Faithful group action and Galois correspondence

Question

Let $L|K$ be a finite field extension and let $H$ be a normal subgroup of the Galois group Gal$(L|K)$. I want to show that the extension Fix$_HL|K$ is a Galois extension, where $M=$Fix$_HL$ is the fixed field of $H$.

From the knowledge of group actions we know that Gal$(L|K)$ acts on $L$ and since $H$ is a normal subgroup of Gal$(L|K)$, it acts on $M$ also. I have also shown that $H=H^{\prime}$, where $H^{\prime}$ is the kernel of the group homomorphism $$\varphi: \text{Gal}(L|K)\to S(M)$$ where $S(M)$ denotes the permutation group of $M$. Thus $Gal(L|K)/H$ acts faithfully on $M$.

Now $[L:K]=[L:M][M:K]\implies \dfrac{[L:K]}{[L:M]}=[M:K]$.

But $\dfrac{[L:K]}{[L:M]}=\dfrac{\#\text{Gal}(L|K)}{\#H}$. I am done if I can show that $\dfrac{\#\text{Gal}(L|K)}{\#H}=\#\text{Gal}(M|K)$. But I could not understand how to show this from the information that $Gal(L|K)/H$ acts faithfully on $M$. Any help is appreciated.

Answer 1

You mean $L/K$ is Galois, $G = Gal(L/K)$, $H$ a normal subgroup, $L^H$ its fixed field.

Moreover $K= L^G$ (you can take $K = L^G$ as the definition of $L/K$ is Galois with Galois group $G$)

For $\sigma \in G$ then $\sigma(L^H) = L^{\sigma H \sigma^{-1}}$. As $H$ is normal then $\sigma H \sigma^{-1} = H$ so $\sigma(L^H) =L^H$ and $\sigma |_{L^H} \in Gal(L^H/K)$.

The restriction to $L^H$ sends $G$ to $G/ \ker(\sigma \to \sigma|_{L^H}) = G/ Gal(L/L^H) = G/H$.

Thus it makes sense to look at the fixed subfield $(L^H)^{G/H} = L^G = K$ which implies $L^H/K$ is Galois with Galois group $G/H$.

If $L/K$ is not Galois then let $H = \{1\}$ to see it cannot work.