MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

Return to Answer

3 added 44 characters in body

[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 (at least for Hausdorff spaces, maybe in general) this is equivalent to the map being closed with compact quasi-compact fibers. In particular, since properness is preserved under composition and $G$ is proper over a point, to prove the Hausdorff $E$ is proper 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

2 added 86 characters in body

[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 map between topological spaces defined to be proper when it is universally closed, and that (at least for Hausdorff spaces, maybe in general) this is equivalent to being closed with compact fibers. In particular, since properness is preserved under composition and $G$ is proper over a point, to prove $E$ is proper 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

1

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 map between topological spaces defined to be proper when it is universally closed, and that (at least for Hausdorff spaces, maybe in general) this is equivalent to being closed with compact fibers. In particular, since properness is preserved under composition and $G$ is proper over a point, to prove $E$ is proper 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