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.