I am currently tutoring a course on introduction to Galois theory. For some reason the professor (whom we also cannot contact, don't ask, I have no idea) decided to define a Galois extension as an extension where $\#\mathrm{Aut}(M/K)=[M:K]$. We have not seen the concept of normal or separable extensions at all (again, don't ask me...). This definition is at least equivalent in the case of finite extensions, but the proof of the main theorem of Galois theory has a problem, which I cannot fix.
In particular we want to show that for any field $L$ between $M$ and $K$ we have that $M/L$ is also Galois. Usually we'd simply prove that $M/L$ is normal and separable, which we cannot do here (since we don't know either of the concepts). Instead we assume otherwise (i.e. $[M:L]>\#\mathrm{Aut}(M/L)$) and then try to decompose $\mathrm{Aut}(M/K)$ as follows:
For a fieldextension $A/B$ and some $\kappa\in\mathrm{Aut}(B)$ define $\mathfrak{A}_{A/B,\kappa}$ as the set of extensions of $\kappa$ to $A$, i.e. $\mathfrak{A}_{A/B,\kappa}\subseteq \mathrm{Aut}(A)$. Then $$\mathrm{Aut}(M/K)=\dot\bigcup_{\kappa\in\mathfrak{A}_{L/K,\mathrm{id}_K}}\mathfrak{A}_{M/L,\kappa}$$ i.e. first consider extensions of the identity on $K$ to $L$ and then further extend them to $M$. Now I have the following problem: Without knowing that $M/L$ is normal, how do we prove that there are no automorphisms on $M$ that don't fix $L$? The remainder of the "proof" is just some counting argument which would give us a contradiction to the tower formula of field extensions, so no more theory in there. I just don't know why the above equality of sets should be true - I mean, it obviously is true, since everything is Galois here and thus normal, but we simply don't know that with the given information. Am I missing something or is this not solvable?
Cheers for any input on the matter.