I am studying Galois Theory from the lecture notes given by our instructor. Here I came across the following theorem which has been left as an exercise for us.
Theorem $:$ Let $L|K$ be a finite Galois extension. Let $G=\text {Gal} \left (L|K \right).$ Let $H$ be a normal subgroup of $G.$ Let $M=\text {Fix}_H(L).$ Then $M|K$ is a Galois extension.
The proof of the above theorem is not given in the lecture notes as I have already said but a hint has been given there (Observe that $G/H$ acts faithfully on $M$). With the help of this hint I myself have managed to write a proof of the above theorem. Here it is $:$
Consider the left action of $G$ on $L.$ Then we have a group homomorphism $\vartheta : G \longrightarrow \text {Sym}\ (L)$ defined by $\tau \mapsto \sigma_{\tau}$ where $\sigma_{\tau}$ is a bijection of $L$ given by $\sigma_{\tau} (x) = \tau(x),\ x \in L.$ Since $H$ is a normal subgroup of $G$ so $G$ acts on $\text {Fix}_H L = M$ and $M$ is invariant under the left action of $G$ on $L$ i.e. $\sigma \left (M \right ) \subseteq M$ for all $\sigma \in G.$ So we have a homomorphism $\varphi : G \longrightarrow \text {Sym}\ (M)$ defined by $\tau \mapsto \vartheta(\tau)|_{M} = \sigma_{\tau}|_M = \tau|_M.$ Let $\widetilde H = \text {Ker}\ \varphi.$ Then it is easy to see that $\text {Fix}_H (L) = \text{Fix}_{\widetilde H} L.$ So by Fundamental Theorem of Galois Theory it follows that $H=\widetilde H.$ Let us consider the action of $G/H$ to $M$ given by homomorphism $\psi : G/H \longrightarrow \text {Sym}\ (M)$ defined by $\sigma H \mapsto \sigma|_{M}, \sigma \in G.$ Then $\text {Ker}\ \psi = \widetilde H/H = \{H \},$ since $\widetilde H = H.$ So $G/H$ acts faithfully on $M.$ That means $G/H \cong \psi \left (G/H \right ) \subseteq \text {Gal}\ \left (M|K \right ).$ Hence $\#\ G/H \leq \#\ \text {Gal}\ \left (M|K \right ).$ Now we need to prove that $M|K$ is a Galois extension i.e. we need to show that $\#\ \text {Gal}\left (M|K \right ) = [M:K].$ Since $L|K$ and $L|M$ both are Galois extensions it follows that $\#\ \text{Gal} \left (L|K \right) = [L:K]$ and $\#\ \text {Gal} \left (L|M \right ) = \#\ H = [L:M]$ (by Fundamental Theorem of Galois Theory). Now $$\begin {align*} [M:K] & = \frac {[L:K]}{[L:M]} \\ & = \frac {\#\ \text {Gal} \left (L|K \right )} {\#\ \text {Gal} \left (L|M \right )} \\ & = \frac {\#\ G} {\#\ H} \\ & = \#\ G/H \\ & \leq \#\ \text {Gal} \left (M|K \right ). \end{align*}$$ So $[M:K] \leq \#\ \text {Gal} \left (M|K \right).$ On the other hand by Dedekind-Artin Theorem for finite field extensions it also follows that $\#\ \text {Gal} \left (M|K \right ) \leq [M:K].$ Combining these two inequalities we have $\#\ \text {Gal} \left (M|K \right ) = [M:K],$ as claimed.
Does my proof hold good? If there is any flaw crept in my proof would anybody please let me know this? I will be glad to rectify it. Any suggestion or comment regarding this will be highly appreciated.
Thanks for your valuable time for reading.