I'm having a hard time proving this statement. Can anyone point me in the right direction, or outline a possible proof? The claim was made in a brief remark in Fesenko and Vostokov's book on Local Fields (pg. 97), after introducing Fontaine and Wintenberger's field of norms.
Here's what I want to understand...
Let $K$ be a local field, $\overline K$ a separable closure of $K$, and put $G_M:= \text{Gal}(\overline K/M)$ for any intermediate extension $K\subset M\subset \overline K$. Prove that if $L/K$ is a Galois extension of finite degree contained in $\overline K$ and $u\in \mathbb{R}_{\geq -1}$ then $G_K^{u}G_L$ is open in $G_K$ if and only if $\text{Gal}(L/K)^{u}$ is open in $\text{Gal}(L/K)$.
Okay, I think I might have figured it out. The problem is equivalent to showing that $G_K^{u}G_L$ has finite index in $G_K$ if and only if $\text{Gal}(L/K)^{u}$ has finite index in $\text{Gal}(L/K)$, so this is what I'll do. Using Herbrand's theorem (which is also true for infinite Galois extensions), we know that $$\bigg(\frac{G_K}{G_L}\bigg)^u \cong \frac{G_K^uG_L}{G_L} $$ since $G_L$ is normal in $G_K$ by Galois theory. So now
$$\frac{\text{Gal}(L/K)}{\text{Gal}(L/K)^{u}}\cong \frac{\big(G_K/G_L\big)}{\big(G_K/G_L\big)^u}\cong \frac{\big(G_K/G_L\big)}{ \big(G_K^uG_L/G_L)}\cong \frac{G_K}{G_K^uG_L},$$
where the first isomorphism is by Galois theory, the second by Herbrand's theorem, and the third by the 3rd isomorphism theorem.