profinite spaces coming from profinite groups This is probably well-known:
Does every nonempty profinite space occur as the underlying space of a profinite group? If not, which conditions have to be imposed?
- Is every profinite group isomorphic to the Galois group of some Galois extension? (yes, see the comments)
 A: (Note: This was intended to be a comment to unknown (google)'s answer - but as I'm new here I can't post comments.)
As Pete L. Clark points out, unknown (google)'s answer is false as stated.
However, this is only because of the omission of the word "infinite".
A correct statement is:
An infinite profinite group $G$ is homeomorphic to $\{0,1\}^{w(G)}$,
where $\{0,1\}$ is the $2$-point discrete space,
and $w(G)$ is the weight of $G$.
This is Theorem 9.15 (pages 95-98) of "Abstract Harmonic Analysis I" by Edwin Hewitt and Kenneth A. Ross. (Hewitt and Ross actually state the result using the minimum cardinality of a local base at $1_G$, rather than the weight of $G$, but these are equal for infinite profinite groups.)
Notice that the case of countable weight is an immediate consequence of the usual characterisation of the Cantor set.
A: It's not hard to prove Waterhouse's theorem that all profinite groups
are Galois groups.
Note first that each quotient of a Galois group by a normal
closed subgroup is a Galois group, and as each profinite group is
the quotient of a product of finite groups by a closed subgroup
(this follows from the inverse limit construction),
we reduce to the case where $G$ is a product of finite groups.
Now each closed subgroup of a Galois group is a Galois group,
so by Cayley's theorem we may reduce to the case where
$$G=\prod_{i\in I}S_{m_i}$$
is a product of finite symmetric groups.
Finally we can write down an extension with this Galois group. Let $k$ be
any field, and $L$ be the field generated over $k$ by algebraically
independent indeterminates $X_{i,j}$ for $i\in I$ and $1\le j\le m_i$.
Then $G$ acts ion $L$ by letting the $i$-th factor $S_{m_i}$
permute the $X_{i,j}$. Let $K$ be the fixed field of the action of $G$. Then $L/K$ has
Galois group $G$ (in the Krull topology).
A: In the book Abstract Harmonic Analysis I be E. Hewitt and K.A. Ross - which I do not have at hand it is shown that the underlying space X of every compact, totally disconnected group is homeomorphic to the Cartesian power $\{0,1\}^c$, where c is the weight of the topological space X.
