Profinite completions and inverse limits

Question

The following questions are mainly curiosities; I recently came across the definitions of inverse limits and profinite completions, and below are two questions about them I could not answer myself. I don't think the questions are of any practical importance, but I would nonetheless like to get to the bottom of them so as to understand the concepts better.

Question 1: The profinite completion $\hat G$ of a group $G$ is defined as the inverse limit of its finite quotients ordered by inclusion and equipped with the canonical projection maps. If $G$ is finite, the profinite completion is isomorphic to $G$ itself; indeed, the inverse limit will consist of finite sequences of compatible elements of $G$, which are determined by the first entry, which is an element of the quotient $G / \{e\}$.

My question regards the case where $G$ is infinite. Is it possible to find a group $G$ with only a finite number of normal subgroups? In this case, the completion $\hat G$ will be isomorphic to the biggest quotient $G / N$, where $N \trianglelefteq G$. As $G$ is infinite, $G$ itself will not be a finite quotient in this case. Of course $\hat G$ is supposed to contain $G$, so I am not sure if this case exists, however I have not been able to come up with an argument why not.

Question 2: This question concerns the cardinality of inverse limits. Given an inverse system $(G_i, \phi_{ij})$ for $i,j$ in some infinite indexing set $I$, the inverse limit $\displaystyle\lim_{\longleftarrow} G_i$ is a subgroup of $\prod_{i \in I} G_i$. This product is always uncountable (assuming the $G_i$ are non-trivial), as Cantor's diagonal argument shows. My question is whether or not there are non-trivial examples of the inverse limit being countable. Again, my intuition says this can't happen, but I'm missing the tools to prove this.

Any help would be appreciated!

Answer 1

1. Yes, for example there are infinite simple groups, such as $PSL_2(\mathbb{R})$. The profinite completion of any such group is trivial. More generally, a group $G$ does not always embed in its profinite completion; this is true iff $G$ is residually finite. (You forgot the condition that the normal subgroups have finite index.)

2. No, it's either finite or uncountable. There are several ways to show it: one is to use the existence of a Haar measure of total measure $1$ on any compact (Hausdorff) abelian group, which is impossible if the group is countable.