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I am a bit confused on how to go about this problem:

Let G be a finite group and $p$ the smallest prime dividing its order. Assume $P$ is a $p$-Sylow subgroup which is cyclic. Show $P$ is contained in the center of its normalizer.

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up vote 3 down vote accepted

Let $N$ be the normalizer of $P$, and let $N$ act on $P$ by conjugation; this induces a homomorphism $N\to\mathrm{Aut}(P)$; since $P$ is abelian, it is contained in the kernel of that map, so we have that $N/P$ maps into $\mathrm{Aut}(P)$. But what is the order of the automorphism group of a cyclic group of order $p^n$, and what is the order of $N/P$? What does that tell you about the action of $N$ on $P$?

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You have to show that if an element $g$ of $G$ normalises $P$, then in fact it commutes with all the elements in $P$, i.e. if the conjugation by $g$ preserves the set, then it actually preserves all the elements.

Now, if $g$ normalises $P$, then conjugation by $g$ gives you an automorphism of $P$. What is the automorphism group of a cyclic $p$-group? (Hint: an automorphism is determined by what it does to a generator.) Don't forget the restriction, that $p$ is the smallest prime divisor of $|G|$.

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Since P is abelian, when acting upon P by conjugation from P, say you have p,q in P, then pqp^{-1} = q. Thus P is indeed contained in the kernel of N -> Aut(P), and hence induces a map N/P -> Aut(P). Examine the order of Aut(P) and this allows you to conclude that the image of N/P in Aut(P) must be trivial. From this you can conclude the desired result.

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