# Faithful group action and Galois correspondence

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.

• anything still unclear ? with my answer I meant that $|G|=[L:K]$ in an obfuscated way to say $K = L^G$, the same for your other relations. If $L/K$ is not Galois and $H$ is not normal let $N = \bigcap \sigma H \sigma^{-1}$ you'll obtain the tower $L/L^N/L^H/L^G/K$ Apr 9, 2019 at 22:39

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.