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

Question

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.

Answer 1

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 $$\tau(\alpha)=\tau(\tilde\sigma(\alpha))=\sigma(\tau(\alpha)).$$

Answer 2

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 $n \in T$