Proof of a result on closed subgroups of Galois group Let $M\supseteq K$ be an algebraic normal extension. Then its Galois group is profinite. I have been told that in this hypothesis, if $H\leq G$, then $H''=H\iff \bar H=H$, where $H'=\mathrm{Fix}(H)=\{\alpha\in M:\alpha^g=\alpha\,\,\forall g\in H\}$ and $H''=\{g\in G:\alpha^g=\alpha\,\,\forall\alpha\in H'\}$, and finally $\bar H$ is the topological closure of $H$ in $G$, which makes sense since $G$ is a topological space. I was wondering where I could find a proof of this. I would have tried googling, but the best query I could think of was topological closure of subgroup of Galois group and coincidence of it with the subgroup tha fixes all elements fixed by all of the group which is definitely too long to produce any sensible result with GOogle. So I came here to ask. Does anyone know a good reference available online (please avoid Google Books)? Or could someone write a proof in an answer? Thx.
 A: Since you know that $G = \mathop{\mathrm{Gal}}(M/K)$ is profinite, you know that the open subgroups are exactly those of the form $\mathop{\mathrm{Gal}}(M/L)$ where $L/K$ is a finite subextension of $M/K$. I'll now prove the statement:

For any subextension $L/K$ of $M/K$, the subgroup $\mathop{\mathrm{Gal}}(M/L)\leq G$ is closed.

Proof: Choose any $\sigma\in G - \mathop{\mathrm{Gal}}(M/L)$. We'll show there is an open neighbourhood of $\sigma$ in $G$ which does not meet $\mathop{\mathrm{Gal}}(M/L)$, so that the complement of $\mathop{\mathrm{Gal}}(M/L)$ is open in $G$. Let $\alpha\in L$ be such that $\sigma\alpha \neq \alpha$, possible because $\sigma$ by definition does not fix $L$. Set $H = \mathop{\mathrm{Gal}}(M/K[\alpha]) \leq G$, which is open as $K[\alpha]/K$ is a finite extension. For any $h\in H$, $\sigma h(\alpha) = \sigma\alpha\neq \alpha$, so $\sigma H\cap \mathop{\mathrm{Gal}}(M/L)$ is empty and $\sigma H$ is the desired open neighbourhood of $\sigma$. $\square$
The following is essentially the converse to the above statement:

If $H\leq G$ is closed, then $H = \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))$.

Proof: The inclusion $H\leq \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))$ follows from the definition of fixing field. As $H\leq G$ is closed, $G - H\subseteq G$ is an open subset. By the definition of the profinite topology, for any $\sigma\in G - H$, there exists a finite extension $L/K$ such that $\sigma\mathop{\mathrm{Gal}}(M/L)\subseteq G - H$. Thus, $\sigma\notin H\mathop{\mathrm{Gal}}(M/L)$. By enlarging $L$ (shrinking $\mathop{\mathrm{Gal}}(M/L)$), we may assume that $L/K$ is finite Galois. Let $H|_L\leq \mathop{\mathrm{Gal}}(L/K)$ be the image of $H$ under the canonical isomorphism
  $$\phi:\mathop{\mathrm{Gal}}(M/K)/\mathop{\mathrm{Gal}}(M/L)\to \mathop{\mathrm{Gal}}(L/K).$$
Similarly, the restriction $\sigma|_L\in \mathop{\mathrm{Gal}}(L/K)$ of $\sigma$ to $L$ is the image of $\sigma$ under $\phi$. As $\sigma\notin H\mathop{\mathrm{Gal}}(M/L)$, $\sigma|_L\notin H|_L$, so there is $\alpha\in \mathop{\mathrm{Fix}}(H|_L)$ (fixed field of $H|_L$ inside $L$) such that $\sigma|_L(\alpha) = \sigma(\alpha)\neq \alpha$. As $\mathop{\mathrm{Fix}}(H)\supseteq \mathop{\mathrm{Fix}}(H|_L)$, this shows that $\sigma\notin \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))$. Thus, $G - H\subseteq G -  \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))$. $\square$
Now, let's prove the statement you're asking about:

If $H\leq G$, then $\mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H)) = \overline{H}$.

Proof: By our first statement, $\mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))\leq G$ is closed. As $H\leq \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))$, this shows that $\overline{H}\leq \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))$.
As $H\leq \overline{H}$, we have $\mathop{\mathrm{Fix}}(H)\supseteq \mathop{\mathrm{Fix}}(\overline{H})$, so $\mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(H))\leq \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(\overline{H}))$. By the second statement, $\overline{H} = \mathop{\mathrm{Gal}}(M/\mathop{\mathrm{Fix}}(\overline{H}))$, which gives the other inclusion. $\square$
