Let $G$ be an infinite profinite group, so $$G=\lim_{\longleftarrow}G/N$$ where $N$ runs through the open normal subgroups. I have two questions:

  1. Is $G$ of Haar measure zero in the compact group $\prod_NG/N$?
  2. What is the relation between the Haar measure of a subset $E$ of $G$ and the numbers $\frac{|EN/N|}{|G/N}$, the size of the image of $E$ in $G/N$?
  • $\begingroup$ You probably want $G$ connected? Otherwise, just take a finite simple group. $\endgroup$
    – abx
    Dec 23, 2019 at 11:51
  • 4
    $\begingroup$ @abx Profinite group is just a totally disconnected compact group. Sure I mean an infinite profinite group as I would edit. $\endgroup$ Dec 23, 2019 at 11:57
  • $\begingroup$ Yes, I think that 1 has a positive answer (it's a good exercise). $\endgroup$
    – YCor
    Dec 23, 2019 at 14:43
  • $\begingroup$ For 2, there's an obvious inequality (for $E$ measurable, $\lambda(E)\le\inf_N|EN/N|/|G/N|$) which for $E$ dense of measure zero is clearly not an equality. $\endgroup$
    – YCor
    Dec 23, 2019 at 14:47
  • $\begingroup$ @YCor I choose this title because of my first question. $\endgroup$ Dec 23, 2019 at 15:01

1 Answer 1


1) The measure of a closed subgroup $H$ of a profinite group $G$ is $\frac{1}{\vert G:H \vert}$. So $G$ has measure zero in $\prod G/N$ if and only if it has infinite index. This way you should be able to show that $G$ always has measure zero in $\prod G/N$.

2) As Yves mentioned, you always have the inequality $\mu(S) \le \inf \frac{\vert NS/N\vert}{\vert G/N \vert}$ for any measurable subset $S$. In fact the right-hand side is the measure of the closure of $S$. If $S$ is closed it is an equality. If $G$ is the profinite completion of an abstract countable group $\Gamma$, then $\mu(\Gamma)=0$, but the closure of $\Gamma$ is $G$.


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