If $G$ is a finite subgroup of $\Gamma(L:K), [L:K] = pq,$ and $[G^{\dagger} : K] > 1$ where $p, q$ are distinct primes, prove that $G$ is cyclic.
This was a problem on one of my exams and I proceeded to try the following: $[G^{\dagger} : K] = p, q, pq$ by the Tower Law, so $|G| = |\Gamma(L:K)|/|G^{\dagger} : K] = pq/|G^{\dagger} : K] = 1, p, q,$ and from here $G$ is cyclic by Lagrange's Theorem or otherwise.
However, this assumes $|\Gamma(L:K)| = [L:K]$ and $G^{\dagger *} = G,$ which requires normality of $L:K.$ Without normality, you can still say $[L:G^{\dagger}] = 1, p, q,$ and that $G^{\dagger *} \subseteq \Gamma(L:G^{\dagger}).$ Under the assumption of normality, $|\Gamma(L:G^{\dagger})| = [L : G^{\dagger}]$ and we may finish as before since $G \subseteq G^{\dagger *}.$
However, not every extension of prime degree is normal (consider $\mathbb{Q}(2^{1/3}) : \mathbb{Q}$ for example), so you can't prove $L:G^{\dagger}$ is normal, and I didn't notice the mistake in time.
There's a similar problem with $[L:K]$ being merely prime, but there's no way I'd have come up with the solution there and generalized it within the time limit of the exam.
Note: For a subfield $H, H^*$ is the group of $H$-automorphisms of $L$ and for a subgroup $G, G^{\dagger}$ is the fixed field of $G.$ $\Gamma(L:K)$ is the group of all $K$-automorphisms of $L.$ Forget Galois, you do not know Galois.
Be careful, in your note you mentioned that Aut(L/K) the Galois group of L over K; it is when the extension is Galois (normal and separable), but not always. By defining Aut(L/K) as the Galois group, you're automatically assuming L/K is Galois and you're done, as you said.