I am reading Galois Theory 4th edition by Ian Stewart and the proof of one part of the Galois correspondence is circular.
$L : K$ is a finite normal field extension inside $\mathbb{C}$ with Galois group $G, \mathscr{F}$ is the set of intermediate fields, and $\mathscr{G}$ is the set of subgroups of $G.$ For $M \in \mathscr{F}, M^*$ is the group of $M$-automorphisms of $L$ and for $H \in \mathscr{G}, H^†$ is the fixed field of $H.$
(2) The maps $∗$ and $†$ are mutual inverses, and set up an order-reversing one-to one correspondence between $\mathscr{F}$ and $\mathscr{G}.$
We need to show that $M^{*†} = M, H^{†*} = H,$ which the book proceeds to do as follows:
Suppose that $M$ is an intermediate field, and let $[L : M] = d.$ Then $|M^∗| = d$ by Theorem 10.5. On the other hand, if $H$ is a subgroup of $G$ of order $d,$ then $[L : H^†] = d$ by Corollary 11.11. Hence the composite operators $∗†$ and $†∗$ preserve $[L : M]$ and $|H|$ respectively. From their definitions, $M \subseteq M^{∗†}$ and $H \subseteq H^{†∗}.$ Therefore these inclusions are equalities.
Theorem 10.5: Let $G$ be a finite subgroup of the group of automorphisms of a field $K,$ and let $K_0$ be the fixed field of $G.$ Then $[K : K_0] = |G|.$
Corollary 11.11. If $L : K$ is a finite normal extension inside $\mathbb{C},$ then there are precisely $[L : K]$ distinct $K$-automorphisms of $L.$ That is, $|\Gamma(L : K)| = [L : K]$
If $M$ is the fixed field of $M^*,$ then Theorem 10.5 indeed gives us $|M^*| = [L : M] = d.$ However, this is equivalent to $M^{*†} = M,$ which is what we are trying to show.
Supposing that $[L:H^†]$ is a normal extension, Corollary 11.11 gives us $[L:H^†] = |\Gamma(L : H^†)| = |H^{†*}|,$ but we want this to equal $|H| = d$ and hence need to show $H = H^{†*}.$
As it stands, all we really know is that $|M^*| \le d$ and $[L:H^†] \ge d.$
We start with a field $L \leq M \leq K$, and we want to show that $M^{\star\dagger} = M$. Suppose that $[L:M] = d$.
The first thing we do is apply Corollary 11.1, this tells us that $|M^{\star}| = d$ (the book says they are applying Theorem 10.5 here, but I believe this is a mistake).
Then we apply Theorem 10.5 (again the book mistakenly here says to apply Corollary 11.1) to the group $M^{\star}$, which lets us conclude that $$[L:M^{\star\dagger}] = |M^{\star}| = d = [L:M].$$
We tie this together by remembering how the $\star$ and $\dagger$ operations are defined; $M^{*}$ is a group that fixes everything in $M$, and $M^{\star\dagger}$ is a field $L \leq M^{\star\dagger} \leq K$ that contains precisely the elements fixed by every element of $M^{\star}$, so certainly $M \leq M^{\star\dagger}$. Since they are field extensions of $L$ with the same index, they must be equal.