I want to show that the following are equivalent for a compact topological group $G$:

  1. $G$ is the inverse limit of finite groups $G_i$.

  2. There's a family $\left\{N_i\right\}$ of open normal subgroups that intersect trivially and $G$ is the inverse limit of the (finite) groups $G/N_i$.

  3. There's a family $\left\{H_j\right\}$ of open subgroups that intersect trivially.

Now, I know how to do all the implications except 1->2 or 1->3. The reason is that if $G$ has the trivial topology, where the only open and closed sets are the empty set and the whole group, then the only open subgroup would be all of $G$, which would prevent the other two properties from holding. Is there a reason a compact topological group can't have the trivial topology?

  • 2
    $\begingroup$ Are you assuming that group is Hausdorff (or has any separation properties)? A Hausdorff topological group has the trivial topology iff it is the trivial group. $\endgroup$ – Iian Smythe Jul 7 '14 at 22:35
  • $\begingroup$ Hmm, I think I'm assuming that the finite groups we're taking the inverse limit over each have the discrete topology. Does that help? $\endgroup$ – Nishant Jul 7 '14 at 22:57
  • $\begingroup$ Yes, an inverse limit of Hausdorff groups $G_i$ is a (closed) subgroup of the product $\prod_i G_i$, and since products of Hausdorff spaces and subspaces of Hausdorff spaces are Hausdorff, this shows that the inverse limit is Hausdorff. See Lemma 2.5 in math.ucdavis.edu/~osserman/classes/250C/notes/profinite.pdf. $\endgroup$ – Iian Smythe Jul 7 '14 at 23:00
  • $\begingroup$ And then I can take the $N_i$ to be all of $G$ except one of the factors in the subgroup of the direct product? $\endgroup$ – Nishant Jul 7 '14 at 23:04
  • $\begingroup$ I'd have to think about it, I'm rusty with inverse limits (I had to look up that link). What does follow from what I said is that if $G$ has the trivial topology, then it is the group $\{e\}$, and conditions 2 and 3 in your question are trivial. $\endgroup$ – Iian Smythe Jul 7 '14 at 23:10

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