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## Are extensions of profinite groups profinite?

Assume $X$, $E$ and $G$ are topological groups and $1\to X\to E\to G\to 1$ a short exact sequence of continuous group homomorphisms. Under which of these conditions is $E$ a profinite group?

(i) $G$ profinite, $X$ finite

(ii) $G$ profinite, $X$ pro-$p$ for some prime $p$

(iii) $G$ profinite, $X$ profinite

In those case where $E$ is profinite, is there a way to write down the inverse system for $E$ explicitly in terms of the inverse system of $G$ (and $X$)

Thank you very much.

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The answer to all three questions is no in general. You need to assume in addition that $G$ carries the quotient topology from $E$. Otherwise, starting from any such exact sequence with $E$ compact and $G$ infinite, you can endow the compact group $E$ with the topology inherited from the embedding into $E\times G_d$ where $G_d$ is $G$ with the discrete topology; let $E'$ be the resulting group; then $1\to X\to E'\to G\to 1$ is an exact sequence of continuous group homomorphisms, $X$ and $G$ are compact but $E'$ is not. Here $X$ can even be the trivial group, so $E'=G_d$.

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 Dear Yves: In my answer I explicitly mention where I am assuming the OP meant for $G$ to have the quotient topology and $X$ the subspace topology (so our answers aren't incompatible!). I will edit to bring up those hypotheses at the outset. – kreck Jan 2 at 18:10 @Kreck I noticed this just after posting! this was the natural hypothesis to do indeed. – Yves Cornulier Jan 2 at 18:50

[EDIT: I assume $X$ is given the subspace topology and $G$ the quotient topology.]

The answer to (iii) (and (i) and (ii)) is "yes". Is this not treated in the book "Profinite groups" (which I've never looked at)? I'm less sure about the "explicit" request, since (even for finite $X$) it seems a bit hard to "see" an open subgroup of $E$ that omits a given nontrivial $x \in X$ (then we could shrink it a bit to also be normal and intersect with preimages in $E$ of open normal subgroups of $G$ to get what you want).

Here is the proof for (iii) in the affirmative.

Step 1 (Hausdorff nonsense): Firstly, since $G$ is (I presume) given the quotient topology of $E$ and that is Hausdorff, its identity is closed and hence $X$ is closed in $E$. I assume $X$ is meant to have the subspace topology and so since its identity point is closed (as $X$ is Hausdorff) it follows that the identity of $E$ is a closed point. Thus, since $E$ is a topological group, it follows that $E$ is Hausdorff. That was pretty boring, and perhaps you were assuming $E$ to be Hausdorff at the outset. Quomodocumque.

Step 2 (reformulation): Now we recall that among Hausdorff topological groups, the profinite ones are precisely those that are compact and totally disconnected (i.e., only non-empty connected subsets are points). This is proved in Montgomery-Zippin and elsewhere I presume. So we just have to check that $E$ inherits each such property separately from $X$ and $G$, and these are elementary as follows.

For the total disconnectedness it is equivalent to say that the only connected closed subgroup is the trivial one (since the connected component of the identity point is closed and visibly a subgroup). But such a subgroup of $E$ has trivial image in the totally disconnected $G$ and so lies in the totally disconnected $X$ and thus is trivial, so $E$ is totally disconnected.

Step 3 (properness): Next, we verify the compactness. There may be a clever way to see it using open covers or nets, but I don't see such an argument offhand (since I don't recall in what generality one knows that quotient maps between topological groups admit local continuous cross-sections), so here is a direct argument using topological properness (for Hausdorff spaces) in the sense of Bourbaki. Maybe the argument can be done more efficiently; I just give what comes to mind at the moment.

Recall that a separated continuous map between topological spaces defined to be proper when it is universally closed (in the category of all topological spaces), and that this is equivalent to the map being closed with quasi-compact fibers. In particular, since properness is preserved under composition and $G$ is proper over a point, to prove the Hausdorff $E$ is compact it suffices to show that $f$ is proper.

More specifically, since $f$ has compact fibers (translates of $X$), we just have to show that $f$ is closed. That is, if $C$ is a closed set in $E$ then we want to show that $f(C)$ is closed in $G$. Since $G$ has the quotient topology from $E$, this means that $f^{-1}(f(C))$ is closed in $E$. To prove this closedness, we'll use compactness of $X$ in another way.

The map $X \times E \rightarrow E \times_G E$ defined by $(x,e) \mapsto (xe, e)$ is a topological isomorphism (respecting 2nd projections), so if $C$ is a closed set in $E$ then $X \times C$ goes over to a closed set in $E \times_G E$, and this closed set is $S := f^{-1}(f(C)) \times_{f(C)} C$. Note that its image under the first projection to $E$ is $f^{-1}(f(C))$. But $E \times_G E$ is also identified with $E \times X$ respecting first projections (via $(e,x) \mapsto (e, ex)$, say), and this first projection is proper since $X$ is compact Hausdorff. In particular, this first projection is a closed map, so $f^{-1}(f(C))$ is closed in $E$ because of the closedness of $S$ in the fiber product. That completes the proof of compactness of $E$.

QED

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