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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)

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  • $\begingroup$ Waterhouse proved that all profinite groups occur as galois groups, see en.wikipedia.org/wiki/Profinite_group $\endgroup$
    – dke
    Commented Jun 30, 2010 at 14:18
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    $\begingroup$ The answer to the first question is no: at a minimum, the automorphism group of the space must act transitively, ruling out a space like {1, 1/2, 1/3, ..., 0}. $\endgroup$
    – Reid Barton
    Commented Jun 30, 2010 at 14:21
  • $\begingroup$ @dke: Thanks. Also, the proof is very short! I've edited my question. @Reid: Ok. Let's assume that $Aut(X)$ acts transitively on $X$. Are more conditions needed? $\endgroup$
    – Martin Brandenburg
    Commented Jun 30, 2010 at 14:33

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(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.

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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).

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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.

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  • $\begingroup$ Thanks, Francois. I will learn this editor. Peter Plauman $\endgroup$
    – user7207
    Commented Jun 30, 2010 at 15:12
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    $\begingroup$ False: consider e.g. the finite discrete group of order $3$. This is of course a trivial counterexample; off the top of my head I don't see why this shouldn't be true so long as the profinite group is infinite. Could someone check the source? $\endgroup$
    – Pete L. Clark
    Commented Jun 30, 2010 at 17:16

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