My question is based on John M Howie's 'Fields and Galois Theory', chapter 7.
Let $L:K$ be a finite field extension. Let $E$ be a subfield of $L$ including $K$, and let the maps $\Gamma$ and $\Phi$ be defined as follows:
$$\Gamma(E) =\{\varphi \in \text{Aut }L : \varphi(z)=z \text{ for all } z \in E\}\\\Phi(H) = \{x \in L : \varphi(x)=x \text{ for all } \varphi \in H \}$$
Now my understanding is that the following relations hold for all finite extensions $L:K$, not just when $L:K$ is normal and separable:
$$H\subseteq\Gamma(\Phi(H))\\H_1\subseteq H_2 \implies \Phi(H_1)\supseteq\Phi(H_2)$$
Where $H_1$ and $H_2$ are subgroups of $\text{Gal}(L:K)$. This immediately implies
$$\Phi(H)\supseteq\Phi(\Gamma(\Phi(H)))$$
It is also true that the relation $E\subseteq\Phi(\Gamma(E))$ holds even when $L:K$ is not normal or separable. Therefore if we substitute $E$ for $\Phi(H)$ this gives
$$\Phi(H)\subseteq\Phi(\Gamma(\Phi(H)))$$
Combining the two results above gives $\Phi(H)=\Phi\Gamma\Phi(H)$. Now my understanding is that the relation $[L:\Phi(H)]=\lvert H \rvert$ also holds for all finite extensions - again, not just normal separable extensions. Therefore we have
$$\lvert H\rvert=[L:\Phi(H)]=[L:\Phi\Gamma\Phi(H)]=\lvert \Gamma\Phi(H)\rvert$$
Combining this with the result that $H\subseteq\Gamma\Phi(H)$ gives
$$H=\Gamma\Phi(H)$$
$\textbf{Question:}$ none of the above required $L:K$ to be Galois, so the result would appear to be true for all finite extensions. Is this actually the case? And if so, why could we not run a similar argument to show that $E=\Phi\Gamma(E)$ for all finite extensions - which is clearly not the case (since, for example, when considering the non-normal extension $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ we have $\Phi(\Gamma(\mathbb{Q}))=\mathbb{Q}(\sqrt[3]{2})$).