Understanding the terms in a terse-style written exercise about $G$-sets

Question

This exercise is in Neukirch's Algebraic Number Theory chapter on Abstract Class Field Theory.

I don't think knowing the definition of a profinite group is helpful here so before you leave just keep in mind it is a certain type of topological group.

Exercise.

Let $G$ be a profinite group and $B(G)$ the category of finite $G$-sets, i.e., of finite sets $X$ with a continuous $G$-operation. Show that the connected, i.e., transitive $G$-sets in $B(G)$ are, up to isomorphism, the sets $G/G_{K}$, where $G_{K}$ is an open subgroup of $G$, and $G$ operates via multiplication on the left.

My question(s)

I will enumerate the questions for reference. These look easy, but I want to make sure.

1. First of all, $X$ has discrete topology, right?
2. The term "transitive $G$-sets" stands for $G$-sets $X$ such that $G$ acts transitively on $X$?
3. If $X$ has discrete topology, then the only connected $G$-sets $X$ are the singletons. So I guess this means (1) should be false. Which topology on $X$, then? Arbitrary topology?
4. The word 'isomorphism' refers to bijections which commute with the $G$ action?
5. Why the notation $B(G)$ for such category?

Answer 1

1. Yes.

2. Yes.

3. This is not how the term connected is being used. I read it to mean that $G$ acts transitively on $X$. (So if I draw a graph with nodes labelled by elements of $X$ and arrows $x\to y$ if there is $g\in G$ with $x=g.y$, the graph is connected.)

4. An isomorphism $\phi:X\to Y$ in the category of $G$-sets is a $G$-equivariant bijection (i.e. $\phi(g.x)=g.\phi(x)$).

5. I have no idea. Maybe $B$ is for "bounded" as the $G$-sets are finite.

Now, you are right that the topology is not really important, it is really a group theory question. All the topological adjectives follow from the fact that the action map $$a:G\times X\to X$$ is continuous (and $X$ is discrete).

Fix $x\in X$. Since $G$ acts transitively on $X$, the map $$\phi: G\to X,\;\;\;g\mapsto g.x$$ is surjective. The fibers of this map are of the form $\phi^{-1}(g.x)=gG_x$, where $G_x=\{h\in G\mid h.x=x\}$ is the stabilizer of $X$. It follows that there is a bijection $G/G_x\to X$, $gG_x\mapsto g.x$.

To see that $G_x$ is open, it is enough to note that it is the preimage of the open set $\{x\}$. To see that the bijection above is $G$-equivariant, one computes $$\phi(g.(hG_x))=\phi((gh)G_x)=(gh).x=g.(h.x)=g.\phi(h).$$