A group is profinite if and only if it is a Stone space

Question

A profinite group is an inverse limit of an inverse system of discrete finite groups. Alternatively, a profinite group is a topological group that is also a Stone space. Under the second, axiomatic definition it's obvious that a profinite group $G$ is a Stone space.

My professor made the claim that a group $G$ is profinite if and only if $G$ is a Stone space. The forward implication, as I mentioned, seems trivial. I need some help figuring out what is meant to be shown in the opposite direction. It is true that a topological space $X$ being a Stone space is equivalent to $X$ being homeomorphic to a projective limit of an inverse system of finite discrete spaces. My thought is that the reverse implication means to show that if $G$ is a Stone space, then the finite discrete spaces in the projective limit may be taken to be finite discrete groups.

If this is the case, how would someone go about proving such a thing? Otherwise, can someone lay out this duality in a more sensible way, and indicate the direction to go for studying profinite groups, and more generally pro-objects in a category?

Answer 1

Suppose $G$ is a topological group that is a Stone space, and let $g\in G$ be different from $1$. Since $G$ is a Stone space, there is a clopen set $U\subseteq G$ such that $1\in U$ and $g\not\in U$ (the basic open sets given by the presentation of a Stone space as an inverse limit of finite discrete spaces are clopen). If $\mu:G\times G\to G$ is the multiplication map, then $\mu^{-1}(U)$ is an open set containing $\{1\}\times U$. Since $U$ is compact, this implies $\mu^{-1}(U)$ contains $V\times U$ for some open set $V\subseteq G$ containing $1$. Let $W=V\cap V^{-1}$ and let $H\subseteq G$ be the subgroup generated by $W$. Note that $H\subseteq U$, since every element of $H$ is a product of elements of $W$ and $W\times U\subseteq\mu^{-1}(U)$ so any product of elements of $W$ will remain in $U$. Also, $H$ is open, since for any $h\in H$ the open neighborhood $hW$ of $h$ is contained in $H$. The coset space $G/H$ is then discrete and thus finite by compactness. The action of $G$ on $G/H$ then gives a continuous homomorphism $G\to S$ where $S$ is the symmetric group on $G/H$ (with the discrete topology). The kernel of this homomorphism is contained in $H$, and so in particular $g$ is not in the kernel.

To sum up we have now shown that for any $g\in G$ different from $1$, there is a continuous homomorphism from $G$ to a finite discrete group whose kernel does not contain $g$. Now consider the collection $I$ of all open normal subgroups of $G$ ordered by reverse inclusion. There is a natural inverse system of finite discrete groups indexed by $I$, with $K\in I$ getting sent to the quotient $G/K$. Let $L$ be the inverse limit of this system. The quotient maps $G\to G/K$ induce a continuous homomorphism $f:G\to L$ by the universal property. Since the quotient maps $G\to G/K$ are all surjective, the image of $f$ is dense, and thus $f$ is surjective since $G$ is compact and $L$ is Hausdorff. Also, from what we showed before, for every $g\in G$ different from $1$ there is some $K\in I$ such that the quotient map $G\to G/K$ does not map $g$ to $1$, so $f$ is injective. Thus $f$ is bijective, and thus a homeomorphism since $G$ is compact and $L$ is Hausdorff. Thus $G$ is an inverse limit of finite discrete groups.

(There is not any significantly simpler way to prove this result, and in particular there is no easy "abstract nonsense" way to turn the inverse system of finite sets into an inverse system of finite groups. In particular, for instance, there are other finitary algebraic structures besides groups for which this result is not true. See this answer for more about that.)