Galois correspondence for pro-p extensions I am experiencing persistent confusion about the Galois correspondence when restricted to pro-$p$ extensions of a field $K$ (of characteristic $0$), $p$ a prime.
My naive expectation is that, because pro-$p$ extensions are `well-behaved' (closed under forming subgroups, quotients, extensions and products), the usual Galois correspondence should follow through, i.e. pro-$p$ extensions of $K$ correspond (bijectively) with quotients of $G_K(p)$, the maximal pro-$p$ quotient of the absolute Galois group of $K$, or equivalently $Gal(K(p)/K)$ where $K(p)$ is the maximal pro-$p$ extension of $K$.
This expectation leads me to think that if $F/K$ is a finite pro-$p$ extension, then $G_F(p)$ can be realised as a subgroup of $G_K(p)$ with the quotient being isomorphic to $Gal(F/K)$, by analogy to the case with arbitrary Galois extensions. However, this seems to be wrong, because work by Demushkin-Labute-Serre showed that if we e.g. take $K=\mathbb{Q}_2$ and $F=\mathbb{Q}_2(\sqrt{-1}))$, then the pro-$2$ groups can be computed, with explicit generators and relations. Crucially, the rank of $G_K(2)$ here equals 3, while that of $G_F(2)$ equals 4, so surely $G_F(2)$ cannot be realised as a subgroup of $G_K(2)$. Rather, $G_K(2)$ looks like a quotient of $G_F(2)$.
What is the general relationship between $G_F(p)$, $G_K(p)$ and $Gal(F/K)$ for an arbitrary pro-$p$ extension $F$ of $K$? Why does my naive expectation fail, or why was my expectation confused in the first place?
 A: There is much vagueness in your definitions/assertions about the Galois correspondence in a profinite extension $L/K$. Let us start again from the fundamental notions :
(1) To speak of $Gal(L/K)$, you should assume (as practically everybody does) that the extension $L/K$ is Galois. In your query about the relationship between $G_F(p), G_K(p)$ and $Gal(F/K)$, the first two pro-$p$-extensions are automatically Galois by maximality, but I think that  you  assume implicitly that $F/K$ is Galois.
(2) The profinite Galois correspondence states that, for a profinite Galois extension $L/K$ with group $G$, there is a bijection (built in the usual way) between the set of all the sub-extensions $F/K$ of $L/K$ and the set of all closed subgroups of $G$. This is a genuine limitation because, practically by definition, $G$ is the projective limit of the quotients $G/U$, where $U$ runs through all open normal subgroups of $G$. Recall that a subgroup $H$ of $G$ is open iff $H$ is closed and has finite index in $G$. Moreover, if $G$ is topologically of finite type (i.e. $G$ admits a subroup of finite type which is dense), then its subgroups of finite index are open.
(3) In the last question (which worries you because of a contradiction), you don't need the subextension $F/K$ to be Galois, only that it has finite degree. You take for granted the erroneous assertion that the "rank"of a subgroup $H$ of $G$ (even of finite idex) should be less than that of $G$. But:
(a) We must first define the "rank" $d(G)$ of a pro-$p$-group $G$. In view of the last sentence of (2), $d(G)$ can be defined, if $G$ is topologically of finite type, as the minimal number of topological generators of $G$ if it is finite, $\infty$ otherwise. Finiteness does not always occur, even in examples coming from number theory. The pro-$p$-analog of Burnside's basis theorem for (finite) $p$-groups tells us that $d(G)$ is the $\mathbf F_p$-dimension of $G/[G,G]G^p$ viewed as a $\mathbf F_p$ vector space. When $G= G_K(p)$ and $K$ is a local $p$-adic field, local class field theory allows to show that $d(G)$ is finite (actually explicit formulas are available). When $K$ is a number field, $d(G_K(p))$ is no longer finite in general, but global class field theory again allows to show that finiteness occurs for certain quotients of $G_K(p)$ defined by imposing adequate ramification conditions. For all this, see e.g. H. Koch's book "Galois Theory of $p$-extensions", chap.9-11.
(b) Assuming that $G$ is a pro-$p$-group with $d(G)$ finite and $H$ a subgroup of finite index, the explicit formulas alluded to in (a) give $d(H)$. For example, if $K$ is a local field of degree $n$ over $\mathbf Q_p$, then $d(G_K(p))=n+1+\epsilon$, where $\epsilon =1$ (resp.$0$) if $K$ contains (resp. does not contain) a primitive $p$-th root of unity, and analogously $d(G_F(p))$. It follows that $d(G_F(p))-d(G_K(p))=n([F:K]-1)$.
A: I don't know what you mean by "products" under "well-behaved," but pro-$p$ extensions aren't closed under compositum, so they're not maximally nice. As a simple example, let $K$ be the splitting field of $x^3 - 2$ over $\mathbb{Q}$. $K$ has subextensions $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\sqrt[3]{2} \omega)$ of degree $3$ and their compositum is $K$, which has degree $6$. The corresponding group-theoretic fact is that the intersection of subgroups of $p$-power index need not have $p$-power index.
On the other hand I also don't know what you mean by the rank of a nonabelian group. For any of the meanings I can think of that might plausibly be relevant, it's not true that if $G$ is a subgroup of $H$ then the rank of $G$ must be less than or equal to the rank of $H$.
