Fix an algebraic closure, $\overline{\mathbb{Q}}$ for the rationals and consider the set, $B\_p$, of all places of $\overline{\mathbb{Q}}$ over a fixed (possibly infinite) prime, $p$, of $\mathbb{Q}$. Let $G\_\mathbb{Q}=\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.


1: Let $G\_\mathbb{Q}$ act on $B\_p$ in the natural way. Then what is the structure of the stabilizer, $H\_v$ of $v$, for $v|p$ an element of $B\_v$?


Infinite primes are just embeddings of $\overline{\mathbb{Q}}$ into $\mathbb{C}$, so the case $p=\infty$ is really asking for the stabilizer of the action of left multiplication in $G\_\mathbb{Q}$, and for finite $p$ it should be the inverse limit--over finite, Galois extensions of $\mathbb{Q}$--of the decomposition groups of primes $\mathfrak{p}|p$. However, actually getting ones hands on the structure of the stabilizers in question seems to be a hard question, at least no one I've talked to so far seems to know much more than the obvious things stated here.


2: Is is perhaps easier to describe the cosets $G/H\_v$? If so, what should they look like?



Place stabilizers for the absolute Galois Group
If $p$ is a rational prime, then choosing a prime $v$ of $\overline{\mathbf{Q}}$ lying over $p$ amounts to choosing an embedding $i:\overline{\mathbf{Q}}\hookrightarrow\overline{\mathbf{Q}}\_p$. This gives rise to a map $\varphi:G\_{\mathbf{Q}\_p}\rightarrow G\_{\mathbf{Q}}$defined as follows: given $s$ in the source, $\varphi(s)$ is the unique automorphism of $\overline{\mathbf{Q}}$ such that $i\circ\varphi(s)=s\circ i$. This is continuous. Its image is the decomposition group (i.e. the stabilizer of) $v$, $G\_v\subseteq G\_\mathbf{Q}$. The kernel is the Galois group of $\overline{\mathbf{Q}}\_p$ over $\mathbf{Q}\_pi(\overline{\mathbf{Q}})$, which is trivial by Krasner's lemma. So you have an isomorphism $G\_{\mathbf{Q}\_p}\cong G\_v$ (it is a homeomorphism because it is bijective with compact source and Hausdorff target). 


The case of Archimedean primes is identical. In particular, $G\_v$ for $v$ Archimedean has order $2$. When people talk about "choosing a complex conjugation" in $G\_{\mathbf{Q}}$, they are referring to the choice of an embedding $\overline{\mathbf{Q}}\rightarrow\mathbf{C}$ which gives rise to an injection $\mathrm{Gal}(\mathbf{C}/\mathbf{R})\hookrightarrow G\_{\mathbf{Q}}$, and the image of the unique non-trivial element of the source is the ``complex conjugation." 


Whenever you have a, say, $\ell$-adic, Galois representation $\rho:G\_\mathbf{Q}\rightarrow\mathrm{GL}\_d(\mathbf{Q}\_\ell)$, and a theorem talks about the local structure at $p$ of $\rho$, it means the restriction of $\rho$ to a decomposition group for a prime above $p$, which, by the paragraph above, can be identified with $G\_{\mathbf{Q}\_p}$. So a representation of $G\_\mathbf{Q}$ gives rise to representations of $G\_{\mathbf{Q}\_p}$ for all $p$ by restriction...at least after choosing a decomposition group, which is unique up to conjugacy. 


EDIT: This is in response to the question posed in the comments. The reason the map $\varphi$ (which depends on $i$) is well-defined is because $\overline{\mathbf{Q}}$ is a normal extension of $\mathbf{Q}$, so the image of any embedding of $\overline{\mathbf{Q}}$ into $\overline{\mathbf{Q}}\_p$ is the same (the subfield of elements algebraic over $\mathbf{Q}$). So, given $s\in G\_{\mathbf{Q}\_p}$, the embeddings $s\circ i$ and $i$ have the same image, so $i^{-1}\circ s\circ i$ makes sense, and is an element of $G\_\mathbf{Q}$. This is $\varphi(s)$.


Since $\varphi$ is a homomorphism, it's enough to check continuity at the identity. Take a finite extension $F$ of $\mathbf{Q}$ in $\overline{\mathbf{Q}}$, so $U=G\_F\leq G\_\mathbf{Q}$ is a typical neighborhood of the identity. Suppose $\varphi(s)\in G\_F$. We want to prove that there is an open subgroup $U^\prime$ of $G\_{\mathbf{Q}\_p}$ with $s\in U^\prime$ and $\varphi(U^\prime)\subseteq G\_F$. Let $F^\prime=i(F)\mathbf{Q}\_p\subseteq\overline{\mathbf{Q}}\_p$. This is a finite extension of $\mathbf{Q}\_p$. Since $\varphi(s)=i^{-1}\circ s\circ i$ fixes $F$, $s$ fixes $i(F)$, and therefore, since $i(F)$ generates $F^\prime$ over $\mathbf{Q}\_p$ and $s$ is $\mathbf{Q}\_p$-linear, $s$ fixes $F^\prime$. Conversely anything in $U^\prime=G\_{F^\prime}$ has image under $\varphi$ in $G\_F$, so $s\in U^\prime\leq\varphi^{-1}(U)$, as desired.


These arguments are totally general. They show that, if $k\hookrightarrow K$ is a map of fields and $i:k\_s\hookrightarrow K\_s$ is a choice of map of separable closures lifting $k\hookrightarrow K$ , then we get a continuous homomorphism $G\_K\rightarrow G\_k$. This homomorphism is not always injective though, as the injectivity of $\varphi$ above used a property particular to that setup (Krasner's lemma).


