How to show that a Group is not profinite?

Question

I tried to show that $\mathbb{Z}$ is not profinite. So I know that a topological group $G$ is profinite if $G\cong \underleftarrow{\lim}~ G/H$, with $H$ all open normal subgroups of $G$. And I know that there is no surjective map $\mathbb{Z}\longrightarrow\underleftarrow{\lim}~\mathbb{Z}/n\mathbb{Z}$. So $\mathbb{Z}$ is not profinite with the profinite topology. But how to show that there is no other topology that $\mathbb{Z}$ is profinite?

Answer 1

Your reference (the book Introduction of Profinite Groups and Galois Cohomology by Ribes) defines a profinite group to be a compact Hausdorff totally disconnected topological group. I'll argue directly from this definition that any profinite group is either finite or has cardinality $\geq 2^{\aleph_0}$.

Suppose $G$ is a profinite group. If $G$ has an isolated point, then every point of $G$ is isolated (since for any $a\in G$, the map $x\mapsto ax$ is a homeomorphism). But then $\{\{a\}\mid a\in G\}$ is an open cover of $G$, so $G$ is finite by compactness.

Now suppose $G$ has no isolated points (such a space is called perfect). It's a standard fact that a perfect locally compact Hausdorff space has cardinality $\geq 2^{\aleph_0}$. But the proof is a bit easier for a compact Hausdorff totally disconnected space, so I'll outline the argument in this case.

We build a complete binary tree of non-empty clopen sets in $G$ by associating to each finite binary sequence $w$ a clopen set $C_w\subseteq G$, such that for any $w$, writing $w0$ and $w1$ for the sequences obtained by extending $w$ by a $0$ and a $1$, respectively, we have $C_{w0}\subseteq C_w$, $C_{w1}\subseteq C_w$, and $C_{w0}\cap C_{w1} = \emptyset$.

We start by associating the empty sequence to all of $G$. Now given $C_w$, since $G$ has no isolated points, $C_w$ has at least two points, $x$ and $y$. Since $G$ is Hausdorff and totally disconnected, there are two complementary clopen sets such that one contains $x$ and the other contains $y$. We define $C_{w0}$ and $C_{w1}$ by intersecting these two clopen sets with $C_w$.

Now for any infinite binary sequence $s$, we can take the intersection of the clopen sets corresponding to the finite initial segments of $s$: $\bigcap_{n\in \mathbb{N}} C_{s\restriction n}$. By compactness, this intersection is non-empty, so we can pick a point $x_s$ in the intersection. This gives an injective map from the set of infinite binary sequences (equivalently, paths through the complete binary tree we constructed above) to $G$, which shows that $|G|\geq 2^{\aleph_0}$.

Answer 2

An inverse limit of finite groups is either finite or uncountable. $\mathbb Z$ is neither of those.

For the proof, consider an inverse sequence of finite groups: $$\cdots \xrightarrow{f_i} G_i \xrightarrow{f_{i-1}} \cdots \xrightarrow{f_2} G_2 \xrightarrow{f_1} G_1 \xrightarrow{f_0} G_0 $$

Case 1: Suppose there exists $I$ such that for all $i \ge I$ the map $f_i$ is an isomorphism. Then the inverse limit is isomorphic to $G_I$ and hence is finite.

Case 2: Suppose for all $I$ there exists $i \ge I$ such that the map $f_i$ is not an isomorphism. Thus one can find a subsequence $i_1 < i_2 < i_3 < \ldots$ such that $f_{i_k}$ is $n_k$-to-one where $n_k \ge 2$. Now one uses a Cantor-type diagonalization argument to prove that the inverse limit is uncountable.

ADDED: As pointed out in the comment of @AlexKruckman, you can also get exact cardinality $2^{\aleph_0}$, but that's more than you asked and so is really another question.