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 $P$ ?
Normalizer of the normalizer of the sylow $p$-subgroup
-
4This is the first exercise on page $82$ of N. Jacobson's Basic Algebra I (e2), if anyone was wondering. – 2017-06-28
5 Answers
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.
-
0Nice solution! +1 – 2018-04-11
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, $N_G(M)\subseteq N_G(P)$.
-
4@qwr Don’t know how that happened 7 years ago... – 2018-10-28
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 $P\unlhd N(N(P))$.
Hint: P is a normal Sylow p-subgroup of $N_G(P)$...
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 $N_G(P)\subseteq N_G(N_G(P))$.