# Normalizer of the normalizer of the sylow $p$-subgroup

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 $P$ ?

• This was a question on the 1995 Columbia Algebra qual, if anyone was wondering. Jan 5, 2013 at 2:48
• This is an exercise from Fraleigh's A First Course In Abstract Algebra (7e), if anyone was wondering. Nov 20, 2016 at 3:16
• This is the first exercise on page $82$ of N. Jacobson's Basic Algebra I (e2), if anyone was wondering. Jun 28, 2017 at 17:25
• This is exercise 5.8 in Isaacs $\textit{Algebra A Graduate Course}$, if anyone was wondering.
– user644428
Oct 18, 2019 at 0:59
• This is exercise 2.12.17 from Herstein's Topics in Algebra, if anyone was wondering. Nov 25, 2019 at 14:45

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.

• Can I ask why y∈N(P) ?
– 최선웅
Oct 17, 2017 at 16:35
• $P$, $xPx^{-1}$ are Sylow subgroups of $N(P)$. By Sylow's theorem, they are conjugate in $N(P)$ : there is $y\in N(P)$ such that $yPy^{-1} = xPx^{-1}$ Nov 5, 2017 at 10:34
• Nice solution! +1
– RFZ
Apr 11, 2018 at 16:34

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, $$N_G(M)\subseteq N_G(P)$$.

• The notation is strange - shouldn't it be $N_G(P)$?
– qwr
Oct 28, 2018 at 10:13
• @qwr Don’t know how that happened 7 years ago... Oct 28, 2018 at 18:24
• I'm sorry. I wanted to upvote but by mistake I downvoted your answer. Now I can't change it. May 9, 2022 at 6:57

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 $N_G(P)$...

Another proof

We have that $$P\in\text{Syl}_p(\mathbf{N}_G(P))$$ and $$\mathbf{N}_G(P)\trianglelefteq \mathbf{N}_G(\mathbf{N}_G(P))$$. By Frattini's Argument: $$\mathbf{N}_{\mathbf{N}_G(\mathbf{N}_G(P\,))}(P)\cdot\mathbf{N}_G(P)=\mathbf{N}_G(P)=\mathbf{N}_G(\mathbf{N}_G(P)),$$ because $$\mathbf{N}_{\mathbf{N}_G(\mathbf{N}_G(P\,))}(P)\subseteq \mathbf{N}_G(P)$$.

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 $N_G(P)\subseteq N_G(N_G(P))$.

There are many answers already but I'll give anoter one (I didn't see it somewhere above).

It is based on the following proposition:

Let $$G$$ be a finite group and $$P$$ a Sylow $$p-$$subgroup of $$G$$. If $$N_G(P)\leq H\leq G$$ then $$H=N_G(H)$$.

Proof: Let $$g\in N_G(H)$$. Then $$P\leq H\Rightarrow gPg^{-1}\subseteq gHg^{-1}=H$$. So $$P,gPg^{-1}$$ are Sylow $$p-$$subgroups of $$H$$. Hence they are conjugate, i.e. there exists $$h\in H$$ s.t. $$P=hgPg^{-1}h^{-1}\Rightarrow hg\in N_G(P)\leq H\Rightarrow g\in H\Rightarrow N_G(H)\subseteq H\checkmark$$.

Now, apply the proposition for $$H=N_G(P)$$.