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I am a bit confused about this exercise. Is there someone who can help?

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.

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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 $n \in T$

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    $\begingroup$ 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 $ghg^{-1}(x)=x$ $\endgroup$
    – Hello
    Sep 30 '20 at 12:00
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    $\begingroup$ $L^H$ is the set of element of $L$ fixed by $H$ $\endgroup$
    – 1123581321
    Sep 30 '20 at 12:01
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    $\begingroup$ 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? $\endgroup$
    – Hello
    Sep 30 '20 at 12:09
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    $\begingroup$ Yes of course it is ok! $\endgroup$
    – 1123581321
    Sep 30 '20 at 12:14
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    $\begingroup$ 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 $g'n(x)=n(x)$? $\endgroup$
    – Hello
    Oct 1 '20 at 9:04
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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 $$\tau(\alpha)=\tau(\tilde\sigma(\alpha))=\sigma(\tau(\alpha)).$$

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    $\begingroup$ But why do you look at \tau \in G? Aren't we supposed to look at \sigma\in G $\endgroup$
    – Hello
    Sep 30 '20 at 10:29
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    $\begingroup$ 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. $\endgroup$ Sep 30 '20 at 10:39
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    $\begingroup$ So when you write \tau, you mean \sigma? $\endgroup$
    – Hello
    Sep 30 '20 at 10:59
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    $\begingroup$ No. $\tau$ is just an arbitrary element of $N_G(H)$. Similarly, $\sigma$ is an arbitrary element of $H$. $\endgroup$ Sep 30 '20 at 11:04
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    $\begingroup$ But we have that $\sigma \in G$? $\endgroup$
    – Hello
    Sep 30 '20 at 11:08

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