# Let $L/F$ be Galois with Galois group $G$, and let $K$ be a subfield corresponding to the subgroup $H$. Then $N_{G}(H)=\{\sigma\in G:\sigma K=K\}$

The exercise is

Let $$L/F$$ be a Galois extension with Galois group $$G$$, and let $$K$$ be an intermediate field corresponding to the subgroup $$H$$ of $$G$$. Show that the normalizer $$N_{G}(H)$$ consists of those $$\sigma \in G$$ for which $$\sigma K=K$$.

A hint is given which is: To say that $$\sigma K=K$$ is the same as saying that $$\sigma \alpha \in K$$ for every $$\alpha \in K$$.

I know that I have to show $$N_G(H)=\{\sigma\in G \mid \sigma\cdot K=K\}$$. Which is the same as showing $$\sigma \alpha \in K \iff \sigma h \sigma^{-1} =h.$$

But I don't understand how I should do that.

Let $$N=N_G(H)=\{g\in G| \ gH=Hg\}$$ and $$T=\{g\in G|\ gK=K\}$$

• We show that $$T\subseteq N$$:

Let $$g\in T, h\in H,\ x\in K=L^H$$. Then $$ghg^{-1}(x)=g(hg^{-1}(x))$$ and since $$gK=K$$ it is $$g^{-1}(x)\in K$$ and $$hg^{-1}(x)=g^{-1}(x)$$ since $$h\in H$$ and $$K=L^H$$. Then $$ghg^{-1}(x)=gg^{-1}(x)=x$$ hence $$g\in N$$

• We show that $$N\subseteq T$$:

Let $$n\in N, g'\in H,x\in K=L^H$$. Then $$g'n(x)=n[n^{-1}g'n](x)$$. Since $$N=N_G(H)$$ it is $$n^{-1}g'n\in H$$ thus $$ng'n^{-1}(x)=x$$ so $$g'n(x)=n(x)$$ so $$n(x)$$ is in the subfield fixed by $$H$$ which is $$K$$. Hence $$n \in T$$

• When you show T\subseteq N. Then I don't understand what you mean by $K=L^H$, what is $L^H$. And how do you get that $hg^{-1}(x)=g^{-1}(x)$. And how do you get that $g\in N$, just because you have shown that $ghg^{-1}(x)=x$ Sep 30 '20 at 12:00
• $L^H$ is the set of element of $L$ fixed by $H$ Sep 30 '20 at 12:01
• Okay, thank you, I'll look at it again now. Is it okay that I ask you again, if I still don't understand it? Sep 30 '20 at 12:09
• Yes of course it is ok! Sep 30 '20 at 12:14
• I have tried to look at it again, and in the second part where you show $N \in subseteq T$ I still don't understand $ng'n^{-1}(x)=x$ but I realize if $ng'n^{-1}(x)=x$ then $ng'n^{-1}(x)=Id$ which might imply that $ng'(x)=n(x)$ but then I don't understand how you get $g'n(x)=n(x)$? Oct 1 '20 at 9:04

By definition, $$N_G(H)=N=\{\tau\in G\mid \tau H=H\tau\}$$. Let $$\tau\in N,\sigma\in H$$ and $$\alpha\in K$$. To show $$\tau(\alpha)\in K$$, it suffices to show that $$\sigma(\tau(\alpha))=\tau(\alpha)$$. Now by hypothesis, $$\sigma\tau=\tau\tilde\sigma$$ for some $$\tilde\sigma\in H$$. Thus $$\tau(\alpha)=\tau(\tilde\sigma(\alpha))=\sigma(\tau(\alpha)).$$

• But why do you look at \tau \in G? Aren't we supposed to look at \sigma\in G Sep 30 '20 at 10:29
• The only reason I started out with $\tau$ was that it only has three letters to type. Don't worry too much about it. Can you follow the argument? If there's any problem, please let me know. Sep 30 '20 at 10:39
• So when you write \tau, you mean \sigma? Sep 30 '20 at 10:59
• No. $\tau$ is just an arbitrary element of $N_G(H)$. Similarly, $\sigma$ is an arbitrary element of $H$. Sep 30 '20 at 11:04
• But we have that $\sigma \in G$? Sep 30 '20 at 11:08