroup of $N(P)$, and in fact is normal in $N(P)$; that means that $P$ is the only Sylow $p$-subgroup of $G$ that is contained in $N(P)$.

Now, suppose $g\in G$ normalizes $N(P)$ (that is, $g\in N(N(P))$). Then $g^{-1}N(P)g = N(P)$. We also know that $g^{-1}Pg$ is a Sylow $p$-subgroup of $G$, and since $P\subseteq N(P)$, then $g^{-1}Pg\subseteq g^{-1}N(P)g = N(P)$.

What does that tell you about $g^{-1}Pg$? Conclusion?

Note that this particular result does not require you to assume $G$ is nilpotent.

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I got it~ Many thanks~ – ShinyaSakai Mar 13 '11 at 18:47

A fairly simple way to do this is to note that a normal Sylow-subgroup is always characteristic. Thus, since you can make a subnormal chain from the Sylow-subgroup to the group itself, this Sylow-subgroup must in fact be normal in each subgroup in the chain and thus in the entire group.

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This is really an easy way of thinking. Thank you very much~ – ShinyaSakai Mar 13 '11 at 18:48

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