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If $P$ is a Sylow $p$-subgroup of $G$, how do I prove that normalizer of the normalizer $P$ is same as the normalizer of $P$ ?

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    This is the first exercise on page $82$ of N. Jacobson's Basic Algebra I (e2), if anyone was wondering.2017-06-28

5 Answers 5

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We have the following: $P\leq N(P)\leq N(N(P))$. We see that $P$ is also a Sylow $p$-group of $N(P)$ and of $N(N(P))$. If $x\in N(N(P))$, then $xPx^{-1}\leq xN(P)x^{-1}=N(P)$, and since all Sylow $p$-subgroups are conjugate, we have that there exists $y\in N(P)$ such that $xPx^{-1}=yPy^{-1}$. But since $y\in N(P)$, we have that $yPy^{-1}=P$, and so $xPx^{-1}=P$. This shows that $x\in N(P)$, and they must be the same.

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    Nice solution! +12018-04-11
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Let $M= N_G(P)$. Clearly, $M\subseteq N_G(M)$.

Now, notice that $P$ is normal in $M$, so it is the unique Sylow $p$-subgroup of $M$. Therefore, if $x\in N_G(M)$, then since $xPx^{-1}$ is a Sylow $p$-subgroup of $xMx^{-1}=M$, then $xPx^{-1} = P$, because $P$ is the only Sylow $p$-subgroup of $M$. That means that $x\in N_G(P) = M$. Therefore, $N_G(M)\subseteq N_G(P)$.

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    @qwr Don’t know how that happened 7 years ago...2018-10-28
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Hints ($N(H)$ denotes the normalizer of a subgroup $H\le G$ in $G$):

1) Show that $P$ is the only Sylow $p$-subgroup of$N(P)$. Remember that they are all conjugate in $N(P)$.

2) If $P$ and P' are different Sylow $p$-subgroups, show that $N(P)$ and N(P') are A) conjugate in $G$, B) different.

3) Show that $P$ is the only Sylow $p$-subgroup of $N(N(P))$.

4) Show that $P\unlhd N(N(P))$.

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Hint: P is a normal Sylow p-subgroup of $N_G(P)$...

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Let $N=N_G(P)$. Let $x\in N_G(N)$, so that $xNx^{-1}=N$. Then $xPx^{-1}$ is a Sylow $p$-subgroup of $N\leq G$. Since $P$ is normal in $N$, $P$ is the only Sylow $p$-subgroup of $N$. Therefore $xPx^{-1}=P$. This implies $x\in N$. We have proved $N_G(N_G(P))\subseteq N_G(P)$.

Let $y\in N_G(P)$ Then certainly $yN_G(P)y^{-1}=N_G(P)$, so that $y\in N_G(N_G(P))$. Thus $N_G(P)\subseteq N_G(N_G(P))$.