Are absolute Galois groups compact topological groups

Let $T$ be a finite set of primes and let $K$ be the maximal extension of $\mathbf{Q}$ unramified outside $T$.

We have three Galois groups:

$G_{\mathbf{Q}} = \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$

$G_{T} = \mathrm{Gal}(K/\mathbf{Q})$

and for any prime number $p$

$G_p = \mathrm{Gal}(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)$

Are these compact topological groups?

Also, are there any canonical maps between these groups? I think $G_T$ maps to $G_p$ if $p$ is in $T$. Is that correct?

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They're profinite groups: inverse limits of finite groups. The inverse limit is a closed subspace of the direct product, which is compact by Tychonoff (or an easier case of it). –  Dylan Moreland Apr 17 '12 at 21:00

There is, however, a nice topology that can be defined on Galois groups called the Krull topology. Let $E/F$ be a Galois extension and $\mathrm{Gal}(E/F)$ its Galois group. The Krull topology on $\mathrm{Gal}(E/F)$ has as basis for its closed sets the subgroups of $\mathrm{Gal}(E/F)$ which fix some finite intermediate extension of $F$ in $E$ (together with all right and left cosets of such subgroups). With the Krull topology, $\mathrm{Gal}(E/F)$ is a compact topological group. In fact, it is a profinite group, i.e. it is Hausdorff, compact, and totally disconnected, or equivalently it is the inverse limit of discrete finite groups.
To answer the second part of your question: $G_T$ is obviously a quotient of $G_{\mathbf{Q}}$, and one can view $G_p$ as a subgroup of $G_{\mathbf{Q}}$, but this is only uniquely defined up to conjugation in the latter group -- to define it but one has to make a non-canonical choice (one has to choose, compatibly, a prime above $p$ in every number field).
Composing these one obtains a map $G_p \to G_T$. If $p \notin T$ then this map is rather dull (it factors through a rather small quotient of $G_p$). If $p \in T$ then the map is very interesting: it has recently been shown by Chenevier and Clozel that it is injective if $T$ is large enough (I think $|T| \ge 2$ suffices).